FORUM GEOMETRICORUMA Journal on Classical Euclidean Geometry and Related Areas published by Department of Mathematical Sciences Florida Atlantic University b b b FORUM GEOM Volume 6 2006 http://forumgeom.fau.edu ISSN 1534-1178 Editorial Board Advisors: John H. Conway Julio Gonzalez Cabillon Richard Guy Clark Kimberling Kee Yuen Lam Tsit Yuen Lam Fred Richman Princeton, New Jersey, USA Montevideo, Uruguay Calgary, Alberta, Canada Evansville, Indiana, USA Vancouver, British Columbia, Canada Berkeley, California, USA Boca Raton, Florida, USA Editor-in-chief: Paul Yiu Boca Raton, Florida, USA Editors: Clayton Dodge Roland Eddy Jean-Pierre Ehrmann Chris Fisher Rudolf Fritsch Bernard Gibert Antreas P. Hatzipolakis Michael Lambrou Floor van Lamoen Fred Pui Fai Leung Daniel B. Shapiro Steve Sigur Man Keung Siu Peter Woo Orono, Maine, USA St. John’s, Newfoundland, Canada Paris, France Regina, Saskatchewan, Canada Munich, Germany St Etiene, France Athens, Greece Crete, Greece Goes, Netherlands Singapore, Singapore Columbus, Ohio, USA Atlanta, Georgia, USA Hong Kong, China La Mirada, California, USA Technical Editors: Yuandan Lin Aaron Meyerowitz Xiao-Dong Zhang Boca Raton, Florida, USA Boca Raton, Florida, USA Boca Raton, Florida, USA Consultants: Frederick Hoffman Stephen Locke Heinrich Niederhausen Boca Raton, Floirda, USA Boca Raton, Florida, USA Boca Raton, Florida, USA Table of Contents Khoa Lu Nguyen and Juan Carlos Salazar, On the mixtilinear incircles and excircles, 1 Juan Rodr´ıguez, Paula Manuel and Paulo Semi˜ao, A conic associated with the Euler line, 17 Charles Thas, A note on the Droz-Farny theorem, 25 Paris Pamfilos, The cyclic complex of a cyclic quadrilateral, 29 Bernard Gibert, Isocubics with concurrent normals, 47 Mowaffaq Hajja and Margarita Spirova, A characterization of the centroid using June Lester’s shape function, 53 Christopher J. Bradley and Geoff C. Smith, The locations of triangle centers, 57 Christopher J. Bradley and Geoff C. Smith, The locations of the Brocard points, 71 Floor van Lamoen, Archimedean adventures, 79 Anne Fontaine and Susan Hurley, Proof by picture: Products and reciprocals of diagonals length ratios in the regular polygon, 97 Hiroshi Okumura and Masayuki Watanabe, A generalization of Power’s Archimedean circles, 103 Stanley Rabinowitz, Pseudo-incircles, 107 Wilson Stothers, Grassmann cubics and desmic structures, 117 Matthew Hudelson, Concurrent medians of (2n + 1)-gons, 139 Dieter Ruoff, On the generating motions and the convexity of a well-known curve in hyperbolic geometry, 149 Mowaffaq Hajja, A very short and simple proof of “the most elementary theorem” of euclidean geometry, 167 S´andor Kiss, The orthic-of-intouch and intouch-of-orthic triangles, 171 Kurt Hofstetter, A 4-step construction of the golden ratio, 179 Eisso J. Atzema, A theorem by Giusto Bellavitis on a class of quadrilaterals, 181 Wladimir Boskoff and Bogdan Suceav˘a, A projectivity characterized by the Pythagorean relation, 187 Bogdan Suceav˘a and Paul Yiu, The Feuerbach point and Euler lines, 191 Eric Danneels, A simple perspectivity, 199 Quang Tuan Bui, Pedals on circumradii and the Jerabek center, 205 Bernard Gibert, The Simmons conics, 213 Nikolaos Dergiades, A synthetic proof and generalization of Bellavitis’ theorem, 225 Huub van Kempen, On some theorems of Poncelet and Carnot, 229 Charles Thas, The Droz-Farny theorem and related topics, 235 ˇ Zvonko Cerin, On butterflies inscribed in a quadrilateral, 241 Eric Danneels, On triangles with vertices on the angle bisectors, 247 Matthew Hudelson, Formulas among diagonals in the regular polygon and the Catalan numbers, 255 Khoa Lu Nguyen, A note on the barycentric square root of Kiepert perspectors, 263 Clark Kimberling, Translated triangle perspective to a reference triangle, 269 James L. Parish, On the derivative of a vertex polynomial, 285 Alexei Myakishev, On two remarkable lines related to a quadrilateral, 289 Max A. Tran Intersecting circles and their inner tangent circle, 297 K. R. S. Sastry, Two Brahmagupta problems, 301 Floor van Lamoen, Square wreaths around hexagons, 311 Jean-Pierre Ehrmann, Some geometric constructions, 327 Amy Bell, Hansen’s right triangle theorem, its converse and a generalization, 335 Paul Yiu, Some constructions related to the Kiepert hyperbola, 343 Author Index, 359 b Forum Geometricorum Volume 6 (2006) 1–16. b b FORUM GEOM ISSN 1534-1178 On Mixtilinear Incircles and Excircles Khoa Lu Nguyen and Juan Carlos Salazar Abstract. A mixtilinear incircle (respectively excircle) of a triangle is tangent to two sides and to the circumcircle internally (respectively externally). We study the configuration of the three mixtilinear incircles (respectively excircles). In particular, we give easy constructions of the circle (apart from the circumcircle) tangent the three mixtilinear incircles (respectively excircles). We also obtain a number of interesting triangle centers on the line joining the circumcenter and the incenter of a triangle. 1. Preliminaries In this paper we study two triads of circles associated with a triangle, the mixtilinear incircles and the mixtilinear excircles. For an introduction to these circles, see [4] and §§2, 3 below. In this section we collect some important basic results used in this paper. Proposition 1 (d’Alembert’s Theorem [1]). Let O1 (r1 ), O2 (r2 ), O3 (r3 ) be three circles with distinct centers. According as ε = +1 or −1, denote by A1ε , A2ε , A3ε respectively the insimilicenters or exsimilicenters of the pairs of circles ((O2 ), (O3 )), ((O3 ), (O1 )), and ((O1 ), (O2 )). For εi = ±1, i = 1, 2, 3, the points A1ε1 , A2ε2 and A3ε3 are collinear if and only if ε1 ε2 ε3 = −1. See Figure 1. The insimilicenter and exsimilicenter of two circles are respectively their internal and external centers of similitude. In terms of one-dimensional barycentric coordinates, these are the points r2 · O1 + r1 · O2 , (1) ins(O1 (r1 ), O2 (r2 )) = r1 + r2 −r2 · O1 + r1 · O2 . (2) exs(O1 (r1 ), O2 (r2 )) = r1 − r2 Proposition 2. Let O1 (r1 ), O2 (r2 ), O3 (r3 ) be three circles with noncollinears centers. For ε = ±1, let Oε (rε ) be the Apollonian circle tangent to the three circles, all externally or internally according as ε = +1 or −1. Then the Monge line containing the three exsimilicenters exs(O2 (r2 ), O3 (r3 )), exs(O3 (r3 ), O1 (r1 )), and exs(O1 (r1 ), O2 (r2 )) is the radical axis of the Apollonian circles (O+ ) and (O− ). See Figure 1. Publication Date: January 18, 2006. Communicating Editor: Paul Yiu. The authors thank Professor Yiu for his contribution to the last section of this paper. 2 K. L. Nguyen and J. C. Salazar A3− A2− O1 A2+ A3+ A1− O+ O− O2 A1+ O3 Figure 1. Lemma 3. Let BC be a chord of a circle O(r). Let O1 (r1 ) be a circle that touches BC at E and intouches the circle (O) at D. The line DE passes through the midpoint A of the arc BC that does not contain the point D. Furthermore, AD · AE = AB 2 = AC 2 . Proposition 4. The perspectrix of the circumcevian triangle of P is the polar of P with respect to the circumcircle. Let ABC be a triangle with circumcenter O and incenter I. For the circumcircle and the incircle, r·O+R·I = X55 , ins((O), (I)) = R+r −r · O + R · I = X56 . exs((O), (I)) = R−r in the notations of [3]. We also adopt the following notations. A0 point of tangency of incircle with BC A1 intersection of AI with the circumcircle A2 antipode of A1 on the circumcircle Similarly define B0 , B1 , B2 , C0 , C1 and C2 . Note that (i) A0 B0 C0 is the intouch triangle of ABC, (ii) A1 B1 C1 is the circumcevian triangle of the incenter I, On mixtilinear incircles and excircles 3 (iii) A2 B2 C2 is the medial triangle of the excentral triangle, i.e., A2 is the midpoint between the excenters Ib and Ic . It is also the midpoint of the arc BAC of the circumcircle. 2. Mixtilinear incircles The A-mixtilinear incircle is the circle (Oa ) that touches the rays AB and AC at Ca and Ba and the circumcircle (O) internally at X. See Figure 2. Define the B- and C-mixtilinear incircles (Ob ) and (Oc ) analogously, with points of tangency Y and Z with the circumcircle. See [4]. We begin with an alternative proof of the main result of [4]. Proposition 5. The lines AX, BY , CZ are concurrent at exs((O), (I)). Proof. Since A = exs((Oa ), (I)) and X = exs((O), (Oa )), the line AX passes through exs((O), (I)) by d’Alembert’s Theorem. For the same reason, BY and CZ also pass through the same point. A2 Y A A B1 B0 Ob Oc Z Ca O I C1 Ba C0 Ca Oa B C Ba I Oa A0 B X C X A1 Figure 2 Lemma 6. (1) I is the midpoint of Ba Ca . (2) The A-mixtilinear incircle has radius ra = (3) XI bisects angle BXC. See Figure 3. Figure 3 r cos2 A 2 . Consider the radical axis a of the mixtilinear incircles (Ob ) and (Oc ). Proposition 7. The radical axis a contains (1) the midpoint A1 of the arc BC of (O) not containing the vertex A, (2) the midpoint Ma of IA0 , where A0 is the point of tangency of the incircle with the side BC. the homothetic center of the triangles is the point J which divides the segment OI in . Let Mb and Mc be the midpoints of IB0 and IC0 respectively. L. Note that the triangles A1 B1 C1 and Ma Mb Mc are directly homothetic. Since A1 B1 C1 is inscribed in the circle O(R) and Ma Mb Mc in inscribed in the circle I(2r ). Ac and A1 are collinear. and that of (Oa ) and (Ob ) is the line C1 Mc . Theorem 8. Similarly. and is therefore a point on the radical axis of (Ob ) and (Oc ). Since Ab . Proof. C. This shows that A1 is on the radical axis of (Ob ) and (Oc ). the radical axis of (Ob ) and (Oc ) is the line A1 Ma .4 K. Z. the radical axis of (I) and (Oc ) passes through the midpoints of IA0 and IB0 . and the circle (Ob ) touches the same two lines at Cb and Ab . Nguyen and J. (Oc ) is the point J which divides OI in the ratio OJ : JI = 2R : −r. Salazar Y A B1 Bc Cb B0 C1 Ob Mb C0 Z Oc I Mc Ma B Ac Ab A0 C A1 Figure 4. the radical axis of these two circles is the line joining the midpoints of Cb C0 and Ab A0 . (Ob ). A1 . (2) Consider the incircle (I) and the B-mixtilinear incircle (Ob ) with common ex-tangents BA and BC. A1 Ac · A1 Z = A1 B 2 = A1 C 2 = A1 Ab · A1 Y . (1) By Lemma 3. I. By Proposition 7. Proof. The radical center of (Oa ). Also. so are Y . Ab . the radical axis of (I) and (Ob ) passes through the midpoints of IA0 and IC0 . Since the circle (I) touches BA and BC at C0 and A0 . Cb are collinear. Then the radical axes of (Oc ) and (Oa ) is the line B1 Mb . It follows that the midpoint of IA0 is the common point of these two radical axes. (I)). This is the triangle center X57 in [3]. Proof. where J is the radical center of the mixtilinear incircles. BY . the lines AX . Y . Theorem 10. Let T be the homothetic center of the excentral triangle Ia Ib Ic and the intouch triangle A0 B0 C0 . the ratio See Figure 5. The A-mixtilinear excircle (Oa ) can be easily constructed by noting that the polar of A passes through the excenter Ia . If the mixtilinear excircles touch the circumcircle at X . similarly for the other two mixtilinear excircles. Theorem 9. CZ are concurrent at ins((O). the reflection of I in O. The mixtilinear excircles The mixtilinear excircles are defined analogously to the mixtilinear incircles.On mixtilinear incircles and excircles 5 Y A B1 Bc Cb B0 C1 Ob Mb C0 Z Oc Mc J O I Ma B Ac Ab A0 C A1 Figure 5. Comparison with (3) shows that J is the reflection of T in O. Since the excentral triangle has circumcenter I . The polar of A with respect to (Oa ) passes through the excenter Ia . 2 (3) Remark. except that the tangencies with the circumcircle are external. . See Figure 6. Similarly for the other two polars of B with respect to (Ob ) and C with respect to (Oc ). r OJ : JI = R : − = 2R : −r. The radical center of the mixtilinear excircles is the reflection of J in O. OT : T I = 2R : −r. Z respectively. 3. Since B3 C3 is parallel to Ib Ic (both being perpendicular to the bisector AA1 ). C0 C3 respectively. Since B4 and C4 are the midpoints of the diagonals B0 B3 and C0 C3 . I. and A2 is the midpoint of Ib Ic . and the excentral triangle is the medial triangle of A3 B3 C3 . The ratio of homothety is clearly 4R−r R . This means that the line B4 C4 is the radical axis of (I) and (Oa ). Ib . the line B4 C4 also contains the midpoints of B0 Ba and C0 Ca . . Consider the isosceles trapezoid B0 C0 B3 C3 . Let A4 . which is the radical center of the mixtilinear excircles. which are on the radical axis of (I) and (Oa ). Let A3 B3 C3 be the triangle bounded by these three polars. Similarly. Salazar Ob A Oc Y Z I B C X Oa Figure 6. We claim that A4 B4 C4 is homothetic to A2 B2 C2 at a point J . This is homothetic to the intouch triangle A0 B0 C0 at r . L. Nguyen and J. Note also that I is the circumcenter of A3 B3 C3 (since it lies on the perpendicular bisectors of its three sides). Ca of the circle (Oa ) with AC and AB. with ratio of homothety − 4R If A4 is the midpoint of A0 A3 . and B3 C3 contains the points of tangency Ba . See Figure 7.6 K. B0 B3 . C. similarly for B4 and C4 . then A4 B4 C4 is homothetic to A3 B3 C3 with ratio 4R−r 4R . Since Ia Ib A3 Ic is a parallelogram. B4 . C4 be the midpoints of A0 A3 . it is also the midpoint of A3 Ia . Ic are the midpoints of C3 A3 and A3 B3 . Ia is the midpoint of B3 C3 . This is the radical center J of the mixtilinear incircles. Theorem 11 below gives further details of this circle. and similarly for B5 and C5 . Apollonian circles Consider the circle O5 (r5 ) tangent internally to the mixtilinear incircles at A5 . the homothetic center is the point J which divides IO in the ratio J I : J O = 4R − r : 2R. It follows that A4 is on the radical axis of (Ob ) and (Oc ). the ratio 2R : −r. The reflection of in O divides OI in Equivalently. Clearly. . B5 . and an easier construction. We call this the inner Apollonian circle of the mixtilinear incircles. See Figure 8. It can be constructed from J since A5 is the second intersection of the line JX with the A-mixtilinear incircle. and circumradii R and 4R−r 2 . Since these two triangles have circumcenters O and I. This means that the radical center of the mixtilinear excircles is the homothetic center of the triangles A2 B2 C2 and A4 B4 C4 . A2 also lies on the same radical axis.On mixtilinear incircles and excircles 7 A3 Ob A4 Ib A2 A Oc Ic Y Z B0 C0 I J C2 C4 B B2 C A0 C3 X B4 Ba Ia Ca Oa B3 Figure 7. 4. OJ J I (4) J : = −2R : 4R − r. C5 respectively. Qa . which is common to AA5 . (5) OI 2R − r Since P = exs((O5 ). Qa . (I)). A5 from the circles (O5 ). X ). (Oa )). (2) A. and R + r5 OO5 = . Thus. (I). L. (Oa ). (3) A. It follows that OJ : JI : JO5 = 2R : −r : −2r5 . (1) Triangles A5 B5 C5 and ABC are perspective at ins((O). (Oa ). (6) OI R+r . (I)) are the same point. (Oa ). and 2(R − r5 ) OO5 = . ins((O). From Theorem 8. X . (1) Let P = exs((O5 ). It follows that P and ins((O). (O5 )). C. and Qa = exs((O5 ). (O5 )). (4) A. BB5 and CC5 . (I)). Nguyen and J. (I). For the same reason. (Oa ). we have OJ : JO5 = R : −r5 . (Oa ). OP : P O5 : P I = R : r5 : r. (2) The inner Apollonian circle of the mixtilinear incircles has center O5 divid3Rr . it is also ins((O). the lines BB5 and CC5 contain the same two points. (I)) = ins((O). (I)) from the circles (O). Theorem 11. See Figure 9. (I)) (along with Qa . P from the circles (O5 ). (I). A5 . Therefore the lines AA5 contains the points P and ins((O). OJ : JI = 2R : −r. P from the circles (O5 ). (2) Now we compute the radius r5 of the circle (O5 ). Salazar Y A A5 O Z O5 J C5 B5 B C X Figure 8.8 K. ing the segment OI in the ratio 4R : r and radius r5 = 4R+r Proof. (I)). As J = exs((O). The following triples of points are collinear by d’Alembert’s Theorem: (1) A. 4R 4R+r and O5 divides OI in the ratio OO5 : O5 I = 4R : r. If the lines J X .On mixtilinear incircles and excircles 9 A Qa A5 P I O5 Oa C B X X Oa Figure 9. (1) Triangles A6 B6 C6 and ABC are perspective at exs((O). C6 respectively. then the circle A6 B6 C6 is tangent internally to each of the mixtilinear excircles. J Y . ing the segment OI in the ratio −4R : 4R + r and radius r6 = R(4R−3r) r . Consequently. B6 . 3Rr Comparing (5) and (6). we easily obtain r5 = 4R+r . (2) The outer Apollonian circle of the mixtilinear excircles has center O6 divid. Theorem 12 below gives an easier construction without locating the radical center. J Z intersect the mixtilinear excircles again at A6 . Theorem 12. OO5 OI = The outer Apollonian circle of the mixtilinear excircles can also be constructed easily. (I)). For the same reason BB6 and CC6 pass through the same point. (Ob )) on the radical axis of (O) and (O6 ). and Oa Ob concur at S = exs((Oa ). we conclude that X A6 Y B6 is cyclic. A B is parallel to A6 B6 . It follows . (O6 )). Y B6 C6 Z is cyclic. For the same reason. and the center of homothety is P = exs((O). and P I : P O = r : R. and KX ·KA6 = KY · KB6 . B C and C A are parallel to B6 C6 and C6 A6 respectively. C. CC6 intersect the circumcircle (O) at A . Now: SA · SB = SX · SY = SA6 · SB6 . (O6 )) = exs((I). (I)). Nguyen and J. BB6 . (O)). Let AA6 . by d’Alembert’s Theorem. the line A6 X passes through K = ins((O). (Oa )). AB. (O6 )). (1) Since A6 = exs((Oa ). We claim that K is the radical center J of the mixtilinear excircles. X Y . B . (Oa )) and X = ins((Oa ).10 K. Proof. A6 B6 . (I)). ABA6 B6 is cyclic. Since ∠BAA6 = ∠BB A = ∠BB6 A6 . (O6 )). B6 Y and C6 Z pass through the same point K. Also. the triangles A6 B6 C6 and ABC are perspective at P = exs((I). Therefore the triangles A B C and A6 B6 C6 are directly homothetic. C respectively. Similarly. and KY · KB6 = KZ · KC6 . it is also exs((O). (2) Since A6 = exs((O6 ). Since P = exs((O). See Figure 10. Since SX ·SY = SA6 ·SB6 . L. Salazar B6 Ob C6 A Oc Z C B Y O6 J P B C A X Oa A6 Figure 10. By Proposition 2. (O6 )) and A = exs((I). Since SA · SB = SA6 · SB6 . (O6 )). by d’Alembert’s Theorem. the line AA6 passes through P = exs((O6 ). showing that K = ins((O). By Pascal’s Theorem for the six points. then both Q and S lie on the line connecting the intersections XY ∩ X Y and Y Z ∩ Y Z. A simple application of Ceva’s Theorem shows that the AX . OJ : IO =2R : 2R − r. Then. J O : J O6 = R : −r6 . The intersection point of these three lines is called the cyclocevian conjugate of P . Construct the circle through X. The same reasoning shows that if A B C and X Y Z are perspective at a point S. For example. CA. 2R−r Corollary 13. Clearly. Similarly. Since OJ + J O6 = OO6 . p. (O6 )) is the radical center J of the mixtilinear excircles. The cyclocevian conjugate Let P be a point in the plane of triangle ABC. Hence. the intersections of the lines X A and Y B lies on the line connecting Q to the intersection of XY and X Y . Note also P O : P O6 = R : r6 . Let Q be a point on the line P P ◦ . BY . OO6 : IO =r6 − R : R − r. Y . Y .On mixtilinear incircles and excircles 11 that KX · KA6 = KY · KB6 = KZ · KC6 . Y . 5. (P ◦ )◦ = P . Z . CZ are concurrent. We prove two interesting locus theorems. Since Q lies on P P ◦ . the centroid and the orthocenters are cyclocevian conjugates. X . B . r Since K = ins((O). it follows that X A . Y . the perspector is also on the same line. We denote this point by P◦ . Z on the sidelines BC. lies on P P ◦ . we have the following relations. Proof. which is the line P P ◦ . IO5 · IO6 = IO2 . we obtain O6 = and J = Remark. CA. AB respectively. r r6 ·O+R·O6 R+r6 (4R+r)O−4R·I . 2R − r 2R − r R−r This gives: r6 = (4R−r)O−2R·I 2R−r R(4R−3r) . (O6 )) = are the same point. For Q on P P ◦ . which according to Pappus’ theorem (for Y . for example. we have 2r6 r6 − R 2R + = . See. with traces X. [2. Z. Y . A . . and Gergonne point is the cyclocevian conjugate of itself. X. Theorem 14. Y ). Y . A B C and X Y Z are perspective at a point on P P ◦ . AB again at points X . Z and C. J O6 : IO =2r6 : 2R − r. Z. and P P ◦ are concurrent. Y B . The radical circle of the mixtilinear excircles has center J and radius R (4R + r)(4R − 3r). This circle intersects the sidelines BC. The locus of Q whose circumcevian triangle with respect to XY Z is perspective to X Y Z is the line P P ◦ .226]. = sin A AC sin C CB sin B BA Therefore. Some further results We establish some further results on the mixtilinear incircles and excircles relating to points on the line OI. By Theorem 14. For example. Theorem 15. then the circumcevian triangle of O (with respect to the medial triangle) is perspective with the orthic triangle at the nine-point center N . Nguyen and J. 6. A B C is perspective with ABC if and only if it is perspective with X Y Z . Salazar A A Y Y Z Z S P O P Q ◦ C B B X X C Figure 11. First note that sin Z ZA sin A AY sin ZAA sin Z X A sin Z XA = = · · sin A X Y sin A Y Y sin ZAA sin A Y Y sin A AY sin Y XA sin BAA AA A Y sin BAX = · . if P = G. If Q = O. the locus of Q is the line P P ◦ . The locus of Q whose circumcevian triangle with respect to XY Z is perspective to ABC is the line P P ◦ . .12 K. then P ◦ = H. C. · · = ZA AA sin XAC sin A XZ sin A AC It follows that sin Z X A sin X Y B sin Y Z C · · sin A X Y sin B Y Z sin C Z X sin BAA sin ACC sin CBB sin Y XA sin ZY B sin XZC · · · · = sin A XZ sin B Y X sin C ZY sin A AC sin C CB sin B BA sin BAA sin ACC sin CBB · · . Proof. L. The line P P ◦ is the Euler line. the circumcenter. and B1 C2 ∩ B2 C1 are collinear. Proposition 18. We first show that D = B1 Y ∩C1 Z lies on the line AA1 .On mixtilinear incircles and excircles 13 A A Y Y Z O Z P B P S Q ◦ C B X X C Figure 12. if P = ins((O). The line OI is the locus of Q whose circumcevian triangle with respect to A1 B1 C1 (or XY Z) is perspective with DEF . (2) with X Y Z at the centroid of the excentral triangle. Y . In particular. The triangle A2 B2 C2 is perspective (1) with XY Z at the incenter I. A1 B1 C1 is the circumcevian triangle of I with respect to triangle DEF . Theorem 16. Corollary 17. Applying Pascal’s Theorem to the six points Z. I = B2 Y ∩ C2 Z. the triangle DEF is perspective with A1 B1 C1 at I. C1 . if E = C1 Z∩A1 X and F = A1 X∩B1 Y . . then the perspector Q divides the same segment in the ratio OQ : QI = (1 + t)R : −2tr. Since B1 C2 and B2 C1 are parallel to the bisector AA1 . Remark. B1 . the locus of P whose circumcevian triangle with respect to A1 B1 C1 is perspective with XY Z is a line through I. C2 on the circumcircle. The line OI is the locus of P whose circumcevian triangle with respect to A1 B1 C1 is perspective with XY Z. Triangle XY Z is formed by the second intersections of the circumcircle of A1 B1 C1 with the side lines of DEF . Now. B2 . If P divides OI in the ratio OP : P I = t : 1 − t. By Theorem 14. since O is one such point. the points D = B1 Y ∩ C1 Z. (I)). See Figure 13. Equivalently. the homothetic center of the excentral and intouch triangles. this perspector is T . (The circumcevian triangle of O with respect to A1 B1 C1 is perspective with XY Z at I). Proof. This is indeed the line IO. it follows that D lies on AA1 . Theorem 19. Proof.14 K. The last column gives the indexing of the triangle centers in [2. 3]. the radical axis of the mixtilinear incircles (Ob ) and (Oc ). L. Remark. This point divides OI in the ratio 4R − r : −4r and has homogeneous barycentric coordinates a(b + c − 5a) b(c + a − 5b) c(a + b − 5c) : : . The triangles A7 B7 C7 and XY Z are perspective at a point on the line OI. . Let A7 be the second intersection of the circumcircle with the line a . X and Ia are collinear. and it intersects B2 C2 at its midpoint X . In other words. Therefore X Ia is a median of triangle X Ba Ca . the excenter Ia is the midpoint Ba Ca . Nguyen and J. See Figure 14. Summary We summarize the triangle centers on the OI-line associated with mixtilinear incircles and excircles by listing. A2 . So do B2 Y and C2 Z. (1) follows from Lemma 6(b). hence the centroid. the line A2 X contains a median. Since A2 B2 Ia C2 is a parallelogram. X . (2) Referring to Figure 7. of the excentral triangle. for various values of t. Similarly define B7 and C7 . Salazar D A2 Y A B1 Ob C1 Oc Z I C2 B B2 C A1 Figure 13. the points which divide OI in the ratio R : tr. C. b+c−a c+a−b a+b−c 7. (I)) X56 s−a perspector of ABC and XY Z perspector of ABC and A6 B6 C6 a homothetic center of excentral X57 s−a and intouch triangles a2 (b2 + c2 − a2 − 4bc) radical center of mixtilinear X999 incircles a2 (b2 + c2 − a2 + 8bc) center of Apollonian circle of mixtilinear incircles a2 f (a. c) radical center of mixtilinear excircles a2 g(a.On mixtilinear incircles and excircles 15 Y A A7 B1 Ob C1 Oc O Z J Oa B7 B C C7 X A1 Figure 14. b. c) center of Apollonian circle of mixtilinear excircles a(3a2 − 2a(b + c) − (b − c)2 ) centroid of excentral triangle X165 a(b+c−5a) perspector of A7 B7 C7 and XY Z b+c−a The functions f and g are given by . b. t 1 −1 2R − 2R+r − 12 1 4 − 4R−r 2r − 4R+r 4r − 4R r 4R − 4R−r first barycentric coordinate a2 (s − a) Xn X55 ins((O). (I)) perspector of ABC and X Y Z perspector of ABC and A5 B5 C5 a2 exs((O). unpublished manuscript. 1991. Triangle centers and central triangles. [5] P. 106 (1999) 952–955. Math. g(a. L. 1998. Mendoza. available at http://faculty.com . Kimberling. Nguyen and J. c) =a5 − a4 (b + c) − 2a3 (b2 − bc + c2 ) + 2a2 (b + c)(b2 − 5bc + c2 ) + a(b4 − 2b3 c + 18b2 c2 − 2bc3 + c4 ) − (b − c)2 (b + c)(b2 − 8bc + c2 ).16 K. Mixtilinear incircles. Texas. Urb. Amer. Houston.. Estado Bol´ıvar. Puerto Ordaz 8015. C. Kimberling. 1920. Gabay reprint. Yiu.. 6th ed. Venezuela E-mail address: caisersal@yahoo. c) =a4 − 2a3 (b + c) + 10a2 bc + 2a(b + c)(b2 − 4bc + c2 ) − (b − c)2 (b2 + 4bc + c2 ). Encyclopedia of Triangle Centers.html. References [1] F.edu/ck6/encyclopedia/ETC.evansville. Paris. Mixtilinear incircles II. USA E-mail address: treegoner@yahoo. b. Khoa Lu Nguyen: 806 Candler Dr. G.com Juan Carlos Salazar: Calle Matur´ın No C 19. Exercices de G´eom´etrie. Salazar f (a. 129 (1998) 1–285. Monthly. 77037-4122. [4] P. b. Yiu. [3] C..-M. [2] C. Congressus Numerantium. has a prescribed Euler line.b Forum Geometricorum Volume 6 (2006) 17–23. The authors are extremely grateful to P. They are the points x y −x2 + 1 . y). . Paula Manuel. The Euler conic Given two points A and B and a real number m. We show that this locus is a conic through A and B. and Paulo Semi˜ao Abstract. Yiu for his help in the preparation of this paper. Without loss of generality. with A and B fixed. 1. we assume a Cartesian coordinate system in which A = (−1. its slope is given by: −3x2 − y 2 + 3 . for example. b b FORUM GEOM ISSN 1534-1178 A Conic Associated with Euler Lines Juan Rodr´ıguez. [2]. we study the locus of a point C for which the Euler line of triangle ABC has slope m. and with center (if it exists) at the midpoint of AB. . The main properties of such an Euler conic are described. (3) . y = 0. GH = 3 3y (2) is parallel to the Euler line. Communicating Editor: Paul Yiu. and write C = (x. The vector −−→ 2x −3x2 − y 2 + 3 . We also give a construction of a point C for which triangle ABC. When the Euler line is not vertical. the coordinates of the vertex C satisfy the equation 3x2 + 2mxy + y 2 = 3. We study the locus of a point C for which the Euler line of triangle ABC with given A and B has a given slope m. m= 2xy Therefore. The centroid G and the orthocenter H of a triangle can be determined from the coordinates of its vertices. This is a conic through A and B. and H = x. 0). Publication Date: January 24. x. 2006. 0) and B = (1. (1) G= 3 3 y See. Semi˜ao This clearly represents a conic. . if m = ± 3. m 1 The characteristic polynomial being λ2 − 4λ − (m2 − 3). We call this the Euler conic associated with A. respectively. the orthocenter. Suppose m2 < 3. 0. eigenvector eigenvalue √ √ 2 (√m + 1 + 1. or √ (2) (2 + √m2 + 1)(x cos α − y sin α)2 + (2 − m2 + 1)(x sin α + y cos α)2 = 3. . with A or B. if m = ± 3. It clearly has center at the origin. equation (3) can be rewritten in the form √ (1) (mx √ + y)2 = 3. the centroid. Classification of the Euler conic The Euler conic is an ellipse or a hyperbola according as m2 < 3 or m2 > 3. a triangle ABC with C on the ellipse. (1) The pairs (±1. and the circumcenter coincide. Manuel and P. It degenerates into a pair of straight lines when m2 = 3. the midpoint M of AB. (2) The pairs 0. =. itis not√possible to define the triangle ABC.18 J. or −1 according as m >. 0) are always solutions of (3) and correspond to the singular cases in which the vertex C coincides. ± −sgn(m) · 2 3−m 3 − m2 where sgn(m) = +1. Figure 1 shows the Euler ellipse for m = 34 . Proposition 1. −m) 2 − m2 + 1 Thus. or < 0. sin α = . Rodr´ıguez. ε= √ m2 + 1 + 2 The foci are the points √ √ 2 2 3( m + 1 − 1) 3( m + 1 + 1) . and consequently. where √ √ m2 + 1 + 1 m2 + 1 − 1 √ √ cos α = . B and slope m. m) 2 + √ m2 + 1 ( m2 + 1 − 1. Its axes are the eigenvectors of the matrix 3 m . P. ± 3 are also solutions of (3) and correspond to the trivial case when the triangle ABC is equilateral. 2 m2 + 1 2 m2 + 1 Remarks. and its Euler line of slope m. In this case. The Euler conic is an ellipse with eccentricity √ 2 m2 + 1 . 2. its eigenvectors with corresponding eigenvalues are as follows. Suppose m2 > 3.A conics associated with Euler lines 19 C H O A M B Figure 1 Proposition 2. The Euler conic is a hyperbola with eccentricity √ 2 m2 + 1 . . ± sgn(m) · 2 m −3 m2 − 3 where sgn(m) = +1 or −1 according as m > 0 or < 0. ε= √ m2 + 1 − 2 The foci are the points √ √ 2 2 3( m + 1 + 1) 3( m + 1 − 1) . The asymptotes are the lines . the Euler conic degenerates into a pair of parallel lines. and for √ m = − 3 in Figures 4A and 4B. The slope of the Euler line of the triangle ABC is √ m = ± 3. and its Euler line of slope m. if and only if. √ Examples of triangles for m = 3 are shown in Figures 3A and 3B. Corollary 3. one of angles A and B is 60◦ or 120◦ . √ When |m| = 3. y = (−m ± m2 − 3)x. whose equations are: √ y = −mx ± 3. a triangle ABC with C on the hyperbola. Figure 2 shows the Euler hyperbola for m = 12 5 . . P. Manuel and P.20 J. Rodr´ıguez. Semi˜ao C O H A B M Figure 2 C O H C O A B M A B M H Figure 3A Figure 3B C O H C O A M B A M B H Figure 4A Figure 4B . it is the origin of the Cartesian system. Lemma 4. It can therefore be constructed as the intersection of this line with the circle (4). the Euler line is the perpendicular bisector of AB. 0) and B = (1. The circumcenter is the intersection of the Euler line with the line x = 0. It therefore has an equation of the form y = mx + k. We shall henceforth assume that the Euler line is not perpendicular to AB. C lies on the line y = mx + 3k. If G is the centroid. (4) Let M be the midpoint of AB. the perpendicular bisector of AB. Triangles with given Euler line In this section we find the cartesian coordinates of the third vertex C in order that a given line be the Euler line of the triangle ABC with vertices A = (−1. k). Since G lies on the line y = mx + k. The Euler line of triangle ABC is perpendicular to AB if and only if AB = AC. The circumcircle is x2 + (y − k)2 = k2 + 1 or x2 + y 2 − 2ky − 1 = 0. It is the point O = (0. the vertex C is such that M C : M G = 3 : 1. y = mx + 3k y = mx + k C O C G G A H M H Figure 5 B . In this case. 0).A conics associated with Euler lines 21 3. In the hyperbolic and degenerate cases m2 ≥ 3.22 J. The lines y = mx + 3k are tangent to the Euler ellipse (3) at the points 3 + m2 −2m . m2 < 3. or > −4. Rodr´ıguez. Semi˜ao Proposition 5. P. In the elliptic case. The number of points C for which triangle ABC has Euler line y = mx + k is 0. . 1. =. or 2 according as (m2 − 3)(k2 + 1) <. For m2 < 3 and k = ± m . There are two such triangles if and only if 2 +1 . there are always two such triangles. k2 < m 3−m2 2 +1 Corollary 6. Manuel and P. there is a unique triangle ABC 3−m2 whose Euler line is the line y = mx + k. ± . . the midpoint of AB. C O H A M Figure 6 B . (m2 + 1)(3 − m2 ) (m2 + 1)(3 − m2 ) 2 +1 . The other one Figure 6 shows the configuration corresponding to k = m 3−m2 can be obtained by reflection in M . Nauka. Univ. 1994. Fac. Quarteira. Univ. Postnikov.A conics associated with Euler lines 23 References [1] A. de Matem´atica. Fac.a Laura Ayres. do Algarve. [2] M. 1987. Portugal E-mail address:
[email protected] Paula Manuel: Escola Secund´aria/3 Dr. Lectures in Geometry. Faro. Portugal E-mail address: paula manuel@sapo. de Ciˆencias e Tecnologia. de Ciˆencias e Tecnologia.pt Paulo Semi˜ao: Dep. Faro. Analytic Geometry. Mir publishers. Juan Rodr´ıguez: Dep. do Algarve. Pogorelov. de Matem´atica.pt . Portugal E-mail address: psemiao@ualg. . call this pair the ODF-lines through P with respect to ABC. and the line through P . the DF-lines through P are the pairs of conjugate lines in the involution I determined by the lines through P that are tangent to the parabolas of the pencil of parabolas inscribed in ABC. l ) is a pair of DF-lines if and only if l and l are tangent lines of a parabola inscribed in ABC (see also [5]). Erhmann and F. Thus. P B and the line through P parallel with CA (and P C and the line trough P parallel with AB) are also conjugate Publication Date: January 30. and that the anticomplementary triangle A B C of ABC is the triangle whose medial triangle is ABC ([4]). CA in Y and Y . l ) is a pair of conjugate diameters of the conic CP with center P . Proof. circumscribed at the anticomplementary triangle A B C of ABC. and (P C. the ODF-lines through P are the axes of this conic. there passes just one orthogonal pair of DF-lines with respect to ABC. A pair (l. Through a general point P . and AB. 2006. parallel with BC. Y Y . . Considering the tangent lines through P at the three degenerate inscribed parabolas of ABC. We give a simple characterization of the Droz-Farny pairs of lines through a point of the plane. CA. Communicating Editor: J. Chris Fisher. Theorem. are conjugate diameters of the conic CP . Since A is the midpoint of B C . Recall that the medial triangle A B C of ABC is the triangle whose vertices are the midpoints of BC. b b FORUM GEOM ISSN 1534-1178 A Note on the Droz-Farny Theorem Charles Thas Abstract. and AB in Z and Z in such a way that the midpoints of the segments XX . it also follows that (P A. In [3] J-P. it follows immediately that P A. (P B. if and only if (l. and B C is parallel with BC. In particular. van Lamoen prove a projective generalization of the Droz-Farny line theorem. line through P parallel with BC).b Forum Geometricorum Volume 6 (2006) 25–28. line parallel through P with AB). They say that a pair of lines (l. l ) of lines is a pair of DF-lines through P with respect to ABC. They then prove that (l. and ZZ are collinear. l ) is a pair of DF-lines through a point P with respect to a given triangle ABC if they intercept the line BC in the points X and X . are three conjugate pairs of lines of the involution I. line parallel through P with CA). In the same way. which is a point of the nine-point circle of ABC (the center of any circumscribed rectangular hyperbola is on the nine-point circle). H. where IA . . associated with ABC. C. H. Its axes are the ODF-lines through X115 with respect to the medial triangle A B C . Thas diameters of CP .. It is homothetic to (and has the same axes of) the Steiner ellipses of the medial triangle A B C and of the anticomplementary triangle A B C of ABC. It has center X115 with trilinear coordinates (bc(b2 − c2 )2 . with respect to the medial triangle A B C of ABC. . and its center is the focus of the Kiepert parabola (inscribed parabola with directrix the Euler line of ABC). and to A B C ). trilinear coordinates (bc(b − c)2 (b + c − a). and its center is Kimberling center X125 with trilinear coordinates (bc(b2 +c2 −a2 )(b2 −c2 )2 . which is Kimberling . the incenter I of ABC. C. the circumcenter O. The Stammler hyperbola of ABC has trilinear equation (b2 − c2 )x21 + (c2 − a2 )x22 + (a2 − b2 )x23 = 0. . IC .. It is also the Feuerbach hyperbola of the tangential triangle of ABC. B. Since two pairs of corresponding lines determine an involution. we can characterize the axes of any circumscribed ellipse or hyperbola of ABC as the ODF-lines through its center with regard to the medial triangle A B C of ABC. 3. The Jerabek hyperbola of ABC (the isogonal conjugate of the Euler line of ABC) is the rectangular hyperbola through A.... The axes of this hyperbola are the ODF-lines through X125 . The Kiepert hyperbola of ABC is the rectangular hyperbola through A. Since any two orthogonal diameters of a circle are conjugate. The Steiner ellipse of ABC is the circumscribed ellipse with center the centroid G of ABC. and through the Spieker center (the incenter of the medial triangle of ABC). As a corollary of this theorem. . and the symmedian point K. 4. the centroid G of ABC. The Feuerbach hyperbola is the rectangular hyperbola through A. . the excenters IA ... the orthocenter H. It is the rectangular hyperbola through the incenter I. 2.. we find by this special case the classical Droz-Farny theorem: Perpendicular lines through H are DF-pairs with respect to triangle ABC. IB . Its axes are the ODF-lines through F ..)) . with respect to the medial triangle A B C of ABC. IB . C.26 C. Examples 1. Remark that the orthocenter H of ABC is the center of the circumcircle of the anticomplementary triangle A B C . B. this completes the proof. with center the Feuerbach point F (at which the incircle and the nine-point circle are tangent.) on the nine-point circle.). IC are the excenters of ABC). And in the same way we can construct the axes of any circumscribed ellipse or hyperbola of any triangle. These axes are the ODF-lines through G with respect to ABC (and to A B C . 5. . the circumcenter O and the Lemoine (or symmedian) point K of ABC. the Mittenpunkt (the symmedian point of the excentral triangle IA IB IC . B. R ) determine an involution I on C. which is often called the Wallace or the Steiner hyperbola. respectively). ac. determined by the conjugate pairs (P A. C ∩ P B = R.. The conic with trilinear equation a2 (b2 − c2 )x21 + b2 (c2 − a2 )x22 + c2 (a2 − b2 )x23 = 0 is the rectangular hyperbola through the incenter I. C ∩ lb = R . with regard to the medial triangle of IA IB IC .A note on the Droz-Farny theorem 27 a center X110 with trilinear coordinates (b2 −c 2 . ab). and C) of the conic through A. It is also circumscribed to the anticomplementary triangle A B C (recall that the trilinear coordinates of A . we give a construction for the ODF-lines through a point P with respect to the triangle ABC.. and through the centroid G of ABC. C ∩ la = Q . on the circumcircle of ABC. Each line through T intersect the circle C in two conjugate points of I . Remark that center X110 is the fourth common point (apart from A. . ac. (1) A biographical note on Arnold Droz-Farny can be found in [1]. It follows that the axes of this hyperbola. and C are (−bc. . line la through P parallel to BC) and (P B. the line through T and through the center of C intersects C in two points S and S . ca . The axes of this Stammler hyperbola are the ODF-lines through X110 . i. (2) A generalization of the Droz-Farny theorem in the three-dimensional Euclidean space was given in an article by J. C. In particular.e. which is the nine-point circle of the excentral triangle IA IB IC . −ab). and the circumcircle of ABC. (3) Finally. Remarks. . with center QQ ∩RR = T . IB . and with center the symmedian point K of ABC. IC . B. 6. a point of interSteiner point X99 with trilinear coordinates (b2bc −c2 c2 −a2 a2 −b2 section of the Steiner ellipse and the circumcircle of ABC. −ac. which has trilinear equation a(−a2 + b2 + c2 )x2 x3 + b(a2 − b2 + c2 )x3 x1 + c(a2 + b2 − c2 )x1 x2 = 0. (bc. ab ). for the orthogonal conjugate pair of lines through P of the involution I in the pencil of lines through P . Bilo [2]. Remark that the circumcircle of ABC is the nine-point circle of A B C and also of IA IB IC . through the excenters IA . and (bc. are the ODF-lines through the Steiner point X99 with regard to ABC.. line lb through P parallel to CA): intersect a circle C through P with these conjugate pairs: C ∩ P A = Q. B . ab). Its center is the . then (Q. and also with regard to the medial triangle of the excentral triangle IA IB IC . B.). Q ) and (R.. such that P S and P S are the orthogonal conjugate pair of lines of the involution I. 4 (2004).be . Charles Thas: Department of Pure Mathematics and Computer Algebra. Thas. [4] C. On ellipses with center the Lemoine point and generalizations. Ehrmann and F. 1. Krijgslaan 281 . Forum Geom. Forum Geom.28 C. Ayme. Belgium E-mail address: pt@cage. Mathesis. 4 (2004). Bilo. A purely synthetic proof of the Droz-Farny line theorem. 1 – 7.S22. [3] J-P. Congressus Numerantium. A projective generalization of the Droz-Farny line theorem. 255 – 259. 225 – 227. 56 (1947).ugent. Nieuw Archief voor Wiskunde... 219 – 224. Kimberling. 129 (1998) 1 – 285. [5] C. University of Ghent. M. Thas References [1] J-L. Triangle centers and central triangles. ser 4. [2] J. nr. 11 (1993). Generalisations au tetraedre d’une demonstration du theoreme de Noyer-Droz-Farny. van Lamoen. B-9000 Ghent. b are also homothetic and two adjacent flanks. E D K A L J F C G B H I Figure 1. This construction was used in a recent simple proof. I drew in Figure 2 these sixteen quadrilaterals together with their names. To every cyclic quadrilateral corresponds naturally a complex of sixteen cyclic quadrilaterals. The radical axes of the various pairs of circumcircles. though. but with reversed orientation. More precisely. 1. b b FORUM GEOM ISSN 1534-1178 On the Cyclic Complex of a Cyclic Quadrilateral Paris Pamfilos Abstract. bottom-flank b = BIJA and left-flank l = DKLA of q respectively. right-flank r = CGHB. To spare words. t. resulting by constructing other quadrangles on the sides of q. considered to be basic in the study of the whole complex. Besides. CDEF : top-flank of ABCD The sides of the new align with those of the old. from their construction. t are antihomothetic with respect Publication Date: February 6. Here we consider a simple figure. gives the previous basic Figure 1. All these quadrilaterals are cyclic and share the same angles with q. Introduction Consider a generic convex cyclic quadrilateral q = ABCD. flanks l. like r. The principle is to construct the quadrilateral q = CDEF . only the main flanks are similar to q. the various circumcenters and anticenters combine to interesting configurations. on a side of and similar to q. there are some other flanks. It seems though that the figure is interesting for its own. repeating the procedure three more times with the other sides of q. r are homothetic. Communicating Editor: Paul Yiu.b Forum Geometricorum Volume 6 (2006) 29–46. created by the extensions of the sides of q and the extensions of sides of its four main flanks. by Antreas Varverakis [1]. For convenience I call the quadrilaterals: top-flank t = CDEF . 2006. In addition to these main flanks. similar to q. . of the minimal area property of cyclic quadrilaterals. Here are studied some of these. In general. Later create the big flank denoted below by q∗ . rt) = (q. ltr lt t rt lqr l q r lbr lb b rb q* tlb tqb trb Figure 2. Not all of them are different though. a pair of symbols in parentheses. the total number of radical axes. and the symbol in brackets. will denote the circumcenter. Each side coincides with the radical axis of four pairs of circles of the complex. like (rt. [rt]. Pamfilos to their common vertex. Intersections of lines of the complex . will denote the radical axis of the circumcircles of the corresponding flanks. here C. For example the radical axes (t. Radical Axes By taking all pairs of circles. rb) = (tqb. (rt). rb). The same happens with every side out of the eight involved in the complex. Finally. The cyclic complex of q 2. will denote the circumcircle. will denote the anticenter of the corresponding flank. appears to be 120. rqb) coincide with line BC. In order to study other identifications of radical axes we need the following: U D V G X B J X* W Y Z U* Figure 3. The symbol in braces.30 P. involved. the symbol in parentheses. r) = (b. {rt}. The various sides are radical axes of appropriate pairs of circles and there are lots of coincidences. I call this figure the cyclic complex associated to the cyclic quadrilateral. Analogously t. Indeed. Y. V. The first of the next two figures shows the centers of the small flanks. from the intersection of these two circles with {lb} we see that Y is on their radical axis. tqb). b are similar and their similarity center is U . and certain parallelograms created by them. X ∗ are intersections of opposite sides of q∗. V. (3) Line XY Z coincides with (lbr.On the cyclic complex of a cyclic quadrilateral 31 Lemma 1. Centers The centers of the cyclic complex form various parallelograms. q) = (trb. Other radical axes Proposition 2. r are similar and their similarity center is X. Z are intersections of opposite sides of other flanks of the complex. b). l. This implies easily the properties of the lemma. lb) = (t. b) = (lqr. The proof is a trivial consequence of the similarity of opposite located main flanks. U ∗ . The second gives a panorama of all the sixteen centers together with a parallelogramic pattern created by them. Referring to Figure 3. Referring to Figure 4. . V G U D W {rb} B {lbr} {b} J X Y Z Figure 4. l) = (tqb. (2) Line U U ∗ coincides with (lt. W. rb) = (ltr. The following lines are common radical axes of the circle pairs: (1) Line XX ∗ coincides with (lt. trb). W are aligned on two parallel lines. lbr). Y. t) = (q∗ . Besides triangles XDY . Thus. (4) Line U V W coincides with (tlb. V DU are anti-homothetic with respect to D and triangles XGZ. Z and U. Points X. lqr). I show that line XY Z is identical with (lbr. from the intersection of the two circles {lbr} and {b} with circle {rb} we see that Z is on their radical axis. 3. Hence line ZY X coincides with the radical axis (lbr. U are intersections of opposite sides of q. Similarly. r) = (lb. q) = (ltr. W JU are similar to the previous two and also anti-homothetic with respect to B. points X. rb) = (tlb. The other statements are proved analogously. Namely those that have sides the medial lines of the sides of the flanks. b) = (rt. rt) = (l. b). r) = (q∗ . implying the equality of horizontal sides of the two left parallelograms. (lqr)(r). The parallelity of the other sides is proved analogously. in the first figure. (rb)(b)(t)(lt). In the second figure (lt)(t). all. (l)(q)(b)(rb) and (lqr)(r)(rb)(lbr). Since the labeling is arbitrary. (tqb)(q∗ ) is orthogonal to the axis U V W . The proof of the various parallelities is a consequence of the coincidences of radical axes. In addition. the centers of the flanks build equal parallelograms with parallel sides: (lt)(t)(tqb)(tlb). Referring to Figure 5. The parallelogramic pattern is symmetric with respect to the middle M of segment (q)(q∗) and the centers (tqb). The symmetry about M is a simple consequence of the previous considerations. is parallel to XY Z. from the parallelograms. (q). (ltr)(rt)(trb)(q∗). An importand case. after lemma-1. (lqr). Recall that the anticenter is the symmetric of the circumcenter with respect to the centroid of the quadrilateral. as. and the previous remarks imply that all parallelograms shown are equal. (q)(r). (b)(rb) are parallel because. (lbr)(rb) are orthogonal to A∗ D∗ . (lqr) are collinear. which. The equality of the parallelograms results by considering other implied parallelograms. For example. Similarly (ltr)(rt). See also Court [3] and the miscelanea (remark after Proposition 11) below. (l)(q). There are other interesting quadrilaterals with vertices at the centers of the flanks. (tlb)(tqb). . Characteristically it is the common intersection point of the orthogonals from the middles of the sides to their opposites. (q∗ ). Pamfilos B* (lt) (l) (rb) (t) (q) (b) (t) (ltr) D (tqb) C (tlb) (q) (l) (q*) (lqr) A B (b) (rb) (lbr) (lt) (rt) (r) (rb) C* A* (rt) (trb) (r) (rb) D* Figure 5. hence the collinearity. (q). Panorama of centers of flanks Proposition 3. related directly to q.32 P. in the second figure. (rb)(b) are all orthogonal to AD. any main flank can be considered to be the left flank of the complex. coinciding with line BC. (q∗ ). follows that the lengths are equal: |(tqb)(q)|=|(q∗ )(lqr)|. Some properties of the anticenter are discussed in Honsberger [2]. for example. sides (t)(rt). are orthogonal to the corresponding radical axis. Both lines (tqb)(q) and (q ∗ )(lqr) are orthogonal to the axis XY Z of the previous paragraph. (q∗)(trb). is that of the collinearity of the centers (tqb). For example the next proposition relates the centers of the main flanks to the anticenter of the original quadrilateral q. Then repeat the procedure and continuing that way fill all the area of the angle AQD with flanks. One could construct further flanks. (1) follows from the fact that triangles F T C and CSB are similar isosceli. All these having alternatively their anticenters and . since the trapezia DAHG and CBLK are similar and inversely oriented with respect to the sides of the angle AQD. Thus. passing through Q.On the cyclic complex of a cyclic quadrilateral 33 R E F T=(tf) D K U=(lf) Q L A G C S=(rf) M O B H V=(bf) J I Figure 6. tightly related to the anticenters of q and its left and right flanks (Figure 7). it coincides with their intersection. Referring to the previous figure. coincides with the anticenter M of q. formed by the centers of the main flanks. M. One sees easily that triangles U XM and M Y S are similar and points U. which is a diagonal of the quadrangle with vertices at the anticenters of the flanks. The same is true for lines QX and QY . consider the parallelograms p = OY M X. p∗∗ = U XM ∗∗ Y and p∗ = SY ∗ M ∗ Y . The last assertion follows along the same arguments. S are aligned. Referring to Figure 6. X ∗ and Y ∗ being the middles of the respective sides. T V of the quadrilateral T SU V . Thus the anticenter M of q lies on line U S. Analogously it must lie also on the line joining the two other circumcenters. Remark. (3) The intersection point of the diagonals of the quadrilateral formed by the anticenters of the main flanks coincides with the circumcenter O of q. Y. O is on the line M ∗ M ∗∗ . left from the left and right from the right flank. To see that property (2) is valid. the following properties are valid: (1) The circumcircles of adjacent main flanks are tangent at the vertices of q. This is again obvious. from the similarity of parallelograms U X ∗ M ∗∗ X and SY M ∗ Y ∗ of Figure 7. (2) The intersection point of the diagonals U S. the lines QM and QO are symmetric with respect to the bisector of angle AQD. Proposition 5. The properties are immediate consequences of the definitions. Anticenter through the flanks Proposition 4. X. W T W* C D U O φ A φ ξ ϖ Z* V S B Z Figure 8. Z∗ . (1) Referring to Figure 8. The proof follows immediately from the similarity of triangles U AO and OBS in Figure 8. of the centers of the main flanks. M ∗ . Z∗ . whose sides are parallel to those of ABCD. Pamfilos Q K C D U G M X* S X L M** A Y O Y* M* B H Figure 7. r being the circumcradius of q and 2φ. gives for |OZ| = r · sin(ω). W. W. M ∗∗ of the flanks circumcenters on the two lines QO and QM and the centers of their sides on the two lines OX and OY . The quadrilateral ST U V . Z. (2) The distances of the vertices of q and q are equal: |ZB| = |AZ ∗ | = |DW | = |CW ∗ | = r · cos(φ + ξ). its incenter coincides with the circumcenter O of q and its radius is r · sin(φ + ξ). Analogous formulas hold for the other segments |OZ∗ | = |OW | = |OW ∗ | = r · sin(ω). Anticenters M. . W ∗ are vertices of a cyclic quadrilateral q . The angle ω = φ + ξ.34 P. The circumscriptible quadrilateral ST U V Proposition 6. Z. W ∗ being the projections of O on the sides of ST U V . 2ξ being the measures of two angles at O viewing two opposite sides of q. Remarks. is circumscriptible. U and passing from B. The next figure gives a panoramic view of the sixteen anticenters (in blue). [tlb][tqb][b][lb]. Referring to Figure 9. one can construct the cyclic quadrangle ABCD. B* (lt) [t] (t) [lt] A* [rt] [ltr] C D [l] [tlb] (q*) [q] [tqb] (lqr) (tqb) (q) [lqr] [q*] [r] [trb] A B [b] (lb) [lb] [rb] (b) [lbr] D* C* Figure 9. similar to the previous one for the centers. A similar argument shows that the other sides are also parallel and also proves the statement about the angles. having centers of its flanks the vertices of ST U V . [lqr] are collinear. [ltr][rt]. [q∗][trb][rb][lbr] and [ltr][rt][r][lqr]. Anticenters of the cyclic complex Proposition 7. define with their intersections on lines AB and CD the right and left flank of ABCD. Simply take on the sides of ST U V segments |ZB| = |AZ∗ | = |DW | = |CW ∗ | equal to the above measure. [q]. [q∗][trb]. For example. Then it is an easy exercise to show that the circles centered at S. [lb][b]. [lbr][rb] are all parallel since they are orthogonal to BC or its parallel B∗ C ∗ . D respectively. [l][q]. The parallelogramic pattern is symmetric with respect to the middle M of segment [q][q∗] and the anticenters [tqb]. C and A. The proof is similar to the one of Proposition 2. segments [lt][t]. together with the centers (in red) and the centroids of flanks (white). To prove the equality of parallelograms one can . [lqr][r].On the cyclic complex of a cyclic quadrilateral 35 (3) Given an arbitrary circumscriptible quadrilateral ST U V . the anticenters of the flanks build equal parallelograms with parallel sides: [lt][t][q][l]. 4. [tlb][tqb]. Besides the angles of the parallelograms are the same with the corresponding of the parallelogramic pattern of the centers. [q∗ ]. Anticenters The anticenters of the cyclic complex form a parallelogramic pattern. [lqr]. which shows the equality of horizontal sides of the two left parallelograms. [q]. Miscelanea Here I will mention only a few consequences of the previous considerations and some supplementary properties of the complex. in view of the parallelities proven so far. The only point where another kind of argument is needed is the collinearity assertion. The details can be completed as in Proposition 2. I omit the calculations.36 P. 5. [q∗ ]. as. Figure 10 shows how this can be done. Pamfilos D C G F A [q] [lqr] E [tqb] H I J B Figure 10. [tqb] are collinear. related to the definition of the three anticenters under consideration. D A C B Figure 11. F. J are middles of sides of flanks. Collinearity of [tqb]. E. for example. giving short hints for their proofs or simply figures that serve as hints. [lt][t][tqb][tlb]. Barycenters of the flanks . H. [lqr] use again implied parallelograms. G. it suffices to show that points [q]. For this. I. It suffices to calculate the ratios and show that |EF |/|EG| = |JI|/|JH|. The alignement of the four middles along the sides of ABCD is due to the equality of angles of cyclic quadrilaterals shown in Figure 14. The symmetry center is the middle of the line of (q)(q ∗ ).On the cyclic complex of a cyclic quadrilateral 37 Proposition 8. are again parallelograms. Referring to Figure 13. by considering the circumcircles of appropriate triangles. formed by parallel diagonals of the four parallelograms. the line joining the intersection points of opposite sides of q. c=((x)+[x])/2 (x) [x] Figure 12. . The equality of the parallelograms follows from the equality of corresponding parallelograms of centers and anticenters.e. Thus. this is a consequence of the corresponding results for centers and anticenters of the flanks and the fact that linear combinations of parallelograms ci = (1−t)ai +tbi . G E H F B A C D M L X K Y N Z J I P O U V W Figure 13. the parallelograms of the main flanks are homothetic to each other and their homothety centers are aligned by three on a line. bj denote the vertices of parallelograms. Indeed. Lines of middles of main flanks This is due to the fact that the main flanks are antihomothetic with respect to the vertices of q. Proposition 10. the quadrilateral (t)(r)(b)(l) of the centers of the main flanks is symmetric to the quadrilateral of the centers of the peripheral flanks (tlb)(ltr)(trb)(lbr). the centers of the sides of the main flanks are aligned as shown and the corresponding lines intersect at the outer diagonal of q i. Later assertion can be reduced to the well known one for similarity centers of three circles. Here t = 1/2. Referring to Figure 15. The barycenters of the flanks build the pattern of equal parallelograms of Figure 11. where ai . Linear combinations of parallelograms Proposition 9. since the corresponding barycenter is the middle between center and anticenter. the quadrilateral [t][r][b][l] of the anticenters of the main flanks is symmetric to the quadrilateral of the anticenters of the peripheral flanks [tlb][ltr][trb][lbr]. Referring to Figure 16. Indeed. Pamfilos B H C G E D A F Figure 14. is also orthogonal to its symmetric A∗ D∗ . The orthogonal from the middle of one side of q to the opposite one. The symmetry center is the middle of the line of [q][q ∗ ]. Symmetric quadrilaterals of centers There is also the corresponding sort of dual for the anticenters. resulting by replacing the symbols (x) with [x]: Proposition 11. which is parallel to AD (Figure 17). the symmetry of center and anticenter about the barycenter leads to a simple proof of the chararacteristic property of the anticenter. Since A∗ D∗ is a chord .38 P. By the way. Equal angles in cyclic quadrilaterals (ltr) (t) (l) (tlb) [q*] (q*) Z [q] (q) (lbr) (r) (trb) (b) Figure 15. consider the symmetric A∗ B ∗ C ∗ D∗ of q with respect to the barycenter of q. to AD say. Harmonic bundle of radical axes .q*)=(q.q*) (q.lqr)=(tqb.lqr) (q. Anticenter’s characteristic property The following two propositions concern the radical axes of two particular pairs of circles of the complex: (lqr. Symmetric quadrilaterals of anticenters of the symmetric of the circumcircle.On the cyclic complex of a cyclic quadrilateral 39 [ltr] [t] [q*] [tlb] [l] [trb] [r] [q] [lbr] [b] Figure 16. the orthogonal to its middle passes through the corresponding circumcenter. A D* B M D C A* Figure 17.q*) Figure 18. which is the anticenter.tqb) Y B* E D K L J A C* {lqr} {q*} (lqr) (q*) F C (q) B (tqb) {q} V A* G H D* I {tqb} (tqb. the radical axes (lqr. Pamfilos Proposition 12. r) is parallel to the radical axis (tlb. The similarities are: d b a (d) (b) (a) (c) c Figure 20. Proposition 13. A complex similar to the cyclic one (r) . b and two other circles c. Namely. lqr) and (q. when considering two arbitrary circles a.lbr) (tlb) (t) (lbr) (t. Common tangents of main flanks 6. The figure generates a complex of quadrilaterals which I call a generalized complex of the cyclic quadrilateral. similar to the cyclic complex. q∗ ) are parallel to the previous two and define with them a harmonic bundle of parallel lines. q ∗ ). lqr) = (tqb. the common tangent (t.40 P. resulting in another context.r) (r) Figure 19. Generalized complexes There is a figure. q∗ ) = (q. There are many similarities to the cyclic complex and one substantial difference. Analogous statements hold for the common tangents of the other pairs of adjacent main flanks. (tlb. This is shown in Figure 20. which prepares us for the discussion in the next paragraph. tqb) and (q. Referring to Figure 18. Referring to Figure 19. d tangent to the first two. Besides the radical axes (tqb. lbr). From our discussion so far. join H with D extend it and define I on the medial line of side DA. In Figure 21. the right cyclic-complex-flank r of q has been also constructed and it is different from the flank created by the general procedure. Join it to B. departing from the big flank q ∗ . (5) The same parallelogramic patterns appear for circumcenters. on the medial line of side AB of q = ABCD. tangent at the vertices of q. uniquely distinguished between the various generalized complexes. discussed so far. start with a point. for example. created by the other intersection points of the sides of q with the circles. it is not possible to construct q by elementary means. In fact. q being cyclic. F say. in general. the properties of the complex. to the flanks. Figure 21 depicts the parallelogrammic pattern for the circumcenters (in red) and the anticenters (in blue). (2) The centers of the circles form a circumscriptible quadrilateral with center at the circumcenter of q. . G. The following lemma deals with a completion of the figure handled in lemma-1. hence defining the configuration of the previous remark. one could use the above remarks to construct infinite many generalized complexes (Figure 22) having q as their central quadrilateral. (4) Adjacent flanks are antihomothetic with homothety centers at the vertices of q.On the cyclic complex of a cyclic quadrilateral 41 (1) The points of tangency of the four circles define a cyclic quadrilateral q. Circumcenters and anticenters of the general complex Having a cyclic quadrilateral q. (3) There are defined flanks. The only difference is that the central cyclic quadrilateral q is not similar. by the property of having its flanks similar to the original quadrilateral q. H and I. Join G with C extend and define the intersection point H with the medial of CD. created in this way. extend F B and define its intersection point G with the medial of BC. anticenters and barycenters. it is clear. The inverse problem The inverse problem asks for the determination of q. Finally. at points F. could have been proved in this more general setting. (d) (a) (b) r (c) Figure 21. correspondingly. The answer is in the affirmative but. 7. that the cyclic complex is a well defined complex. Thus. in general. implies that there are four circles centered. The generalized setting U X F E V K G C D N H M B A I L J Y W X* R Z S U* Figure 23. DXA. . Line X∗ U ∗ is defined by the intersection points of the opposite sides of q∗ . Lines XZY and U V W Lemma 14. The figure has the properties: (1) Lines XZY and U V W are parallel and intersect line X∗ U ∗ at points S. (2) Triangles Y KX. U AX. U IV are similar. R trisecting segment X∗ U ∗ . ZCX.42 P. U CV . (3) Angles V U D = CU X and ZXH (4) The bisectors of angles V U X. = GXU . Referring to Figure 23. Pamfilos H C D I G A B F Figure 22. lines XZY and U V W are defined by the intersection points of the opposite sides of main flanks of q. U XZ are respectively identical with those of DU C. quadrangle. X respectively. From trisecting points U. V X intersecting the sides of q∗ at W. with end-points the intersections of opposite sides of q ∗ . Define the parallelograms QW ZT.On the cyclic complex of a cyclic quadrilateral 43 (5) The bisectors of the previous angles intersect orthogonaly and are parallel to X ∗ W and Y U ∗ Z. IKJE as shown. (1) The central quadrilateral q has angles equal to q∗ . (2) The flanks are always similar to each other. consider a quadrilateral q∗ = ABCD and trisect the segmet QR. T and S. U ∗ M of angles V (1) is obvious. create ABCD with the required properties. S two parallel lines. V draw two arbitrary parallels U Y. Referring to Figure 24. not necessarily cyclic. (3) and (4) is a consequence of (2). The lemma suggests a solution of the inverse problem: Draw from points R. RSY X and through their intersections and the intersections with q∗ define the central quadrilateral q = EF N I and its flanks F EHG. the bisectors X∗ M . F P ON. two adjacent being anti-homothetic . so that the parallelograms X∗ W XV and U ∗ Y U Z. Z A Y M K L J I N E B W O A* D H F X G P C T S B* Q U V R Figure 24. Inverse problem Proposition 15. The angles of the flanks are complementary to those of q∗ . Next proposition investigates a similar configuration for a general. The orthogonality of (5) is a general property of cyclic quadrilaterals and the parallelity is due to the fact that the angles mentioned are opposite in parallelograms. since lines U V W and XZY are diagonals of the parallelograms X ∗ W XV and U ∗ Y U Z. (2) is also trivial since these triangles result from the extension of sides of similar quadrilaterals. namely q and its main flanks. N M LI. with their sides intersections. producing four complexes in general. Pamfilos with respect to their common vertex. applied to a cyclic quadrilateral. Remarks. Thus the given q∗ is identical with the big flank of q as required. having flank-1. (1) is trivial and (2) follows from (1) and an easy calculation of the ratios of the sides of the flanks. defining the ratio k1 . a simple calculation shows that points Y. they show a way to produce a complex out of any quadrilateral. depending on which pair of adjacent sides of q we start it. Then repeat with analogous constructions for the two remaining flanks. by continuity it passes through 1. It is easy to see that this procedure. Thus. condition (3) of the aforementioned proposition suggests a unified approach for general quadrilaterals that produces the cyclic complex. To prove (3) consider point S varying on side BC of q∗ . we land. the two parallels and the whole configuration. in general. to another complex. In fact. and this independently from the pair of adjacent sides of q. Thus defining k1 = |DC|/|BC| and starting with flank-2. In other words. Draw V B ∗ parallel to AB.44 P. defined through them. For S varying on BC. These are defined inductively. for which the corresponding central quadrilateral q has side-lengths-ratio |EF |/|F N | = |ON |/|OP |. In particular. produces its cyclic complex. constructed through the condition |BG|/|GH| = k1 etc. (1) Figure 24 and the related Proposition 14 deserve some comments. whose cyclic complex has corresponding big flank the given one. (2) The second remark is about the results of Proposition 8. different from the previous one. In that case. |BC| |F B| ) to construct flank-2 BGHC. point Z |EF | |ON | moves on the hyperbola from A∗ to infinity and the cross ratio r(S) = |F N | : |OP | varies increasing continuously from 0 to infinity. B∗ being the intersection point with BC. First. Thus. becomes dependent on the location of point S on BC. Proposition 16. when applied to cyclic quadrilaterals. Define the two parallels and in particular V X by joining V to S. The suggested procedure can be carried out as follows (Figure 25): (a) Start from the given general quadrilateral q = ABCD and construct the first flank ABF E using the restriction |AE|/|EF | = |BC|/|AB| = k1 . the condition of the equality of ratios implies that the constructed by the proposition central quadrilateral q is similar to the main flanks. (3) There is a particular direction of the parallels. on the centroids or barycenters of the various flanks. As point S moves from A∗ towards B ∗ on segment A∗ B ∗ . (b) Use appropriate anti-homotheties centered at the vertices of q to transplant the flank to the other sides of q. the procedure has an element of arbitrariness. use the anti-homothety (B. not necessary a cyclic one. They remain valid for the complexes defined . The proof follows immediately by applying (3) of the previous proposition to the given cyclic quadrilateral q∗ . Given a circular quadrilateral q∗ = A∗ B ∗ C ∗ D∗ there is another circular quadrilateral q = ABCD. More precisely. For general quadrilaterals though the complex depends on the initial choice of sides defining k1 . the hyperbola containing Z intersecting BC at point A∗ . Z vary on two hyperbolas (red). the complex and corresponding big flank of the big flank etc. or. Proposition-9. The figure below shows the barycenters for a general complex. Barycenters for general complexes An easy approach is to use vectors. (3) Although Proposition 15 gives an answer to the existence of a sort of soul (q) of a given cyclic quadrilateral (q∗ ). I leave the details as an exercise. since circumcenters and anticenters are not available in the general case. Several questions arise in this context. shows that even general quadrilaterals have souls. a more elementary constuction of it is desirable. constructed with the procedure described in (1). The proof though has to be modified and given more generally. (4) One is tempted to look after the soul of a soul.. Proposition-14. K L J I B* D C B A A* H G C* E F D* Figure 26. can also be carried over to the general case. Flanks in arbitrary quadrilaterals through the previously described procedure. stepping inversely. producing something similar to the original after a finite number of repetitions? (b) which are the limit points.On the cyclic complex of a cyclic quadrilateral 45 B* I K C* D L 4 J 3 A 1 E A* C 2 B F H G D* Figure 25. in combination with the first remark. for the sequence of souls? . with some minor changes. such as (a) are there repetitions or periodicity. p. p. Crete.uoc. [2] R.. Forum Geom. Barnes & Noble 1970.gr . Amer. 1994. 131. Math. Pamfilos References [1] A. A maximal property of cyclic quadrilaterals. [3] N. 35. College Geometry. Episodes in Nineteenth and Twentieth century Euclidean Geometry. Paris Pamfilos: Department of Mathematics. Court. 63–64. 5(2005). University of Crete. Honsberger. Greece E-mail address:
[email protected] P. Varverakis. A. Assoc. It is well known that the tangents at A. The case of non-pivotal isocubics is also considered.b Forum Geometricorum Volume 6 (2006) 47–52. b b FORUM GEOM ISSN 1534-1178 Isocubics with Concurrent Normals Bernard Gibert Abstract. It is well known that the tangents at A. These three normals are concurrent if and only if (pvw + qwu + ruv)(a2 qru + b2 rpv + c2 pqw) = 0. i. C concur at a point. M and its Ω−isoconjugate M∗ are collinear. (SC qu + (SC + SA )pv)x + (SC pv + (SB + SC )qu)y = 0. 2006.. up x − wr z = 0. B. the locus of point M such as P .e. 1. and by LΩ the line which is the Ω−isoconjugate of the circumcircle. namely. (SB ru + (SA + SB )pw)x + (SB pw + (SB + SC )ru)z = 0. 2 These have barycentric equations CΩ : pyz + qzx + rxy = 0. 2This line is also the trilinear polar of the isotomic conjugate of the isogonal conjugate of Ω. and LΩ : b2 c2 a2 x + y + z = 0. also passes through the same point. 1 We characterize the pivotal cubics whose normals at the vertices A. being respectively the lines − vq y + wr z = 0. u(rv2 − qw2 )x + v(pw2 − ru2 )y + w(qu2 − pv 2 )z = 0. These normals are the lines nA : nB : nC : (SA rv + (SA + SB )qw)y + (SA qw + (SC + SA )rv)z = 0. Pivotal isocubics Consider a pivotal isocubic pK = pK(Ω. B. B. − up x + vq y = 0. . Communicating Editor: Paul Yiu. 1The tangent at P . concur at P ∗ = up : vq : wr . This paper studies the situation where the normals at the same points concur. This has equation ux(ry 2 − qz 2 ) + vy(pz 2 − rx2 ) + wz(qx2 − py 2 ) = 0. Let us denote by CΩ the circumconic with perspector Ω. C and P to pK. P ) with pole Ω = p : q : r and pivot P . C to a pivotal isocubic concur. p q r Publication Date: February 13. the conjugate diameter of the line P P ∗ in CΩ . The pivotal cubic pK(Ω. equivalently. P and two other points E1 . E2 of LΩ and the circumcircle. we define the orthopolar of M with respect to the curve as the perpendicular at M to the polar line of M . C are parallel since P∗ lies on the line at infinity. i. We find that the locus of point Q such that the orthopolar of Q contains X is the union of the line at infinity and the circumconic . The orthopolar The tangent tM at any non-singular point M to any curve is the polar line (or first polar) of M with respect to the curve and naturally the normal nM at M is the perpendicular at M to tM . the normal at P is parallel to these three normals. or (2) P lies on LΩ . we may ask whether there are other points on pK such that the normal passes through X. C. antipode of P∗ on the circumcircle. For any point M not necessarily on the curve. See Figure 1. in (1). Obviously. B. equivalently. the tangents at A. nA nB nP nC A E2 P S C B E1 CΩ Figure 1.Isocubics with concurrent normals 53 Theorem 1. E2 lying on the polar line of P ∗ in CΩ . B. See Figure 2. The cubic pK meets CΩ at A. P ∗ lies on the circumcircle and the normals concur at X.. 2. In (2). P ∗ lies on the line at infinity. Hence the normals are also parallel and “concur” at X on the line at infinity. C concurrent if and only if (1) P lies on CΩ . In Theorem 1(1) above. P ) has normals at A. Theorem 1(1): pK with concurring normals More precisely. P ∗ lies on the circumcircle. B.e. These two points are isoconjugates. pK passes through the (not always real) common points E1 . In Theorem 1(2). X523 ) where four real normals are drawn from the Tarry point X98 to the curve. B. Theorem 1(2): pK with concurring normals passing through P and P ∗ . Let M be a point and m its trilinear polar meeting the sidelines of ABC at U . Hence. Denote by nK0 the corresponding cubic without xyz term. V . Non-pivotal isocubics Lemma 2. Let us now consider a non-pivotal isocubic nK with pole Ω = p : q : r and root3 P = u : v : w. The locus of the point of concurrence is the Darboux cubic. The perpendiculars at A. ux(ry 2 + qz 2 ) + vy(pz 2 + rx2 ) + wz(qx2 + py 2 ) = 0. .. This cubic has equation : ux(ry 2 + qz 2 ) + vy(pz 2 + rx2 ) + wz(qx2 + py 2 ) + kxyz = 0.54 B. 3. Figure 3 shows pK(X2 . CW concur if and only if M lies on the Thomson cubic. the locus of point Q such that the orthopolar of Q contains X is now a circum-cubic (K) passing through P∗ and therefore having six other (not necessarily real) common points with pK. C to the lines AU . BV . there are no other points on the cubic with normals passing through X. i. V . W lying on the trilinear polar of the root. Gibert nA A X B C nC P P* E1 E2 LΩ nB Figure 2. the isoconjugate of the line P P ∗ .e. 3An nK meets again the sidelines of triangle ABC at three collinear points U . W . these tangents form a triangle perspective to ABC whose perspector is P ∗ . Xi ) with concurring normals at A. Theorem 1(2): Other normals to pK It can easily be seen that the tangents tA. it is clear that pK2 is the Ω−isoconjugate of the Thomson cubic. B. . or 2 2 2 (2) P lies on the pivotal isocubic pK2 with pole Ω2 = ap2 : qb2 : rc2 and pivot P2 = ap2 : bq2 : cr2 . V . The following table gives a selection of such cubics pK2 . The normals of nK0 at A. Each line of the table gives a selection of nK0 (Ω. pK1 is the pK with pivot the root P of the nK0 which is invariant in the isoconjugation which swaps P and the isogonal conjugate of the isotomic conjugate of P. By Lemma 2. C are concurrent if and only if (1) Ω lies on the pivotal isocubic pK1 with pole Ω1 = a2 u2 : b2 v 2 : c2 w2 and pivot P . B. B. tC do not depend of k and pass through the feet U . W of the trilinear polar of P ∗ 4. Hence it is enough to take the cubic nK0 to study the normals at A. Its vertices are the harmonic associates of P ∗ . 4In other words.Isocubics with concurrent normals 55 nC tA A X99 nB W U C B X98 tC V (K) tB nA Figure 3. Theorem 3. tB. C. C. Gibert. 77. 200. 34. 312 2. 75. 55. 219. The feet of these normals lie on another circum-cubic labeled (K) in the figure. 347. 278 2. 6. 25. 19.html. 57. [3] C. Special Isocubics in the Triangle Plane. 76. The two other normals are labelled OR and OS. 346 4. 81. [4] C. X76 ). 1895 2. 204. B. 273. 69. 33. Encyclopedia of Triangle Centers. 393 1. X25 . 318. 31. 2199 1. Figure 4 shows nK0 (X1 . 342 2. 42100 . Gibert. 63.gibert/index. 1014.St Etienne. 269. 9. Kimberling. 64. the normals at A. 8. 184. all the isogonal nK0 with concurring normals must have their root on the Thomson cubic. Bernard Gibert: 10 rue Cussinel. [2] B. 31. France E-mail address: bg42@wanadoo. 208.html . 76. 221.-P. the locus of M whose orthopolar passes through F being also a nodal circumcubic with node F . 280. 264. In the special case where the non-pivotal cubic is a singular cubic cK with singularity F and root P . The corresponding nodal cubic passes through the points O. available at http://perso. 69. 8.edu/ck6/encyclopedia/ETC.wanadoo. Gibert Ω X1 X2 X3 X4 X9 X25 K175 X31 K346 X32 X55 X56 X57 X58 X75 Cubic K034 K184 K099 Ω2 P2 X2 X75 X76 X76 X394 X69 X2052 X264 X346 X312 X2207 X4 X32 X1 X1501 X6 X220 X8 X1407 X7 X279 X85 X593 X86 X1502 X561 Xi on the curve for i = 1. 84. 63. 3. 75. 55. 85. 1969 For example. Cubics in the Triangle Plane. 129 (1998) 1 – 285. X1073 . 19. 273. 58. 281 1. 312. 2. 57. 604. Congressus Numerantium. available at http://perso. 279 21. 7. 40. 92. X1384 . 394 2. 2192 6. In Figure 5. 329. 7.evansville. 1790 75. 7. Similarly. 271. 56. Triangle centers and central triangles. 9. 48. 41. 77.fr/bernard. 264.wanadoo. 27. 85. 56. Kimberling. 92. available at http://faculty. Ehrmann and B. 78. References [1] J. 8. 561.56 B. 4. X75 ) with normals concurring at O. 189. 285. all the isotomic nK0 with concurring normals must have their root on K184 = pK(X76 .html. X1617 . 318. 32. Furthermore. It is possible to draw from O six other (not necessarily all real) normals to the curve. there are two other points on cK with normals passing through F . 253. 25.gibert/files/isocubics. 86. cK has singularity at O and its root is X394 . 222. 6. 20.fr/bernard. C concur at F if and only if F lies on the Darboux cubic.fr . 304. 78. X75 ) with normals concurring at O tA A R S C B O nB nC tB nA tC Figure 5. nK0 (X1 .Isocubics with concurrent normals 57 Darboux cubic Lemoine axis tA V’ (K) A W’ K nC C B U’ O W nB trilinear polar of X75 V tC nA tB Figure 4. A cK with normals concurring at O . By the SAS similarity theorem and by the geometric interpretation of the quotient of two complex numbers. we define a (non-degenerate) triangle to be any ordered triple (A. C) of distinct complex numbers. In this note. then P is the centroid. We say that triangles ABC and A B C are similar if A − B : A − B = B − C : B − C = C − A : C − A . and CC are the cevians through P . where three different proofs are given. June A. This has already appeared as Theorem 7 in [1]. then triangles ABC and A B C are similar if and only if P is the centroid of ABC. Communicating Editor: Paul Yiu. [5]. A−C A − C A−B the shape of triangle ABC and she A−C studied properties and applications of this shape function in great detail in [4]. According to this definition. and as a problem in the Problem Section of the Mathematics Magazine [2]. Identifying the Euclidean plane with the plane of complex numbers. B. we use this shape function to prove that if P is a point inside triangle ABC. this is equivalent to the requirement that A − B A−B = . The notion of triangle shape is used to give another proof of the fact that if P is a point inside triangle ABC and if the cevian triangle of P is similar to ABC in the natural order. b b FORUM GEOM ISSN 1534-1178 A Characterization of the Centroid Using June Lester’s Shape Function Mowaffaq Hajja and Margarita Spirova Abstract.b Forum Geometricorum Volume 6 (2006) 53–55. there are in general six different triangles having the same set of vertices. and we write it as ABC if no ambiguity may arise. . Our proof is an easy consequence of two lemmas that may prove useful in other contexts. and [6]. Lester called the quantity Publication Date: February 21. BB . and if AA . A generalization to d-simplices for all d is being considered in [3]. The first-named author is supported by a research grant from Yarmouk University. 2006. Proof. (ii) u − 12 v − 12 + v − 12 w − 12 + w − 12 u − 12 ≤ 0. if and only if P is the centroid.. i. we obtain (x + y)S 2 − 2yS + (y + z) = 0. Then (i) uvw ≤ 18 . or xy + yz + zx > 0.e. as desired. 4 because uvw = (1 − u)(1 − v)(1 − w).54 M. . Proof. Let ABC be a non-degenerate triangle. Let uvw = p. x + y = 0 and the discriminant 4y 2 − 4(x + y)(y + z) = −(xy + yz + zx) of (2) is negative. by the AM-GM inequality ≤ 2 2 2 1 . Let S = A−B A−C be the shape of ABC. and CC be the cevians through an interior point P of triangle ABC. Hajja and M. y. = 8 with equality if and only if u = 12 . note that 1 1 1 1 1 1 v− + v− w− + w− u− u− 2 2 2 2 2 2 3 = (uv + vw + wu) − (u + v + w) + 4 1 = 2uvw − . Then using the cevian condition uvw = (1−u)(1−v)(1−w). and z be real numbers such that x(A − B)2 + y(B − C)2 + z(C − A)2 = 0. v : 1 − v. with equality if and only if u = v = w = 12 . and let x. xy + yz + zx > 0. we see that u(1 − u) v(1 − v) w(1 − w) p = u + (1 − u) v + (1 − v) w + (1 − w) . Suppose that the cevians through an interior point P of a triangle divide the sides in the ratios u : 1 − u.. To prove (ii). Let AA . with equality if and only if u = v = w = 12 .. if and only if P is the centroid. We now use Lemmas 1 and 2 and the shape function to prove the main result. BB . Dividing (1) by (A − C)2 .Spirova Lemma 1. Otherwise. and w = 12 . Now use (i). Then triangles ABC and A B C are similar if and only if P is the centroid of ABC. (2) Since ABC is non-degenerate. Theorem 3.e. i. This proves (i). then y = 0 and hence x = y = z = 0. S is not real. Lemma 2. v = 12 . Thus if x + y = 0. (1) Then either x = y = z = 0. i. and w : 1 − w.e. 30–54. Sci. A. A. Triangles I: Shapes. and CC through P divide the sides BC. 53 (1997). Lester. Math. A − B A−B = . One direction being trivial. Characterization of the centroid of a simplex. either u = v = w = 12 . Triangles II: Complex triangle coordinates. 164 Sofia.. J. Irbid. we assume that A B C and ABC are similar. Yarmouk University. References [1] S.bg . Suppose that the cevians AA .e. u− 2 2 2 By Lemma 1. respectively. Triangles III: Complex triangle functions. i. Aequationes Math. in which case P is the centroid. 52 (1996). u− 2 2 2 2 2 2 in which case Lemma 2(ii) is contradicted. it follows that they have equal shapes. Aequationes Math. Aequationes Math. (3) A−C A − C Substituting the values A = (1 − u)B + uC. University of Sofia. 68. 52 (1996). [6] J. 78 (2005). Mag. Jordan..edu. B = (1 − v)C + vA. [4] J. or 1 1 1 1 1 1 v− + v− w− + w− u− > 0. Since ABC and A B C are similar. E-mail address: spirova@fmi. Abu-Saymeh and M. [5] J.A characterization of the centroid using June Lester’s shape function 55 Proof.. 5 Yames Bourchier. and w : 1 − w. Lester.jo Margarita Spirova: Faculty of Mathematics and Informatics. 4–35. BB . Math. Mowaffaq Hajja: Mathematics Department. Hajja. and AB in the ratios u : 1 − u. C = (1 − w)A + wB in (3) and simplifying. A. 36 (2005) 889–912. Ed. [2] M. and we prove that P is the centroid. 215–245. Bulgaria. [3] M. Hajja. Hajja and H. This completes the proof. In search of more triangle centres. in preparation. CA.uni-sofia. Tech. Lester. Problem 1711. Internat. E-mail address: mhajja@yu. Martini. v : 1 − v. we obtain 1 1 1 (A − B)2 + v − (B − C)2 + w − (C − A)2 = 0. . The Mittenpunkt is also confined to and ranges freely over another punctured disk. . O. Smith Abstract. We are not able to show that O. Sp and Na are collinear and spaced in the ratio 2 : 1 : 3 it follows from Guinand’s theorem [4] that the Spieker center and Nagel point are confined Publication Date: February 27. b and c. G and the first Fermat point F determine the reference triangle by using complex numbers. G and K determine a. The orthocentroidal circle of a non-equilateral triangle has diameter GH where G is the centroid and H is the orthocenter. Communicating Editor: Paul Yiu. it is helpful to rescale by insisting that the distance GH is constant. G. In order to say that I ranges freely over all points of this punctured open disk. 1. We show that. Fermat (F ) and Gergonne (Ge ) points. b and c. the Euler line may rotate and the distance GH may change. Since I. We also show that the circumcenter. The orthocentroidal circle SGH has diameter GH. G and Ge determine a. We show that the Fermat. As the triangle ABC varies. 2006. Euler showed [3] that O. this can be readily achieved by dividing by the distance GH or OG as convenient. and the second Fermat point is confined to and ranges freely over the exterior of the orthocentroidal circle. It is also helpful to imagine that the Euler line is fixed. b b FORUM GEOM ISSN 1534-1178 The Locations of Triangle Centers Christopher J. where G is the centroid and H is the orthocenter of triangle ABC. The Morleys [8] showed that O.b Forum Geometricorum Volume 6 (2006) 57–70. Later Guinand [4] showed that I ranges freely over the open disk DGH (the interior of SGH ) punctured at the nine-point center N . using the same disk DGH but punctured at its midpoint J rather than at the nine-point center N . but we conjecture that they do. Recently Smith [9] showed that both results can be achieved in a straightforward way. Here O denotes the circumcenter and I the incenter. Introduction All results concern non-equilateral non-degenerate triangles. b and c of triangle ABC. that I can be anywhere in the punctured disk follows from Poncelet’s porism. centroid and symmedian point determine the sides of the reference triangle ABC. Bradley and Geoff C. In this paper we are able to prove similar results for the symmedian (K). Gergonne and symmedian points are confined to. This work involved showing that certain cubic equations have real roots. b and c as roots. and range freely over the interior disk punctured at its center. and a formula for IG2 means that the position of I in DGH enables one to write down a cubic polynomial which has the side lengths a. G and I determine the sides a. We show that the Feuerbach point must lie outside the circle SGH . Thus we now know how each of the first ten of Kimberling’s triangle centers [6] can vary with respect to the scaled Euler line. This last result was also announced by Stevanovic. Bradley and G. The orthocentroidal disk This is the interior of the circle on diameter GH and a point X lies in the disk if and only if ∠GXH > π2 . exactly one of the Brocard points is in DGH . and each in conjunction with O and G determines the triangle’s sides. The two Brocard points enjoy the Brocard exclusion principle. It will lie on the boundary if and only if ∠GXH = π2 . We give an areal descriptions of the orthocentroidal circle. In what follows we initially use Cartesian vectors with origin at the circumcenter O. 2. . [2] and [7] for general geometric background. we have |x| = |y| = |z| = 1 (2) . Stevanovic. this suggests that the neighborhood of the Euler line may harbor more secrets than was previously known. together with assertions that the symmedian and Gergonne points (and others) must lie in or outside the orthocentroidal disk were made in what amount to research announcements on the Yahoo message board Hyacinthos [5] on 27th and 29th November 2004 by M. J. a result foreshadowed by a recent internet announcement. the nine-point circle and the polar circle of the triangle. We realize that some of the formulas in the subsequent analysis are a little daunting. G and M are collinear and spaced in the ratio 2 : 1 it follows that M ranges freely over the open disk on diameter OG with its midpoint deleted. . Smith to. then both Brocard points lie on the circle SGH . and we have had recourse to the use of the computer algebra system DERIVE from time to time. OB = y. though his results do not yet seem to be in published form. If it is isosceles. Our results were found in March 2005 though we were unaware of Stevanovic’s announcement at the time. We have also empirically verified our geometric formulas by testing them with the CABRI geometry package. These conditions may be combined to give XG · XH ≤ 0. Since Ge .58 C. when algebraic formulas and geometric reality co-incide to 9 decimal places it gives confidence that the formulas are correct. We recommend this technique to anyone with reason to doubt the algebra. OC = z and. and range freely over. R. certain punctured open disks. We offer this article as a verification of this remark. C. with OA = x. Additionally we observe that the orthocentroidal circle forms part of a coaxal system of circles including the circumcircle. The fact that the (first) Fermat point must lie in the punctured disk DGH was established by V´arilly [10] who wrote . If triangle ABC is not isosceles. (1) with equality if and only if X is on the boundary. taking the circumcircle to have radius 1. We suggest [1]. This result. The locations of triangle centers 59 and a4 + b4 + c4 − 2a2 (b2 + c2 ) (3) 2b2 c2 with similar expressions for z · x and x · y by cyclic change of a. (5) . v. The sum is taken over cyclic changes. This follows from cos 2A = 2 cos2 A − 1 and the cosine rule. b and c. y. but as components in the x. w) lies in the disk DGH is (b2 + c2 − a2 )u2 + (c2 + a2 − b2 )v 2 + (a2 + b2 − c2 )w2 −a2 vw − b2 wu − c2 uv < 0 (4) and the equation of the circular boundary is SGH ≡(b2 + c2 − a2 )x2 + (c2 + a2 − b2 )y 2 + (a2 + b2 − c2 )z 2 − a2 yz − b2 zx − c2 xy = 0. We take X to have position vector ux + vy + wz . w + u. u+v+w not as areals. cyclic where we have used (2). u+v+w Multiplying by (u + v + w)2 we find that condition (1) becomes {(v + w − 2u)(v + w)} cyclic + [(w + u − 2v)(u + v) + (w + u)(u + v − 2w)]y · z ≤ 0. cyclic Dividing by (a+b+c)(b+c−a)(c+a−b)(a+b−c) the condition that X(u. The polar circle has equation SP ≡ (b2 + c2 − a2 )x2 + (c2 + a2 − b2 )y 2 + (a2 + b2 − c2 )z 2 = 0. we obtain 2(u2 + v 2 + w2 − vw − wu − uv)(a2 b2 c2 ) cyclic + (u2 − v 2 − w2 + vw)[a2 (a4 + b4 + c4 ) − 2a4 (b2 + c2 )]. u + v) . v. Next. simplifying and using (3). w). Now y · z = cos 2A = ((v + w − 2u). z frame and 3XG = XH = (v + w. (w + u − 2v). (u + v − 2w)) . The circumcircle has equation SC ≡ a2 yz + b2 zx + c2 xy = 0. u+v+w so that the unnormalised areal co-ordinates of X are simply (u. Theorem 3. the nine-point circle and the polar circle of the triangle. 2c2 ). b2 + c2 − a2 . Smith The nine-point circle has equation SN ≡(b2 + c2 − a2 )x2 + (c2 + a2 − b2 )y 2 + (a2 + b2 − c2 )z 2 − 2a2 yz − 2b2 zx − 2c2 xy = 0. . E and F is the Hagge circle of K). The incenter lies in DGH . We have established the following result. Substituting u = a2 . with similar expressions for points E and F by cyclic change. We offer a second proof. One Brocard point lies in DGH and the other lies outside SGH . c except a = b = c. Theorem 2. Evidently SGH − SC = SP and SN + 2SC = SP . However.60 C. b2 + c2 − a2 ) with similar expressions for E and F . Hence relative to triangle D E F all three areal co-ordinates of K are positive so K is in the interior of triangle D E F and hence inside its circumcircle. Since IGNa are collinear and IG : GNa = 1 : 2 it follows that Nagel’s point is outside the disk. E and F respectively. v = b2 . We are done. w = c2 in the left hand side of equation (4) we get a4 b2 + b4 c2 + c4 a2 + b4 a2 + c4 b2 + a4 c2 − 3a2 b2 c2 − a6 − b6 − c6 and this quantity is negative for all real a. m and n that l2 m l3 + m3 + n3 + 3lmn ≥ sym with equality if and only if l = m = n. The line AK with areal equation c2 y = b2 z meets the circumcircle of ABC at D with co-ordinates (−a2 . J. The symmedian point lies in the disc DGH . Bradley and G. Proof. Theorem 1. This follows from the well known inequality for non-negative l. or they both lie simultaneously on SGH (which happens if and only if the reference triangle is isosceles). The orthocentroidal circle forms part of a coaxal system of circles including the circumcircle. e and f denote the vector positions D . Let d . The reflection D of D in BC has co-ordinates (a2 . it is instructive to verify these facts by substituting relevant areal co-ordinates into equation (5). and we invite the interested reader to do so. It is easy to verify that these points lie on the orthocentroidal disk by substituting in (5) (the circle through D . 2b2 . C. b. It is possible to prove the next result by calculating that JK < OG directly (recall that J is the midpoint of GH). It is clear that s = (2b2 + 2c2 − a2 )d + (2c2 + 2a2 − b2 )e + (2a2 + 2b2 − c2 )f but 2b2 + 2c2 − a2 = b2 + c2 + 2bc cos A ≥ (b − c)2 > 0 and similar results by cyclic change. but it is easier to use the equation of the orthocentroidal circle. We will find a cubic polynomial which has roots a2 . v. q. The idea is to express the formulas for OK2 . b2 a2 . m. q. GK 2 and JK 2 in terms of u = 2 a + b2 + c2 . w) = (a2 b2 . b2 . r) are proportional to the powers of the Brocard points with respect to SGH with the same constant of proportionality. = cyclic abc so It is well known that the circumradius R satisfies the equation R = 4[ABC] . w) denote the left hand side of equation (4). We first note some equations which are the result of routine calculations. The fact that the Fermat point lies in the orthocentroidal disk was established recently [10] by V´arilly. but they have the same denominator when normalised. Proof. Gergonne’s point lies in the orthocentroidal disk DGH . 3.The locations of triangle centers 61 Proof. If the sum of these powers is zero we shall have established the result. b2 c2 . but performing the usual trick of putting a = m + n. n > 0 we get the required inequality (after division by 8) to be m2 n 2(m3 n3 + n3 l3 + l3 m3 ) > lmn sym where the final sum is over all possible permutations and l. v. which is excluded. Equality holds if and only if a = b = c. This is precisely what happens when the calculation is made. v 2 = a2 b2 + b2 c2 + c2 a2 and w3 = a2 b2 c2 . c = l + m where l. r) = (a2 c2 . 16[ABC]2 = (a + b + c)(b + c − a)(c + a − b)(a + b − c) (2a2 b2 − a4 ) = 4v 2 − u2 . Now l3 (m3 + n3 ) > l3 (m2 n + mn2 ) and adding two similar inequalities we are done.1. The symmedian point. c2 a2 ) and the other has unnormalised areal co-ordinates (p. v. The determination of the triangle sides. c2 b2 ). w) and f (p. Theorem 4. 3. m. Let f (u. G and K. v = (a + b − c)(b + c − a). n not all equal. One Brocard point has unnormalised areal co-ordinates (u. b = n + l. w = (b + c − a)(c + a − b) and the left hand side of (5) becomes −a5 (b + c) + 4a4 (b2 − bc + b2 ) − 6b3 c3 + 5a3 (b2 c + bc2 )2 −18a2 b2 c2 + cyclic which we want to show is negative. c2 given the positions of O. It follows that f (u. This is not immediately recognisable as a known inequality. Put u = (c + a − b)(a + b − c). We have r p (6) OG2 = − . These are p. (8) GK 2 = x + 6y − 3s and 2 = 1 − x/(9z) or 9zOK 2 = (9z − x)JK 2 . Now u. q/p. y. J. Note that 9w3 OK 2 = JK 2 (u3 + 9w3 − 4uv 2 ) or u(4v 2 − u2 ) JK 2 = 1 − . Bradley and G. r/q. We have (6) OG2 = z − x/9. 16[ABC]2 (4v 2 − u2 ) 1 a6 + 3a2 b2 c2 − a4 b2 R2 OG2 = 2 2 2 9a b c sym R2 = cyclic = u3 9w3 + − 4uv 2 w3 u a2 + b2 + c2 2 = − = R . (a2 + b2 + c2 )2 9u2 48[ABC]2 4(u3 + 9w3 − 4uv 2 )(u2 − 3v 2 ) 2 2 . r/p2 = x. 4v2 − u2 = q and w3 = r. (7) OK 2 = z − 3s. OK 2 9w3 We simplify expressions by putting u = p. v and w are known (9) OK JK 2 . s respectively. z. q 9 Now u2 − 3v 2 = − 34 (4v 2 − u2 ) + 14 u2 = − 34 q + 14 p2 so ( 1 p2 − 3 q) (p2 − 3q)r r 3r = − 2 =r OK = 4r 4 2 4 = 2 p q p q q p 2 Also 6v2 − u2 = 32 (4v 2 − u2 ) + 12 u2 = GK 2 = 3q 2 + 3 1 − 2 q p (7) p2 2 q 3r p(3q/2 + p2 /2) − 27r p + − 2 = 2 9p 18 6p p (8) pq OK 2 (9) =1− 2 JK 9r We now have four quantities that are homogeneous of degree 1 in a2 . Smith a2 b2 c2 w3 = . b2 and c2 . + u2 (4v 2 − u2 ) 4 (b2 + c2 ) − 15a2 b2 c2 − 6 3a a cyclic cyclic cyclic (a2 + b2 c2 )2 = 6uv 2 − u3 − 27w3 .62 C. − 9(4v 2 − u2 ) (4v 2 − u2 ) 9 9 By areal calculations one may obtain the formulas 2 OK = 4R2 a4 − a2 b2 4w3 (u2 − 3v 2 ) . C. where xs = r/p = yz. = JK =OG 1 − 2 (a + b2 + c2 )2 9u2 (4v 2 − u2 ) GK 2 = = The full details of the last calculation will be given when justifying (14). In pages 206-208 [8] the Morleys show that O. However by our hypothesis we have a homotopy of paths from γ to δ which does not involve z0 being on an intermediate path. Embed the scaled disk in the complex plane. b and c. a rigorous argument is available via complex analysis. it follows that X has winding number 1 (with suitable orientation) with respect to J as we move through a Poncelet cycle. By choosing the neighbourhood of the boundary sufficiently small. J in the scaled orthocentroidal disk. G and F determine triangle ABC using complex numbers. Gergonne or symmedian point. when the starting triangle is recovered. We assume that O. So the small path also has winding number 1 with respect to J. We will show that when r approaches R/2 (as we approach the equilateral configuration) a Poncelet cycle will keep X arbitrarily close to. Moving point A towards the original point B by sliding it round the circumcircle takes us continuously through a family of triangles which are pairwise not directly similar (by angle and orientation considerations) until A reaches the original B. and a variable triangle ABC which has the given circumcircle and incircle. One might think it obvious that every point in the scaled punctured disk must arise as a possible X on a closed path intermediate between a path sufficiently close to the edge and a path sufficiently close to the deletion. and consider the configuration of Poncelet’s porism for triangles.e. Filling the disk Following [9]. save that its vertices have been relabelled. b2 and c2 and therefore a. G and the (first) Fermat point are given. 4. a fixed incircle. we fix R and r. Moving through triangles by sliding A to B in this fashion we call a Poncelet cycle. 3. winding number +1) near the boundary and δ be an anticlockwise path (winding number also 1) close to the puncture. since we have not eliminated the possibility of exotic paths. . Therefore 1 dz dz 1 = = 0. This diagram contains a fixed circumcircle. Let γ be an anticlockwise path (i. passage through a Poncelet cycle takes X round a closed path arbitrarily close to the boundary of the orthocentroidal disk scaled to have constant diameter. 1= 2πi γ z − z0 2πi δ z − z0 Thus 1 = 0 and we have the required contradiction.2. There are technical difficulties for those who seek them. Moving the ratio r/R from close to 0 to close to 1/2 induces a homotopy between the ‘large’ and ‘small’ closed paths. We will show shortly that for X the Fermat.The locations of triangle centers 63 unambiguously and hence the equations determine a2 . The Fermat point. However. The function defined by 1/(z − z0 ) is meromorphic in DGH and is analytic save for a simple pole at z0 . Then F determines and is determined by the second Fermat point F since they are inverse in SGH . Suppose (for contradiction) that the complex number z0 represents a point between the wide path γ and the tight path δ which is not a possible location for X. but never reaching. It is clear from the theory of Apollonius circles that there is IJ < ε/2. GJ This shows that for sufficiently small r. √ By choosing r < R/8 say. c2 ). The GPAT [9] asserts that 2 =2 σI2 + KI 2 + σK abc ab + bc + ca ≤ 4Rr. b2 . (10) GJ R R 3/6 ε For any ε > 0. assuming that ε < 1 the winding number of K about J will increase by 1. Since I lies in the critical disk we have OH > OI so OH 2 > OI 2 = R2 − 2Rr. there is K1 > 0 so that if 0 < r < K1 .64 C. the path of K in the scaled critical disk (as the triangle moves through a Poncelet cycle) will be confined to a region at most ε from the boundary. weighted once by (a. c) and (a2 . Observe that we are dividing by GJ to scale the orthocentroidal disk so that it has fixed radius. Recall that a passage round a Poncelet cycle induces a path for I in the scaled critical disk which is a circle of Apollonius with defining points O and N with ratio IO : IN = 2 R R−2r . Now we have √ 2 Rr 3r KI =4 < √ . c > 0 so σI2 . σS2 > 0. We consider the mean square distance of the vertex set to itself. GJ GJ 2 Adding equations (11) and (12) and rearranging we deduce that (11) (12) KJ . C. A similar result holds for the Gergonne point Ge . Smith 5. because that it is what happens to I. we force 9GJ2 = OH 2 > 3R2 /4 so GJ > R 3/6. b. and J moves in proximity to I. c2 ) respectively. K2 } we have ε IJ < GJ 2 so by the triangle inequality 1− and SI/GJ < ε 2 KJ ε IJ < + . b. K2 > 0 such that if 0 < r < K2 . Close to the edge The areal coordinates of the incenter and the symmedian point of triangle ABC are (a. K1 . c) and once by (a2 . J. 1−ε < . then 1 − GJ Now choosing r such that 0 < r < min{R/8. In what follows we fix R and investigate what we can achieve by choosing r to be sufficiently small. Bradley and G. Note that a. Moreover. The Gergonne point must therefore be inside the incircle. a + b + c a2 + b2 + c2 √ It follows that SI < 2 Rr. then KI GJ < 2 . This is the intersection of the Cevians joining triangle vertices to the opposite contact point of the incircle. b2 . b. The rest of the argument proceeds unchanged. For the purposes of the following calculation only. then Ge I ε GJ < 2 . Now Ge I < r. m and n can be read off. The symmedian point.The locations of triangle centers 65 As before we consider the case that R is fixed. there is a possibly new K1 > 0 so that if 0 < r < K1 . 6. 9a2 b2 c2 (a2 4P10 + b2 + c2 )2 a4 b2 ) . we will normalize so that R = 1. We get a new version of equation (10) which is √ r 2r 3 Ge I < √ (13) = GJ R R 3/6 For any ε > 0. We have a2 b2 c2 KJ 2 =a2 b2 c2 l2 + m2 + n2 + 2mny · z cyclic 2 2 2 2 2 2 =(l + m + n )(a b c ) + mn(a2 (a4 + b4 + c4 ) − 2a4 (b2 + c2 )) cyclic 4P10 = 2 9(a + b2 + c2 )2 where P10 = a10 − 2a8 (b2 + c2 ) + a6 (b4 + 4b2 c2 + c4 ) − 3a4 b4 c2 . Near the orthocentroidal center 6. We proceed as in the argument for the symmedian point. We have a2 x + b2 y + c2 z OK = a2 + b2 + c2 so 2 2 2 2 a2 cyclic (2b + 2c − a )x − 2 KJ = x = 3 a + b2 + c2 3(a2 + b2 + c2 ) cyclic =lx + my + nz where l. cyclic Now OG2 = 1 9a2 b2 c2 a6 + 3a2 b2 c2 − sym cyclic so we define Q6 by OG2 = We have 9a2 b2 c2 OG2 = Q6 and 9a2 b2 c2 KJ 2 = Q6 .1. J. if the points XA . When r approaches R/2. The Gergonne point. XB and XC arise . A F E I Ge XA B D H C Figure 1 Let AD meet IH at XA .66 C. G and H are fixed. Therefore in the orthocentroidal disk scaled to have diameter 1. the points XA .2. It follows that IXA : XA H approaches 1 : 2. BXB and CXC which meet at the Gergonne point of the triangle. so the triangle ABC approaches (but does not reach) the equilateral. Consider the three rays AXA . KJ 2 4P10 Now a computer algebra (DERIVE) aided calculation reveals that Q6 (a2 + b2 + c2 ) − 4P10 = Q6 3(a + b − c)(b + c − a)(c + a − b)(a + b + c) It follows that 48[ABC]2 KJ 2 = 1 − . KJ/OG approaches 0. (14) OG2 (a2 + b2 + c2 )2 Our convenient simplification that R = 1 can now be dropped. Bradley and G. the symmedian point approaches the center J of the circle. since the ratio on the left hand side of (14) is dimensionless. Fix the circumcircle of a variable triangle ABC. Drop a perpendicular ID to BC. Smith so that Q6 (a2 + b2 + c2 )2 OG2 = . As ABC approaches the equilateral. C. H approaches O so HA approaches OA = R. 6. In particular. If we rescale so that points O. Similar results hold for corresponding points XB and XC . these three rays become more and more like the diagonals of a regular hexagon. We consider that the case that r approaches R/2. As the triangle approaches the equilateral. Triangles IDXA and HAXA are similar. XB and XC all converge to a point X on IH such that IX : XH = 1 : 2. Theorem 5. F approaches the midpoint of JK. Thus Ge converges to J. Each of the Gergonne and symmedian points are confined to. 6. so Ge approaches the point J which divides N H internally in the ratio 1 : 2. Thus the symmedian and Gergonne points are confined to the orthocentroidal disk. = √ FK 4 3[ABC] (15) From this it follows that as we approach the equilateral limit. .3. The Fermat point. and so we may take this angle to be obtuse. By continuity we have proved the following result. It follows that F performs closed paths arbitrarily close to the boundary.The locations of triangle centers 67 CF AD BE I XB Ge XA XC H Figure 2 (without loss of generality) in that order on the directed line IH. but I approaches N . However we have shown that K approaches J in the scaled diagram. Thus in the scaled diagram Ge approaches X. and range freely over the orthocentroidal disk punctured at its center. Neither of them can be at J for non-equilateral triangles (an easy exercise). then ∠XA Ge XC is approaching 2π/3. An analysis of areal co-ordinates reveals that the Fermat point F lies on the line JK between J and K. Therefore Ge is inside the circle on diameter XA XC . so in the scaled diagram F can be made arbitrarily close to K (uniformly). so F approaches J. the area [ABC] approaches 0. the center of the orthocentroidal circle. as well as wide passages arbitrarily near its boundary (as moons of I). make tight loops around its center J. As r approaches 0 with R fixed. and a2 + b2 + c2 JF . This is linear in Jx and Kx .68 C. Theorem 6. when expressed in terms of the reference triangle sides. We suppress this remark in the rest of our explanation. The first areal component Kx of K is a2 a2 + b2 + c2 and the first component Jx of J is a4 − 2b4 − 2c4 + a2 b2 + a2 c2 + 4b2 c2 . K are collinear (as is well known) but also that by the section theorem. 48[ABC]2 Hence the first co-ordinate of F is proportional to √ 8 3Kx [ABC](a2 + b2 + c2 ) + 96Jx [ABC]2 . the trigonometry involves a sign change dependent on the region in which F lies. It follows that J. JF/F K is given by (15). 3(a + b + c)(b + c − a)(c + a − b)(a + b − c) One can now calculate the areal equation of the line JK as (b2 − c2 )(a2 − b2 − bc − c2 )(a2 − b2 + bc − c2 )x = 0 cyclic To calculate the areal co-ordinates of F we first assume that the reference triangle has each angle less that 2π/3. of course. Since J is the midpoint of GH the first areal coordinate of J is a4 − 2b4 − 2c4 + b2 c2 + c2 a2 + a2 c2 . and the other co-ordinates are obtained by cyclic change. In such a case. The Fermat point cannot be at J in a non-equilateral triangle because the second Fermat point is inverse to the first in the orthocentroidal circle. and one can use trigonometry to obtain a formula for the areal coordinates which. C. J. not all of the areal co-ordinates are positive. Bradley and G. Fermat’s point is confined to. turns out to be correct for arbitrary triangles. but the final formula for the co-ordinates remains unchanged. The first normalized areal co-ordinate of H is (c2 + a2 − b2 )(a2 + b2 − c2 ) 2b2 c2 + 2c2 a2 + 2a2 b2 − a4 − b4 − c4 and the other co-ordinates are obtained by cyclic changes. and ranges freely over the orthocentroidal disk punctured at its center and the second Fermat point ranges freely over the region external to SGH . Smith Here is an outline of the areal algebra. In the latter event. BF and CF meet at equal angles. One can either invoke the charlatan’s principle of permanence of algebraic form. or analyze what happens when a reference angle exceeds 2π/3. F . The unnormalized first areal co-ordinate of F turns out to be √ 8 3a2 [ABC] + 2a4 − 4b4 − 4c4 + 2a2 b2 + 2a2 c2 + 8b2 c2 . We have therefore established the following result. In this case the rays AF. . Toland for an illuminating conversation about the use of complex analysis in homotopy arguments. The point Fe is always outside the orthocentroidal circle. V´arilly. Statics and the moduli space of triangles. Math. (n. Math.evansville. [5] http://www. 18(R − 2r) 6 Corollary 8. P. 5 (2005) 181–190. Monthly. Bradley Challenges in Geometry. Let J be the center of the orthocentroidal circle.The locations of triangle centers 69 7. 2R − 4r 6 Now I must lie in the orthocentroidal disk so IO/3 < OG and therefore JFe2 > OG2 + 3rR(R − 2r) rR − = OG2 . [3] L. available at http://faculty. Mag. Kimberling.. We thank J. rR 3r 2 2 2 JFe = OG + OG − . Lalesco. and N be its nine-point center. Morley & F.html. Location of incenters and Fermat points in variable triangles. Smith. Novi Comm. Guinand. 3 We have IN = R/2 − r. Acknowledgements. Tritangent centers and their triangles Amer. Petropolitanae 11 (1765). Theorem 7. M. Math. The Feuerbach point Let Fe denote the Feuerbach point. (16) JFe = OG 2R − 4r 6 This leaves the issue in doubt so we press on. serie prima.groups. La G´eom´etrie du Triangle. 26 (ed. Greitzer Geometry Revisited. 7 (2001) 12–129. Speiser). L. F. Coxeter and S.edu/ck6/encyclopedia/ETC.com/group/Hyacinthos. Euler. Jacques Gabay. America. C. by A. 2005.325) 139-157. Scie. In [9] it was established that IJ 2 = OG2 − 2r (R − 2r). reprinted 1987. The positions of I and Fe reveal that the interior of the incircle intersects both DGH and the region external to SGH non-trivially. [10] A. We may apply Stewart’s theorem to triangle JFe N with Cevian JI to obtain rR 2R − r 2 2 − . Vol. J. Forum Geom. Encyclopedia of Triangle Centers. Paris 2e e´ dition. References [1] C. [6] C. [8] F. Solutio facili problematum quorundam geometricorum difficillimorum..yahoo. Morley Inversive Geometry Chelsea. Acad. . Assoc. OUP. Proof. New York 1954. 1967. V. 91 (1984) 290–300. [7] T. [9] G. IFe = r and JN = OG/2. [4] A. S. [2] H. reprinted in Opera omniaa. University of Bath. Smith: Department of Mathematical Sciences.70 C.uk . Smith Christopher J. University of Bath. Bradley and G. Bath BA2 7AY. J. England E-mail address: G. C.C. Bradley: c/o Geoff C. England Geoff C.Smith@bath. Bath BA2 7AY.ac. Smith: Department of Mathematical Sciences. In [2] and [5] we demonstrated that scaling the orthocentroidal circle (on diameter GH) to have fixed diameter and studying where other major triangle centers can lie relative to this circle is a fruitful exercise. (See Figure 1). 2006. the centroid and the Brocard points. Bradley and Geoff C. We now address the Brocard points. . We consider non-equilateral triangles ABC. ±π/6) are restored to Γ. and is confined to. Include the boundary when using (2) but exclude it when using (1). as do each of Γ1 and Γ2 if their deleted points are filled in. We give the locus of points which can arise as a Brocard point of specified type. We show that the set of points of the plane which can be either type of Brocard point consists of the interior of the orthocentroidal disk. We use the Euler line as the reference ray. We show that ABC is determined by the locations of the circumcenter. curves and regions by means of polar coordinates (r. and describe this region and its boundary in polar terms. Publication Date: February 28. Let Γ2 be the reflection of Γ1 in the Euler line. 0) and radius 1 is r 2 − 4r cos θ + 3 = 0.b Forum Geometricorum Volume 6 (2006) 71–77. Delete the unique point Z in the interior which renders GJZ equilateral. Introduction For geometric background we refer the reader to [1]. 2. b b FORUM GEOM ISSN 1534-1178 The Locations of the Brocard Points Christopher J. It is easy to verify that if the points ( 3. Communicating Editor: Paul Yiu. and is confined to. The main theorem Theorem. Let Γ = Γ1 ∪ Γ2 so Γ consists of the set of points inside or on (2) for any θ. To fix ideas. We will have occasion to use polar co-ordinates with origin the circumcenter O. (2) Let Γ1 denote the region enclosed by the closed curve defined by (2) for θ > 0 and (1) for θ ≤ 0. 1. (a) One Brocard point ranges freely over. (1) This circle is enclosed by the curve defined by r 2 − 2r(cos θ + 1) + 3 = 0. and the other ranges freely over. In some circumstances the location of one Brocard point will suffice. save that G and H are removed from the boundary and two points are deleted from the interior√(the points Z such that GJZ is equilateral). We will describe points. We fix and scale to constant size the orthocentroidal disk of a variable non-equilateral triangle ABC. the equation of the orthocentroidal circle with center J = (2. (b) The set of points which can be occupied by a Brocard point is Γ. then it becomes convex. Γ2 . θ). [3] and [4]. Smith Abstract. Γ1 . with OG of length 1. (6) . G and one of the following items determines the sides of triangle ABC and which of the (generically) two possible orientations it has. First we gather some useful information. (d) The data consisting of O. Bradley and G. Smith Z O G J H Figure 1 (c) The points which can be inhabited by either Brocard point form the open orthocentroidal disk. This result can be obtained by substituting the areal co-ordinates of these points into the areal equation of this circle [2]. (1) the locations of the Brocard points without specifying which is which. Proof. (4) the location of a Brocard point of unspecified type on the orthocentroidal circle. JΩ2 + JΩ2 = 2OG2 . J. In [2] we established that 48[ABC]2 JK 2 = 1 − OG2 (a2 + b2 + c2 )2 (3) where [ABC] denotes the area of this triangle. (2) the location of one Brocard point of specified type provided that it lies in the orthocentroidal disk.72 C. (3) the location of one Brocard point of unspecified type outside the closed orthocentroidal disk together with the information that the other Brocard point lies on the same side or the other side of the Euler line. (5) OG2 The sum of the powers of Ω and Ω with respect to the orthocentroidal circle (with diameter GH) is 0. It is well known [6] that cot ω = (a2 + b2 + c2 ) 4[ABC] (4) so JK 2 = 1 − 3 tan2 ω. C. we scale distances so that the length OG is 1 and we rotate as necessary so that the reference ray OG points in a fixed direction. and we will be able to conjure up triangles ABC with convenient properties by specifying the location of K. but reflection symmetry in the Euler line means that we need not study the region inhabited by Ω separately. then a triangle exists which gives rise to this configuration and its sides are determined [2]. The angle α is just the directed angle ∠KOG and ω can be read off from JK2 /OG2 = 1 − 3 tan ω 2 . and the Brocard points are on this circle. Recalling that the length OG is 1 we obtain r 2 − 4r cos ω cos α + 3 = 0. the point Ω is diagonally to the left and Ω diagonally to the right.The locations of the Brocard points 73 √ We can immediately conclude that JΩ. and are mutual refections in the Brocard axis OK. We adopt the convention. so ∠JOΩ = ω − α. α − ω). The positions of the Brocard points are determined by (5). Observe that if we specify the locations of O. Let ∠KOJ = α. Consider the possible locations of Ω for a specified α + ω. Since the product of the roots is 3. . By the cosine rule JΩ2 = 4 + r 2 − 4r cos(ω + α) (8) JΩ2 = 4 + r 2 − 4r cos(ω − α) Now add equations (8) and (9) and use (6) so that (9) and 2OG2 = 8 + 2r 2 − 8r cos ω cos α. From (10) we see that its distance from the origin ranges over the interval 2 cos ω cos α ± 4 cos2 ω cos2 α − 3. (7) and the fact that they are on the Brocard circle. (7) As the non-equilateral triangle ABC varies. JΩ ≤ OG 2. α + ω). This expression (11) is maximized when α = ω and we use the plus sign. (11) Next suppose that α > 0. We work with a non-equilateral triangle ABC. that when standing at O and viewing K. This can be written cos(α + ω) + cos(α − ω) ± ((cos(α + ω) + cos(α − ω))2 − 3. Let r = OΩ = OΩ . we see that the minimum distance also occurs when α = ω and we use the minus sign. Viewed as a directed angle the argument of Y in polar terms would be ∠Ω OJ = α − ω. which seems to have majority support. The other Brocard point Ω has co-ordinates (r. G and the symmedian point K (at a point within the orthocentroidal circle). In the subsequent discussion the points O and G will be fixed. Now start to address the loci of the Brocard points. Now let K be at an arbitrary point of the orthocentroidal disk with J deleted. The Brocard or seven-point circle has diameter OK where K is the symmedian point. (10) We focus on the Brocard point Ω with polar co-ordinates (r. It is well known that the Brocard angle manifests itself as ω = ∠KOΩ = ∠KOΩ . G and a Brocard point X. If Ω is on the orthocentroidal circle then ABC is isosceles and the position of Ω is known. If X is in the disk. In Figure 2. We know JΩ so from (6) we know JΩ . Also 0 < ω < π/6. but it is easy to check that r2 − 2(cos θ + 1)r + 3 = 0 encloses the relevant orthocentroidal semicircle and so is the envelope of the places which may be occupied by at least one Brocard point.74 C. We also know OΩ = OΩ. Next suppose that α = 0. Smith Let θ = ∠ΩOG for Ω on the boundary of the region under discussion. Suppose that we are given O. When K is on the Euler line (so ABC is isosceles) equation (10) ensures that √ Ω is on the orthocentroidal circle. These are enveloped by the curve (2). Thus the points (1. The region Γ1 is shaded. C. then the Brocard point must be Ω. Now the Brocard circle and O are determined. Brocard points come in two flavours. so the antipodal point K to O is known. and (c) is a formality. we need to be told which it is. and if θ < 0. If X is outside the open orthocentroidal disk on the side θ > 0. However. so θ = ω + α = 2α = 2ω. Note that K cannot occupy J. Note that it follows that Ω is on the boundary precisely when Ω is on the Euler line because the argument of Ω is α − ω. and the point Ω free to range on the axis side of the curve defined by r 2 − 4r cos θ + 3 = 0 for θ = α < 0. then it must be Ω . We are done. Notice that this is the equation of the boundary of the orthocentroidal circle (1). If Ω is in the open orthocentroidal disk then only one of the two candidate positions for Ω lies inside the set of points over which Ω may range so the location of Ω is determined. This time equation (11) tells another story. in [2] we showed that the triangle sides and its orientation may be recovered from O. Thus locus of the boundary when θ > 0 is given by (10). In this region the boundary is not attained. Suppose without loss of generality we know the Brocard point is Ω. We have proved (a). our construction ensures that every point in Γ1 arises a possible location for Ω. J. The endpoints of the GH √ interval are on the edge of our region. ( 3. This is equation (2). Using standard trigonometric relations this transforms to r 2 − 2r(cos θ + 1) + 3 = 0 for θ > 0. 0). The expression is maximized (and minimized as before) when ω = 0 which is illegal. Finally we address (d). π/6) and (3. 0) do not arise as possible locations for Ω. We say that this is a forbidden point of Ω. . The more interesting exclusion is that of ( 3. if Ω is outside the closed orthocentroidal disk then the two candidates for the location of Ω are both inside the open orthocentroidal disk and we are stuck unless we know on which side of the Euler line Ω can be found. The reflection of this last curve gives the unattained boundary of Ω when θ > 0. Now. so by intersecting two circles we determine two candidates for the location of Ω . Moreover. Now suppose that α < 0. we illustrate the loci of the Brocard points for triangles with various Brocard angles ω. G and K. Bradley and G. Then (b) follows by symmetry. π/6). We fix OK and increase the argument of K. First consider K near G. Now let K make one orbit. and . Now K sails along the Euler line and the three points come close again together as K approaches H. both points rejoining the circle almost immediately. When reaching the Euler line near G. Ω just outside and Ω just inside orthocentroidal circle. and the two Brocard points nestle close to K in roughly the same configuration until K passes H at which point the Brocard points change roles. For the second half of the passage of K just inside the circle it is Ω which is just inside the circle and Ω which is just outside. and Ω swerves inside. starting with positive arguments. Ω dives inside the circle and Ω moves outside. Ω leaves the orthocentroidal disk and heads towards the boundary. Triangle ABC is isosceles. As K approaches J each Brocard point approaches its forbidden point. A qualitative description We present an informal and loose qualitative description of the movements of Ω and Ω as we steer K around the orthocentroidal disk. In this case Ω swerves round its forbidden point outside the circle. Suppose that the swerve is on the side θ > 0. Though their paths cross. the Brocard points park symmetrically on the circle with Ω having positive argument. Both Brocard points are close to K. mutual reflections in the Euler line and Ω has positive argument. Let K make a small swerve round J to rejoin the Euler line on the other side. with small positive argument. Next let K be at an arbitrary legal position on the Euler line. Both Brocard points also move in the same general direction. All the time ω stays small.The locations of the Brocard points 75 Z O G J H Z Figure 2 3. the Brocard points do not actually meet of course. Now move K along the Euler line towards J. the Brocard points move round the circle. just inside the circle. C. To get the other half of the boundary we must exchange the roles of Ω and Ω and this yields x3 z 2 + y 3 x2 + z 3 y 2 = xyz(x2 + y 2 + z 2 ) so the locus is a quintic in areal co-ordinates. Then K is free to range over a circle with center J and radius KJ where KJ2 = 1 − 3 tan2 ω because of (3). Relabel by replacing r sec θ2 by r and then replacing the remaining occurence of 2θ by θ. but the points G and H must be removed by special fiat. y = a2 b2 . J. The process will unwrap as K reverses direction until it arrives back on the Euler line. so Ω and Ω will each move round circles. When K reaches a certain critical point. Discussion We can obtain an areal equation of the boundary of Γ using the fact that one Brocard point is on the boundary of Γ exactly when the other is on the Euler line. The Brocard point (a2 b2 . but not the information as to whether the Brocard point is Ω or Ω . The three points come close together as K approaches the unattainable circle boundary. Equation (10) exhibits an intriguing symmetry between α = ∠KOG and ω = ∠ΩOS which we will now explain. but Ω reverses direction and plunges back towards K. A final journey of note for K is obtained by fixing the Brocard angle ω. If this Brocard point is Ω. Now OK = r sec θ2 . The description is therefore the union of two curves. 4. Now SΩ O is a right angle so in this particular sweep the position of Ω is the perpendicular projection on the Euler line of the position of K. We derive (12) as follows. Another interesting tour which K can take is to move with positive arguments and starting near G along the path defined by the following equation: r 2 − 4r cos θ + 3 + 3 tan2 θ = 0 (12) This is the path of critical values which has Ω moving on the boundary of Γ1 and Ω on the Euler line. As K takes this circular tour through the moduli space of directed similarity types of triangle. z = b2 c2 is then given by x3 y 2 + y 3 z 2 + z 3 x2 = xyz(x2 + y 2 + z 2 ). Take (2) and express θ in terms of θ/2 and multiply through by sec2 2θ . we call the location . Bradley and G. The equation of the Euler line is (b2 + c2 − a2 )(b2 − c2 )x+ (c2 + a2 − b2 )(c2 − a2 )y + (a2 + b2 − c2 )((a2 − c2 )z = 0. The locus of the other Brocard point x = c2 a2 . Ω reaches the boundary and at the same instant. The direct similarity type of triangles OKΩ and OKΩ will not change. we make the same journey through triangles as when a triangle vertex takes a trip round a Neuberg circle. Ω crosses the Euler line. Smith Ω chases K.76 C. b2 c2 . c2 a2 ) lies on the Euler line if and only if a6 c2 + b6 a2 + c6 b2 = a4 b4 + b4 c4 + c4 a4 . Suppose that we are given the location Y of a Brocard point within the orthocentroidal circle. Now K keeps moving towards Ω followed by Ω . Lalesco. The symmetry in (10) is explained. Let the respective Brocard angles be ω1 and ω2 . so ∠K2 Y O = ∠K1 Y O = π/2 and therefore Y lies on the line segment K1 K2 . Ω1 ) which are linked via their common Brocard point in the orthocentroidal disk. Also OΩ1 = OY = OΩ2 . Paris 2e e´ dition. Forum Geom.uk . Using equation (6) we conclude that the lengths JΩ1 and JΩ2 are equal. α” description of Ω2 which follows from equation (10). There are a pair of triangles determines by by the quadruple (O. exchanging the roles of left and right in the whole discussion (the way in which we have discriminated between the first and second Brocard points). W. We anticipate that there may be interesting geometrical relationships between these pairs of non-isosceles triangles. “Brocard Points. References [1] [2] [3] [4] [5] [6] C. 6 (2006) 57–70. Assoc.The locations of the Brocard points 77 of the other Brocard point Ω1 and the corresponding symmedian point K1 . La G´eom´etrie du Triangle. L. However.. J. C.” From MathWorld–A Wolfram Web Resource. Forum Geom. we have ω = ω2 and α = ω1 . H. Statics and the moduli space of triangles. Greitzer Geometry Revisited. Ω ) using the values (O. England Geoff C. G. Smith: Department of Mathematical Sciences. 5 (2005) 181–190. OUP. J. http://mathworld. Ω2 .com/BrocardPoints. G. Jacques Gabay. Weisstein.C. reprinted 1987. Smith: Department of Mathematical Sciences. England E-mail address: G. Bradley and G.. G. Now ∠Ω2 OΩ1 = ∠Ω2 OG + ∠GOΩ1 = 2ω1 + 2ω2 .ac.html Christopher J. the resulting description of Ω1 would have ω = ω1 and α = ω2 . University of Bath. M.wolfram. Coxeter and S. G. Y. Smith. S. Bath BA2 7AY. Math. C. E. We have two Brocard circles. 1967. Y ) and (O. However. Bradley Challenges in Geometry. Therefore Ω1 and Ω2 are mutual reflections in the Euler line. Bradley: c/o Geoff C. America. Ω. Thus in the “ω. Smith The locations of triangle centers.Smith@bath. 2005. Bath BA2 7AY. T. C. University of Bath. On the other hand if the Brocard point at Y is Ω we call the location of the other Brocard point Ω2 and the corresponding symmedian point K2 . the Euler line is the bisector of ∠Ω2 OG so ∠K2 OG = ω1 and ∠GOK1 = ω2 . . In this paper Publication Date: March. Their radius is rA = r1rr2 . (O ) and C(2rA ) (E in [6]) and the common tangent of (O). mutually tangent in the points A. B and C as shown in Figure 1 below. In their overview [4] Dodge et al have expanded the collection of Archimedean circles to huge proportions. Okumura and Watanabe [6] have added a family to the collection. 3] we know that these Archimedean twins are not only twins.b Forum Geometricorum Volume 6 (2006) 79–96. (O2 ) and (O). 1 Recently Power has added four Archimedean circles in a short note [7]. We explore the arbelos to find more Archimedean circles and several infinite families of Archimedean circles. may seem surprising. (O2 ). which all pass through O. its relevance is immediately apparent when we realize that the common tangent of (O1 ). The arbelos with its Archimedean circles It is well known that Archimedes has shown that the circles tangent to (O). Communicating Editor: Paul Yiu. having radii r1 . (O1 ) and CD and to (O). consisting of three semicircles (O1 ). (W21 ) and the incircle (O3 ) of the arbelos meet on AB. 2006. 29 individual circles and an infinite family of Woo circles with centers on the Schoch line. Preliminaries We consider an arbelos. . and have given new characterizations of the circles of Schoch and Woo. See Figure 1. being a curvilinear triangle. (O2 ) and CD respectively are congruent. To define these families we study another shape introduced by Archimedes. The use of tangent lines in the arbelos. Thanks to [1. but that there are many more Archimedean circles to be found in surprisingly beautiful ways. b b FORUM GEOM ISSN 1534-1178 Archimedean Adventures Floor van Lamoen Abstract. 2. 1. the salinon. 1The circles of Schoch and Woo often make use of tangent lines. r2 and r = r1 + r2 . M D W2 M2 W1 E M M1 A O1 C O O O2 B Figure 1. A similar statement is true for the tangent to B(C) at its intersection with (W14 ) and (O). r1 r2 ) 2. 2 r1 r2 ) B (r1 + r2 . (O2 ) = (CB). which are found by intersecting A(C) and B(C) with (O) and then taking the smallest circles through the respective points of intersection tangent to CD. see Figure 2. r12 and thus the making use of the right triangle O1 O2 K1 . Thus. The Archimedean circles (Wn ) are those appearing in [4]. factor 2 applied to (O1 ). and (O) = (AB) on the same side of the line AB.1. (O2 ) and (O ). r2 ) r +r M (0. The relation is clear when we realize that A(C). √2 E (r1 − r2 . M. Then. 0) C (r1 − r2 . A second characterization of the points K1 and K2 . r1 + r2 ) 2 1 2 M r1 −r 2 . P (r) P (Q) (P ) (P Q) (P QR) circle with center P and radius r circle with center P and passing through Q circle with center P and radius clear from context circle with diameter P Q circle through P . We note that the tangent to A(C) at the point of intersection of A(C). B(C) and (O) can be found by a homothety with center C. (K1 ) and (K2 ). By symmetry the radius of (K2 ) equals rA as well. By symmetry it is easy to find from these the Archimedean circles (K1 ) and (K2 ) tangent to CD. 0) M2 (r 1 . Here are the coordinates of some basic points associated with the arbelos. and r = r1 +r2 . To prove the correctness.0 √ D (r1 − r2 . 0) O1 (−r2 . It is convenient to introduce a cartesian coordinate system. . We shall call the semicircle (O ) = (O1 O2 ) (on the same side of AB) the midway semicircle of the arbelos. We adopt the following notations. r1) 2 O r1 −r 2 . New ones are labeled (Kn ). Q. and is also tangent to (K1 ). these are often interpreted as semicircles.80 F. (W13 ) and (O) passes through B. r2 . with O as origin. A (−(r1 + r2 ). is that these are the points of intersection of (O1 ) and (O2 ) with (O ). R In the context of arbeloi. van Lamoen we introduce some new Archimedean circles and some in infinite families. 0) O2 (r1 . an arbelos consists of three semicircles (O1 ) = (AC). New Archimedean circles 2. of radii r1 . The circles (K1 ) and (K2 ) are closely related to (W13 ) and (W14 ). The common tangent at C to (O1 ) and (O2 ) meets (O) in D. let T be the perpendicular foot of K1 on AB. we find O1 T = r1 +r 2 radius of (K1 ) is equal to O1 C − O1 T = rA . 0) M1 (−r2 . The Archimedean circle (W8 ) has C as its center and is tangent to the tangents O1 K2 and O2 K1 to (O1 ) and (O2 ) respectively. clearly equivalent. Paul Yiu has noted that the circle with center on CD tangent to (O) and (O ) is Archimedean. Recall that this is the circle that passes through C and the points . Indeed.Archimedean adventures 81 W14 D W13 K2 K1 A O1 O C O (W8 ) O2 B Figure 2.2. (K4 ) and (K5 ). The Archimedean circle (K3 ) 2.3. In private correspondence [10] about (K1 ) and (K2 ). if x is the radius of this circle. There is an interesting similarity between (K3 ) and Bankoff’s triplet circle (W3 ). (K3 ). D K3 A O1 C O O O2 B Figure 3. The Archimedean circles (K1 ) and (K2 ) 2. then we have 2 r − r 2 r 1 2 +x − (r − x)2 − (r1 − r2 )2 = 2 2 and that yields x = rA . M1 and M2 be the midpoints of the semicircular arcs of (O). This line M1 M M2 is an angle bisector of the angle formed by lines AB and the common tangent of (O). a fact that seems to have been unnoted so far. · O C = O F3 = r 2 2(r12 + r22 ) 2 − rA and from this we see that O1 F3 : F3 O2 = r12 : r22 . (O ). 3 . The tangency can be shown by using Pythagoras’ Theorem in trianr 2 +r 2 gle CO W3 . There are many ways to find the interesting point T3 . and the circle (BM2 O1 ) intersects the semicircle (O1 ) at X. M. 2 2r r This also gives us in a simple way the new Archimedean circle (K4 ): the circle tangent to AB at O and to (O ) is Archimedean by reflection of (W3 ) through the perpendicular to AB in O . and F3 be the perpendicular foot of T3 on AB. M . Then r (r2 − r1 )r 2 . Z and T3 are collinear. But this means that the points C. Let me give two. This also gives us the circle (K5 ) tangent to (O ) and tangent to the parallel to AB through Z is Archimedean by reflection of (W3 ) through T3 . See Figure 4. yielding that O W3 = 12r 2 . (O3 ) and (W21 ). 11]. van Lamoen of tangency of (O1 ) and (O2 ) with the incircle (O3 ) of the arbelos. see [9. This leaves for the length of the radius of (O ) beyond W3 : r1 r2 r r12 + r22 − = . (O1 ) and (O2 ) respectively. T3 is the second intersection of (O ) with line M1 M M2 . X and Y are the points of tangency 2For simple constructions of (O ). The Archimedean circles (K4 ) and (K5 ) Let T3 be the point of tangency of (O ) and (W3 ). Corollary 3] shows the same for the angle bisector of ∠AZB. (2) The circle (AM1 O2 ) intersects the semicircle (O2 ) at Y . D Z M2 O3 M K5 T3 M1 A O1 W3 F3 K4 C O O O2 B Figure 4. and hence the angle bisector of ∠O1 T3 O2 passes through C.82 F. (1) Let M . If Z is the point of tangency of (O3 ) and (O) then [9. (W3 ) has its center on CD and is tangent to (O ). 2 Just like (K3 ). so that O1 T3 : T3 O2 = r1 : r2 . apart from M . Similarly one finds (K7 ). This circle is easily constructed by noticing that the common tangent of A(C) and (K6 ) passes through O2 . The Archimedean circles (K6 ) and (K7 ) To prove that (K6 ) is indeed Archimedean. let S6 be the intersection of AB and 1 and AS6 = 2(r1 −rA ). D T7 K7 T6 K6 A S6 O1 C O O S7 O2 B Figure 6. the radical axis of A(C) and (O).Archimedean adventures 83 of (O3 ) with (O1 ) and (O2 ) respectively. then K6 is the intersection of the mentioned radical axis and AT6 . each of these circles intersects the midway semicircle (O ) at T3 . See Figure 5. Now. Consider again the circles A(C) and B(C). Construction of T3 2. D M M2 O3 M T3 M1 A X Y W3 C O1 O O O2 B Figure 5. The circle with center K6 on the radical axis of A(C) and (O) and tangent to A(C) as well as (O ) is the Archimedean circle (K6 ). See Figure 6. then O S6 = 2rA + r2 −r 2 . (K6 ) and (K7 ). Let T6 be the point of tangency.4. Of course by similar reasoning this holds for (K11 ) as well. C and the point of tangency of (O) and the incircle (O3 ) are concyclic. O. To see this note that the radius of (M M1 M2 ) equals r12 +r22 . while this follows for (K9 ) as well by symmetry. The points M . van Lamoen If x is the radius of (K6 ). Let A and B be the reflections of C through A and B respectively. We have 4r1 − 2rA − x 4r1 + r2 = r2 x implying indeed that x = rA . Now the circle M (C) meets (O) in two points. Now we know the sides of triangle O M K10 . M1 . The circle with center on AB. The smallest circles (K12 ) and (K13 ) through these points tangent to AB are Archimedean as well. And (K8 ) is Archimedean. M2 . then 2 r2 − r1 2 r1 + r2 + x − 2rA + = (2r1 − x)2 − 4(r1 − rA )2 2 2 which yields x = rA . the center of their circle. The Archimedean circles (K8 ) and (K9 ) 2.84 F. M. For justification of the simple construction note that AS6 2r1 − 2rA 2r1 AT6 = = = = cos ∠AO2 T6 .6. Four more Archimedean circles from the midway semicircle. tangent to the tangent from A to (O2 ) and to the radical axis of A(C) and (O) is the circle (K8 ). See Figure 7. Let x be the radius of (K8 ). The altitude from K10 has 2 √ (r12 +4r1 r2 +r22 )(r12 +r22 ) r 2 +r 2 and divides O M indeed in segments of 12r 2 and length 2r rA . The circles with K10 and K11 as centers and tangent to AB are Archimedean. Similarly we find (K9 ) from B and (O1 ) respectively. This circle (M ) meets (O ) in two points K10 and K11 . the mid-arc circle is M . . A AK8 O1 C C O O O2 K9 B B Figure 7.5. (K8 ) and (K9 ). This can be seen by applying homothety with center C with factor 2 to the points K10 and K11 . cos ∠S6 K6 A = AK6 2r1 − rA 2r1 + r2 AO2 2. B(C) and (O) are images of (O1 ). In triangle √ COD we see that CD = 2 r1 r2 and thus by similarity of COD and F14 AK14 . The Archimedean circles (K14 ) and (K15 ) An additional property of (K14 ) and (K15 ) is that these are tangent to (O) exter2r 2 nally.Archimedean adventures 85 M D Z M2 O3 M M1 K12 K10 K11 O A C O1 K13 B O2 Figure 8. This shows that A(C). just as is Bankoff’s quadruplet circle (W4 ). (O2 ) and (O ) after homothety through C with factor 2. It is easy to see that the semicircles A(C). See Figure 8. the circles (K14 ) and (K15 ) tangent internally to A(C) and B(C) respectively and both tangent to d at the opposite of (O1 ). To see this note that. using linearity. (K11 ). are Archimedean circles. the distance from A to d equals r1 . (K12 ) and (K13 ) The image points are the points where M (C) meets (O) and are at distance 2rA from AB.7. (O2 ) and (O ). The Archimedean circles (K10 ). so AK14 = r1 (2rr1 +r2 ) . As a result. 2. (O2 ) and (O ). K15 W4 K14 E A A O1 C O O O2 B B Figure 9. Let F14 be the perpendicular foot of K14 on AB. (K14 ) and (K15 ). B(C) and (O) have a common tangent parallel to the common tangent d of (O1 ). In the same way it is shown and now we see that K14 F14 A 14 (K15 ) is tangent to (O). Note that O1 K15 passes through the point of tangency of d and (O2 ). λ) to (O) and (O1 ) to get the circles (Ω) and (Ω1 ). (K16 ) and (K17 ). and CD. Using the Pythagorean theorem in triangle U U Ω and U U Ω1 we find (λr1 + ρ)2 − ((λ − 2)r1 + ρ)2 = (λr − ρ)2 − (λr − 2r1 + ρ)2 . and (K17 ) tangent to C(B). which also lies on O1 (D). By symmetry. OF14 = r + r r2 2 + OF 2 = (r + r )2 . See Figure 10. 2. As special members of this family we add (K16 ) as the circle tangent to C(A). K14 F14 = r r2 3 2 2 r2 − r1 r + 4r1 r2 + 4r1 r2 − r13 AK14 = 2 . van Lamoen we have √ √ 2 r1 r2 2r1 r1 r2 (2r1 + r2 ) AK14 = . Apply the homothety h(A. M. and CD. In particular.86 F. We leave details to the reader. K17 K16 A O1 D C O O O2 Figure 10.8. The Archimedean circles (K16 ) and (K17 ) B . this shows that the twin circles of Archimedes are members of a family of Archimedean twin circles tangent to CD. (W6 ) and (W7 ) of [4] are limiting members of this family. See Figure 9. This yields ρ = rA . and U the perpendicular foot of U on AB. A(B). Let U (ρ) be the circle tangent to these two circles and to CD. B(A). Then |U Ω| = λr − 2r1 + ρ and |U Ω1 | = (λ − 2)r1 + ρ. See Figure 12. (O2 (λ + 1)) and (O1 (λ + 1)O2 (λ + 1)) enclose Archimedean circles (K14 (λ)). This is seen best by the well known observation that the two points of tangency of (O1 ) and (O2 ) with their common tangent together with C and D are the vertices of a rectangle. one of Bankoff’s surprises [2]. that is by attaching to (O) a semicircle (O ) to be the two inner semicircles of a new arbelos. From the reasoning of §2. Of course the centers W4 (λ) lie on a line perpendicular to AB. the radius of the incircle (O3 ) of the original arbelos. so that the centers K14 (λ) run through a line as well. passing through C and with radii λr1 and λr2 respectively. Theorem 1]. By homothety the point of tangency of (K14 (λ)) and (O1 (λ + 1)) runs through a line through C.Archimedean adventures 87 3. r 2 − r1 r2 surprisingly equal to r3 . (K15 (λ)) and (W4 (λ)). By repetition this yields infinite families of Archimedean circles. as derived in [4] or in generalized form in [6. This line and a similar line containing the centers of K15 (λ) are perpendicular. See Figure 11. then we must have r1 r2 rr = r r + r which yields rr1 r2 .7 it is clear that the common tangent of (O1 (λ)) and (O2 (λ)) and the semicircles (O1 (λ + 1)). Extending circles to families There is a very simple way to turn each Archimedean circle into a member of an infinite Archimedean family. The result is that we have three families. . When (O ) is chosen smartly. r = D K2 O3 K2 K1 K1 O A O1 C O O O2 B Figure 11 Now let (O1 (λ)) and (O2 (λ)) be semicircles with center on AB. If r is the radius of (O ). this new arbelos gives Archimedean circles exactly of the same radii as Archimedean circles of the original arbelos. van Lamoen K15 (λ) W4 (λ) K14 (λ) W4 E O1 (λ) O1 (λ + 1) A O O1 C O O2 B O2 (λ) O2 (λ + 1) Figure 12 4. See Figure 13. Ot (tr1 + (t − 1)r2 .88 F. 0 r12 −r22 +(2t−1)r1 r2 . Through these we draw a semicircle (Ot. This tangent passes through E.0 r Ct Ot. 3 In this case we call the region bounded by the 4 semicircles the t-salinon of the arbelos. This we do by starting with a point Ot that divides O1 O2 in the ratio O1 Ot : Ot O2 = t : 1 − t. In this paper we will not refer to these results.2 (r1 + (2t − 1)r2 . 0) Ot.2 Ct. Here are the coordinates of the various points. 0) At ((2t − 2)r1 − r2 . as the resulting figure is not really like Archimedes’ salinon.4 ) = (Ct. 0 The radical axis of the circles (x − tr1 − (t − 1)r2 )2 + y 2 = ((1 − t)r1 + tr2 )2 (Ot ) : 3If t < 1 2 we can still find valid results by drawing (Ot.1 ) opposite to the other semicircles with respect to AB. Our method of generalization is to translate the two basic semicircles (O1 ) and (O2 ) and build upon them a (skew) salinon. The family of salina and the arbelos are to have common tangents. These two semicircles have a semicircular hull (Ot ) = (At Bt ) and meet AB as second points in Ct.2 ) with radius rt = (1 − t)r1 +tr2 .2 if and only if t ≥ 12 .1(2tr1 − r2 . Assume r1 < r2 . 0) Ct. 0) Bt (r1 + 2tr2 . Then we create semicircles (Ot. We do this in such a way that to each arbelos there is a family of salina that accompanies it. see [6.4 (t + 2 )r1 + (t − 1 12 )r2 . .1 and Ct.1 ) and (Ot.4 ) on the same side of AB as the other semicircles.2 ) with radii r1 and r2 respectively.2 (r − 2)r2 .1 is on the left of Ct. A family of salina We consider another way to generalize the Archimedean circles in infinite families. 0) 1 + (2t 1 Ot. the external center of similitude of (O1 ) and (O2 ).1 Ot. 0) 1 Ot t − 2 r. and meets AB in N . 0) Ct.2 respectively. Ct. M. Theorem 8]. so that it is tangent to d. We create a semicircle (Ot ) = (Ot.1 ((2t − 1)r1 − r2 . 1Ot B Ot.4 Ot Ct A O1 At C Ot. r So is the radical axis of x2 + y 2 = r 2 (O) : and (Ot ) : 1 x− t− 2 2 2 1 1 2 r +y = r2 .2 tangent to both (Ot ) and t and to (O1 ) and (O2 ) respectively are Archimedean.1. we define points the points Ct on AB and Dt on (O).2 Bt Figure 13. 1 − t r1 + t + 2 2 On this common radical axis t . 5.1 O Ct. Wt.1 and Wt. A t-salinon and (O ) : r1 − r2 x− 2 2 + y2 = r2 4 is the line t : x= (2t − 1)r1 r2 + r12 − r22 . This can be proven with the above coordinates: The semicircles Ot ((1 12 −t)r1 + (t + 12 )r2 − rA )) and O1 (r1 + rA ) intersect in the point 2r1 (2t − 2)r1 r2 + r12 − r22 . rA (3 − 2t)(2t + 1 + ) . The twin circles of Archimedes.2 O2 Ct.1 r r2 . See Figure 13.Archimedean adventures D 89 Dt E Ot. We can generalize the well known Adam and Eve of the Archimedean circles to adjoint salina in the following way: The circles Wt. Archimedean circles in the t-salinon 5. r r1 which lies rA right of t .2 ) Two properties of the twin circles of Archimedes can be generalized as well.1 .1 lies on O2 (A).1 Ot.2 O2 Ct.2 E Wt. Similarly we find for the intersection of Ot ((1 12 − t)r1 + (t + 12 )r2 − rA )) and O2 (r2 + rA ) 2 2r2 r1 + 2r1 r2 t − r22 Wt.2 − O2 ) O2 + r2 + rA (1 + 2t)r1 r22 ((3 − 2t)r1 + 2r2 ) 2r12 − r22 + 2tr1 r2 .2 . .2 ) and (O2 ) is r2 (Wt. These circles are real if and only if 12 ≤ t ≤ 32 . rA (2t + 1)(3 − 2t + ) . Dt r r while the point of contact R2 of (Wt. p. while Wendijk [8] and d’Ignazio and Suppa in [5.1 O Ct. M. • the point Nt. To verify this note that Dt has coordinates r1 r2 ((3 − 2t)r1 + 2r2 )(2r1 + (2t + 1)r2 ) r12 − r22 + (2t − 1)r1 r2 .4 Ot Ct A O1 At C Ot. 236] ask in a problem to show that the point N2 where O2 W2 meets CD lies on O2 (A) (reworded). We can generalize these to • the circle A(Dt ) passes through the point of tangency of (Wt.2 ) and (O2 ). D Dt Wt.2 where O2 Wt.1 ) and (Wt.2 meets the perpendicular to AB through Ct.90 F. = 2r1 + r2 2r1 + r2 . Of course similar properties are found for Wt. . Dodge et al [4] state that the circle A(D) passes through the point of tangency of (O2 ) and (W2 ). See Figure 14.1Ot B Ot.2 Bt Figure 14. The Archimedean circles (Wt. van Lamoen which lies indeed rA left of t . 1 .3 ) 5. When L travels linearly. then the radical axis of (K) and the circle (L) tangent to 1 and 2 moves linearly as well. This shows that this circle (Wt. Let L be a point travelling through the line P K. These circles are real if and only if 12 ≤ t ≤ 1. See Figure 15. Dt Wt. To generalize Bankoff’s quadruplet circle (W4 ) we start with a lemma. R2 )2 = 2r1 (2r1 + (1 + 2t)r2 ). (Wt.1 C Ct. Dt )2 = d(A. Lemma 1. (1 + 2t)r1 ((3 − 2t)r1 + 2r2 ) so that this point lies indeed on O2 Wt. The Archimedean circle (Wt.1 O Ot Ot O2 Ot. (Wt.3 ). 5.2. we see that the point Nt.2 . using the tangency of (W3 ) to (O ) shown above.3 ) is an Archimedean circle and generalizes the Bankoff triplet circle (W3 ). When u is the radius of this circle we have (r t − u)2 − O tCt2 = u2 − CCt2 (t − 1)r12 + tr22 2 ) = u2 − ((2t − 1)rA )2 ((1 − t)r1 + tr2 − u)2 − ( r which yields u = rA . Let (K) be a circle with center K and 1 and 2 be two tangents to (K) meeting in a point P . Furthermore. The speed relative to the speed of L depends only on the angle of 1 and 2 .3. .2 = 2tr1 − r2 .4 ). on O2 (A) and on the perpendicular to AB through Ct. Consider the circle through C tangent to (Ot ) and with center on Ct Dt .2 = O2 + 2r1r+r2 (R2 − O2 ) has coordinates Nt.2 B Bt Figure 15.3 A At O1 Ot.Archimedean adventures 91 Straightforward verification now shows that d(A. (Wt. As a result of Lemma 1.4 ).1 O Ct. (O) and (O ) can be replaced by Ct Dt .92 F. (Wt. 0).7 ). Of course this means that the circles (K12 ) and (K13 ) are found as members of the family (Wt. d. (x − x2 )2 + y 2 = v 2 x22 .e. To see this we note that N is external center of similitude of (Ot.1 ) and (Ot.2 O2 Ct.4 W4 E Ot.1Ot B Ot. The Archimedean circle (Wt.2 ).4 ). it is now clear why the the radical axes of (O) and (Ot ) and of (O ) and (Ot ) coincide for all t. an Archimedean circle (Wt. See Figure 16.14 ). With v = sin φ the circles (K) and (L) are given by (x − x1 )2 + y 2 = v 2 x21 . Without loss of generality let 1 and 2 be lines making angles ±φ with the x−axis and let K(x1 . Whereas (W13 ).13 ) and (Wt. 2 The radical axis of these is x = 1−v 2 (x1 + x2 ) = 1 2 cos2 φ(x1 + x2 ). van Lamoen Proof. (W14 ). This shows that the common tangent of (O) and (Ot ) is parallel to the common tangent of (O1 ) and (O2 ). (Wt. (K1 ) and (K2 ) are defined in terms of intersections of (semi-)circles or radical axes.2 Bt Figure 16. and d is the tangent from N to (Ot ). 0). It is easy to check that the slope of the line through the midpoints of the semi1 and thus equal to te slope of the line circular arcs (O) and (Ot ) is equal to r2 −r r through M1 and M2 . M. The generalization of (W6 ) and (W7 ) as Archimedean circles with . also the common tangent of (O ) and (Ot ). just as the famous example of Bankoff’s quadruplet circle (W4 ).4 ) We note that the the Archimedean circles (Wt.6 ).4. with Lemma 1 their generalizations are obvious: CD. Another consequence is that the greatest circle tangent to d at the opposite side of (O1 ) and (O2 ) and to (Ot ) internally is. (Ot ) and (Ot ). 5. D Dt Wt.4 ) can be constructed easily in any (skew) salinon without seeing the salinon as adjoint to an arbelos and without reconstructing (parts of) this arbelos. Let P be the origin for Cartesian coordinates. i. L(x2 .4 Ot Ct A O1 At C Ot. See Figure 17. .14 . r12 + r22 r1 + r22 xLt = . The tangents from A to (Ot.6 ).6 ) and (Wt.6 C Ot.2 ) and from B to (Ot.13 ) and (Wt.1 O Ct. As a result we have two stacks of three families. .13 ) and (Wt. Tt. Tt.13 r(2r1 + (1 − 2t)r2 ) = .2 Bt Figure 17. 2 . A(C) and (Ot ) meet and similarly define Tt.14 meets AB r((2t − 1)r1 + 2r2 ) .13 0 − yTt.14 r r The slope of the tangent to A(C) in Tt.13 and the x−coordinate of the point Rt where this tangent meets AB is equal to xRt = xTt.13 A O1 At Ct Wt. Wt.14 ). The Archimedean circles (Wt.13 r r √ r12 +(2t+1)r1 r2 −r22 r2 −r1 ((4t2 −4t−3)r1 −(8t+4)r2 ) .2 Ot. (Wt.13 with respect to the x−axis is equal to sA = − xA − xTt.1 ) are tangent to (Wt. (Wt.7 O2 Ct. This is seen by straightforward calculation.13 − yTt.Archimedean adventures 93 center on AB and tangent to CtDt keep having some interest as well.13 ).1Ot B Ot.14 D Dt E Wt. sA 2r1 + (2t − 1)r2 Similarly we find for the point Lt where the tangent to B(C) in Tt.13 be the point where (Wt.14 ) We zoom in on the families (Wt. Let Tt.7 ). (2t − 1)r1 − 2r2 The coordinates of Z are given by 2 r (r1 − r2 ) 2rr1 r2 . which is left to the reader.7 ) respectively. We find that √ r12 +(2t−3)r1 r2 −r22 r1 −r2 ((8t−12)r1 +(4t2 −4t−3)r2 ) .4 Ot Wt. Then the distance between Mt. A(C) and B(C) is equal to 2rA . We note that the midpoints Mt . Mt. See Figure 19.1 Ot. Since the distance between d and the common tangent of (O). Let the tangent to A(C) through R meet A(C) in P1 and similarly find P2 on B(C). (Ot.2 ) 5. that are centers of Archimedean circles (Kt.2 Kt. The lemma can also help to generalize for instance (K8 ) and (K9 ). Let K be a circle through Z with center on AB. we notice that for instance A(2r1 − rA ) and Ot (rt + rA ) have a common tangent parallel to d as well.1 A At O1 Ot. Remark. Let k be the line through the midpoint of P1 P2 perpendicular to AB.6. But that implies that we can use Lemma 1 on their intersection (and radical axis). The Archimedean circles (Kt. Mt. This leads for instance to easy generalizations of (K3 ) and (K7 ).1 ) and (Kt.2 .2 and Mt. (Of course the parameter t does not really play a role here. but for reasons of uniformity we still use it in the naming).1 C Ct. More corollaries. (Ot.5. Theorem 2.11 ). Dt Kt.1 and Ot.2 ) and (Ot.13 ) and (Wt. (Ot ).94 F.1 O Ot Ot O2 Ot. 5. denote by u the distance between Ot. We end the adventures with a sole salinon.1 and Mt.10 ) and (Kt.2 B Bt Figure 18. Their circle.1 ).14 ).4 of the semicircular arcs (Ot ). M.4 ) are concyclic. Then the smallest circles through P1 and P2 tangent to k are Archimedean. the mid-arc circle (O5 ) of the salinon and the circle (Ot. To verify this. Similar characterizations can be found for (Kt.2 ). This circle meets AB in two points L on the left hand side and R on the right hand side. van Lamoen By straightforward calculation we can now verify that ZL2t + ZRt2 = Lt Rt2 and we have a new characterization of the families (Wt. More precisely they are vertices of a rectangle. We go back to Lemma 1. Archimedean circles from the mid-arc circle of the salinon.1 .2 equals u2 + (r2 − r1 )2 and the distance from O to AB .2 ) meet in two points.1 ) and (Kt. but then some more work has to be done. We leave this to the reader. 1Ot B Ot. The Archimedean circles (Kt.1 Kt.7 ) Mt D Dt Mt.3 E Ot.4 Ot Ct A C Ot.2 Bt Figure 19. See Figure 21.4 Figure 20.1 O1 At O Ct.11 ) 2 equals r1 +r 2 .2 E O5 Mt.2 Bt Mt.4 ) and (O5 ) intersects (Ot ) in two points at a distance of 2rA from AB.2 ) and (O5 ) to AB.7 Kt.4 Ot O Ct.1Ot B Ot.10 Ct A O1 At C Ot.2 O2 Ct.11 Ot. We can generalize (K12 ) and (K13 ) in a similar which leads to x = rr11+r 2 way. then 2 r1 + r2 u2 u2 + (r2 − r1 )2 = −x − x2 − 4 2 4 r2 = rA .2 O2 Ct. To see this we note that the circle with center on Ot Mt in the pencil generated by (Mt.Archimedean adventures D 95 Dt Kt.1 Ot.3 ) and (Kt.1 Kt. .10 ) and (Kt. The Archimedean circles (Kt. If x is the distance from the intersections of (Ot. Those ubiquitous Archimedean circles. Math. Are the twin circles of Archimedes really twins?.1 O Ct. reprinted in College Math. Nov. [3] L. Wendijk and T. 72 (1999) 202–213. Forum Geom.12 Ot.. [11] P. Dodge. 5 (2005) 75–96. 108 (2001) 770. Il Problema Geometrico.1Ot B Ot. Guy and R.1 Kt. The Netherlands E-mail address:
[email protected]. The marvelous arbelos. dal Compasso al Cabri. M. 1999. Power. [8] A. A mere coincidence. Yiu. Math. Bankoff. ed. Willibrordcollege.2 O2 Ct.2 Bt Mt.4 Ot Ct A O1 At C Ot.13 ) References [1] L. Forum Geom. Watanabe.. [9] P.K.12 ) and (Kt.4 Figure 21. d’Ignazio and E. Woo and P. Suppa.. Mag. [2] L. Forum Geom. Bankoff. Simple constructions of the incircle of an arbelos. 47 (1974) 214–218. Los Angeles City College. Private correspondence. Woodrow. Yiu. T. Mathematical Association of America. Floor van Lamoen: St. P.96 F. Okumura and M. Math. 1994. Mag.. 1 (2001) 133–136. The Archimedean circles (Kt. Fruitlaan 3.. Problem 10895. Journal. 23 (1992) 106. Forum Geom.W. 1954. Elegant geometric constructions. van Lamoen Mt D Dt Mt.Y.13 Kt. [5] I. R. Monthly.2 E O5 Mt. [7] F. solution. 5 (2005) 133–134. Some more Archimedean circles in the arbelos. [10] P. 247–253. in The lighter Side of Mathematics. The Archimedean circles of Schoch and Woo. [4] C. Schoch. Yiu. 4462 EP Goes. Mathematics Newsletter. Bankoff. [6] H. Woo. Amer. 4 (2004) 27–34. Hermann. 2001. Interlinea Editrice.nl . 110 (2003) 63–64.. T´eramo. k) = 1. i=1 and that for the reciprocal of diagonal length ratios when gcd(n. . We illustrate the formula for the product of diagonal length ratios min[k. Although the length of di equals that of dn−i . where s = min{j > 0 : kj ≡ ±1 mod n}. 2006.b Forum Geometricorum Volume 6 (2006) 97–101. d3 d2 d1 kπ n d1 d4 d5 dk π n d8 d6 Figure 1. . .h. Introduction Consider a regular n-gon.n−h] rh rk = r|k−h|+2i−1 .2 shows us that the the length of dk to that of dj is jπ sin n Publication Date: March 13. Figure 1. In particular. Number the diagonals d1 . .1 for n = 9) including the sides of the polygon as d1 and dn−1 . d2 . we shall use all n − 1 subscripts since this simplifies our formulae concerning the diagonal lengths. The ratio of sin kπ n . s 1 = rk(2j−1) . rk j=1 1. . b b FORUM GEOM ISSN 1534-1178 Proof by Picture: Products and Reciprocals of Diagonal Length Ratios in the Regular Polygon Anne Fontaine and Susan Hurley Abstract.2 Ratios of the lengths of the diagonals are given by the law of sines. dn−1 (as shown in Figure 1.1 d7 Figure 1. These “proofs by picture” link the geometry of the regular n-gon to formulae concerning the arithmetic of real cyclotomic fields. Communicating Editor: Paul Yiu.n−k. 4 illustrate what happens when the nonagon is enlarged so that l1 = s4 . Products of diagonal length ratios It is an exercise in the algebra of cyclotomic polynomials to show that min[k... Steinbach names it the diagonal product formula and makes use of it to derive a number of interesting properties of the diagonal lengths of a regular polygon.. This ratio of sines will be denoted sin πn rk . since rk = ssk1 = sl11 and rh = llh1 . from the second that l2 = .2 Figures 2. consider two regular n-gons with the side of the larger equal to some diagonal of the smaller.h. The summation formula for the diagonals can be visualized by projecting the left edge of the larger polygon onto each of its other edges in turn. In order to understand the geometry of the diagonal product formula.n−k. 2. We observe from the first pair of nonagons that l1 = s4 s1 = s4 . i=1 This formula appears in Steinbach [1] for the case where h + k ≤ n. .n−1 for the larger.n−1 for the smaller polygon and {li }i=1.1-2.h. . It is not hard to extend the formula to cover all n − 1 values of h and k. rk is simply the length of the k-th diagonal.. Note that if the length of the side d1 = 1. so that l1 = sk for some k..n−k.1 Figure 2.n−h] rh rk = r|k−h|+2i−1. l1 l2 s5 s3 s4 Figure 2. Denote the diagonal lengths by {si }i=1.n−h] lh = s|k−h|+2i−1. the product rk rh becomes slh1 and when we multiply through by s1 the diagonal product formula becomes min[k. In this case. Hurley sin kπ n ratio of the length of dk to that of d1 is .. each of the larger diagonal lengths is expressible as a sum of the smaller ones. Fontaine and S. i=1 In other words.98 A. . The sums are l1 =s2 s1 = s2 . l2 =s2 s2 = s1 + s3 . the diagonal length ratio rk is a unit in the ring of integers of the real subfield of Q[ξ]. l3 =s2 s3 = s3 + s4 . When n and k are both even. Knowing that this was the case. and from the last that l4 = s4 s4 = s1 + s3 + s5 + s7 . This is exactly what the diagonal product formula predicts. ξ a primitive n-th root of unity.4 s4 s2 = s3 + s5 .3 Figure 2. but the same strategy of projecting onto each side of the larger polygon in turn still works. n) = 1. n) = 1 and i ≤ n2 forms a basis for this field [2].2 s2 s4 Figure 3. so that ri = the length of the i-th diagonal.1 s3 s1 Figure 3. 2). we searched for a formula to express r1k as an integral linear combination of diagonal length ratios.Proof by picture 99 l3 s6 s4 s7 s5 l4 s3 s2 s1 Figure 2. l3 l2 l1 s2 Figure 3. k) = (6. This time we found the picture first. the polygons do not have the same orientation. Figure 3 shows the case (n. Reciprocal of diagonal length ratios Provided gcd(k.3 3. The set of ratios {ri } with gcd(i. from the third that l3 = s4 s3 = s2 + s4 + s6 . We assume that the polygon has unit side. as in Figure 4. rk j=1 Once we had discovered this formula. n − 1. .2 for (n. . the positive contributions to the sum occur as we move in one direction with respect to diagonal k. k) = (7. 2). until one arrives at a vertex adjoining the starting point. 3). will give the desired reciprocal. r2 1 r3 3π 8 π 8 r4 Start 1 r6 Figure 4. . . we have not been able to locate them. which is s 1 = rk(2j−1) . . Hurley First note that r1k is equal to the length of the line segment obtained when diagonal k intersects diagonal 2 as shown in Figure 4. we realized that ri = −r2n−i would have the correct sign to produce the simplest formula for the reciprocal. where s = min{j > 0 : kj ≡ ±1 mod n}. one can set off along diagonal k and zigzag back and forth.2 Following the procedure outlined above. Fontaine and S. Summing the lengths of the diagonals parallel to diagonal k. Also as we zigzag our way through the polygon. and would appreciate any lead in this regard. we found it that it was a messy but routine exercise in cyclotomic polynomial algebra to verify its truth. k) = (8.1 1 r2 End Figure 4. allowing ri = sin nπ to be n defined for i > n.1 for (n. with positive or negative sign according to the direction of travel. Although we are almost certain that these formulas for manipulating the diagonal length ratios must be in the classical literature. In order to express this length in terms of the ri . notice that. For example follow the diagonals shown in Figure 4. alternately parallel to diagonal k and in the vertical direction. the head of each directed diagonal used in the sum is k vertices farther around the polygon from the previous one.2 to see that r12 = r2 + r6 − r4 .100 A. i = 1. while the negative sin iπ contributions occur as we move the other way. In fact. Golden fields: a case for the heptagon. 12211. 70 (1997) 22–31. 12180.edu . Troy. New York. USA Susan Hurley: Department of Mathematics.Proof by picture 101 References [1] P. Mag. Math.. Introducion to Cyclotomic Fields. 2nd edition. New York. New York. Anne Fontaine: Hudson Valley Community College. C. Steinbach. Washington. 1982. Loudonville. USA E-mail address: HURLEY@siena. Siena College. Springer-Verlag. [2] L. . β and γ form an arbelos with inner semicircles α and β with diameters P A and P B respectively. Communicating Editor: Paul Yiu. Frank Power a+b [2] has shown that for “highest” points Q1 and Q2 of α and β respectively.b Forum Geometricorum Volume 6 (2006) 103–105. Publication Date: March 20. and intersect the line AB at points P1 and P2 respectively (see Figures 2 and 3). Let Q be the intersection of the circle γ and the perpendicular of AB through P . and let δ be a circle touching γ at the point Q from the inside of γ. Let three semicircles α. Circles with radii t = are called Archimedean circles. We generalize these Archimedean circles. Theorem. b b FORUM GEOM ISSN 1534-1178 A Generalization of Power’s Archimedean Circles Hiroshi Okumura and Masayuki Watanabe Abstract. . (2) The circle touching the circles γ and α at points different from A and the line P1 Q1 from the opposite side of B and the circle touching the circles γ and β at points different from B and the line P2 Q2 from the opposite side of A are congruent with common radii (1 − k)t. The tangents of δ perpendicular to AB intersect α and β at points Q1 and Q2 respectively. the circles touching γ and the line OQ1 (respectively OQ2 ) at Q1 (respectively Q2 ) are Archimedean (see Figure 1). 2006. We generalize the Archimedean circles in an arbelos given by Frank Power. Let a and b be the radii of the circles α ab and β. (1) The radii of the circles touching the circle γ and the line OQ1 (respectively OQ2 ) at the point Q1 (respectively Q2 ) are 2(1 − k)t. γ Q1 α A O1 O β P Q2 O2 B Figure 1 We denote the center of γ by O. The radius of δ is expressed by k(a + b) for a real number k satisfying 0 ≤ k < 1. Hence its radius is (1 − k) times of the radii of the twin circles of Archimedes. Q1 and the center of this circle. Let x be the radius of one of the circles touching γ and the line OQ1 at Q1 . we get (a + b − x)2 = x2 + (a − b)2 + 4kab Solving the equation for x. Watanabe Q γ Q1 α Q2 β A O P1 O1 P δ O2P2 B Figure 2 Proof. p. While |P1 Q1 |2 = |P P1 ||P1 A| = 2ka(2a − 2ka) = 4k(1 − k)a2 . (1) Since |P P1 | = 2ka. . Hence |OQ1 |2 = ((b − a) + 2ka)2 + 4k(1 − k)a2 = (a − b)2 + 4kab. 108]. Okumura and M. 2(1−k)ab a+b = 2(1 − k)t. |OP1 | = (b − a) + 2ka. The other case can Q γ Q1 α Q2 β A P1 O1 O P δ O2P2 B Figure 3 (2) The radius of the circle touching α externally and γ internally is proportional to the distance between the center of this circle and the radical axis of α and γ [1.104 H. we get x = be proved similarly. From the right triangle formed by O. Non-Archimedean twin circles in the arbelos. 13 (2005) no. Watanabe. 460-1 Kamisadori Maebashi Gunma 371-0816. The statement (2) is a generalization of the twin circles of Archimedes. Coolidge. Chelsea.A generalization of Power’s Archimedean circles 105 The Archimedean circles of Power are obtained when δ is the circle with a diameter OQ. Q2 and P coincide..jp Masayuki Watanabe: Department of Information Engineering. 5 (2005) 133–134. Japan E-mail address: watanabe@maebashi-it. Plus. Maebashi Institute of Technology.1. The twin circles with the half the size of the Archimedean circles in [4] are also obtained in this case. L. Maebashi Institute of Technology. Forum Geom.ac. Hiroshi Okumura: Department of Information Engineering. Japan E-mail address: okumura@maebashi-it. In this case the points Q1 . 60–62. 4 (2004) 229–251.ac.(in Bulgarian). Forum Geom.jp . 460-1 Kamisadori Maebashi Gunma 371-0816. and we get the circle with radius 2t touching the line AB at P and the circle γ by (1) [3]. Okumura and M. [2] Frank Power. Math. [3] H. Some more Archimedean circles in the arbelos. A treatise on the circle and the sphere.. 1971 (reprint of 1916 edition). The twin circles of Archimedes in a skewed arbelos. References [1] J. [4] H. which are obtained when δ is the point circle. Watanabe. Okumura and M. New York. . (Sb ). I discovered that the result would remain true if the circumcircle was replaced with any circle in the plane of ABC. and Z.b Forum Geometricorum Volume 6 (2006) 107–115. and CZ meet in a point. Communicating Editor: Paul Yiu. Let (S) be any circle in the plane of triangle ABC. This paper generalizes properties of mixtilinear incircles. and Z. and CZ are concurrent (Figure 1). Let the points of tangency of (Sa ). Introduction A mixtilinear incircle of a triangle ABC is a circle tangent to two sides of the triangle and also internally tangent to the circumcircle of that triangle. The point of concurrence is the external center of similitude of the incircle and the circumcircle. 1 Y A Y A Z Sb Sc Z B B C X Sa C X Figure 1 Figure 2 I wondered if there was anything special about the circumcircle in Proposition 1. respectively. and (Sc ) each tangent internally to (S). 1This is the triangle center X in Kimberling’s list [5]. 56 . Then it is shown that the lines AX. Paul Yiu discovered an interesting property of these mixtilinear incircles. (Sb ). and (Sc ) with (S) be X. b b FORUM GEOM ISSN 1534-1178 Pseudo-Incircles Stanley Rabinowitz Abstract. 2006. Y . In 1999. and (Sa ) is inscribed in angle BAC (similarly for (Sb ) and (Sc )). 1. If the points of contact of the mixtilinear incircles of ABC with the circumcircle are X. After a little experimentation. Suppose there are circles (Sa ). BY . Y . Publication Date: March 26. BY . Proposition 1 (Yiu [8]). then the lines AX. Then lines AX. Y . When we say that a circle is inscribed in an angle ABC. Suppose that there are three circles. Figure 2 shows an example where (S) surrounds the triangle and the three circles are all internally tangent to (S). Let (S) be the circle that is externally tangent to all three circles. (Sb ). Pseudo-excircles There is another way of viewing Figure 3a. we mean that the circle − −→ −−→ is tangent to the rays BA and BC. Let (S) be any circle in the plane of ABC. Let triangle ABC be the triangle that circumscribes these three circles (that is. and (Sc ). Let the points of tangency of (S) with the three circles be X. Figure 3c shows an example where (S) surrounds the triangle and the three circles are all externally tangent to (S). we can start with the other three circles. Y . (Sc ) are inscribed in ∠BAC. A A X Y X Z Y B C Figure 3a. each internally (respectively externally) tangent to (S). suppose (Sa ). Then we get the following proposition. There are many configurations that meet the requirements of Theorem 2. the three circles are inside the triangle and each side of the triangle is a common external tangent to two of the circles). (Sb ).108 S. ∠ABC. Proposition 3. BY . Figure 3b shows an example where (S) intersects the triangle. BY . (Sb ). Rabinowitz Theorem 2. The point P is the external (respectively internal) center of similitude of the incircle of ABC and circle (S). Pseudo-excircles Z B C Figure 3b. . each external to the other two. Let the points of tangency of (Sa ). and Z (Figure 3a). Furthermore. and CZ are concurrent. and (Sc ) with (S) be X. Let there be given three circles in the plane. Figure 3a shows an example of pseudo-excircles where (S) lies inside the triangle. Then the lines AX. and CZ are concurrent at a point P . A pseudo-excircle of the triangle is a circle that is tangent to two sides of the given triangle and externally tangent to the given circle. Given a triangle and a circle. (Sa ). Instead of starting with the triangle and the circle (S). Definitions. and Z. respectively. ∠ACB respectively (Figure 2). a pseudo-incircle of the triangle is a circle that is tangent to two sides of the given triangle and internally tangent to the given circle. Pseudo-incircles 109 A A Z Y Z Y C B X X B C Figure 3c. Pseudo-excircles Figure 4. Pseudo-incircles Figure 4 also shows pseudo-incircles (as in Figure 2), but in this case, the circle (S) lies inside the triangle. The other three circles are internally tangent to (S) and again, AX, BY , and CZ are concurrent. This too can be looked at from the point of view of the circles, giving the following proposition. Proposition 4. Let there be given three mutually intersecting circles in the plane. Let triangle ABC be the triangle that circumscribes these three circles (Figure 4). Let (S) be the circle that is internally tangent to all three circles. Let the points of tangency of (S) with the three circles be X, Y , and Z. Then lines AX, BY , and CZ are concurrent. We need the following result before proving Theorem 2. It is a generalization of Monge’s three circle theorem ([7, p.1949]). a Iab b Ica Ibc Eab c Eca Figure 5. Six centers of similitude Ebc 110 S. Rabinowitz Proposition 5 ([1, p.188], [2, p.151], [6]). The six centers of similitude of three circles taken in pairs lie by threes on four straight lines (Figure 5). In particular, the three external centers of similitude are collinear; and any two internal centers of similitude are collinear with the third external one. 2. Proof of Theorem 2 Figure 6 shows an example where the three circles are all externally tangent to (S), but the proof holds for the internally tangent case as well. Let I be the center of the incircle. A Sa X S P Y I Z Sc Sb B C Figure 6 Consider the three circles (S), (I), and (Sc ). The external common tangents of circles (J) and (Sc ) are sides AC and BC, so C is their external center of similitude. Circles (S) and (Sc ) are tangent externally (respectively internally), so their point of contact, Z, is their internal (respectively external) center of similitude. By Proposition 5, points C and Z are collinear with P , the internal (respectively external) center of similitude of circles (S) and (I). That is, line CZ passes through P . Similarly, lines AX and BY also pass through P . Corollary 6. In Theorem 2, the points S, P , and I are collinear (Figure 6). Proof. Since P is a center of similitude of circles (I) and (S), P must be collinear with the centers of the two circles. 3. Special Cases Theorem 2 holds for any circle, (S), in the plane of the triangle. We can get interesting special cases for particular circles. We have already seen a special case in Proposition 1, where (S) is the circumcircle of ABC. Pseudo-incircles 111 3.1. Mixtilinear excircles. Corollary 7 (Yiu [9]). The circle tangent to sides AB and AC of ABC and also externally tangent to the circumcircle of ABC touches the circumcircle at point X. In a similar fashion, points Y and Z are determined (Figure 7). Then the lines AX, BY , and CZ are concurrent. The point of concurrence is the internal center of similitude of the incircle and circumcircle of ABC. 2 A Y Z B C X Figure 7. Mixtilinear excircles 3.2. Malfatti circles. Consider the Malfatti circles of a triangle ABC. These are three circles that are mutually externally tangent with each circle being tangent to two sides of the triangle. Corollary 8. Let (S) be the circle circumscribing the three Malfatti circles, i.e., internally tangent to each of them. (Figure 8a). Let the points of tangency of (S) with the Malfatti circles be X, Y , and Z. Then AX, BY , and CZ are concurrent. Corollary 9. Let (S) be the circle inscribed in the curvilinear triangle bounded by the three Malfatti circles of triangle ABC (Figure 8b). Let the points of tangency of (S) with the Malfatti circles be X, Y , and Z. Then AX, BY , and CZ are concurrent. 2This is the triangle center X in Kimberling’s list [5]; see also [4, p.75]. 55 112 S. Rabinowitz A A X X Z Y Z Y B C B Figure 8a. Malfatti circumcircle C Figure 8b. Malfatti incircle 3.3. Excircles. Corollary 10 (Kimberling [3]). Let (S) be the circle circumscribing the three excircles of triangle ABC, i.e., internally tangent to each of them. Let the points of tangency of (S) with the excircles be X, Y , and Z. (Figure 9). Then AX, BY , and CZ are concurrent. The point of concurrence is known as the Apollonius point of the triangle. It is X181 in [5]. See also [4, p.102]. Y A A Z Y Z C B B X C X Figure 9. Excircles Figure 10. Nine-point circle If we look at the circle externally tangent to the three excircles, we know by Feuerbach’s Theorem, that this circle is the nine-point circle of ABC (the circle that passes through the midpoints of the sides of the triangle). Pseudo-incircles 113 Corollary 11 ([4, p.158]). If the nine point circle of ABC touches the excircles at points X, Y , and Z (Figure 10), then AX, BY , and CZ are concurrent. 4. Generalizations In Theorem 2, we required that the three circles (Sa ), (Sb ), and (Sc ) be inscribed in the angles of the triangle. In that case, lines ASa , BSb , and CSc concur at the incenter of the triangle. We can use the exact same proof to handle the case where the three lines ASa , BSb , and CSc meet at an excenter of the triangle. We get the following result. Theorem 12. Let (S) be any circle in the plane of ABC. Suppose that there are three circles, (Sa ), (Sb ), and (Sc ), each tangent internally (respectively externally) to (S). Furthermore, suppose (Sa ) is tangent to lines AB and AC; (Sb ) is tangent to lines BC and BA; and (Sc ) is tangent to lines CA and CB. Let the points of tangency of (Sa ), (Sb ), and (Sc ) with (S) be X, Y , and Z, respectively. Suppose lines ASa , BSb , and CSc meet at the point J, one of the excenters of ABC (Figure 11). Furthermore, assume that sides BA and BC of the triangle are the external common tangents between excircle (J) and circle (Sa ); similarly for circles (Sb ), and (Sc ). Then AX, BY , and CZ are concurrent at a point P . The point P is the external (respectively internal) center of similitude of circles (J) and (S). The points J, P , and S are collinear. J P X S A Y Z B C Figure 11. Pseudo-excircles Figure 12 illustrates the case when (S) is tangent internally to each of (Sa ), (Sb ), (Sc ). 114 S. Rabinowitz X S A B Y Z C Figure 12. Pseudo-incircles Examining the proof of Theorem 2, we note that at each vertex of the triangle, we have found that the line from that vertex to the point of contact of the circle inscribed in that angle and circle (S) passes through a fixed point (one of the centers of similitude of (S) and the incircle of the triangle). We note that the result and proof would be the same for a polygon provided that the polygon had an incircle. This gives us the following result. Theorem 13. Let A1 A2 A3 . . . An be a convex n-gon circumscribed about a circle (J). Let (S) be any circle in the interior of this n-gon. Suppose there are n circles, (S1 ), (S2 ), . . . , (Sn ) each tangent externally to (S) such that for i = 1, 2, . . . , n, circle (Si ) is also inscribed in angle Ai−1 Ai Ai+1 (where A0 = An and An+1 = A1 ). Let Xi be the point of tangency of circles (Si ) and (S) (Figure 13). Then the lines Ai Xi , i = 1, 2, . . . , n are concurrent. The point of concurrence is the internal center of similitude of the circles (S) and (J). In order to generalize Theorem 2 to three dimensions, we need to first note that Proposition 5 generalizes to 3 dimensions. Theorem 14. The six centers of similitude of three spheres taken in pairs lie by threes on four straight lines. In particular, the three external centers of similitude are collinear; and any two internal centers of similitude are collinear with the third external one. Proof. Consider the plane through the centers of the three spheres. This plane passes through all 6 centers of similitude. The plane cuts each sphere in a circle. Thus, on this plane, Proposition 5 applies, thus proving that the result holds for the spheres as well. Theorem 15. Let T = A1 A2 A3 A4 be a tetrahedron. Let (S) be any sphere in the interior of T (or let (S) be any sphere surrounding T ). Suppose there are four Pseudo-incircles 115 A1 X1 A5 A2 X5 X2 X4 A4 X3 A3 Figure 13. Pseudo-excircles in a pentagon spheres, (S1 ), (S2 ), (S3 ), and (S4 ) each tangent internally (respectively externally) to (S) such that sphere (Si ) is also inscribed in the trihedral angle at vertex Ai . Let Xi be the point of tangency of spheres (Si ) and (S). Then the lines Ai Xi , i = 1, 2, 3, 4, are concurrent. The point of concurrence is the external (respectively internal) center of similitude of sphere (S) and the sphere inscribed in T . The proof of this theorem is exactly the same as the proof of Theorem 2, replacing the reference to Proposition 5 by Theorem 14. It is also clear that this result generalizes to En . References [1] [2] [3] [4] [5] [6] [7] [8] [9] N. Altshiller-Court, College Geometry, 2nd edition, Barnes & Noble. New York: 1952. R. A. Johnson, Advanced Euclidean Geometry, Dover Publications, New York: 1960. C. Kimberling, Problem 1091, Crux Math., 11 (1985) 324. C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1–285. C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html. R. Walker, Monge’s theorem in many dimensions, Math. Gaz., 60 (1976) 185–188. E. W. Weisstein, CRC Concise Encyclopedia of Mathematics, 2nd edition. Chapman & Hall. Boca Raton: 2003. P. Yiu, Mixtilinear incircles, Amer. Math. Monthly, 106 (1999) 952–955. P. Yiu, Mixtilinear incircles, II, preprint, 1998. Stanley Rabinowitz: MathPro Press, 10 Cliff Road, Chelmsford, Massachusetts, 01824. USA E-mail address:
[email protected] . The geometry associates with each such cubic a cubic of type pK. we show that each cubic of type nK but not of type cK has a parent of type pK. R) : rx(wy 2 − vz 2 ) + sy(uz 2 − wx2 ) + tz(vx2 − uy 2 ) = 0. nK(W. that is circumcubics which are invariant under an isoconjugation. We have nK0 (W. We are interested in isocubics. Throughout. not only do (some) points of a desmic structure lie on a cubic. We may view W as the image of G under the isoconjugation.b Forum Geometricorum Volume 6 (2006) 117–138. and that each cubic of type pK has three sisters of type nK0 . The point R = r : s : t is known as the pivot of the cubic pK. The constant k in the latter equation is determined by a point on the curve which is not on a sideline. each cubic of type pK has infinitely many child cubics. One important subclass of type nK occurs when k = 0. We call this the parent cubic. 2006. so the geometry must be rather different. k) : rx(wy 2 + vz 2 ) + sy(uz 2 + wx2 ) + tz(vx2 + uy 2 ) + kxyz = 0. and the root of the cubic nK W = u : v : w is the pole of the isoconjugation. #3991 and follow-up. Introduction In Hyacinthos. We write the coordinates of P as p : q : r. but just three of type nK0 . R. They showed that each cubic of type nK0 has an associated sister of type pK. we meet many familiar cubics. There we learn that an isocubic has an equation of one of the following forms : pK(W. Communicating Editor: Bernard Gibert. Along the way. The key is the existence of a desmic structure associated with parent and child. We show that each cubic of type nK which is not of type cK can be described as a Grassmann cubic. The theory of isocubics is beautifully presented in [1]. This means that the isoconjuagte of X = x : y : z is ux : yv : wz . Here. b b FORUM GEOM ISSN 1534-1178 Grassmann cubics and Desmic Structures Wilson Stothers Abstract. R) : rx(wy 2 + vz 2 ) + sy(uz 2 + wx2 ) + tz(vx2 + uy 2 ) = 0. Our results do not appear to extend to cubics of type cK. but also that they actually generate the cubic as a locus. Another interpretation of k appears below. we use barycentric coordinates. On the other hand. . Ehrmann and others gave a geometrical description of cubics as loci. This extends work of Wolk by showing that. Publication Date: April 3. 1. and every cubic of type pK has infinitely many children. The B. S). as befits a polar object. (b) The N KW -transform of P is N K(W. C P ∩ AB are collinear}. P ) = pu(r 2 v + q 2 w) : qv(p2 w + r 2 u) : rw(q 2 u + p2 v). this has equation pC(W. Notice that. k) with W = f 2 : g2 : h2 . Let ∆ = ABC be the reference triangle. (a) GP (∆ ) = {P : the triangle with vertices A P ∩ BC. the A-harmonic of X is the point XA with barycentrics −x : y : z. Definition 1. in apparently unrelated contexts. XB and XC . R) : rx(hy − gz)2 + sy(f z − hx)2 + tz(gx − f y)2 = 0. pC(W. Definition 2. 2. These are defined in terms of isogonal conjugation. then P K(W. The equation has the form cK(#F. Note that P K(W. Also. let P ∗ denote the W-isoconjugate of P. and k = −2(rgh + shf + tf g). then R is on pC(W. B is not on CA and C is not on AB. if S is on pC(W. We also remark that the equation for the polar conic vanishes identically when W is the barycentric square of R. Suppose that W is a fixed point. Grassmann cubics associated with desmic Structures Definition 3.118 W. These are defined in terms of the root R and node F = f : g : h. it is convenient to regard the class of pK(W. Observe that XA XB XC is the anticevian triangle of X. In such a case. XC also lie on pC(W. R) itself. Hence. P ) is the perspector of the circumconic through P and P∗ . We shall need the general case for W -isoconjugation. N K(W. the intersection of the tripolars of P and P ∗ (if P = P ∗ ). (b) GN (∆ ) = {P : A P ∩ BC. such as R. B P ∩ CA. In the preamble to X(2081) in [5] we meet Gibert’s P K. R). R. R). (a) For a point X with barycentrics x : y : z. each subclass having a common polar conic. R) is undefined if W = R2 . We require a further concept from [1]. It is the nK(W. Note that P K(W. R) contains the four fixed points of the isoconjugation. From these definitions. if X is on pC(W. P ) occurs several times in our work. for fixed W . B P ∩ CA. Also. R) : (vt2 − ws2 )x2 + (wr 2 − ut2 )y 2 + (us2 − vr 2 )z 2 = 0. if W = u : v : w and P = p : q : r. P ). R). that of the polar conic of R for pK(W. the crosspoint of P and P∗ . so F is a fixed point of the isoconjuation. so is determined by one other point.P).and N K-transforms. P K(W. (a) The P KW -transform of P is PK(W. C P ∩ AB is perspective with ∆}. R) then XA . Let ∆ = A B C where A is not on BC. P ) = pu(r 2 v − q 2 w) : qv(p2 w − r 2 u) : rw(q 2 u − p2 v). R) as consisting of just pK(W. . the class of cubics of type pK splits into disjoint subclasses. (b) The harmonic associate of the triangle P QR is the triangle PA QB RC .and C-harmonics. are defined analagously. Stothers Another subclass consists of the conico-pivotal isocubics. In our notation. XB . As we shall see. X(20)) is K004 in Gibert’s catalogue. B. and others where they can be identified. BBP and CCP . then AP BP CP = A P ∩ BC = 0 : a2 x − a1 y : a3 x − a1 z. Under the properties listed is the fact that it is the locus of points whose pedal triangle is a cevian triangle. . = B P ∩ CA = b1 y − b2 x : 0 : b3 y − b2 z. Lemma 2. if P = x : y : z. C and A . = C P ∩ AB = c1 z − c3 x : c2 z − c3 y : 0. If ∆ = A B C has A = a1 : a2 : a3 . there are other examples where suitable triangles are listed.Grassmann cubics and desmic structures 119 We note that GN (∆) is a special case of a Grassmann cubic. B = b1 : b2 : b3 . The Darboux cubic pK(K.Gibert’s K216 it is mentioned that it is GN for a certain triangle. Proof. There are three examples in the current (November 2004) edition of [2]. If ∆ = A B C has A = a1 : a2 : a3 . The determinant is formed from the previous one by changing the sign of one entry in each row. X(5)) . B . (b) A B C is triply perspective with ABC if and only if a1 b2 c3 = a2 b3 c1 = a3 b1 c2 . This is another determinant condition. (b) GN (∆ ) has equation (a2 x − a1 y)(b3 y − b2 z)(c1 z − c3 x) − (a1 z − a3 x)(b1 y − b2 x)(c2 z − c3 y) = 0. We observe that the perspectivity in (a) is equivalent to the concurrence of AA . C . (a) GP (∆ ) has equation (a2 x − a1 y)(b3 y − b2 z)(c1 z − c3 x) + (a1 z − a3 x)(b1 y − b2 x)(c2 z − c3 y) = 0.Gibert’s K001. The given equality expresses the condition for the intersection of AA and BB to lie on CC . C = c1 : c2 : c3 . X(30)) . The corresponding GN is then the union of the line at infinity and the circumcircle. This is GP (A B C ). Part (b) follows by noting that A B C is triply perspective with ABC if and only if each of A B C . where the vertices are the infinite points on the altitudes. In the discussion of the cubic nK(K. The condition in GP (∆) is equivalent to the concurrence of AAP . Lemma 1. and that the corresponding GP is the Neuberg cubic. It is easy to see that. BB and CC . then. pK(K. We will also require a condition for a triangle A B C to be perspective with ABC. B = b1 : b2 : b3 . (c) Each locus contains A. This gives (a). B C A and C A B is perspective with ABC. This gives (b). (a) A B C is perspective with ABC if and only if a2 b3 c1 = a3 b1 c2 . C = c1 : c2 : c3 . a degenerate cubic. The condition in GN (∆) is equivalent to the vanishing of the determinant of the coefficients of these points. Once we have the equations. then. Proof. (c) is clear. The equation in Lemma 1(b) has this property if and only if (a2 b3 c1 − a3 b1 c2 )D = 0. and C not on AB. with A not on BC. Suppose that A = a1 : a2 : a3 . From the equation in Lemma 1(b). The equation in Lemma 1(a) has k = 0 if and only if a2 b3 c1 = a3 b1 c2 . This establishes (a). so P is on GN (∆ ). Suppose that GN (∆ ) is non-degenerate. also perspective with ABC. the intersection of P A with BC lies on L. so ∆ and hence GN (∆ ) are degenerate. B not on CA. . B and C. B not on CA. and P C with AB. we are led to consider a further triangle. the other condition is equivalent to the perspectivity of the triangles. as do those of P B with CA. Suppose that A B C is perspective to ABC. These two conditions are equivalent to triple perspectivity by Lemma 2(b). and C not on AB. with A not on BC. Then (c) GN (∆ ) is of type nK if and only if ∆ is perspective with ABC. (b) If ∆ is degenerate. B = b1 : b2 : b3 . Our work so far leads us to consider triangles A B C perspective to ABC. as in (c). k = a2 b3 c1 + a3 b1 c2 − 2a1 b2 c3 . By Lemma 2(a). with a1 b2 c3 = 0. This turns out to be the desmic mate of A B C . we require a2 b3 c1 − a3 b1 c2 = 0. C = c1 : c2 : c3 . (d) GN (∆ ) is of type nK0 if and only if ∆ is triply perspective with ABC. For any P on L. (b) If ∆ is degenerate. Thus. (a) We begin by observing that equation (1) gives a cubic of type pK if and only if f1 g1 f1 + f2 g2 h2 = 0 and k = 0. Theorem 3. Now suppose that the locus GN (∆ ) is non-degenerate. then A . B and C . (c) Equation (1) gives a cubic of type nK if and only if f1 g1 f1 − f2 g2 h2 = 0. B and C lie on a line L. Now. so that it must be degenerate. (d) For a cubic of type nK0 . Looking at our loci. (1) This has the correct form for the cubic to be of type pK or nK. then GN (∆ ) is degenerate. D = 0 if and only if A . Theorem 4.120 W. We also require that k = 0. these intersections are collinear. By Lemma 2(a). Now the locus contains the line L. B and C are collinear. A Maple calculation shows that the other condition for a pK is satisfied if the triangles are perspective. Stothers We observe that each of the equations in Lemma 1 has the form x(f1 y 2 + f2 z 2 ) + y(g1 z 2 + g2 x2 ) + z(h1 x2 + h2 y 2 ) + kxyz = 0. this is the condition for the triangles to be perspective. so we are led to consider desmic structures which include the points A. For a triangle ∆ = A B C . (a) GP (∆) is of type pK if and only if ∆ is perspective with ABC. Proof. Let the perspector be P1 = p11 : p12 : p13 . where D is the determinant of the matrix whose rows are the coordinates of A . 2(p21 p22 p23 − p11 p12 p13 )). (g) GP (A B C ) = GP (A B C ). S) is a harmonic range. (g). R = p11 − p21 : p12 − p22 : p13 − p23 . R. two vertices and the intersection with the given tripolar. (b) GP (A B C ) = pK(W. C must be as described. Suppose that a point X on B C lies on the locus. Let A = p11 : p22 : p23 . (i) Once again. . The common locus includes P1 . with perspector S. S). (j) P1 and P2 lie on pK(W. P2 . the coordinates of the vertices A . (j) This is a routine verification. we get the same equations. this can be checked by calculation.(h) First. The fact that the given points lie on the respective loci are simply verifications. R). C = p11 : p12 : p23 . the tripolar of R. B . we have A = p21 : p12 : p13 .(c) These are simply verifications using the equations in Lemma 1. P2 and S. (h) GN (A B C ) = GN (A B C ). A B C and A B C have common perspectrix. S = p11 + p21 : p12 + p22 : p13 + p23 . (i) ABC. (b). with perspector P2 = p21 : p22 : p23 . (e) A B C is perspective with A B C . We can also argue geometrically. B = p21 : p12 : p23 . then X = BC ∩B C . (f) The coordinates of the points make this clear. C . These are the W -isoconjugates of A . R. The identification of the perspectrix uses the fact that the cubic meets the each sideline of ABC in just three points. Then XB and XC are B C .Grassmann cubics and desmic structures 121 (a) Suitably normalized. (d) This follows at once from the coordinates of A . (f) (P1 . (e) This requires the calculations that the lines A A . we note that. so this must be the common line. B B and C C all pass through S. (d) A B C is perspective with ABC. if we interchange the roles of P1 and P2 . This shows that we must have a common perspectrix. Proof. (a) Since we have the perspector. Let W = p11 p21 : p12 p22 : p13 p23 . But X is on B C . C = p21 : p22 : p13 . B . (c) GN (A B C ) = nK(W. B and C . so X = BC ∩B C . The common locus includes the intersections of the tripolar of R with the sidelines. If X also lies on B C . Then XA must meet BC on this line. B = p11 : p22 : p13 . the sum and difference of those of P1 and P2 . Note that the conditions on P are necessary to ensure that S can be expressed in the stated from. k) and pK(W. pK(W. and the corresponding nK as a Grassmann cubic. we have a desmic structure with the twelve points described as vertices A. As P is on pK(W. P2 . W = u : v : w. we say that the perspectors are normalized. (1) Is every cubic of type pK a locus of type GP associated with a desmic structure? (2) Is every cubic of type nK a locus of type GN associated with a desmic structure? The answer to (1) is that. S). We then show that this choice does lead to a description of the cubic as a Grassmann cubic. and R as the harmon of the structure. Suppose that P is a point on pK(W. respectively. C = mp : mq : nw/r. R. S). put A = nu/p : mq : mr. B . For the second question. B. C. RB ). Notice that the desmic structure in not determined by its perspectors. There are two obvious questions. First. this has locus GP (A B C ) = pK(W. The first result uses the idea of A-harmonics introduced in the Introduction. k) are A. The answer to (2) is more complicated. If P = p : q : r. S is on P P ∗ . This identifies the vertices of the structure. perspectors P1 . and is not S or its W -isoconjugate. B. B. at the intersections of the tripolar of R with AB and AC. with mn = 0. For each other cubic of type nK. When P2 is suitably scaled. Many authors describe S as the desmon. C . we use the notation of Theorem 4. Then S = mP + nP ∗ . R. A . He observed that the twelve points all lie on pK(W. A . This description of a desmic structure with vertices including A. we shall write X∗ for the W -isoconjugate of a point X. Stothers In the notation of Theorem 4. and at two further points. S) which is not fixed by W isoconjugation. B. C and perspector P1 = P with locus GP = pK(W. and P2 as the coperspector. S. C and intersections of nK(W. What may be new is the fact that the other vertices may be used to generate this cubic as a locus. k) with the cubics pK(W. we show that a cubic of type nK has at most one suitable desmic structure. For brevity. Throughout. provided neither perspector is on a sideline. RC ). each point on a pK is a perspector of a suitable desmic structure. we need to scale P2 so that the barycentrics of the desmon and harmon are. This is the class of conico-pivotal isocubics. n. From Theorem 4. C . Proof. If we choose barycentrics for P1 . Theorem 6. there is a unique desmic structure. B = mp : nv/q : mr. The normalization is determined by a single vertex. B . Theorem 5. S). for some constants m. RA ) touch at B and C. (a) The vertices of the desmic structure on nK(W. with six exceptions. S). . we proceed in two stages. intersect at A. (b) nK(W.122 W. R. RA ). C is discussed by Barry Wolk in Hyacinthos #462. pK(W. There is a class of nK which do not possess a suitable desmic structure. Then there is a unique desmic structure with vertices A. Some refer to P1 as the perspector. we require two perspectors interchanged by W -isoconjugation. These are clearly W -isoconjugate. The intersections with the sidelines in each case include the stated meetings with the tripolar of R. and this involves the product of the first coordinates. there are two unaccounted for. From Theorem 4. with two on each of the associated pK(W. The discriminant vanishes precisely when nK(W. R). C. 2vrt/s or 2wrs/t. The point A = p21 : p12 : p13 is on nK(W. RA ) touch at B and C. the six points do constitute a suitable desmic structure. . Part (b) is largely computational. The tangents at B and C coincide. in any other case. Now consider the loci derived from the perspectors P1 and −P2 . We observe that the vertices of the desmic structure can be derived from the normalized versions of the perspectors P1 and P2 . R) and nK(W. Thus (a) holds. In the proof. Provided W is not on a sideline. For part (c). k) and pK(W. there is at most one desmic structure defining the cubic. B and C. Then we need only verify that the cubics meet twice at F . we postpone the discussion of these cases to an Appendix. Thus. Also from Theorem 6. then cK(#F. there are at most six candidates. for some k . RC ) gives the other cases listed. X = A. k ). k). The normalization is such that R = P1 −P2 and S = P1 +P2 . R. The structure is a degenerate kind of desmic structure. (d) If the final two points in (b) coincide. and twice at F. Obviously the cubics meet at A.Grassmann cubics and desmic structures 123 (c) If W = f 2 : g2 : h2 . RA ). k) is of type cK. R. there is no structure if the cubic is a cK(#F. R. Solving the equations for nK(W. R). We need to investigate the cases where either of the final solutions in Theorem 6(b) lie on a sideline. z}. so there are two intersections here. RX ). R. Looking at other pK(W. and meet at A. To describe a cubic of type nK as a Grassmann cubic as above. Part (d) relies on a Maple calculation. we need six vertices not on a sideline. we assume that we can choose an intersection of nK(W. we get the known points and the solutions of a quadratic. Since two cubics have a total of nine meets. where F is such that W = F 2 . Proof. They still contain a unique structure which can be used to generate the cubics as Grassmann loci. for then the “six” points and the perspectors are all F . The quadratic equation in (d) has constant term zero precisely when k = 2vrt/s. k) and pK(W. Now F is clearly on pK(W. S. B. RB ). Rather than interrupt the general argument. Part (e) uses the same computation. we use the result of (b) to get seven intersections. so we need the discussion of the Appendix to tidy up the remaining cases. the cubics are pK(W. RA ) not on a sideline. R) and pK(W. R). at the intersections of the tripolar of R with AB and AC. where W = u : v : w. R. But F is a double point on cK(#F. RA ) as only the first term is affected by the sign change. R). We show that. RA ) for {y. so that AA = −p21 : p12 : p13 is on pK(W. Then A is on pK(W. R = r : s : t. (e) If either of the final points in (b) lie on a sideline. then nK(W. then k = 2ust/r. k) = cK(#F. pK(W. After Theorem 6. We can solve for p11 in terms of p12 . . So far. This gives three relations in p11 and p21 in terms of p12 and p13 . We also require that u : v : w = p11 p21 : p12 p22 : p13 p23 . Provided that we can choose A also on C. and k = 2rst. But then k = 2ust/r = 2vrt/s = 2wrs/t. there are just two such choices of A . Suppose that C = pK(W. RA ). R.124 W. We require perspectors P1 = p11 : p12 : p13 . we saw that each cubic C of type nK which is not of type cK is the Grassmann cubic associated with a unique desmic structure. Proof. This amounts to two linear equations which can be used to solve for p22 and p23 in terms of p11 . RC ). Theorem 5 shows that a cubic C of type pK contains infinitely many desmic structures. so we have just one suitable desmic structure. S). each defining a cubic of type nK. 3. k ). p12 . p13 and p21 . It follows that there is a unique desmic structure unless C has the points −u/r : v/s : w/t. p13 and p21 . S) which is not S and not fixed by W -isoconjugation. If a cubic C is of type nK. R. and these are isoconjugate. This involves the equation of the polar conic of S. k). so we get C as a Grassmann cubic. The GP locus is the whole plane. Suppose that C = nK(W. Gibert observes this for K001 and K216. RA ). R. and R = r : s : t. but not of type cK. This uses a Maple calculation. We could equally use a point of intersection of C with pK(W. u/r : −v/s : w/t and u/r : v/s : −w/t. S) with W = u : v : w. and similarly for B . It follows that C is the degenerate cubic (ty + sz)(rz + tx)(sx + ry) = 0. On the other hand. S) : (vt2 − ws2 )x2 + (wr 2 − ut2 )y 2 + (us2 − vr 2 )z 2 = 0. We call each of these cubics a child of C. with R on pC(W. and S = r : s : t. Our first task is to describe the children of a cubic pK(W. p13 = v/s and p21 = w/t. and P2 = p21 : p22 : p23 such that r : s : t = p11 − p21 : p12 − p22 : p13 − p23 . R). It is easy to check that this is given as a Grassmann cubic by the degenerate desmic structure with A = A = −r : s : t. RB ) or with pK(W. We call the cubic C the parent of C. (b) If R is a point of pC(W. He refers to K216 as a sister of K001. then there is a unique child of C of the form nK(W. k) which is not of type cK has parent of the form pK(W. then there is a unique desmic structure which defines C as a Grassmann cubic. we have shown that. k) with W = u : v : w. In Theorem 7. (c) Any cubic nK(W. S and C . then we can reconstruct perspectors which give rise to some nK(W. In [2. (a) Any child of C is of the form nK(W. Parents and children The reader will have noted the resemblance between the equations for the cubics GP (∆ ) and GN (∆ ). R. B . Our calculations also give information on the parents of the family of cubics of type nK with fixed pole and root. S) with S on pC(W. Theorem 8. provided we do not have (after scaling) p12 = −u/r. notes on K216]. This has perspectors. R. k). desmon and harmon equal to R. These are consistent provided A = p21 : p12 : p13 is on pK(W. see §1. and hence with a unique cubic C of type pK. but not on a sideline. so that W = R2 . then k = k directly. Stothers Theorem 7. As we saw in Theorem 6. if A is on pK(W. We now have additional centers on pC(K. As noted above. S). Roots and pivots If we have a cubic nK(W. X(523). From Theorem 4(a). X(15)} gives a cubic of the form nK(K.Grassmann cubics and desmic structures 125 (d) If nK(W. k). and X(30). S) and on pK(W. The pair {X(14). S). we get the required perspectors. S) = 0. C and the four points fixed by W -isoconjugation. R) if and only if R is on pC(W. k). it is noted that it is a rectangular hyperbola passing through I. we can establish them quite simply by “guessing” the pole of RS with respect to pC(W. R. the excenters. Suppose that R = x : y : z. As they are isogonal conjugates. Now these cubics meet at A. X(5). In terms of triangle centers. k) of C arises from a desmic structure. From Theorem 8(c). The information in [2] identifies it as K216. The perspectors P1 . X(396). By Theorem 4(f). R. (c) is really just the observation that S is on pC(W. X(30)) is mentioned in [5] in the discussion of its center. X(523) is on the polar conic. R). We also have the harmonic associates of each of these points! 4. X(74)}. we can reverse the process in (a) to obtain a suitable value for (m/n)2 . R). . S) is a harmonic range. S). R) and pK(W. the most prolific parent seems to be the Neuberg cubic = pK(K. (b) Given such an R. we can identify S from an equation for RS. The pair {O. then it has a parent cubic pK(W. the perspectors must be at infinity. X(30)). X(30). u v w If we eliminate m2 and n2 from these. k). S). we see that its asymptotes pass through X(74) and X(110). Example 1. X(5). Suppose the perspectors are P1 and P2 . Since R is also on the conic. the other infinite point must be X(523). The polar cubic pC(K. the Tixier point. The pair {X(13). Of course. k) defined by a desmic structure. Although the results were found by heavy computations. W is the barycentric product of P1 and P2 . being rectangular. X476. (d) In Theorem 4. we get pC(W. These have already been noted as lying on K001. X(16)} gives a cubic of the form nK(K. (a) We know from Theorem 7 that a child nK(W. then the perspectors are the non-trivial intersections of pK(W. k) has parent pK(W. B. we noted that the perspectors lie on pK(W. X(396). X(395). P2 of desmic structures on K001 must be its isogonal pairs other than {X(30). P2 . There. R. Choosing either root. From Theorem 4(f) (P1 . X(395). they must be the infinite circular points. There must be just two other (non-trivial) intersections. we know that S is on pC(W. Using the information in [5]. k). Since X(523) is not on the cubic. the root of the child cubic must be the mid-point of P1 and P2 . as well as X(1). X(523). so we also have a child of the form nK(K. R. Then we have m2 y 2 − n 2 s 2 m2 z 2 − n 2 t2 m2 x2 − n2 r 2 = = . X(30)). Gibert’s K001. Proof. R). H} gives a cubic of the form nK(K. It follows that there are constants m and n with P1 = mR + nS and P2 = −mR + nS. R) if and only if P . R = r : s : t and k are such that nK(W. Thus the perspectors arise as the intersections of a line and (circum)conic. R) which are of the form nK0 (W. (b) The point S has first barycentric coordinate 4r(−r2vw + s2wu + t2uv) − 4kstu + k2r. the line RS. R) is vt2 − ws2 = (gh − g h )(hg − gh ). It would also allow us to find the perspectors since these are the W -isoconjugate points on the line RS. S is as stated.the Napolean cubic. . k = 2(f g h − f gh). Gibert’s K024. k). T ). S ∗ } are a pair of W -isoconjugates on pK(W. O) . The second Brocard cubic nK0 (K. R. Then we have r = f − f . P K(W. v. It follows that the perspectors of the desmic structure are the intersections of the Brocard axis with the Kiepert hyperbola. Gibert’s K018. Our next result identifies the children of a given pK(W. P ∗ } = {S. t = h − h . This result allows us to find the parent of a cubic. Theorem 9(b) gives the parent as pK(K. S)). and Gibert’s K005. u = f f . t. Thus.e. Theorem 9(a) gives RS as the Brocard axis. k. u. i.. w = hh . S) be the parent of nK(W. Then they define a desmic structure with associated cubic of the form nK0 (W. R). If we discard a symmetric factor the polar of P is the line P1 P2 . We cannot identify the perspectors of the desmic structure. Let pK(W. The coefficient of x2 in pC(W.126 W. X(523)). Suppose that the desmic structure has normalized perspectors P1 = f : g : h and P2 = f : g : h . Suppose that {P. s. The root T is most neatly defined using the generalization of Gibert’s P K-transform. (a) The first barycentric of P is then 2(f f (g − g )(h − h ) − (f g h − f gh)(f − f )) = 2(f g − f g )(f h − f h ). R). R. K). They are complex.the McCay cubic and Gibert’s K003. the expression becomes K(f + f ). Example 3. Theorem 9(b) gives the parent as pK(K. where K is symmetric in the variables. Proof. w. Suppose that W = u : v : w. Stothers Theorem 9. even if we cannot find the perspectors explicitly. The kjp cubic nK0 (K. X(5)) . s = g − g . Theorem 10. S). It turns out that the perspectors must lie on another cubic nK0 (W. P ∗ lie on nK0 (W. (b) We know that S = f + f : g + g : h + h . Example 2. v = gg . k) is defined by a desmic structure. If we substitute the above values for r. (a) The line RS is the polar of P = 2ust − kr : 2vtr − ks : 2wrs − kt with respect to pC(W. Example 4. From [2. where T = ur(vt2 − ws2 ) : vs(wr 2 − ut2 ) : wt(us2 − vr 2 ) = P K(W. P ∗ lie on pK(W. Then there are three pairs and three cubics. Suppose that P = x : y : z. S) and pK(W. two of which are real. In the former case. C = AP ∩ BQ. z. C = AQ ∩ BP. Applying Theorem 10 to the McCay cubic pK(K. H) we get the second Brocard cubic K019 = nK0 (K. If we look at a pair of coordinates in the expression P + mP ∗ = nS. if P is on the McCay cubic. We begin with a discussion of an obvious way of constructing such a structure. Then the point P ∗ is u/x : v/y : w/z. K) . Maple shows that this happens precisely when P (and hence P ∗ ) is on the cubic nK0 (W. there exist constants m. Also.see Example 3. 5.so pK(W. y. These are (yt − zt)yx (zr − xt)xz (xs − yr)xy . S). G) and the Grebe cubic pK(K. P K(W. Example 5. S)). X(647)). Then the normalized forms for the perspectors of the desmic structure are P and mP ∗ . B = CP ∩ AQ. W = u : v : w. S) = P K(W. Also. we can identify the perspectors of the desmic structures. T ). my = . O) and the Orthocubic pK(K. Again these are complex. These are the intersections of the Brocard axis and the Kiepert hyperbola. we meet these last two points again. then P P∗ passes through O. A = BQ ∩ CP. This gives four of the intersections.Grassmann cubics and desmic structures 127 Proof. we get an expression for m in terms of two of x. so must be O. P K(W. R. S = r : s : t. The Thomson cubic pK(K. Note that we get an nK0 when P . k) as mx my mz uvw − xyz . S) and nK0 (W. S∗ ) . They give the cubic nK0 (K. If P P ∗ is not OK. a cubic of type nK0 arises from a desmic structure in which the triangles are triply perspective with the reference triangle ABC. and R = P − mP ∗ . n with P + mP ∗ = nS. . we know that the points of K019 are the foci of inconics with centers on the Brocard axis OK. It turns out that almost all triply perspective structures arise in this way. K019]. B = CQ ∩ AP. then the center is on P P ∗ and on OK. mx = ysw − ztv ztu − xrv xrv − ysu We can now compute the k such that the cubic is nK(W. Suppose that P = p : q : r and Q = u : v : w are points with distinct cevians. X(512)). As P is on pK(W. In §5. (a) Let A = BP ∩ CQ. We will meet this cubic again in §5. 2 xyz We have an nK0 if and only if this vanishes. P and P ∗ are the unique K-isoconjugates on OK. Lemma 11. K) give the cubic nK0 (K. Desmic structures with triply perspective triangles As we saw in Theorem 3. mz = . S∗ ) give rise to the same nK0 .see §5 . Otherwise. S). for example the first vertex is rv so is A .c. Q) has perspectors P1 . c. a). Q} is a bicentric pair. we can solve for u. The cubics associated with the desmic structure D(P. then the desmic structure D(P. Q with P = Q. and hence triply perspective. and perspectors normalized as in (a). h(b. Stothers Then the desmic structure with vertices A. The desmon of the second structure is the P -Hirst inverse of Q. For example. Q are the Q2 -isoconjugate of P and the P 2 -isoconjugate of Q. P2 with functions h(a. C . This has the 1 1 1 : ru : pv . b. This is on BP and CQ. even if 1 1 1 : ru : pv . and (ii) the perspectors are distinct arises from a unique pair P . The equality of the third coordinates follows from the condition f gh = f g h . Proof. . P )) and nK0 (W. If P and Q are triangle centers with functions p(a. r and the coordinates of P1 . c. (b) Each desmic structure including the vertices A.a. q. p(a. These involve ideas introduced in our Definition 2. some cevians coincide. c. if P . c. we can recover the vertices. Theorem 12. then D(P1 . A . c) = p(b. (c) The vertices of the triangles are [h(a. b. As the perspectors are distinct.a)q(c. A B C which are triply perspective. N K(W. P2 are centers. We can compute the equations of the cubics from the information in Lemma 11(a). P = Q. We leave it as an exercise to the reader that. Then the perspectors are P1 = f : g : h and P2 = f : g : h withf gh = f g h (see Lemmata 2 and 3). c) = q(a. P2 } is a bicentric pair. Q) has 1 and h(a. B. P2 . perspectors P1 = qw A B C . vertices named. As further exercises. we can find q and r in terms of p. C . the reader may verify that P . b). c) = p(a. c. B. and 1 1 1 1 1 1 : ru : pv and P2 = rv : pw : qu has triangles ABC. c) and q(a. Theorem 13. (b) Suppose we are given such a desmic structure. C. where W is the isoconjugation which interchanges P and Q. (a) We begin by looking at the desmic structure which is derived from the given P1 and P2 as in Theorem 4. Q as perspectors. the first three vertices are obtained by replacing a coordinate of P1 by the corresponding coordinate of P2 . Q) are pK(W. a. B . P )). b). A . then P1 . Suppose that P = Q. Q). b) and q(a. with perspectors P and Q. as P and Q are centers. and so on. It is easy to see that the triangles doubly perspective. Thus. We find points P and Q which give rise to these as in part (a).128 W. It follows that. h(c. From the coordinates of P1 . (a) perspectors P1 . b. The desmic structure defined in Lemma 11 is denoted by D(P. using two of the coordinates of P2 . Then. b)]. v and w in terms of p. b. b). B . P2 ) has triangle centers P . if {P. c). Note that from the normalized perpsectors. A is rv Definition 4. Q are centers and D(P.b) (b) {P1 . The proof requires only the observation that. P K(W. C in which (i) the triangles are triply perspective. b. where a is a root of k 2 x + 1 = 0. If C = nK(R2 . From a remark following Definition 2. Now observe that this also interchanges P and Q. This is T (R∗ ). a = −1. Q}. the corresponding pK is pK(R2 . we know that C is a Grassmann cubic associated with a desmic structure which has triply perspective triangles. R). we ignored cubics of type nK(R2 . so Q = P∗ . Suppose that C = nK0 (W. These give isoconjugation as that which interchanges P1 and P2 . k). Proof. Suppose that P = Q. so RS is the polar of T (R) ∩ T (R∗ ) = P K(W. and that C = nK0 (W. R. we have coordinates for the desmon S and the harmon R in terms of those of P1 and P2 . the desmic structure and C are degenerate. R) is not of type cK. so we do not have this case. where S is the desmon of the desmic structure. These are the intersections of C(R) and T (R∗ ). (a) From Theorem 13. Definition 5. Q are the cevian points for C. . then the equation shows that R = X and W = X2 . Since they are W -isoconjugates. so this gives precisely the pair {P. and does not have W = R2 . Theorem 15. and R = P K(W. where R = r : s : t. Now R is the pole of T (R) for this conic. (b) The perspectors of D(P. Then (a) P . (a) The cevian points for C are the W -isoconjugate points on T (R∗ ). (b) The Grassmann points for C are the W -isoconjugate points on the polar of P K(W. But we assumed that W = R2 . which is the union of the R-cevians. x + 1+ 2 When k = 2. P ) = P K(W.see Theorem 4. If the perspectors of the structure coincide at X. Q). −6. the Grassmann points are W -isoconjugate and lie on RS. We have the perspectors P1 and P2 of the desmic structure from Lemma 11(a). From Theorem 4. we replace the constant k by k rst. Q). R). Theorem 9 gives S as a point which we recognize as the pole of T (R∗ ) with respect to the conic C(R). Q) with Q = P ∗ . Using our formulae for P1 and P2 . (b) By Theorem 4. P and Q lie on the conic C(R). To make the algebra easier. When k = 2. These are precisely the pair of W -isoconjugates on T (R∗ ). R. In Theorem 14. then C is the Grassmann cubic associated with the desmic structure having perspectors R and aR. From Theorems 3 and 7.Grassmann cubics and desmic structures 129 Proof. R) is associated with the desmic structure D(P. The conic and line have just two intersections. Theorem 14. they also lie on the W isoconjugate of C(R). k rst) is not of type cK. Q) are the Grassmann points for C. we get the stated values of S and R. R) with respect to C(R). this structure is D(P. As C is an nK0 . and N K(K. C have already been identified. so we get the same vertices from either choice. C . but the proof that the triangles obtained from the intersections are perspective with ABC uses special properties of G and K. so that the line T (R∗ ) is the Euler line. B . A = BH ∩ CO. RB . C = AH ∩ BO. This identifies the perspectors as R and aR. with a as above. B = B . it is possible that these have complex coordinates. Here. From the proof of Lemma 11. Stothers Proof. as well as I.130 W. Also. As it is of type nK0 . O) = N K(K. but not of type cK. X(647)). In the “desmic structure” A = A . for example. is K019 in Gibert’s list. G) = P K(K. C . Here. the vertices A . Example 8. the “cubic” equation is identically zero as a = 1. [4]. When we use Maple to solve the equations to identify A and A . Here. X(39)) and nK0 (K. Example 6. −6. P2 are the Brocard points. RC of R2 -isoconjugation. and is again R. B = CH ∩ AO. Gibert’s website shows the points on K020. X(185)) is not (yet) listed in [2]. but the associated desmic structure is well-known. A . The cubics pK(K. The conic C(R) is the Jerabek hyperbola. B . but the equation factorizes as (ty + sz)(rz + tx)(sx + ry) = 0. They are the vertices of the first and third Brocard triangles. Again. together with {P1 . the excenters and X(185). X(385)). a = −1. X(76)}. The third Brocard cubic. X(384)). This configuration is discussed in. The associated GP is pK(K. B . G) = N K(K. nK0 (K. P2 } = {X(32). Take = G. This gives some new points on the cubic: A = BO ∩ CH. the perspectors P1 and P2 . The quadratic has equal roots when k = 2 or −6. the cubic can be described as the Grassmann cubic GN (A B C ) or GN (A B C ). The vertices of the structure are the intersections of medians and symmedians. pK(R2 . nK0 (K. There. C . for example. X(523)). B . Note that here the cubic is not of type cK as it contains three fixed points RA . Also. These cubics do not appear in Gibert’s list. R) is (ty − sz)(rz − tx)(sx − ry) = 0. the desmon is defined. Of course. is K017 in Gibert’s list. . The associated GP has the pivot N K(K. P1 . This happens. K) = X(512). C = AO ∩ BH. Example 7. In the former case. H) = X(185). A . The first Brocard cubic. Note that the two roots are inverse. R∗ is X(648). When k = 2. C = C as a = −1. nK0 (K. but we know that it has the points A . so we do get the nK as a Grassmann cubic. our Lemma 8(a) gives a very simple (geometric) proof of the general case. K) = X(39). Again we get isogonal cubics. Theorem 12(b) applies. the fourth Brocard cubic and K020 in Gibert’s list. It is easy to verify that this choice leads to C. The points P and Q for a cubic of type nK0 are identified as the intersections of a line with a conic. X(512)). The latter gives the cK in the class. we discover them as r : as : at. B = CO ∩ AH. In this case. Q = K in Lemma 11(a). The cubic pK(K. P K(K. These intersect in (the K-isoconjugates) O and H. for the second Brocard cubic. The points P and Q are the Brocard points. Finally. If nK(W. We refer the reader to the proof of Theorem 6(a). The second Brocard cubic nK(K. X(5)) = K005. 6. It is GN (∆). It is of the form nK(K. we introduced the idea of harmonic associates. k ). so these lie on K021. pK(W. pK(W. Q). Harmonic associates and other cubics In the Introduction. X(512)) was introduced in Example 8. A Cabri sketch shows that these do not have real intersections. From Theorem 16. The desmic structure is D(P. X(385)) = K128 is GP (∆ ). We will give a geometrical description of these shortly. k ) has parent K005. Then the desmic structure D with normalized perspectors P1 . Theorem 16. k) with parent pK(K. where ∆ is the harmonic associate of either of these triangles. and (b) cubics nK(W. Corollary 17. X(512)) are the Brocard points. −P2 has (a) vertices the harmonic associates of those of D. We have met these intersections in §4. Suppose that D is a desmic structure with normalized perspectors P1 . R. These are the infinite circular points. Using Theorem 9(b). R. From the Corollary.Grassmann cubics and desmic structures 131 the line is the Brocard axis and the conic is the Kiepert hyperbola. k) = GN (∆). a calculation shows that K067 = nK(K. Thus K067 is the harmonic associate of K216. K) is K024 in Gibert’s list. This was mentioned in §3. R). From Corollary 17. where ∆ is the harmonic associate of ∆. where ∆ is either the first or third Brocard triangle. k ) obtained in this way as the harmonic associate of nK(W. X(512)) = K021 is GP (∆ ). S. X(5). X(30)) = K001. The cubic nK0 (K. This gives us six new points on K128. R) = GP (∆ ). which in turn uses the notation of Theorem 4(a). as do the vertices of ∆ . then pK(W. k ) with parent pK(K. This gives us six points on K067 as harmonic associates of the points identified in [2] as being on K216. . S. S). Let C = K216 of [2]. We met this cubic in Example 3. This amplifies remarks made in the proof of Theorem 6. R. Q are the intersections of the line at infinity and the circumcircle. k). X(385)) = K017 was discussed in Example 7. We begin with a result which relates desmic structures. where ∆ is formed from intersections of medians and symmedians. k). Example 11. the cubic pK(K. Now the Grassmann points for nK0 (K. where P . Example 9. X(30). Proof. and cubics nK(W. Example 10. It is GN (∆). the fifth Brocard cubic pK(K. X(30). From Theorem 8(b). where we identified the perspectors of the desmic structure as the intersections of the Brocard axis with the Kiepert hyperbola. The kjp cubic nK0 (K. We refer to the cubic nK(W. P2 . This gives a pairing of our cubics. K005 has a unique child with root X(30). the harmonic associate is of the form nK(K. The fact that G and x(y − z) : y(z − x) : z(x − y) lie on the cubic is a simple verification using Maple. Suppose that X = x : y : z. These give a triangle with perspector H. The other perspector is K = cX(69). This is given by Theorem 4. X(524). then. The coordinates of W . If X = x : y : z. R. In the Introduction. k). For X = H. Also. the infinite point on GX. the mid-point of X and cX. Note that the first perspector is the complement cX of X and the second is X. Of course. B. The latter are the the intersections of the X(69)cevians with lines through G parallel to the corresponding sidelines. The final part needs an equation for the harmonic associate. the desmic structure D(X) is that with normalized perspectors y + z : z + x : x + y and −2x : −2y : −2z. and this leads to an nK(K. the parent of K022 is pK(O. X(141). and its parent which is the Thomson cubic pK(K. X(20)) is GP (∆). and hence their O-isoconjugates. AB. For X = x : y : z. so K022 is the harmonic associate of the cubic nK(O. The mid-point of X(69) and K is X(141). including the harmonic associates of the second Brocard triangle. we get K216 and K001 and their harmonic associates K067 and K005. but now we know that those points can be used to generate K002 as a locus of type GP . k) passes through G. R. k). The A-harmonic associate is y + z : 2y : 2z. k ) with parent pK(O. where (i) W = x(y + z) : y(z + x) : z(x + y). It is the union of the circumcircle and the line at infinity. There is one further example in [2]. X(141)). C in BC. GN (∆) degenerates. as ∆ is degenerate. We can generalize these to a general X as the result of extending each X-cevian by its own length. Proof. where ∆ has vertices the infinite points on the altitudes. starting from A. The fact that the midpoints lie on K002 is noted in [5]. The harmonic associate ∆ has vertices the mid-points of the altitudes. Theorem 18. k) and pK(W. S). the infinite point of T (X). and their W -isoconjugates. the center of the inconic with perspector X. This will follow from our next result. it is observed that the cubic contains the vertices of the second Brocard triangle. This suggests the following definition. These cubics contain the vertices of D(X(69)). These are the harmonic associates of three vertices of D(X(69)). CA. (ii) R = 2x + y + z : x + 2y + z : x + y + 2z. we get the point y + z : −2y : −2z. In the notes on K022 = nK(O. (iii) S = −2x + y + z : x − 2y + z : x + y − 2z.132 W. We can summarize our results on such structures as follows. The cubics associated with D(X) are nK(W. we mentioned that the Darboux cubic pK(K. The harmonic associate of nK(W. Definition 6. We can replace the vertices of ∆ or . G) = K002. The desmic structures and the points given in Theorem 18 account for most of the known points on K216 and K067. X(524)). R and S follow at once from those of the perspectors and Theorem 4. X(20). Stothers Three of the vertices of the desmic structure for K216 are the reflections of the vertices A. the intersection of the H-cevian at A with the parallel to BC through G. the complement of the isotomic conjugate of X. it includes the midpoints of the sides of ABC. It is also the perspector of E(X) other than X. (b) If pC(W. where X is the isotomic conjugate of the anticomplement of W . and hence the anticomplement of the isotomic conjugate of the anticomplement of the pole. S). the desmic structure E(X) is that with normalized perspectors y + z : z + x : x + y and x : y : z. The last six are the vertices of a defining desmic structure. Yiu derives interesting geometry related to pK(W. . (2) The cubic pK(W. then the Grassmann points are (i) the non-trivial intersections of pK(W. Remarks. it contains the G-Ceva conjugate of W and its W -isoconjugate (the point X). S). neither fixed by W -isoconjugation. G. the mid point of the cevian at A is y + z : y : z. and GP (∆ ) = K004. The case X = H gives W = K. For X = x : y : z. and GP (∆ ) = pK(W. Then GN (∆) = nK(W. k ). We have seen that there are several pairs of cubics of type pK which are loci of type GP from harmonic associate triangles. The harmonic associates are GN (∆ ) = nK(W. From the previous remark. Definition 7. R) = GP (∆) and pK(W. After Theorem 19. In [6]. The center is then the complement of X. R) and pK(W. and hence the G-Ceva conjugate of W . G) clearly contains G and W . R). S). §3. and these are summarized in [1. In the case of ∆ the isogonal points lie on the circumcircle. R = X(20).2]. k = 2((y + z)(z + x)(x + y) − xyz). R). For a given W . R) = pC(W. Most of the result follow from the equations given by Theorem 4. k) and GP (∆) = pK(W.Grassmann cubics and desmic structures 133 ∆ by their isogonal conjugates. (ii) the W -isoconjugate points on RS. there is a unique central pK with pole W . this arises from the desmic structure E(X). we know that if W = G. Theorem 19. (a) There exist harmonic triangles ∆ and ∆ with pK(W. For any point X = x : y : z. Theorem 20. G). R = −x + y + z : x − y + z : x + y − z. Theorem 3.1. so we get GP (∆) = K002. S) = GP (∆ ) if and only if pC(W. It also includes the mid-points of the X-cevians and their W -isoconjugates. the Thomson cubic. Let ∆ be the triangle A B C of E(X). We make the following definition. (1) From [1. suppose that R and S are distinct points. the Darboux cubic. but it is a known result. which degenerates as C(W ) and the line at infinity. Yiu shows that the cubic defined by GP (∆ ) has the given center. Finally. R) is central is quite easy to check. which is a central cubic with center the complement of X. The fact that pK(W. Suppose that X = x : y : z.1. The pivot of the central pK is then the anticomplement of X. We can describe when this is possible. where W = x(y + z) : y(z + x) : z(x + y). R. R) = pC(W. the anticomplement of X.3]. pC(W. The desmic structures are those for K067 and K216. X(239)) with parent K132 = pK(X(1). R) = pC(W. R) = pC(W. (b) The Grassmann points are the same for ∆ and ∆ . Our first two examples come from CL041. we found children of K001 with roots X(395). R.134 W. we showed that K024 = nK0 (K. k ) = GN (∆) for a triangle ∆. K001 and K129b = pK(K. H. In Example 3. and the cevian points are those described in [3] as the Jerabek points. and the cevian points are the Brocard points. R). There are just two non-trivial points of intersection. The Grassmann points are K. Then R is on pC(W. (a) If pK(W. nK(W. and lie on RS. X(239)). S). there is a unique child of pK(W. By Theorem 16. S). S). X(75). X(396)) with Grassmann points X(13). The Grassmann points are the barycentric square and isotomic conjugate of W . Stothers Proof. X(396). If we put W = X(1) in construction CL041. We have chosen examples where at least one of the cubics involved is in [2]. we have the class CL041. S. S) are loci of the given type. We then have K001 and K129a = pK(K. In our notation. the G-Hirst inverse of W . S. The desmic structures appear above. and lie on both cubics. The cevian points are the bicentric pair 1/r : 1/p : 1/q and 1/q : 1/r : 1/p. From Theorem 7. The Grassmann points are W -isoconjugate. K020 and K128 with Grassmann points X(32). X(76). X(15). GP (∆ ) = pK(W. These are complex. X(16). X(523). See the first of Example 10. The parent is pK(W. so these are the Grassmann points. R). We already have some examples of pairs of this kind: K001 and K005 with Grassmann points O. K001 and pK(K. we look at some cubics with different poles. S). X(76). k ). From Example 1. k) by Theorem 16. In the latest edition of [2]. We therefore have K003 and K102 = pK(K. we get nK0 (X(1). where R = p2 − qr : q 2 − pr : r 2 − pq. the Grebe cubic. where S = p2 + qr : q 2 + pr : r 2 + pq. Example 7 is the case where W = K. X(523)) with Grassmann points the infinite circular points. 7. Thus this nK has parent pK(W. Example 12. H. Further examples So far. S). K). In this section. X(894)). K002 and K004 with Grassmann points O. The Grassmann points are the intersection of the Brocard axis and the Kiepert hyperbola. we have the cubic nK0 (W. . giving (ii). S). R) and pK(W. By Theorem 8. then GN (∆ ) = nK(W. By Theorem 8. R) of the form nK(W. This includes cubics derived from W = p : q : r. Now suppose that pC(W. and GP (∆) = pK(W. almost all of our examples have been isogonal cubics. A harmonic associate of K132 is then K323 = pK(X(1). X(395)) with Grassmann points X(14). K) is a child of K003. so that the Grassmann points are the centers X(32). The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector X(37). these meet in G and H.Grassmann cubics and desmic structures 135 Example 13. X(1585)). the anticomplement of X(7) is X(144). X(7) and X(9). The center of the inconic with perspector X(69) is O. X(2). and has parent K202 = pK(X(1). S). X(8). The cubic pK(X(37). The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector O. and has parent pK(O. k ) with parent K070b. X(145). where S is as given in the discussion of CL041. then we get a cubic C = nK(X(37). X(193). k) with parent K168 = pK(O. X(193)). G) does not appear in the current [2]. G) does not appear in the current [2]. and has parent K201 = pK(X(9). X(144)). but contains X(1). The Grassmann points are G and H. is X(1). The next five examples arise from Theorem 19. The shoemaker’s cubics are K070a = pK(H. The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector X(1). The center of the inconic with perspector X(7) is X(1). The complement of X(7). The Grassmann points are X(8) and X(1). X(2). the Mittenpunkt. G). The center of the inconic with perspector X(1) is X(9). The cubic pK(X(9). The Grassmann points are X(7) and X(9). The Grassmann points are X(69) and K. The center of the inconic with perspector X(1) is X(37). X(8)). then we get a cubic C = nK(X(9). X(145)). X(1586)) and K070b = pK(H. Example 15. and has parent K033 = pK(X(37). with λ = ±1. the anticomplement of X(69) is X(193). G). The cubic pK(X(1). Example 16. If we apply Theorem 19 with X = X(1). we get an nK(H. we get nK0 (H. If we put W = H in construction of CL041. X(69). If we apply Theorem 19 with X = X(7). Further examples may be obtained from the page on central pK in [2]. k) with parent pK(X(37). Example 17. k) with parent pK(X(1). G) does not appear in the current [2]. then we get a cubic C = nK(X(1). The Grassmann points are X(1) and X(10). and the cevian points are those described in [3] as the cosine orthocenters. Note that K070a and K070b are therefore harmonic associates. If we normalize G as 1 : 1 : 1 and H as λ tan A : λ tan B : λ tan C. Example 18. X(8) and X(9). The complement of X(69) is K. k) with parent K070a. but contains X(1). The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector X(9). X(144). and an nK(H. X(10) and X(37). k) with parent pK(X(9). X(1585). The complement of X(1) is X(10). but contains X(1). the incenter. the Nagel point. Example 14. If we apply Theorem 19 with X = X(69). the anticomplement of X(8) is X(145). the Nagel point. the Spieker center. is X(9). If we apply Theorem 19 with X = X(8). the anticomplement of X(1) is X(8). . G). X(1585). G). then we get a cubic C = nK(O. the Gergonne point. The complement of X(8). X(2). The Grassmann points are X(393). X(297)) with parent pK(H. As stated in [2]. W ) has parent pK(W. Stothers Example 19. X(206)). let Xt be the tangential of X. k) has parent pK(X(32). X(206). Example 22. the cubic nK0 (X(1). K). We give four examples. at Q. Example 23. The harmonic associate nK(H. we met the kjp cubic K024 = nK0 (K. Gibert’s theorem In private correspondence. From Theorem 19 with X = X(66). is related to the class CL009. the Grassmann points and cevian points may be complex. The cubic nK0 (X(32). K). S) at Qt (twice). This accounts for all three intersections of .and Q are two W -isoconjugate points on the cubic pK(W. V ). but appears in [2] as P 161 . S). G). and to the class CL026. Now pC(P t ) meets pK(W. k). A . The cubic nK0 (X(9). k) has parent K101 = pK(X(1). B .see K161. which contains cubics nK0 (W. B . and at three other points A . X(32)) has parent K160 = pK(X(32). and has parent K161 = pK(X(32). with parent pK(W. X(9). and that this has parent the Grebe cubic K102 = pK(K. C are the vertices of a desmic structure with perspectors P and Q. and pC(X) be the polar conic of X. As above. The harmonic associate nK(X(32). The center of the inconic with perspector X(66) is X(32). X(32)). we get a cubic C = nK(X(32). Example 21. As above. C . This can be verified computationally. This means that the class CL007. The Grassmann points are X(66) and X(206). H). k) with parent K177 = pK(X(32). P 161). W ). which contains cubics pK(W. at P . Then the points A . The parent is the McCay cubic K003 = pK(K. X(1). Suppose that P . k) has parent K181 = pK(H. S). From Theorem 9(b). X(9)). V ). X(1249). where V is the G-Ceva conjugate of W . The harmonic associate will be of the form nK(W. k). In general. The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector X(32). X(1)). O. X(1)). Theorem 21 (Gibert). the general nK0 (W. C . k) meets the sidelines of ABC at A. B . W ). The complement of X(66) is X(206). B. which contains cubics pK(W. H) has parent K159 = pK(H. the anticomplement is not in [5]. Bernard Gibert has noted a further characterization of the vertices of the desmic structures we have used. V. Example 20. For X on pK(W.136 W. O). W ). 8. X(1249)). C and at the intersections with T (R). C . S) at P t (twice). P 161. The harmonic associate nK(X(9). X(9)) has parent K157 = pK(X(9). In Examples 3 and 9. the cubic nK0 (H. The harmonic associate nK(X(1). Appendix We observe that the cubic nK(W. and at three other points A . R. X(1)) has parent pK(X(1). It follows that the harmonic associate of K024 is of the form nK(K. and pC(Qt ) meets pK(W. k) has parent pK(X(9). X(9)). B . and z = ax. and at the intersections of T (R) with AB and AC. Let B = b1 : b2 : b3 . y = −2vx(t+ar) 2aus+k . so we must have c1 = 0. It satisfies the perspectivity conditions. We now examine the locus GN (A B C ) where the vertex A = a1 : a2 : a3 lies on a sideline. This cannot include the cubic nK(W. If a1 = 0. If we look at the equations for nK(W. This establishes the criteria set out in Theorem 6(e). R. 2wrs or k = . The calculation referred to in Theorem 6(e) shows that nK(W. we get k = 2ust r t pK(W. giving y = 0. Then ABC and A B C have perspector B. we find a unique solution. the latter to 0 : 0 : 1. B . From our earlier work. the equation for the locus has some zero coefficients. After scaling this is A = r : −s : 0. R. giving z = 0. If we consider the case k = 2vtr s . C = c1 : c2 : c3 . We should expect to obtain a desmic stucture by taking W -isoconjugates of A . so in the generic case. then we get the expected solutions. C = 0 : −s : t. we know that there is no other way to express these cubics as loci of type GN . k) and pK(W. so we can regard the solutions as limits as a tends to 0 or ∞. We now look at the Maple results in detail. Taking a2 = 0 or b3 = 0 leads to an equation with zero coefficients. or a = − rt . Theorem 7 . then Theorem 7 holds as stated. giving nK(W. z. R. The former leads to r : −s : 0. The equation for a has a nonzero root only when k = 2vrt s . RA ) and solve for y. and at most two such triangles. C = A. When we equate the coefficients of the equations for the locus and nK(W. We cannot have x = 0. R. we still get the same perspectors. there are nine intersections. then there is a triangle ∆ with C = GN (∆). R. we now examine the case a3 = 0. ABC and CRB A have perspector P = r : 0 : t. ory = z = 0. If we write the isoconjugate of X = x : y : z as uyz : vzx : wxy. 2vtr/s. If we replace If we put − rt in the equation for a. C. If we allow this as a desmic. 2vtr/s). If a cubic C is of type nK. On algebraic grounds. CRB A and A B C have perspector RB . We will meet these again below. where a satisfies 2u(2vtr − ks)X 2 + (4uvt2 + 4vwr 2 − 4uvs2 − k2 )X + 2w(2vtr − ks) = 0. The condition for the locus to be an nK becomes a2 b3 c1 = 0. If not. . RB ) or pK(W. k) with k = 2ust/r. but not of type cK. but has only eight distinct points. If we replace B by any point. then the isconjugates are A = C. A little thought shows that for fixed W . RA ) touch at B. there are two further intersections. RA ) by pK(W. the equation for a becomes X = 0. 2wrs/t. k) other than y2 z and xyz. It is a moot point whether this should be termed a desmic structure. R. meet at A. k) and pK(W. k) unless it is the whole plane. R. there are only three such loci. Guided by the above discussion. RC ). C . Then the locus is nK(W. B = r : −s : t = RB . Thus we must have a = 0. B = u/r : −v/s : w/t. Thus we cannot define a cubic as a locus in this case. we clearly get the same results.Grassmann cubics and desmic structures 137 the cubic with each sideline. we can either add these three cubics to the excluded list or reword in the weaker form. A check using Theorem 9(b) shows that the parent is indeed pK(W. starting from B. Stothers To get R as the barycentric difference of perspectors.wanadoo. J.138 W. we get C. This reflects the fact that the other points do not determine B . Episodes in Nineteenth and Twentieth Century Euclidean Geometry. [6] P. Ehrmann and B. Replacing each coordinate of P in turn by the corresponding one from B.evansville.gibert/downloads. Math. Math. 0 : 0 : 0. Special Isocubics in the Triangle Plane. available at http://faculty. Int. Yiu. [3] B. 1996.gla. Gibert.fr/bernard. C . we get A . [2] B. On the other hand.wanadoo.pdf.wanadoo. Gibert. Sci. Technol.gibert/index. Kimberling.uk . as expected. University of Glasgow. we need to scale B to 0 : −s : 0. A. Cubics in the Triangle Plane. Assoc. New Mathematical Library 37. Honsberger. Glasgow. Educ. available from http://perso.edu/ck6/encyclopedia/ETC. Barycentric coordinates and Tucker cubics.. UK E-mail address: wws@maths. Then the sum is RB . RB ) = GP (A B C ). Wilson Stothers: Department of Mathematics.ac. Amer. Thus cubics of this kind are Grassmann cubics for only one triangle rather than the usual two. When we compute the equation of GN (CRB A). Encyclopedia of Triangle Centers. we find that all the coefficients are zero. Gibert. available from http://perso. References [1] J-P.html. [5] C.fr/bernard.fr/bernard. The uses of homogeneous barycentric coordinates in euclidean plane geometry. available from http://perso.html. 31 (2000). 569–578. [4] R.html. RB .gibert/Resources/tucker. Necessarily. 0). this process fails to exhaust the median-concurrent pentagons. this pentagon has a Publication Date: April 10. depicted in Figure 1. such polygons must have an odd number of edges. v1 = (1. One easy source of such polygons is to begin with a regular (2n + 1)-gon centered at the origin and transform the vertices using an affine transformation. A non-affinely regular median concurrent pentagon It is easy to check that for each i (all indices modulo 5). We describe how to regard certain collections of sets of medians as a linear subspace of related collections of sets of lines. This exhausts the triangles as every triangle is an affine image of an origin-centered equilateral triangle. Alternatively. 2006. Communicating Editor: Floor van Lamoen. we show that every set of 2n + 1 concurrent lines are the medians of some (2n + 1)-sided polygon. We exhibit conditions that determine whether a set of 2n + 1 lines are the medians of a (2n + 1)-sided polygon. v3 = (−2. Motivation It is well-known that the medians of a triangle intersect in a common point. b b FORUM GEOM ISSN 1534-1178 Concurrent Medians of (2n + 1)-gons Matthew Hudelson Abstract. Each of these constructions demonstrates a procedure that generates (2n + 4)-degree of freedom families of median-concurrent polygons. We wish to explore which polygons in general have this property. On the other hand. 1 2 (vi+2 + vi+3 ) contains the origin. Furthermore. we derive conditions on n + 1 points so that they can be consecutive vertices of a (2n + 1)sided polygon whose medians intersect at the origin. . Consider the pentagon with the sequence of vertices v0 = (0. and as a consequence. 1. Also. −2). it suffices to check vi+2 + vi+3 are scalar multiples. 1). On the other hand. 2). the line through vi and that vi and single selfintersection whereas a regular pentagon has none and a regular pentagram has five. v4 v0 v2 v1 v3 Figure 1. v2 = (2. this number of degrees of freedom is maximal.b Forum Geometricorum Volume 6 (2006) 139–147. and v4 = (6. if we begin with either a regular pentagon or a regular pentagram. 1). The signed law of sines 1 . denoted θ12 is the angle whose magnitude (in the range (0. 2. Thus. Hudelson so this example cannot be the image under an affine map of a regular 5-gon. Families of polygons constructed by medians 2. Let Di be a point on the positive side of C along i . In the next section.140 M. and with opposite unit vectors is π. We approach this problem by two different means. we will say that B is on the “negative side” of A along . 3 2 θ23 A23 β θ13 α A12 A13 Figure 2. Oriented lines and the signed law of sines. The “signed angle” from 1 to 2 . we will be working with oriented lines. Let 1 and 2 be two non-parallel oriented lines and C be their intersection point. this quantity is denoted d (A. B). we consider families of lines that serve as the medians of polygons and in the section afterwards. the vectors thought of as lying in the z = 0 plane of R3 . Then given points −− → A and B on . The signed angle of two parallel lines with the same unit vector is 0. Switching the orientation of a line switches the sign of the signed length from one point to another on that line. we seek alternative and more comprehensive means to construct median concurrent pentagons specifically and (2n + 1)-gons in general. Switching the orientation of a single line switches the sign of the signed angle between them. The oriented line is defined as the pair ( . If t > 0.1. we associate with it a unit vector v that is parallel with . if t < 0. we begin with a collection of n + 1 consecutive vertices and show how to “complete” the collection with the remaining n vertices. Given a line in R2 . In this section. π)) is that of ∠D1 CD2 and whose sign is that of the cross product v1 × v2 . we can solve AB = tv for t and say that t is the “signed length” from A to B along . the result will also be a (2n + 1)-gon whose medians intersect. v). we will say that B is on the “positive side” of A along . . A13 ) = 2 . Constructing consecutive points via medians Given a point Ak on k . in R2 . Letting α = ∠A23 A13 A12 and β = ∠A12 A23 A13 . and let δi+1 be di+1 (Bi. Point Aij is at the intersection of i and j .i+1 . Let 0 . Bi+1.1 .i+2 ). 2 . A13 is on the positive side of A12 along 1 . As drawn. Any of these orientation switches leaves the LHS and RHS equal. . A23 is on the positive side of A12 along 2 . Let point Ak+2 be the intersection of lines m2 and k+2 . Construct point C on line m1 so that segments Ak B and BC are congruent.Concurrent medians of 2n + 1-gons 141 In Figure 2 we have three oriented lines 1 . Let Bi. k+2 m1 Ak+2 C k+1 B m2 Ak k Figure 3.j be the point of intersection of i and j . By the ordinary law of sines and the above comments about α and β. and A23 is on the positive side of A13 along 3 . we have sin α = sin θ13 and sin β = sin θ23 . sin θ23 sin θ13 Note that if we switch the orientation of 1 . Construct line m2 through C and perpendicular to m1 .2. . 1 . . no two parallel. A23 ) d1 (A12 . A0 ). we call this equation “the signed law of sines. Switching the orientation of 2 changes the signs of the numerator of the RHS and the denominator of the LHS. Let point B be the intersection of line k+1 and m1 . we have d (A12 . choose a point A0 on 0 and let S0 = d0 (B0. then the numerator of the LHS and the denominator of the RHS change signs. Switching the orientation of 3 changes the signs of both denominators. 3 . we construct the point Ak+2 on k+2 (the indices of lines are modulo 2n + 1) as follows and as depicted in Figure 3. Constructing polygons via medians. Construct line m1 through Ak and perpendicular to k+1 .” 2. Finally. 2n be 2n + 1 oriented lines. . Construction 1. and so the equation above is true for oriented lines and signed angles as well. We are interested in the case when we begin with a point A0 on 0 and eventually construct the point A2(2n+1) which will also be on line 0 . When k = 0 and m = 2n + 1.k+1 . Letting m be the line through Ak and Ak+2 and ϕ = θk+1. if we define hk.142 M.2.0.k+1 Sk − δk+1 .p. we have Sk+2 = sin θk.m = gk+2p gk+2p+2 gk+2p+4 .2n+1 = 1. Sk x = sin θk. In general. sin θk+1.k+2 = Sk+2 + δk+1 .k+2 Let gk = sin θk. we have S2(2n+1) = h0. we have by the signed law of sines. and Sk+6 = gk+4 Sk+4 − δk+5 = gk+4 gk+2 gk Sk − gk+4 gk+2 δk+1 − gk+4 δk+3 − δk+5 .0. Hudelson We note that line k+1 intersects segment Ak .2n+1 = 2n k=0 g2k 2n 2n sin θ2k. it follows that the terms in the numerator are a permutation of those in the denominator. sin(θk.i.k+2 Then we have the recurrence Sk+2 = gk Sk − δk+1 . gk+2(m−1) .k+1 + ϕ) Eliminating x.k+1 + ϕ) and x sin θk+1. We define Sk = dk (Bk.m .2k+2 k=0 sin θ2k+1. .2n+1 S0 − h0. sin θ2k+1.2n+1 values.1. The second observation is that S2(2n+1) = S0 + a linear combination of the h0. We notice first that h0.0.k+1 sin(θk.2n+1 δ1 − h0.2k+1 sin θ2k.2k+2 k=0 Since the subscripts in the latter products are all modulo 2n + 1.m δk+1 − hk. This means that h0. .2n+1 δ3 − · · · − δ2(2n+1)−1 .k+1 . sin θk+1.0.m δk+3 − · · · − δk+2m−1 .2. Ak ) and θij to be the signed angle subtended from i to j .m Sk − hk.1. we have Sk+2m = hk. . Ak+2 at its midpoint.2k+1 = = 2nk=0 . This leads to Sk+4 = gk+2 Sk+2 − δk+3 = gk+2 gk Sk − gk+2 δk+1 − δk+3 . y). there are two more degrees of freedom in the choice of the point of concurrency.Concurrent medians of 2n + 1-gons 143 The coefficients of this linear combination are the δ’s. d). the choice of each being a single degree of freedom (for instance. and let m4 be the line through (e. intersecting m2 at (x. We seek a fourth point (u. Constructing the fourth point The point (u. b) and (c. Each line can be chosen by choosing a unit vector. 0) are collinear. . Families of median-concurrent polygons constructed by vertices Suppose we have three points (a. but half the distance from (e. no two parallel. Consider choosing a family of 2n + 1 concurrent lines. b + d) and (0. b). We have shown Proposition 1. Let A be the midpoint of the segment joining (a. Construct the line m2 parallel to m1 that is on the same side of. (a + c. −d). Finally. δ2n )T . (c.b) and the origin. Any set of 2n + 1 concurrent lines. d) and m1 be the line through A and the origin. f + v) and (0. f ) as m1 . The nullspace of the h values will in fact be a codimension 1 subspace of the space of all possible choices of (δ0 . . 3. m4 m2 (c. . 0) (u. A4n+2 = A0 . y). v) such that (u. v) is the intersection of lines m1 and m4 . f ) and (x. A4 . δ1 . Let m3 be the line through (a. An immediate consequence of this is that if for all i we have δi = 0 then S4n+2 = S0 and so we have closed the polygon A0 . Another degree of freedom is the choice of point A0 on 0 . y) (0. The point (u. the angle that vector makes with the vector (1. A2 . . f ) in R2 such that (a. . b) Figure 4. . and (e. b) = (−c. . This is a total of 2n + 4 degrees of freedom for constructing (2n + 1)-gons with concurrent medians. . and (a. (e + u. . v) (a. 0) are also collinear. b). v) can be constructed as follows: Construction 2. in R2 are the medians of some (2n + 1)-gon. v). f ) A (x. d) m1 m3 (e. 0)T ). The case j = 0 is obvious. ∆j. . C = (c. For all j ≥ 0.k = ((ak + ak+1 )bk+1 − (bk + bk+1 )ak+1 ) ∆k. k. (A + C) and A.k+1 = ak+1+n bk+1 − bk+1+n ak+1 ∆k+n.j = ∆n. For 0 ≤ j.k+1 = ∆k+n. ∆i. E) = ad − bc This point satisfies the property that the lines F. provided ad − bc = 0. (E + F ) intersect at the origin. we have be − af .0 which completes the induction. bi ). and suppose that no line joining two consecutive points contains the origin.k = aj bk − bj ak . b + d) and also u(b + d) = v(a + c) b(e + u) = a(f + v). Given point A = (a. xi+1 .0 .144 M. Armed with this.i (xi + xi+1 ). ∆j+n. Isolating k.k = ∆n.i+1 Also. xi+n ). . f ). We formalize this in the following definition. We can recast our definition of the point xi+n+1 using the following definition. Starting at i = 0 we define xi+n+1 = F (xi . Definition 2. Hudelson It must be the case that (u. we have ud − be = vc − af or k(a + c)d − be = k(b + d)c − af.. We address what happens in the sequence x0 .k+1 ∆k+n. C. we use induction to prove: Proposition 2. k= Definition 1. x2 . C. v) = k(a + c. 0 ≤ i ≤ n. Subtracting. we have xi+n+1 = ∆i+n. ad − bc Notice that the fourth point is uniquely determined by the other three. Proof. xi = (ai . Suppose the result is true for j = k. Now. F = F (A. Then ∆k+1+n. suppose we have n + 1 points in R2 . b). . x1 .k (ak bk+1 − bk ak+1 ) = ∆k. and E = (e. E) by the formula be − af (A + C). d). . we define the point F (A. n 1 (∆n−1.n 1 (∆n.j bk = (aj bk − bj ak )b + (a bj − b aj )bk = aj bk b − bj ak b + abj bk − b aj bk = (a bk − b ak )bj = ∆.0 xn + ∆n−1.0 (xi + xi+1 ).k bj .0 xn + ∆n. .n If 1 ≤ i ≤ n.n and so k0 = ∆n−1.n x0 ) + ∆n.0 . we have ∆n. ∆n−1. k.n ∆n−1.i+1 xi+n+1 = We verify a useful property of ∆j. and ∆j.0 xn ∆n−1. We work component-by-component: ∆j.k x + ∆.k xj .0 + ∆n.k : Proposition 3.j ak = (aj bk − bj ak )a + (a bj − b aj )ak = aj bk a − bj ak a + abj ak − b aj ak = (a bk − b ak )aj = ∆. Proof.0 . we calculate x2n + x0 = = = = = ∆n.k aj .0 (xn−1 + xn ) + x0 ∆n−1.n 1 ((∆n.j xk = ∆. Proof. For all 0 ≤ i ≤ 2n.n x0 ) ∆n−1.Concurrent medians of 2n + 1-gons 145 As a consequence. For i = 0. For all j. We can now prove the following: Proposition 4. then by the definition of xn+i .0 xn ) ∆n−1. ki = ∆i−1. ∆j. .k b + ∆.k a + ∆. ∆i. there is a ki such that xi−1 + xi = ki xn+i (all subscripts modulo 2n + 1).0 xn−1 + ∆n.0 xn−1 + ∆n−1.0 xn ) ∆n−1.0 + ∆n.i /∆n. m ). .0 = (xm−2 + xm−1 ) + (xm−1 + xm ) ∆m−2.1 xn + ∆n.1 and so ∆n.0 /(∆m−2.1 xn + ∆n.m (xm−2 + xm−1 ) + ∆m−2.0 x0 = ∆0.m + ∆m−2.1 1 (∆n. we find that the points xj .i−n + ∆i−n−1.0 x0 ) = ∆0.m−1 xm ) + (∆m−1.m + ∆m−1. Each point contributes two degrees of freedom for a total of . But this means that 1 2 (xj+n + xj+n+1 ) is also on the same line.m xm−2 + ∆m−2. x0 .m + ∆m−1.m−1 (xm−1 + xm ) = δ((∆m−1.1 1 = (∆0. kn+1 = xi−1 + xi = xn+m−1 + xn+m ∆n.m−1 )xi+n noting that m − 1 = i − n − 1 ≡ i + n modulo 2n + 1.1 + ∆n. . we have ∆n. x1 .0 x1 + ∆n.0 .0 (x0 + x1 )) ∆0.0 x0 ) = ∆0.m−1 )xm−1 = δ(∆m−2. .1 ∆n. using the symbol δ = ∆n. As a direct consequence. n + 2 ≤ i ≤ 2n. is that the points xi−1 + xi .m + ∆m−2. Hudelson To handle the case when i = n + 1. xj+n + xj+n+1 and the origin are collinear. geometrically. x0 . ∆0.i−n−1) . Alternatively.m + ∆m−2. .i−n−1 ∆i−n−1.i−n What this proposition says. xi+n and the origin are collinear. we set m = i − n and we have. we obtain the following result: Proposition 5. and so the line connecting xj and the midpoint of the segment joining xj+n and xj+n+1 contains the origin.146 M.1 1 (∆0. and so for n + 2 ≤ i ≤ 2n.m−1 ∆m−1. setting i = j + n + 1.1 x0 + ∆n. Here. we can choose n + 1 points to determine a (2n + 1)-gon whose medians intersect at the origin.i−n + ∆i−n−2.m−1 ∆m−1.m = δ(∆m−1. xi+1 or xn . then these points are n + 1 vertices in sequence of a unique (2n + 1)-gon whose medians intersect in the origin. ki = ∆i−n−2. xn such that the origin is not on any line xi .1 For the values of i. Given any sequence of n + 1 points.m−1 )xm−1 ) = δ(∆m−2.0 ∆n.1 + ∆n.0 (∆i−n−2.0 (x0 + x1 ) xn + xn+1 = xn + ∆0. we calculate ∆n. 99164-3113. Washington State University. We cannot hope for more freedom than this. This echoes the final result from the previous section. Pullman. USA E-mail address: hudelson@math. for a total of (2n + 4) degrees of freedom. Two more degrees of freedom are obtained for the point of concurrency. That we cannot obtain further degrees of freedom follows from the previous section as well.Concurrent medians of 2n + 1-gons 147 2n+2 degrees of freedom. Washington. There.wsu. Matthew Hudelson: Department of Mathematics.edu . any set of 2n + 1 concurrent lines (in general position) produced a concurrent-median (2n + 1)-gon. . one obtains a wider. What will be established here is the convexity of the locus Ω of the point P which lies in H and which determines together with A and B an angle ∠AP B of a given fixed size. In this paper the most basic property of the curve. and a line which has points in common with H but avoids segment AB. Then. In hyperbolic geometry on the other hand a set of equivalently defined points P determines a different kind of curve. The fact of the convexity of our curve yields at least one often used by-product: Theorem. As to proving the Lemma itself. Section 50]. Whereas in Euclidean geometry it has to be a rotation. Section 2]). Introduction In the hyperbolic plane let AB be a segment and H one of the halfplanes with respect to the line through A and B. No straight-forward proof could be found. is established. Many thanks to my colleague Dr. . and also [6. b b FORUM GEOM ISSN 1534-1178 On the Generating Motions and the Convexity of a Well-Known Curve in Hyperbolic Geometry Dieter Ruoff Abstract. Exercise 4]. Communicating Editor: Paul Yiu. or. The argument rests on a comparison of the rigid motions that map one of the angles ∠AP B into other ones. In Euclidean geometry this locus is well-known to be an arc of the circle through A and B whose center C determines the (oriented) angle ∠ACB = 2 · ∠AP B. and it seems that a distinction of cases is the only way to proceed. Still. surprisingly. horocyclic rotations. H a halfplane with respect to the line through A and B. it would be desirable if the possibility of an overarching but nonetheless elementary argument would be investigated further. even a translation. In Euclidean geometry the vertices P of those angles ∠AP B of size α that pass through the endpoints A. In hyperbolic geometry. For our convexity proof it appears to be practical to consider the given angle as fixed and the given segment as moving. 2006.79. its convexity. on the other hand. one has to take into account that the motions involved can be rotations. Publication Date: April 17. [1]. the relative position of the centers or axes of our motions can be described in a very simple fashion. as will be shown in the Main Lemma. The evidently greater complexity of the non-Euclidean version of this locus shows itself most clearly when one considers the (direct) motion that carries a defining angle ∠AP B into another defining angle ∠AP B. Chris Fisher for his careful reading of the manuscript and his helpful suggestions. or translations. p. 1.b Forum Geometricorum Volume 6 (2006) 149–166. flatter curve (see Figure 1. B of a given segment trace the arc of a circle. [2. it can in hyperbolic geometry also be a horocyclic rotation about an improper center. with the sought-after convexity proof as an easy consequence. Let AB be a segment. [7. 150 D. Ruoff Figure 1 Then the point X, when running through in H, determines angles ∠AXB that first monotonely increase, and thereafter monotonely decrease in size. Our approach will be strictly axiomatic and elementary, based on Hilbert’s axiom system of Bolyai-Lobachevskian geometry (see [3, Appendix III]). The application of Archimedes’ axiom in particular is excluded. Beyond the initial concepts of hyperbolic plane geometry we will only rely on the facts about angle sum, defect, and area of polygons (see e.g. [2, 5, 6, 8]), and on the basic properties of isometries. To facilitate the reading of our presentation we precede it with a list of frequently used abbreviations. 1.1. Abbreviations. 1.1.1. [A1 A2 . . . Ah . . . Ai . . . Ak . . . An ] for an n-tuple of points with Ai between Ah and Ak for 1 ≤ h < i < k ≤ n. 1.1.2. AB, CD, . . . for segments, and (AB), (CD), . . . for the related open in− −→ −−→ tervals AB − {A, B}, CD − {C, D}, . . . ; AB, CD, . . . for the rays from A −−−→ −−−→ through B, from C through D, . . ., and (A)B, (C)D, . . . for the related halflines − − → −−→ AB − {A}, CD − {C}, . . .; (AB), (CD), . . . for the lines through A and B, C and D, . . . . − → → → 1.1.3. a, b, c, . . . are general abbreviations for lines, − a , b ,− c , . . . for rays in → − − → − → those lines, and ( a ), ( b ), ( c ), . . . for the related halflines. 1.1.4. H(a, B), where the point B is not on line a, for the halfplane with respect to a which contains B, and H(a, B) for the halfplane with respect to a which does not contain B. The improper ends of rays which enter halfplane H through a are considered as belonging to H. 1.1.5. perp(a, B) for the line which is perpendicular to a and incident with B; proj(S, ) for the orthogonal projection of the point or pointset S to . 1.1.6. ABCD for the Lambert quadrilateral with right angles at A, B, C and an acute angle at D. 1.1.7. R for the size of a right angle. Generating motions and convexity in hyperbolic geometry 151 → 1.1.8. a \ b, a \ − p , . . . for the intersection point of the lines a and b, of the → line a and the ray − p ,... . 1.1.9. ·, ··, ◦ (in figures) for specific acute angles with ·· denoting a smaller angle than ·. Remark. In the figures of Section 3, lines and metric are distorted to better exhibit the betweenness features. 2. Segments that join the legs of an angle In this section we compile a number of facts about segments whose endpoints move along the legs of a given angle. All statements hold in Euclidean and hyperbolic geometry alike; the easy absolute proofs are for the most part left to the reader. → − → Let ∠(− a , b ) be an angle with vertex P , and C be the class of segments Aν Bν → − → of length s that have endpoint Aν on leg (− a ) and endpoint Bν on leg ( b ) of this angle, and satisfy the equivalent conditions (1a) ∠P Aν Bν ≥ ∠P Bν Aν , (1b) P Aν ≤ P Bν , → − → (see Figures 2a, b). We will always draw − a , b as rays that are directed downwards → − → and, to simplify expression, refer to P as the summit of ∠(− a , b ). As a result of → a ), and (1a) the segments Aν Bν are uniquely determined by their endpoints on (− → − C can be generated by sliding downwards through the points on ( a ) and finding → − the related points on ( b ). Figure 2a Figure 2b It is easy to see that during this downwards movement ∠P Aν Bν decreases and ∠P Bν Aν increases in size. Due to (1a) the segment A0 B0 which satisfies ∠P A0 B0 ≡ ∠P B0 A0 , P A0 ≡ P B0 is the lowest of class C. 152 D. Ruoff → − → If ∠(− a , b ) < R then the class C contains a segment Ap Bp such that ∠P Ap Bp = R. Note that when Aν moves downwards from P to Ap , Bν moves in tandem down from the point O s units below P to Bp , but that when Aν moves on downwards from Ap to A0 , B0 moves back upwards from Bp to B0 (see Figure 2a). If → − → − → → ∠(− a , b ) ≥ R no perpendicular line to (− a ) meets ( b ) and the points Bν move invariably upwards when the points Aν move downwards (see Figure 2b). → Now consider three segments AB, A1 B1 , A2 B2 ∈ C whose endpoints on (− a) satisfy the order relation [AA1 A2 P ], and the direct motions that carry segment AB to segment A1 B1 and to segment A2 B2 . These motions belong to the inverses of the ones described above and may carry B first downwards and then upwards. As a result there are seven conceivable situations as far as the order of the points B, B1 and B2 is concerned (see Figure 3): (I) (III) (V) (VII) [B2 B1 BP ], [B1 B2 BP ], [B1 BB2 P ], [BB1 B2 P ]. (II) [B1 BP ], B2 = B1 (IV) [B1 BP ], B2 = B, (VI) [BB2 P ], B1 = B, and Figure 3 → − → In case ∠(− a , b ) ≥ R the point B moves solely downwards (see Figure 2b) and − → → we find ourselves automatically in situation (I). On the other hand if ∠(− a, b)<R and A lies on or above Ap both endpoints of segment AB move simultaneously upwards, first to A1 B1 and then on to A2 B2 (see Figure 2a); this means that we are dealing with situation (VII). In Figure 3 the level of the midpoint N1 of BB1 is indicated by an arrow to the left, and the level of the midpoint N2 of BB2 by an arrow to the right. We recognize at once that we can use N1 and N2 instead of B1 and B2 to characterize the above seven situations. Set forth explicitly, a triple of segments AB, A1 B1 , A2 B2 ∈ C with [AA1 A2 P ] can be classified according to the following conditions on the midpoints N1 , N2 of BB1 , BB2 : Generating motions and convexity in hyperbolic geometry (I) (III) (2) (V) (VII) [N2 N1 BP ], [N1 N2 BP ], [N1 BN2 P ], [BN1 N2 P ]. 153 (II) [N1 BP ], N2 = N1 , (IV) [N1 BP ], N2 = B, (VI) [BN2 P ], N1 = B, and 1 Note that always N1 N2 = B1 B2 . The midpoints M1 , M2 of AA1 , AA2 sim2 1 ilarly satisfy M1 M2 = A1 A2 ; here the direction M1 → M2 like the direction 2 A1 → A2 points invariably upwards. Figure 4 Figure 5 In closing this section we deduce two important inequalities which involve the points M1 , M2 , N1 and N2 . From P A ≤ P B, P Ai < P Bi (see (1b)) follows (3) P Mi < P Ni (i = 1, 2), (see Figure 4). In addition, for situations (III) - (VII) in which B2 lies above B1 and N2 above N1 we can establish this. In the right triangles A1 A1 B1 , A2 A2 B2 where A1 = proj(A1 , b), A2 = proj(A2 , b), A1 A1 > A2 A2 , A1 B1 ≡ A2 B2 (= s), and as a result A1 B1 < A2 B2 (see Figure 5). So A1 A2 > B1 B2 , and because A1 A2 > A1 A2 , A1 A2 > B1 B2 . Noting what was said above we therefore have: (4) If N2 lies above N1 then M1 M2 > N1 N2 . 3. The centers of two key segment motions In this section we locate the centers of the segment motions described above. Our setting is the hyperbolic plane in which (as is well-known) three kinds of direct motions have to be considered. The Euclidean case could be subsumed with few modifications under the heading of rotations. 154 D. Ruoff Figure 6 Let µi be the rigid, direct motion that carries the segment AB ∈ C onto the segment Ai Bi ∈ C (i = 1, 2) where Ai lies above A, and let mi = perp(a, Mi ), ni = perp(b, Ni ). If lines mi and ni meet, µi is a rotation about their intersection point Gi , if they are boundary parallel, µi is an improper (horocyclic) rotation about their common end γi , and if they are hyperparallel, µi is a translation along their common perpendicular gi (see e.g. [4, p. 455, Satz 13; Figure 6]). We call Gi , γi or gi the center [Gi ] of the motion µi . For any point X disjoint from the center, (X[Gi ]) denotes the line joining X to the center of µi , namely (XGi ), (Xγi ), or perp(gi , X). The ray from X contained in this line and in the direction of [Gi ] −−−→ will be referred to as the ray X[Gi ] from X towards the center of µi ; specifically −−−→ → −−−−→ → → pi = P [Gi ], − mi = Mi [Gi ] and (if it exists) − ni = for X = P, Mi , Ni we define − −−−−→ Ni [Gi ]. Figure 7 We now show that the center [Gi ] of motion µi must lie in H(a, B). → a ) this is clear; if ni meets a in a point I (see FigIf ni does not intersect (− ure 7) we verify the statement as follows. Segment P I as the hypothenuse of P INi is larger than P Ni and so (see (3)) larger than P Mi . Consequently the Generating motions and convexity in hyperbolic geometry 155 angle ∠P INi = ∠Mi INi is acute, which indicates that ni when entering H(a, B) at I, approaches mi . As a result [Gi ] must lie in H(a, B). Some additional consequences are implied by the fact that the center [Gi ] of either motion µi is determined by a pair of perpendiculars mi , ni to lines a and b (see again Figure 6). If [Gi ] = γi is a common end of mi , ni and thus the → − center of a horocyclic rotation, it cannot be the end of ray b . Similarly, if [Gi ] = gi is the common perpendicular of mi and ni , and thus the translation axis, it is hyperparallel to both of the intersecting lines a, b and, as a result, has no point in common with either; furthermore, a and b, being connected, must belong to the same halfplane with respect to gi . On the other hand, if [Gi ] = Gi is the common point of mi and ni , and thus the rotation center, it is indeed possible that it lies on → − ( b ). The point Gi then is collinear with B and with its image Bi which means that for B = Bi the rotation is a half-turn and Gi coincides with the midpoint Ni of BBi ; in addition Gi should be the midpoint Mi of AAi which is impossible. So B = Bi = Ni = Gi ; conversely, one establishes easily that if any two of the three points B, Bi , Ni coincide, µi is a rotation with center Gi equal to all three. We now assume that our plane is furnished with an orientation (see [3, Section − −→ 20]), and that without loss of generality P lies to the left of ray AB. This ray → − → enters H(a, B) at the point A of (− a ) and H(b, A) at the point B of ( b ). Also −−−→ → − → = − Mi [Gi ] enters H(a, B) at a point of (− a ) and so has P on its left hand m i −−−−→ → − side as well (see Figure 8). As to the ray ni = Ni [Gi ] which (if existing, i.e. for → − [Gi ] = Ni ) originates at the point Ni of ( b ), it has P on its left hand side if and only if it enters H(b, A), i.e. if and only if [Gi ] belongs to H(b, A). Figure 8a Figure 8b → to A on the side of P , A lies to the Because the motion µi carries A across − m i i i − → left and A to the right of mi . Being a direct motion, µi consequently also moves −−−−→ → ni = Ni [Gi ] to Bi on the left hand side B (if B = Ni ) from the right hand side of − which is the side of P iff [Gi ] belongs to H(b, A). In short, motion µi carries B → − upwards on ( b ) iff [Gi ] lies in H(b, A). We gather from the previous two paragraphs that → − • [Gi ] belongs to H(b, A) if Bi and Ni lie above B on ( b ), 156 D. Ruoff → − • to ( b ) if Bi = Ni = B, and → − • to H(b, A) if Bi and Ni lie below B on ( b ). Considering the motions µ1 , µ2 again together we can tell in each of the seven situations listed in (2) where the two motion centers [G1 ], [G2 ] (which both belong to H(a, B)) lie with respect to b. As we shall see, the relative positions of [G1 ], [G2 ] can be described in a way that covers all seven situations: rotating ray −−−→ −−−→ −→ → − a = P A about P into H(a, B) we always pass ray P [G1 ] first, and ray P [G2 ] second. More concisely, −−−→ → → → a ,− p2 ) = ∠AP [G2 ]. MAIN LEMMA (ML). Ray − p1 = P [G1 ] always enters ∠(− Proof. (The essential steps of the proof are outlined at the end.) From (2) and the previous paragraph follows that [G1 ] lies in H(b, A) in sit− → uations (I)-(V), on (b) in situation (VI) and in H(b, A) in situation (VII); [G2 ] → − lies in H(b, A) in situations (I)-(III), on ( b ) in situation (IV) and in H(b, A) in situations (V)-(VII), (see Figure 9). As a result the Lemma follows trivially for situations (IV)-(VI). The other situations are more complex, and their proofs require that the nature of the motion centers [Gi ], (i = 1, 2), which can be a point Gi , end γi or axis gi be taken into account. Thus a pair of motion centers [G1 ], [G2 ] can be equal to G1 , G2 ; G1 , γ2 ; G1 , g2 ; γ1 , G2 ; γ1 , γ2 ; γ1 , g2 ; g1 , G2 ; g1 , γ2 ; g1 , g2 . The arguments to be presented are dependent on the mutual position of P, M1 , M2 → − → on − a and of P, N1 , N2 on b , and are best followed through Figure 9. → − → → → p2 ). We first consider situations (I)-(III) in which ∠(− a , b ) includes (− p1 ) and (− − → → − → − To verify (ML) we have to show that p2 does not enter ∠( a , p1 ), or equivalently → → − → → → p2 ). (This assumes − p1 = − p2 which either follows that − p1 does not enter ∠( b , − automatically or as an easy consequence of the arguments below.) → We begin with the special case that − p1 meets m2 in a point I. In this case → − statement (ML) holds if p2 does not intersect line m2 at I or in a point between → p2 and m2 do M2 and I. Obviously this is so if [G2 ] = γ2 or g2 because then − → − not intersect. If [G2 ] = G2 , p2 and m2 do intersect and we have to show that the intersection point, which is G2 , does not coincide with I or lie between M2 and → p1 in I or between I and P I. We first note that line n1 does not intersect ray − because the intersection point would have to be G1 and so lie on m1 , a line entirely below m2 . As a consequence I, P, M2 , and, if it would lie between M2 and I, also G2 , would all belong to the same halfplane with respect to n1 , namely H(n1 , P ). However this would entail that line n2 which runs through G2 would belong to this halfplane, which is not the case in situations (I) and (II). Thus we have established for those situations that G2 = I, and [M2 G2 I] does not hold, which means (ML) is true. We will present the proof of the same in situation (III) later on. Due to the Axiom of Pasch the point I always exists if P M1 [G1 ] is a proper or improper triangle, i.e. if [G1 ] = G1 , or γ1 . This means that we have so far proved (ML) for the cases G1 , γ2 ; G1 , g2 ; γ1 , γ2 ; γ1 , g2 and in addition for G1 , G2 ; γ1 , G2 in situations (I) and (II). In show that for [G1 ] = γ1 or g1 ray p1 does not enter ∠( b . − fact it is useful to mention here that this statement and its proof can be extended to → include configurations in which − p2 meets n1 in an improper point ι . g1 . γ2 in situations (I) and (II). G2 . γ2 in situation (III) require metric considerations and will be presented later. This means that we have proved (ML) also for g1 . In situation (I) the point I always exists if [G2 ] = G2 or γ2 due to the Axiom of Pasch. g2 in situations (I)-(III). g1 . G2 . G2 . and G1 .Generating motions and convexity in hyperbolic geometry 157 Figure 9 → In the opposite special case that − p2 meets n1 in a point I we can analogously → → − → − p2 ) and (ML) holds. . G2 . In situation (II) with n1 = n2 I exists for [G2 ] = G2 and ι exists for [G2 ] = γ2 because in the first case G2 = I and in the second case γ2 = ι . g1 . The proofs of the remaining cases. namely g1 . γ1 . proving (ML) for g1 . g2 . m2 \ g1 = U. − p2 ). g1 . In addition. g1 . γ1 . G2 . G2 . which because of (5) requires that . n2 intersect in G2 then side N1 G1 of N1 M1 G1 is shorter than side N2 G2 of N2 M2 G2 . γ2 cannot occur. i. m2 and → so avoids − p2 . both in situation (VII) and situations (I) (III). thus causing ∠Ni Mi [Gi ] to be acute. angle ∠(− → − − → − → → (m2 ). At the same time. n1 intersect in G1 and m2 . for [G2 ] = γ2 . → → → a . Ruoff Remark.158 D. see Figure 10) the point W does → − → and (− → → → − → n p1 ) with g1 in the interior of ∠(− a . γ2 . P N2 G2 implies ∠( b . (n2 ) together with ( b ). So. g2 . VII) follows N1 N2 < M1 M2 and so (5) N1 M1 = N1 M2 − M1 M2 < N1 M2 − N1 N2 = N2 M2 .e. g1 with − 1 thus fulfilling (ML). p2 \ g1 = W . − and so settles (ML) in the case of G1 . runs in the interior of ∠(− → → →) in order to meet (− m a . Angle ∠P Mi Mi of the right triangle P Mi Mi is also acute with P above mi . and so (6) ∠N1 M1 [G1 ] < ∠N2 M2 [G2 ]. G2 . This and → → − → → − p1 ) < ∠( b . proj(P. a) < M1 M2 (see (4) and Figure 9. defining V . From (6) follows that Lambert quadrilateral N1 S1 R1 M1 has the smaller angle sum and so the larger area than N2 S2 R2 M2 . A). 2) Turning to situation (VII) we observe that each of the rays − m i → − → intersects ( b ) in a point Mi and approaches ray − ni in H(b. Lemma crosses (− m 2 1 (ML) thus is fulfilled for G1 . n2 \ g1 = V. gi ) = Pi . i. enters quadrilateral P N1 S1 W and leaves it. γ1 . From N1 N2 < M1 M2 and M1 M2 = proj(M1 M2 . g2 . between S1 and W . → If in situation (VII) (in which − p2 lies above m2 . G1 . Taking into account that we have already established (ML) in the case in → which − p1 and m2 meet in a point I and [G2 ] = γ2 or g2 we will assume when → p1 and m2 do not meet. from the fact that P M2 M2 has the smaller area (larger defect) than P M1 M1 follows ∠P M2 M2 > ∠P M1 M1 . if in addition [G1 ] = G1 or γ1 . The main case left is that of g1 . P N1 > P N2 applied to P N1 G1 .− p2 ).− p2 ) includes If − p2 does not intersect m2 . gi ) = Si . i. that the cases γ1 .− pi ) have halfline ( b ) in their interior. g2 that − → taking into account that we have already established (ML) in the case that − p2 and → − n1 meet in a point I and [G1 ] = γ1 or g1 we will assume that p2 and n1 do not meet. from (5) and (6) follows that if m1 . halfline (− p1 ) →). If W exists. The remaining cases of (VII) depend on two metric properties. → (i = 1. γ2 .e. which means ∠Ni Mi [Gi ] is its vertically opposite angle −−−→ → pi = P [Gi ] they both enter H(b. As to the rays − → − → → which means that the angles ∠(− a . g2 and γ1 . g2 . gi ) = Ri .e.− p2 ) not exist n1 lies with ( b ). line n2 which runs between the lines n1 . For use in the following we define proj(Mi . A) at P and Ni lies below mi . Also. and. From (5) and (6) it is clear that if m2 intersects or is boundary parallel to n2 then m1 must intersect n1 . proj(Ni . assuming the points exist. 2) with interior altitude P Pi . b relating to situations (I). consequently (− → → interior of ∠(− a . As ∠S1 V N2 = ∠S1 V S2 of N2 N1 S1 V is acute. Adding the images Mi∗ . Ri∗ of Mi . Moreover ∠Pi P Mi > ∠Pi P Ni as ∠Pi P Mi ≡ ∠Pi P Mi∗ ≤ ∠Pi P Ni together with Pi Ri∗ < Pi Si would imply that P Mi∗ Ri∗ Pi would be a part . [P P2 W ] respectively. W on ray S1 R1 satisfy [N2 S2 V ].Generating motions and convexity in hyperbolic geometry 159 Figure 10 −−−→ N1 S1 > N2 S2 . Ri under reflection in P Pi (as illustrated for i = 2 in Figure 11a) we note that Pi Ri∗ < Pi Si because otherwise we would have P Mi ≡ P Mi∗ ≥ P Ni in contradiction to (3). This means that ∠W = ∠P W V must be acute and −−→ → p1 ) = (P P1 ) must lie with V. (III) contains two pentagons P Mi Ri Si Ni (i = 1. Figure 11a Figure 11b Each of the Figures 11a. N2 in the identical with ∠P W P1 . ∠V in P2 S2 V W is obtuse and ∠P2 + ∠S2 + ∠V > 3R.− p2 ). again confirming (ML). As a result V. as we mentioned. So (7a) Pi Ri < Pi Si . In situation (III) we automatically have V such that [U V S1 ] and W such that [U W V ] is fulfilled. n2 and V W include an acute angle which is congruent to ∠N2 V S1 and lies on the lower side of g1 . Now. ∠P W X as the fourth angle of Lambert quadrilateral XR2 P2 W is acute. i. smaller defect). Consequently ∠M2 U R1 < ∠N2 V S1 . angle ∠P2 W Y in triangle P2 W Y is acute. −−→ In both situations m2 and U R1 include an acute angle which coincides with the fourth angle ∠R1 U M2 of Lambert quadrilateral M2 M1 R1 U and so lies on the upper side of g1 . Finally. together with [R1 P1 S1 ] this extends to [R1 U P1 S1 ]. G2 in situation (III) when G2 lies below g1 . and so does the auxiliary point X = proj(W. ∠P W U < R which establishes (ML) for g1 . in addition to ∠W U γ2 < ∠W V γ2 .e. m2 ) in situation (I). This. (7c) Ri Mi > Si Ni . In view of an earlier Remark we assume that the point U exists and that it satisfies [R1 U P1 ]. ∠N2 V S1 is acute. which leads to the same conclusion. ∠S2 V S1 = ∠S2 V W therefore obtuse and in quadrilateral P2 S2 V W ∠P2 + ∠S2 + ∠V > 3R. P2 and S2 lie below g1 then the intersection point V of n2 and g1 exists and lies between N2 and S2 .160 D. If R2 P2 = XW then ∠P W U < ∠P W X = ∠P P2 R2 = R. proves (ML). At the same time (7a) and this angle inequality would imply that the former quadrilateral would fit into the latter. g1 ) belongs to leg (W )U of ∠P W U . if P1 = −−−→ proj(P. ∠P W U . ∠P W U < R.e. We note that this is the case iff ∠P W U is acute. An analogous argument applies to the case g1 . altogether therefore [R1 U W V S1 ]. Since this is contradictory ∠P2 W U must be larger than the adjacent angle ∠P2 W V . and because ∠P W U < ∠P W X. if R2 P2 lies above XW . The proof of (ML) in situation (III) can be extended with only very minor changes to situation (II). In situation (III). as a consequence ∠W = ∠P2 W V is acute and so is its vertically opposite angle. have the smaller area. as a result ∠P2 W V < R and vertically opposite. for similar −−→ reasons. Also closely related is the case of g1 . γ2 in situation (III). Statement (ML) holds in both situations if [U P1 W ] is fulfilled i. Note that if line m2 intersects g1 in a point U between . g2 in situation (III). In situation → (I) we can similarly assume that − p2 and n1 do not meet which means that the point W exists and that it satisfies [U W S1 ]. (7b) ∠Pi P Mi > ∠Pi P Ni . the inequality ∠γ2 W U ≤ ∠γ2 W V were to hold then line m2 would have to run farther away from line p2 than line n2 in contradiction to (3). (7c). If we also had ∠P2 W U ≤ ∠P2 W V then quadrilateral R2 P2 W U would have a smaller angle sum and larger defect than S2 P2 W V . If R2 P2 lies below XW −−−→ and ray R2 P2 intersects g1 in a point Y .e. This concludes the proof of (ML) in situation (I). if in situation (I) R2 . and so ∠W U R2 < ∠W V S2 on the other side of g1 . In situation (III) we have area M2 M1 R1 U > N2 N1 S1 V because of (4). a relation that can be extended to [R1 U W S1 ]. If here. Ruoff polygon of P Ni Si Pi while not having a larger angle sum (i. As a result of all this in situation (III) the closest connection R2 S2 between m2 and n2 lies below g1 . It is congruent to the vertically opposite angle between m2 −−→ and U W which thus lies on the lower side of g1 . b) and note that ∠IP M2 > ∠IP J (7b) implies (i) M2 I > JI. γ1 .(III). γ1 . G1 . (ML) holds if − p2 \ n1 = I (ι ) with [G1 ] = G1 . So G2 and − → − → − not lie in ∠( a . and (ii) ∠P IM2 < ∠P IJ.(VI) are trivial. and N10 between M2 and M1 . γ1 . N10 lies on M2 M1 and I 0 on M2 I. ∠P1 IM2 > ∠P1 IJ. → (2) In situations (I) . and so due to (7a) with S1 V < S1 P1 . (ML) holds if − p1 \ m2 = I with [G2 ] = G2 . −→ G1 . and in situations (I). This is so because according to (4) and (7c) the existence of a Lambert quadrilateral M2 M1 R1 U with R1 U < R1 P1 implies the existence of N2 N1 S1 V with S1 V < R1 U < R1 P1 . b) would determine a segment N1 N2 > N1 J. i. M2 ) then the point N2 = proj(G2 . → (3) In situations (I) . G2 → and g1 . (II) also with [G2 ] = G2 .(III). G2 of (I). and due to (4) the inequality (iii) M1 M2 > N1 J would result. G2 (this with I = m2 \ − p1 on or above P1 ) of situation (III). and due to (ii) halfline (I 0 )P 0 lies in the interior of ∠P1 IM2 which implies that Pb0 is a proper part of polygon Pa = M2 M1 R1 P1 I in contradiction to → p2 do the fact that Pb0 has the smaller angle sum. γ2 . Figure 12a Figure 12b Call J = proj(I. . M2 ) we have according to (i). We present here the last case (Figure 12b) which is easy and representative also for the proofs of the other two cases (Figure 12a). We now carry the pentagon Pb = JN1 S1 P1 I by an indirect motion to Pb0 = −−−−→ −−→ J 0 N10 S10 P10 I 0 where J0 = M2 . G1 . Due to (7c) ray S10 P10 lies in the interior of −−−−→ ∠M1 R1 P1 .(III).Generating motions and convexity in hyperbolic geometry 161 → R1 and P1 rather than − p1 in a point I between P and P1 then G must lie below g1 (see Figure 11b). Assuming that G2 belongs to H(p1 . g2 of (I) . γ1 . G2 . Summary of the Proof. the larger defect. (iii) that I0 lies between M2 −−−→ and I. γ2 . the point P1 thus lies between U and V . g2 . If G2 would lie in H(p1 . G2 . (1) Situations (IV) . (II). p1 ) and the proof of (ML) is complete. and M2 and N2 meet below g1 . To conclude the proof of (ML) we still have to settle the cases G1 .e. 4. N2 M2 [G2 ] reveals the relative position of [G1 ]. (III). G2 . re-formulate and prove a statement which essentially contains the convexity claim of Section 1. [G2 ] in all but one case. −→ g1 . −→ all cases of (VII) except g1 . γ2 of (I). −→ g1 . Subsequently we discuss the details which make the convexity proof complete. N2 S2 R2 M 2 helps to solve the remaining case. G2 . −→ g1 . (8) The area comparison between a part polygon of N1 S1 P1 P . −→ g1 . P M2 R2 S2 N2 helps to solve the same case as in 5. . (7) The arguments of 6. B1 . Theorem 1. γ1 . P. G2 of (III). P1− and P three points in the same halfplane with respect to the line through A and B such that (8) ∠AP2− B ≡ ∠AP1− B ≡ ∠AP B and (9) ∠BAP2− > ∠BAP1− > ∠BAP ≥ ∠ABP. In the configuration of the points A. (III) the area comparison of P M1 R1 S1 N1 . g2 of (VII). G2 of (III) for G2 below g1 . and Theorem 2 to the statement −−→ that the intersection point C1 of A1 P and r1 (i. Let AB be a fixed segment and P2− . Ruoff −→ g1 . A1 . G1 . P1− .162 D. Then the line r which joins P2− and P separates the point P1− from the segment AB (see Figure 13a). B. g2 of (I). together with two similar comparisons. g2 of (II). g1 . g1 . This allows us to substitute the following equivalent theorem for Theorem 1. the image of C1− under the motion µ1 ) lies between A1 and P . γ2 of (III). (5) In situation (VII) the area comparison of N1 S1 R1 M 1 . For the purpose of re-formulating this theorem we carry the points A. direct motion µ1 into the points −→ −−→ A1 . Theorem 2. P. can be extended to three more cases. G2 with G2 above g1 . and one of M1 R1 P1 P . Remark. P1 and the line r1 as defined above. P1 and the line r1 respectively such that A1 lies on P A and B1 on P B (see Figure 13b). (4) In situation (VII) a direct comparison of N1 M1 [G1 ]. settle the remaining cases of (III). Note that Theorem 1 amounts to the statement that the intersection point −−−→ C1− of ray AP1− and line r lies between A and P1− . P and the line r of this configuration by a rigid. Reinterpretation and solution of the posed problem In the following we formulate. (II). g2 . B. g1 . the line r1 separates the point P from segment AB. (6) In situations (I).e. B1 . ρ2 . We first augment our configuration by the images of another rigid. r1 have a point I in . if r. B). → → p2 which connect At this point we augment our figure further by the rays − p1 . and −→ −→ (because ray P A does not enter ∠ρP ρ2 ) P A enters ∠ρ P ρ2 . ρ2 . B). P2− and the line r into −→ −−→ A2 . −−−→ −→ −−→ A1 P = µ1 (AP1− ) and AP are equally directed. B2 . direct motion µ2 which carries the points A. We note that because r joins P2− and P. P. P2 . and since A2 P = µ2 (AP2− ). ρ. From (10) follows that ρ2 lies on the same side of line r = (P ρ) as A. We note that as an exterior angle of AP2− C1− . r2 joins P and P2 . ∠AC1− ρ > ∠AP ρ . ρ2 ∈ H(a. From ∠BAP2− > ∠BAP1− (see (9)) follows ∠B2 A2 P = µ2 (∠BAP2− ) > µ1 (∠BAP1− ) = ∠B1 A1 P and so [AA1 A2 P ] according to Section 2. ρ1 resp. [G2 ] of the motions µ1 . µ2 and. and that as an exterior angle of AC1− P . P and r2 respectively where A2 lies on P A and B2 on P B. ρ1 and ρ lie together with C1− in H(a.Generating motions and convexity in hyperbolic geometry Figure 13a 163 Figure 13b Proof of Theorem 2. The supplementary angles consequently satisfy (10) ∠AP ρ > ∠AC1 ρ1 > ∠AP ρ2 . ρ2 . − P to the centers [G1 ]. Applying µ2 and µ1 on the two sides of the first and µ1 on the left hand side of the second inequality we obtain ∠A2 P ρ2 > ∠A1 C1 ρ1 and ∠A1 C1 ρ1 > ∠AP ρ . Since ρ lies on the left (right) side of AP2− and of AP1− −−−→ −→ −−→ if and only if it lies on the left (right) side of AP . B. ρ1 . In the following we denote the ends of r by ρ and ρ (with ρ on the same side of a = (AP ) as C1− ) and their images on r1 and −−−→ −−−→ r2 by ρ1 . ∠AP2− ρ > ∠AC1− ρ . ρ. This means P ρ and P ρ1 either have a point I or the ends ρ. Because µ2 maps r and ρ to r2 → and ρ2 . Thus ∠ρ P ρ1 = ∠ρ P A + ∠AP ρ1 > ∠ρ P A + ∠AP ρ2 = ∠ρ P ρ2 . the ray −→ P A enters ∠ρ P [G2 ]. The lower halves of the compared → p1 ∗ angles consequently satisfy ∠ρ I[G1 ] > ∠ρ P [G2 ] which means that neither − → − nor the boundary parallel ray p1 would enter ∠ρ P [G2 ]. ρ2 ∈ H(a. according to (10). Since (− lies together with ρ2 in H(a. again in contradiction to Lemma (ML). B). B). We now show by indirect proof that C1 cannot lie on or above P on a. and we can add to the sentence following (10) that also ρ1 and A lie on the same side of r. and ∠ρ P [G1 ] = 12 ∠ρ P ρ1 > 12 ∠ρ P ρ2 = ∠ρ P [G2 ]. ρ1 . Let us first assume that −→ −−→ P ρ.164 D. Note that ∠ρ Iρ1 is equal to the exterior angle ∠C1 Iρ of P C1 I and so satisfies ∠ρ Iρ1 > ∠P C1 I + ∠C1 P I. This → means that − p1 does not enter ∠ρ P [G2 ] and so does not enter ∠AP [G2 ] in contradiction to Lemma (ML). the ray − p2 is the bisector of angle → − → ∗ p2 ) ∠ρ P ρ2 . by the ray − p1 ∗ connecting I to [G1 ]. on a. −−→ In this case line r intersects both segment C1 A and ray C1 ρ1 which means that A and ρ1 lie on the same side of r. approach −→ −−→ −−→ ray C1 ρ1 when entering H(a. B) (see Section 3) whereas ρ lies in H(a. ρ1 in common. . For C1 = P (see Figure 14a) formula (10) reads: ∠AP ρ > ∠AP ρ1 > ∠AP ρ2 . Ruoff → common. B). or r = (P ρ) and r1 = (P ρ1 ) share a perpendicular line whose intersection point with r lies in H(a. Because ∠P C1 I (= ∠AC1 ρ1 ) > ∠AP ρ2 (see (10)) and because ∠C1 P I ≡ ∠ρ P A we have ∠ρ Iρ1 > ∠AP ρ2 + ∠ρ P A = ∠ρ P ρ2 . ray P ρ of r would. B). P ρ1 meet in I (see Figure 14b). while µ1 maps r and ρ to r1 and ρ1 . Figure 14a Figure 14b −→ If C1 were to lie above P . and (if existing) the ray p1 is the bisector of angle ∠ρ Iρ1 . Figure 14c Figure 14d Finally. P1− . ρ to r1 . As a result the ray − ∠AP [G2 ]. and again fails to enter center [G1 ] of motion µ1 . if r and r1 have the perpendicular line g1 in common (see Figure 14d) → p1 = then µ1 which maps r. This completes the proof of Theorem 2 and of Theorem 1. that P2− . line (P Q) separates the points of arc (P Q) from those of segment AB. To do so we choose a point PX . Figure 15 It should be noted that Theorem 1 contains the assumption (9). By symmetry the Theorem also shows the convexity of the half-arc from B to P0 . P belong to the half-arc of our locus from A to the point P0 on the perpendicular bisector of segment AB (see Figure 15)). without loss of generality .Generating motions and convexity in hyperbolic geometry 165 Should ρ = ρ1 (see Figure 14c) then this common end is at the same time the −→ → p1 = P ρ. As a result − −−−→ −→ P [G1 ] must coincide with P ρ which means it does not enter ∠ρ P ρ2 and so does not enter ∠ρ P [G2 ] and ∠AP [G2 ] in contradiction to Lemma (ML). ρ1 is a translation with axis g1 . In order to establish the convexity of the whole arc we need to confirm the additional fact that for P a point on the first half arc and Q a point on the second. Perron. a relation which is fulfilled if P0 . Neue Begr¨undung der ebenen Geometrie. Math. B. Mathematisch-Naturwissenschaftliche Klasse. La Salle.uregina. Fladt). Nichteuklidische Elementargeometrie der Ebene. Ann. [6] O. VEB Deutscher Verlag der Wissenschaften. which enters ∠P AQ. 1962. 1971. Gans. Stuttgart. Berlin. (P0 Q) belong to H(P Q. Orlando. Canada S4S 0A2 E-mail address:
[email protected]. Seminar on Geometry. Leipzig. 1978. Faculty of Mathematics. [3] D. Hilbert. Nichteuklidische Geometrie. Foundations of Geometry. Inc. Cluj-Napoca (1991). The Open Court Publishing Company. a fact of absolute geometry for which there are many easy proofs.ca . Dieter Ruoff: Department of Mathematics and Statistics.G. A). Miszellen zur hyperbolischen Geometrie. 2. [2] D. which means that our claim follows from [ADX CX ]. An Introduction to Non-Euclidean Geometry. 449-474 (1907). Bitay. Sur les angles inscrits dans un segment circulaire en g´eom´etrie hyperbolique. Illinois. Preprint Nr. B. University of Regina. Academic Press. 1925. Sitzungsbericht 1964. References [1] L.166 D. Obviously ray APX . Florida. [7] O. and so (P0 P ). Simon. meets P Q in a point DX .. Nichteuklidische Geometrie in elementarer Behandlung (bearbeitet und herausgegeben von K. Ruoff on half-arc (P P 0 ). Research Seminars. by Theorem 1 segment APX has a point CX in common with segment P0 P . 64. Teubner Verlagsgesellschaft. Teubner. 157–176. “Babes-Bolyai” University. Perron. Also. Quaisser. Bayerische Akademie der Wissenschaften. This however is a consequence of the fact that P0 has a greater distance from (AB) than P and Q. [5] B. [4] J. Regina. 1973. Hjelmslev. [8] M. and establish that segment APX meets segment P Q between −−−→ A and PX . Klotzek and E. Pedoe calls it the most “elementary” theorem of Euclidean geometry and gives variants and equivalent forms of the theorem and cites references where proofs can be found. [11] and [7. The “only if” part is attributed to Urquhart. Problem 73. [14]. The origin and some history of this theorem are discussed in [9]. The theorem referred to in the title states that if ABB and AC C are straight lines with BC and B C intersecting at D and if AB + BD = AC + C D. see Figure 1. and is referred to by Dan Pedoe as “the most elementary theorem of Euclidean geometry”. [10]. where Professor Pedoe attributes the theorem to C C D A B B Figure 1 the late L. [4]. Unaware of most of the existing proofs of this theorem (e. Urquhart (1902-1966) who discovered it when considering some of the fundamental concepts of the theory of special relativity. [13]..g. then AB + B D = AC + CD . see [2] and [1]. Publication Date: April 24. [8]. the author of this note has published a yet another proof in [5]. 2006. then AB +BD = AC + C D if and only if AB + B D = AC + CD. In view of all of this. and where Professor Pedoe asserts that the proof by purely geometric methods is not elementary. in [3]. M. We give a very short and simple proof of the fact that if ABB and AC C are straight lines with BC and B C intersecting at D. This work is supported by a research grant from Yarmouk University. . b b FORUM GEOM ISSN 1534-1178 A Very Short and Simple Proof of “The Most Elementary Theorem” of Euclidean Geometry Mowaffaq Hajja Abstract. pages 23 and 128-129]).b Forum Geometricorum Volume 6 (2006) 167–169. it is interesting to know that De Morgan had published a proof of Urquhart’s Theorem in 1841 and that Urquhart’s Theorem may be viewed as a limiting case of a result due to Chasles that dates back to 1860. Communicating Editor: Floor van Lamoen. ∠CAD = 2γ.168 M. are triangles with angles Aj = 2αj . . AB + AC sin 2γ + sin 2β sin 2γ + sin 2β p(ABC) =1 + =1+ =1+ BC BC sin 2α sin(2γ + 2β) 2 sin(γ + β) cos(γ − β) cos γ cos β + sin γ sin β =1 + =1+ 2 sin(γ + β) cos(γ + β) cos γ cos β − sin γ sin β 2 2 cos γ cos β = . Bj = 2βj . ∠BDA = 2β. we make no claims that our proof meets the standards set by Professor Pedoe who hoped for a circle-free proof. and where we use the law of sines and the addition formulas for the sine and cosine functions. Clearly our proof does not qualify since it rests heavily on properties of circular functions. and if B1 C1 = B2 C2 . then the perimeter p(A1 B1 C1 ) of A1 B1 C1 is equal to or greater than the perimeter p(A2 B2 C2 ) of A2 B2 C2 according as tan β1 tan γ1 is equal to or greater than tan β2 tan γ2 . Breusch’s lemma [12] states that if Aj Bj Cj (j = 1. we give a much shorter proof based on a very simple and elegant lemma that Robert Breusch had designed for solving a 1961 M ONTHLY problem. we see from Breusch’s Lemma that p(AB D) = p(ACD) ⇐⇒ tan β tan(90◦ − γ ) = tan γ tan(90◦ − β) ⇐⇒ tan β cot γ = tan γ cot β ⇐⇒ tan β tan β = tan γ tan γ ⇐⇒p(ABD) = p(AC D). However. as shown in Figure 2. as desired. This lemma follows immediately from the following sequence of simplifications. Hajja In this note. and ∠C DA = 2γ . 2). where we work with one of the triangles after dropping indices. = cos γ cos β − sin γ sin β 1 − tan γ tan β C C 2γ 2β D 2γ 2β A B B Figure 2 Urquhart’s Theorem mentioned at the beginning of this note follows. together with its converse. Cj = 2γj . Referring to Figure 1. and letting ∠B AD = 2β . immediately. J. Washington.au/Publ/Gazette/1997/Apr97/letters. [11] D. New Mathematical Library 4. Mowaffaq Hajja: Department of Mathematics. in which case the four perimeters are equal. Irbid. [9] D. Yarmouk University. 9 (1982) 100. It states that if D.The most elementary theorem of euclidean geometry 169 The M ONTHLY problem that Breusch’s lemma was designed to solve appeared also as a conjecture in [6. Hajja. page 78].austms. 66 (1973) 643–644. Sokolowsky. Yet more on Urquhart’s theorem. Soc. [6] N. Monthly. Geometry Graphics. An elementary proof of the most “elementary” theorem of Euclidean Geometry. [8] L. 1961. The Mathematics Teacher. 1996. S. Crux Math. [13] K. Pedoe. In contrast with the analogous problem obtained by replacing perimeters by areas and the rich literature that this area version has generated. 2 (1976) 133–134. Washington D. Math. (Eureka). [14] K. 15 (1976) 42–44. Breusch Problem 4964. Jordan E-mail address: mhajja@yu. Amer. C. 2 (1976) 132–133. Urquhart’s quadrilateral theorem. addendum. 68 (1961) 384. http://www. (Eureka). CA. Velleman. ibid. Crux Math. Math. Which Way Did The Bicycle Go? . Deakin. Math. A ‘no-circle’ proof of Urquhart’s theorem. B. The provenance of Urquhart’s theorem. and S. A. 4 (1976) 40–42. Williams. D. Eustice. E. B. Gazette. respectively.edu. [12] E. of a triangle ABC. Ontario Mathematics Gazette. 2 (1976) 163– 170. Sauv´e. Deakin. Aust. [7] J.. 8 (1981). E. Extensions of two theorems by Grossman. and AB. then p(DEF ) ≤ min{p(AF E). 8 (2004) 17–22. [5] M. and F are the midpoints of the respective sides.html [3] D. Kazarinoff. (Eureka).. C. Dolciani Mathematical Expositions 18. p(BDF ). D. (Eureka). Grossman. [10] D.org. Mag. S. ibid. Crux Math. On Urquhart’s elementary theorem of Euclidean geometry. MAA. (Eureka).. 2 (1976) 108–109. solution. Sokolowsky. and Other Intriguing Mathematical Mysteries. 69 (1962) 672–674. References [1] M. A. 2 (1976) 63–67. Pedoe’s formulation of Urquhart’s theorem. and F are points on the sides BC. [4] H. On circumscribable quadrilaterals.. [2] M. Trost and R.. Geometric Inequalities.. Urquhart’s theorem and the ellipse. p(CED)} if and only if D. Crux Math. Williams. The most “elementary” theorem of Euclidean geometry. D. Crux Math.jo . Breusch’s solution of the perimeter version is essentially the only solution that the author was able to trace in the literature. Wagon. Konhauser.. 26. MAA. . Let H be the orthocenter of ABC. Indeed. and the incircle (with inradius r) touching the sidelines BC. SCA = SC · SA . both triangles are homothetic to the reference triangle. Ratios and centers of homothety are found. Introduction We consider a pair of triangles associated with a given triangle: the orthic triangle of the intouch triangle. We shall show that this is true if the given triangle is acute. In this paper. 274] asks if these two triangles are homothetic. F respectively. Let I denote the incenter. See Figure 1. F = CH ∩ AB. H= 1 1 1 : : SA SB SC = (SBC : SCA : SAB ). 2 and SBC = SB · SC . so that D E F is the orthic triangle of ABC. AB at D. 2 SB = c2 + a2 − b2 . 2 SC = a2 + b2 − c2 . we adopt standard notations of triangle geometry. Publication Date: May 1. and denote the side lengths of triangle ABC by a. and the intouch triangle of the orthic triangle. The author thanks Paul Yiu for his help in the preparation of this paper. Conway: I =(a : b : c). b b FORUM GEOM ISSN 1534-1178 The Orthic-of-Intouch and Intouch-of-Orthic Triangles S´andor Kiss Abstract. 1. SAB = SA · SB . so that DEF is the intouch triangle of ABC.b Forum Geometricorum Volume 6 (2006) 171–177. p. . O =(a2 SA : b2 SB : c2 SC ) = (SA (SB + SC ) : SB (SC + SA ) : SC (SA + SB )). b. and certain collinearities are proved. E. where SA = b2 + c2 − a2 . Barycentric coordinates are used to prove that the othic of intouch and intouch of orthic triangles are homothetic. CA. Clark Kimberling [1. and indeed each of them is homothetic to the reference triangle. In this paper we make use of homogeneous barycentric coordinates. We shall also denote by O the circumcenter of ABC and R the circumradius. c. E = BH ∩ CA. 2006. Communicating Editor: Paul Yiu. and let D = AH ∩ BC. Here are the coordinates of some basic triangle centers in the notations introduced by John H. Two pairs of homothetic triangles 2. The homothetic center is the point P1 = (a(−a(s − a) + b(s − b) + c(s − c)) : b(a(s − a) − b(s − b) + c(s − c)) : c(a(s − a) + b(s − b) − c(s − c))) =(a(s − b)(s − c) : b(s − c)(s − a) : c(s − a)(s − b)) b c a : : . for example. Kiss 2. P/Q = x − + + u v w u v w u v w See. Let P and Q be arbitrary points not on any of the sidelines of triangle ABC. 2.2. It is well known that the cevian triangle of P = (u : v : w) is perspective with the anticevian triangle of Q = (x : y : z) at x y x y x y z z z :y − + :z + − . [3.3]. Ib A Ic E F P1 I I O C B D Ia Figure 1. = s−a s−b s−c This is the triangle center X57 in [2]. Perspectivity of a cevian triangle and an anticevian triangle.1. The intouch and the excentral triangles are homothetic since their corresponding sides are perpendicular to the respective angle bisectors of triangle ABC. . The intouch and the excentral triangles. §8.172 S. If ABC is obtuse-angled. V . The orthic triangle and the tangential triangle are also homothetic since their corresponding sides are perpendicular to the respective circumradii of triangle ABC. The orthic and the tangential triangle. 1 See Figures 2A and 2B. U = h1 (A). and H is the incenter of the orthic triangle. F into U . Let h1 be the homothety with center P1 . HE and HF are the angle bisectors of the orthic triangle. = SA SB SC This is the triangle center X25 in [2]. W respectively. This is the image of the altitude Ia A of the excentral triangle under the homothety h1 . E. The ratio of homothety is positive or negative according as ABC is acute-angled and obtuse-angled. The homothetic center is the point P2 = a2 (−a2 SA + b2 SB + c2 SC ) : b2 (−b2 SB + c2 SC + a2 SA ) : c2 (−c2 SC + a2 SA + b2 SB ) =(a2 SBC : b2 SCA : c2 SAB ) 2 b2 c2 a : : . swapping D. . The orthic-of-intouch triangle The orthic-of-intouch triangle of ABC is the orthic triangle U V W of the intouch triangle DEF . the same homothety maps B and C 1This ratio of homothety is 2 cos A cos B cos C. the incenter of the orthic triangle is the obtuse angle vertex.The orthic-of-intouch and intouch-of-orthic triangles 173 2. When ABC is acute-angled. Consider an altitude DU of DEF . A B H A C E O F C P2 F B A E B H P2 B D O C C D A Figure 2A.3. 3. Similarly. In particular. HD . Figure 2B. See Figure 3. It follows that U V W is the image of ABC under the homothety h1 . W =(a(s − b)(s − c) : b(s − c)(s − a) : (a + b)(s − a)(s − b)) = s−a s−b s−c U =((b + c)(s − b)(s − c) : b(s − c)(s − a) : c(s − a)(s − b)) = Proof. Kiss into V and W respectively. s−a s−b s−c c+a c a : : . The vertices of the orthic-of-intouch triangle are b c b+c : : . it has radius r r . Ib A Ic E U F I P1 W V B C D Ia Figure 3. The intouch triangle DEF has vertices D = (0 : s − c : s − b). V =(a(s − b)(s − c) : (c + a)(s − c)(s − a) : c(s − a)(s − b)) = s−a s−b s−c b a+b a : : . . Since the circumcircle of U V W is the nine-point circle of DEF . DE : (s − a)x + (s − b)y − (s − c)z = 0. Proposition 1. FD : (s − a)x − (s − b)y + (s − c)z = 0. F = (s − b : s − a : 0).174 S. The sidelines of the intouch triangle have equations EF : −(s − a)x + (s − b)y + (s − c)z = 0. 2 It follows that the ratio of homothety is 2R . E = (s − c : 0 : s − a). The orthic-of-intouch and intouch-of-orthic triangles 175 The point U is the intersection of the lines AP1 and EF . See Figure 3. The line AP1 has equation −c(s − b)y + b(s − c)z = 0. Solving this with that of EF , we obtain the coordinates of U given above. Those of V and W are computed similarly. Corollary 2. The equations of the sidelines of the orthic-of-intouch triangle are V W : −s(s − a)x + (s − b)(s − c)y + (s − b)(s − c)z = 0, W U : (s − c)(s − a)x − s(s − b)y + (s − c)(s − a)z = 0, U V : (s − a)(s − b)x + (s − a)(s − b)y − s(s − c)z = 0. 4. The intouch-of-orthic triangle Suppose triangle ABC is acute-angled, so that its orthic triangle D E F has incenter H, and is the image of the tangential triangle A B C under a homothety h2 with center P2 . Consider the intouch triangle XY Z of D E F . Under the homothety h2 , the segment A A is swapped into D X. See Figure 4. In particular, h2 (A) = X. For the same reason, h2 (B) = Y and h2 (C) = Z. Therefore, the intouch-of-orthic triangle XY Z is homothetic to ABC under h2 . B A C O E X F B H P2 Z Y D C A Figure 4 Proposition 3. If ABC is acute angled, the vertices of the intouch-of-orthic triangle are 2 b + c2 b2 c2 : : , X =((b2 + c2 )SBC : b2 SCA : c2 SAB ) = SA SB SC 2 a c2 + a2 c2 2 2 2 2 : : , Y =(a SBC : (c + a )SCA : c SAB ) = SA SB SC 2 a b2 a2 + b2 2 2 2 2 : : . Z =(a SBC : b SCA : (a + b )SAB ) = SA SB SC 176 S. Kiss Proof. The orthic triangle D E F has vertices D = (0 : SC : SB ), E = (SC : 0 : SA ), F = (SB : SA : 0). The sidelines of the orthic triangle have equations E F : −SA x + SB y + SC z = 0, SA x − SB y + SC z = 0, F D : D E : SA x + SB y − SC z = 0. The point X is the intersection of the lines AP2 and E F . See Figure 4. The line AP2 has equation −c2 SB y + b2 SC z = 0. Solving this with that of E F , we obtain the coordinates of U given above. Those of Y and Z are computed similarly. Corollary 4. If ABC is acute-angled, the equations of the sidelines of the intouchof-orthic triangle are Y Z : −SA (SA + SB + SC )x + SBC y + SBC z = 0, ZX : SCA x − SB (SA + SB + SC )y + SCA z = 0, U V : SAB x + SAB y − SC (SA + SB + SC )z = 0. 5. Homothety of the intouch-of-orthic and orthic-of-intouch triangles Proposition 5. If triangle ABC is acute angled, then its intouch-of-orthic and orthic-of-intouch triangles are homothetic at the point Q= a(a(b + c) − (b2 + c2 )) b(b(c + a) − (c2 + a2 )) c(c(a + b) − (a2 + b2 )) : : (s − a)SA (s − b)SB (s − c)SC . Proof. The homothetic center is the intersection of the lines U X, V Y , and W Z. See Figure 5. Making use of the coordinates given in Propositions 1 and 3, we obtain the equations of these lines as follows. bc(s − a)SA (c(s − c)SB − b(s − b)SC )x +c(s − b)SB ((b2 + c2 )(s − a)SC − (b + c)c(s − c)SA )y +b(s − c)SC (b(b + c)(s − b)SA − (b2 + c2 )(s − a)SB )z = 0, VY : c(s − a)SA (c(c + a)(s − c)SB − (c2 + a2 )(s − b)SC )x +ca(s − b)SB (a(s − a)SC − c(s − c)SA )y +a(s − c)SC ((c2 + a2 )(s − b)SA − (c + a)a(s − a)SB )z = 0, WZ : b(s − a)SA ((a2 + b2 )(s − c)SB − (a + b)b(s − b)SC )x +a(s − b)SB (a(a + b)(s − a)SC − (a2 + b2 )(s − c)SA )y +ab(s − c)SC (b(s − b)SA − a(s − a)SB )z = 0. UX : It is routine to verify that Q lies on each of these lines. Remark. Q is the triangle center X1876 in [2]. The orthic-of-intouch and intouch-of-orthic triangles 177 Q A X E F F H Z Y W B E U I V D D C Figure 5 6. Collinearities Because the circumcenter of XY Z is the orthocenter H of ABC, the center of homothety P2 of ABC and XY Z lies on the Euler line OH of ABC. See Figure 4. We demonstrate a similar property for the point P1 , namely, that this point lies on the Euler line IF of DEF , where F is the circumcenter of U V W . Clearly, O, F , P1 are collinear. Therefore, it suffices to prove that the points I, O, P1 are collinear. This follows from 1 1 1 = 0, cos A cos B cos C (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) which is quite easy to check. See Figure 1. References [1] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1–285. [2] C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html. [3] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University lecture notes, 2001. S´andor Kiss: 440095 Satu Mare, str. Avram Iancu, nr. 58, sc. A, ap. 8, Romania E-mail address:
[email protected] b Forum Geometricorum Volume 6 (2006) 179–180. b b FORUM GEOM ISSN 1534-1178 A 4-Step Construction of the Golden Ratio Kurt Hofstetter Abstract. We construct, in 4 steps using ruler and compass, three points two of the distances between which bear the golden ratio. We present here a 4-step construction of the golden ratio using ruler and compass only. More precisely, we construct, in 4 steps using ruler and compass, three points with two distances bearing the golden ratio. It is fascinating to discover how simple the golden ratio appears. We denote by P (Q) the circle with center P , passing through Q, and by P (XY ) that with center P and radius XY . H F E C A B C1 C3 G D C2 I Figure 1 Construction. Given two points A and B, construct (1) the circle C1 = A(B), (2) the line AB to intersect C1 again at C and extend it long enough to intersect (3) the circle C2 = A(BC) at D and E, (4) the circle C3 = E(BC) to intersect C1 at F and G, and C2 at H and I. Then GH CH = √ 5+1 2 . Proof. Without loss of generality let AB = 1, so that BC = AE = AH = EH = 2. Triangle AEH is equilateral. Let C4 = H(A), intersecting C1 at J. By symmetry, AGJ is an equilateral triangle. Let C5 = J(A) = J(AG) = J(AB), intersecting C1 at K. Finally, let C6 = J(H) = J(BC). See Figure 2. Publication Date: May 8, 2006. Communicating Editor: Paul Yiu. 180 K. Hofstetter With C1 , C5 , C2 , C6 , following√[1], K divides GH in the golden section. It suffices to prove CH = GK = 3. This is clear for GK since the equilateral triangles AJG and AJK have sides of length 1. On the other hand, in the right triangle ACH, AC = 1 and AH = 2. By the Pythogorean theorem CH = √ 3. C4 H F K J1 C B E D A J C3 C6 C7 C5 C1 G C2 I Figure 2 Remark. CH FH = GH CH = √ 5+1 2 . Proof. Since CH 2 = GH · KH, it is enough to prove that F H = KH. Let C4 intersect C1 again at J1 . Consider the circle C7 = J1 (A). By symmetry, F lies on C7 and F H = KH. Reference [1] K. Hofstetter, A simple construction of the golden section, Forum Geom., 2 (2002) 65–66. Kurt Hofstetter: Object Hofstetter, Media Art Studio, Langegasse 42/8c, A-1080 Vienna, Austria E-mail address:
[email protected] b Forum Geometricorum Volume 6 (2006) 181–185. b b FORUM GEOM ISSN 1534-1178 A Theorem by Giusto Bellavitis on a Class of Quadrilaterals Eisso J. Atzema Abstract. In this note we prove a theorem on quadrilaterals first published by the Italian mathematician Giusto Bellavitis in the 1850s, but that seems to have been overlooked since that time. Where Bellavitis used the functional equivalent of complex numbers to prove the result, we mostly rely on trigonometry. We also prove a converse of the theorem. 1. Introduction Since antiquity, the properties of various special classes of quadrilaterals have been extensively studied. A class of quadrilaterals that appears to have been little studied is that of those quadrilaterals for which the products of the two pairs of opposite sides are equal. In case a quadrilateral ABCD is cyclic as well, ABCD is usually referred to as a harmonic quadrilateral (see [2, pp.90–92], [3, pp.159– 160]). Clearly, however, the class of all quadrilaterals ABCD for which AB · CD = AD · BC includes non-cyclic quadrilaterals as well. In particular, all kites are included. As far as we have able to ascertain, no name for this more general class of quadrilaterals has ever been proposed. For the sake of brevity, we will refer to the elements in this class as balanced quadrilaterals. In his Sposizione del metodo delle equipollenze of 1854, the Italian mathematician Giusto Bellavitis (1803-1880) proved a curious theorem on such balanced quadrilaterals that seems to have been forgotten.1 In this note, we will give an elementary proof of the theorem. In addition, we will show how the converse of Bellavitis’ theorem is (almost) true as well. Our proof of the first theorem is different from that of Bellavitis. The converse is not discussed by Bellavitis at all. 2. Bellavitis’ Theorem Let the lengths of the sides AB, BC, CD and DA of a (convex) quadrilateral ABCD be denoted by a, b, c and d respectively. Similarly, the lengths of the quadrilateral’s diagonals AC and BD will be denoted by e and f . Let E be the point of intersection of the two diagonals. The magnitude of ∠DAB will be Publication Date: May 15, 2006. Communicating Editor: Paul Yiu. 1Bellavitis’ book is very hard to locate. We actually used Charles-Ange Laisant’s 1874 translation into French [1], which is available on-line from the Biblioth`eque Nationale. In this translation, the theorem is on p.26 as Corollary III of Bellavitis’ derivation of Ptolemy’s theorem. . Atzema referred to as α. Quadrilateral Notations With these notations. ∠ADB etc will be denoted by αB . then αB + βC + γD + δA = βA + γB + δC + αD = 180◦ . Finally. or cos (γB + αD ) + cos (δ + α) = cos (αB + γD ) + cos (δ + γ). δC and so on (see Figure 1). The magnitudes of ∠DAC. the magnitude of ∠CED will be referred to as . That is. 1854). The second equality sign does not hold for non-convex quadrilaterals.182 E. J. cos (γB + αD ) − cos (γB − α + αB ) = cos (αB + γD ) − cos (αB − γ + γB ). Note that the convexity condition is a necessary one. with similar notations for the other angles of the quadrilateral. Theorem 1 (Bellavitis. If a (convex) quadrilateral ABCD is balanced. A trigonometric proof of Bellavitis’ Theorem follows from the observation that by the law of sines for any balanced quadrilateral we have sin γB · sin αD = sin αB · sin γD . the following result can be proved. or cos (γB + αD ) − cos (γB − αD ) = cos (αB + γD ) − cos (αB − γD ). D δC δ A d c f A αB αD ε e E γD a βC βA γB C b B Figure 1. . the first factor has to be zero and we are done. Likewise. 2 2 Now. we know that a2 + c2 = b2 + d2 . or 1 1 cos (δC + γB + βA + αD ) · cos (γB + αD − δC − βA ) 2 2 1 1 = cos (αB + βC + γD + δA ) · cos (αB + γD − βC − δA ). will be zero if and only if ABCD is cyclic. 1 1 (δC + γB + βA + αD ) + (αB + βC + γD + δA ) = 180◦ . Finally. in the case that ABCD is cyclic. we may assume that ABCD is not cyclic and that the third term does not vanish. 2 2 It follows that 1 1 cos (δC + γB + βA + αD ) · cos ( + (β + δ)) 2 2 1 1 = − cos (δC + γB + βA + αD ) · cos ( − (β + δ)). Again. In other words. ABCD has to be a kite. it is easy to see that in that case Bellavitis’ theorem is true. we also have cos (δC + βA ) + cos (α + β) = cos (βC + δA ) + cos (α + δ). Clearly. likewise αB + γD − βC − δA = 2 − β − δ. In combination with the initial condition ac = bd. 2 2 or 1 1 cos (δC + γB + βA + αD ) · cos () cos (δ + β) = 0. It is easy to see that any such quadrilateral has the angle property of Bellavitis’ theorem. adding these two equations gives cos (γB + αD ) + cos (δC + βA ) = cos (αB + γD ) + cos (βC + δA ). if neither of the last two factors are equal to zero. however. Therefore. This proves Bellavitis’ theorem. 2 2 This almost concludes our proof. Since cos (α + β) = cos (δ + γ). the second factor only vanishes in case ABCD is orthogonal. this implies that each side has to be congruent to an adjacent side.A theorem by Giusto Bellavitis on a class of quadrilaterals 183 By cycling through. We can safely assume that ABCD is not a kite and that the second term does not vanish either. Bellavitis’ theorem is true. For such quadrilaterals. The last factor. Consequently. note that γB + αD − δc − βA = 360 − 2 − δ − β and. the area of ABC is equal to both abe/4RABC and 12 e · EB · sin with similar expressions for ADC. . Now. The Converse to Bellavitis’ Theorem Now that we have proved Bellavitis’ theorem. the relation between the four circumradii can be rewritten to the form EB ·EC = EA·EC. sin δC · sin βA = sin βC · sin δA + K. where RABC denotes the radius of the circumcircle to the triangle ABC etc. . Consequently. it is only natural to wonder for exactly which kinds of (convex) quadrilaterals the angle sums δC + γB + βA + αD and αA + βC + γD + δA are equal. d sin γD · sin γB We find sin βA · sin βC · sin δC · sin δA = sin αB · sin αD · sin γD · sin γB . b sin αB · sin αD Cycling through twice also gives us sin δC · sin δA b = . we have bd = ac and ABCD is balanced. and BCD. sin γB sin γD K − =− sin αB sin αD sin αB · sin αD and sin δA K sin δC − = sin βC sin βA sin βA · sin βC or K K a b d a − =− − = . So. it follows that sin βA · sin βC d = . Likewise. we find that sin γB · sin αD + K = sin αB · sin γD for some K. We have the following result: Theorem 2. Any (convex) quadrilateral ABCD for which αB + βC + γD + δA = βA + γB + δC + αD = 180◦ is either cyclic or balanced. If K = 0.184 E. Division of each side by abcd and grouping the factors in the numerators and denominators appropriately shows that this equation is equivalent to the equation RABC · RADC = RBAC · RBCD . J. But this means that ABCD has to be cyclic. Atzema 3. Assuming that the two angle sums are equal and working our way backward from the preceding proof. c b sin αB · sin αD d c sin βA · sin βC If K = 0. BAC. Laisant) (Paris: Gauthier-Villars.edu 2This method essentially amounted to a sometimes awkward mix of vector methods and the use of complex numbers in a purely geometrical disguise. The sums αB +βC +γD +δA and βA +γB +δC +αD do not usually show up in plane geometry. Erg¨anzungsb¨ande. Bellavitis.2 Indeed. In fact. however. for those interested in the use of complex numbers in plane geometry. Most of these were definitely wellknown at the time. 1913) [3] M. References [1] G. Maine 04469. Eisso J. These earlier publications. 1906 (= Jahresberichte der Deutschen Mathematiker-Vereinigung. This should not be too hard. University of Maine. Orono. however. 3See the introduction of [1] for a list of references. Bellavitis may have derived the theorem in one of the many papers that he published between 1833. Atzema: Department of Mathematics. Teubner. Exposition de la m´ethode des equipollences (traduit par C-A. . are extremely hard to locate and we have not been able to consult any. the Sposizione features a fair number of (minor) results on quadrilaterals that are proved using the method of equipollences. and 1854.A theorem by Giusto Bellavitis on a class of quadrilaterals 185 4. it is not entirely surprising that this result seems to have been forgotten. when he first published on the method. in which these sums play a role as part of a generalization of Ptolemy’s theorem to arbitrary (convex) quadrilaterals. 1874) [2] W. Jahrhundert.3 Whether the theorem originated with Bellavitis or not.umaine. The Modern Geometry of the Triangle (2nd ed. Simon.) (London: Hodgson. USA E-mail address: atzema@math. Bellavitis was clearly mostly interested in the theorem because it allowed him to showcase the power of his method of equipollences. Conclusion We have not been able to find any references to Bellavitis’ theorem other than in the Sposizione. Ueber die Entwicklung der Elementar-Geometrie im XIX. Gallatly. We do hope to finish up a paper shortly. it might be a worthwhile exercise to rework Bellavitis’ equipollences proof of his theorem to one that uses complex numbers only. Leipzig. This suggests that perhaps our particular result was reasonably well-known at the time as well. I). Alternatively. B. . b Forum Geometricorum Volume 6 (2006) 187–190. or 2 the projectivity would be called elliptic. ∆ABC becomes ∆ACC (see Figure 1) and AM becomes AM . which is what we wanted to prove. Since ∠M AM = π2 . Prove that ∠M AN = π4 . is hiding an elliptic projectivity of focus A.45]. b b FORUM GEOM ISSN 1534-1178 A Projectivity Characterized by the Pythagorean Relation Wladimir G. A Romanian Olympiad problem It is known that any projectivity relating two ranges on one line with more than two invariant points is the identity transformation of the line onto itself. and this means ∠M AN ≡ ∠N AM . N. or hyperbolic (see Coxeter [1. or [3. M. 2006. Therefore. Consider a counterclockwise rotation around A of angle π2 . parabolic. The configuration appears in a problem introduced in the National Olympiad 2001. We study an interesting configuration that gives an example of an elliptic projectivity characterized by the Pythagorean relation. pp. The equality BM 2 + N C 2 = M N 2 transforms into CM 2 + N C 2 = M N 2 = M N 2 . Communicating Editor: Paul Yiu. 1. ∆M AN ≡ ∆N AM (SSS case). This note will point out an interesting and unusual configuration that gives an example of projectivity characterized by a Pythagorean relation. Depending on whether the number of invariant points is 0. By applying this rotation.4143]). First. this is what makes this problem and this geometric structure so special and deserving of our attention. C A M B M N C Figure 1 We present first an elementary solution for this problem. p. since ∆CN M is right in C. We shall show that the metric relation introduced in the problem above. . the angle ∠M AM is right. thus. by Mircea Fianu. in Romania. Boskoff and Bogdan D. similar to the Pythagorean relation. 1. Publication Date: May 22. The statement of the problem is the following: Consider the right isosceles triangle ABC and the points M. we get ∠M AN = π4 . C such that BM2 + N C 2 = M N 2 . Actually. N on the hypothenuse BC in the order B. we would like to recall a few facts of projective geometry. Suceav˘a Abstract. the cross ratio of four ordered lines is the cross ratio determined by the points of intersection with a line L. Suceav˘a 2. mq − np = 0. 3. The points Ai and Bi are called homologous points of the projectivity on L. ÷ = BC BD x3 − x2 x4 − x2 (1) This definition may be extended to a pencil consisting of four ordered lines L1 . and D be four points. B. respectively. Consider a system of coordinates on L such that A. and x4 .248]): (ABCD) = AD x3 − x1 x4 − x1 AC ÷ . i = 1. Theorem 1. we will use this consequence in the proof we present below. We call a projectivity on a line L a map f : L → L with the property that the cross ratio of any four points is preserved. C. is by definition (see for example [5. in this order. as for example in the case of the feet of interior and exterior bisectors associated to the side BD of a triangle M BD (see Figure 2). x3 . p.12. C. D on L. L3 . D. x2 . In fact. Actually. C. M A B Figure 2 C D . and D correspond to x1 . that is (L1 L2 L3 L4 ) = (A1 A2 A3 A4 ) = (B1 B2 B3 B4 ) where Bi = f (Ai ) . A projectivity on L is determined by three pairs of homologous points. L4 . 4. on the line L in the Euclidean plane. The following result is presented in many references (see for example [3]. y= px + q where m. p. Projectivities Let A. Boskoff and B. L2 . A1 A3 A1 A4 ÷ A2 A3 A2 A4 where {Ai } = L ∩ Li . The cross ratio of four ordered points A. n. p. and the relation Bi = f (Ai ) is denoted Ai → Bi . A consequence of this theorem is that two projectivities which have three common pairs of homologous points must coincide. By definition. q ∈ R. The law of sines shows us that the above definition is independent on L. G. B.188 W. the coordinates x and y of the homologous points under a projectivity are related by mx + n . 2.34). Therefore. B. Theorem 4. it is possible that (ABCD) takes the value −1. In formula (1). A projective solution to Romanian Olympiad problem With these preparations. M3 . The rotation of the moving angle ∠hk yields. Their intersections with the line L yield the points M1 . we will say that the homologous of the point C is the point at infinity. a). then point A is not on the line determined by points B and D. the pairs of equal angles: ∠M1 AM2 =∠N1 AN2 = β1 . k2 . 3. for the pencil of rays h1 . h4 and k1 . we are ready to give a projective solution to the initial problem. respectively. In the particular case when C is the midpoint of BD. since M A becomes parallel to BD. we need the following. ∠h4 k4 . ∞D BD For our result. h2 . M2 . To extend the bijectivity of the projectivity presented above. ∠M3 AM4 =∠N3 AN4 = β3 . 0). we have that ∆M BD is isosceles and the external bisector M A is parallel to BD. Indeed. and C(a. ÷ sin β2 sin(β2 + β3 ) This proves the claim that f is a projectivity on L in which the homologous points are M and N. By the law of sines we get that the two cross ratios are equal. The relation BM 2 + N C 2 = M N 2 becomes (x + a)2 + (a − y)2 = (y − x)2 . ∠h2 k2 . We shall also accept the convention BC ∞C ÷ = −1. 0) and N (y. Denote by {M } = h ∩ L and {N } = k ∩ L. let A be a fixed point and L a fixed line such that A is not on L. We have to prove that f : L → L defined by f (M ) = N is a projectivity on the line L determined by the rotation of ∠hk. Consider four positions of the angle ∠hk. both of them having the value sin(β1 + β2 ) sin(β1 + β2 + β3 ) . respectively. N3 . As mentioned in the statement. We consider also M (x. 0). Consider a system of coordinates in which the vertices of the right isosceles triangle are A(0. . ∠h3 k3 . k4 . B(−a. N4 . denoted ∞. N1 . Lemma 2. See Figure 3. in a pair of points related by a projectivity. Consider the rays h and k with origin in A. N2 . Proof. The point A with the property (ABCD) = −1 can be found at the intersection between the external bisector of ∠BM D and the straight line BD. A moving angle with vertex in the fixed point A in the plane intersects a fixed line L. ∠M2 AM3 =∠N2 AN3 = β2 . the moving angle ∠hk with the vertex in A and of constant measure α. denoted consecutively ∠h1 k1 . M4 .A projectivity characterized by the Pythagorean relation 189 Observe that if C is the midpoint of the segment BD. 0). for any C on the straight line BD there exist the point M in the plane (not necessarily unique) such that M C is the interior bisector of ∠BM D. which we attach to the line d. A not on L. h3 . It is sufficient to prove that the cross ratio [M1 M2 M3 M4 ] and [N1 N2 N3 N4 ] are equal. k3 . since by replacing the x−coordinate of B in (2) we get 0. Euclidean and Non-Euclidean Geometries. Coxeter. 2003. California State University. 0) Figure 3 or. M. S. N. Projective Geometry. Sixth Edition.edu . This concludes the proof and the geometric interpretation: the Pythagorean-like metric relation from the original problem reveals a projectivity. (2) a−x This is the equation of a projectivity on the line BC. California 92834-6850.e. G´eom´etrie sup´erieure. E-mail address: bsuceava@fullerton. 0) O(0. H. Bd. Mamaia 124 900527 Constantza. we see that B → O. S. Third Edition. We also have O → C. 1993. 1992. the x−coordinate of O.S. B → O. M. which makes this geometric structure remarkable. 1998. 0)N (y. solving for y. 1981. On the other hand. Therefore. Moscow. Freeman & Co. G.A. J. since ∠BAO = π4 . Suceav˘a: Department of Mathematics. C → ∞. U. H. Furthermore. D.. University Ovidius. C → ∞. Greenberg. Coxeter. Since a projectivity is completely determined by a triple set of homologous points. Mir.ro Bogdan D. Springer-Verlag. a) B(−a. Springer-Verlag. the pair M → N has the property ∠M AN = π4 . 0 → a and a → ∞ express that O → C and. 0) C(a. First. since ∠CA∞ = π4 . Fullerton. represented by the homologous points M → N. Wladimir G. 0) M(x. Third edition. MAA. Ed.190 W. Boskoff and B. Similarly. This projectivity is completely determined by three pairs of homologous points. respectively. Second Edition. M. Boskoff: Department of Mathematics and Computer Science. since ∠CAO = π4 . Suceav˘a A(0. y= References [1] [2] [3] [4] [5] H. i. S. Finally. Consider now another projectivity on BC determined by the rotation about A by π4 (see Lemma 2). Coxeter. ax + a2 . The Real Projective Plane. Non-Euclidean Geometry. Efimov. M. Romania E-mail address: boskoff@univ-ovidius. the two projectivities must coincide. this solution shows that M and N can be anywhere on the line determined by the points B and C. the intersection of the reflections of the line joining the circumcenter and incenter in the sidelines of the intouch triangle. since AY = AZ = AE. then the six points A. The authors thank Jean-Pierre Ehrmann for his interesting remarks on the paper. F respectively. the midpoints of the sides of triangle DY Z. AY = AF . If E and F are the midpoints of DZ and DY . F . we show that the Feuerbach point in the Euler reflection point of the intouch triangle. In Figure 1. E . Y A Z Ha F E E F I B D C Figure 1. b b FORUM GEOM ISSN 1534-1178 The Feuerbach Point and Euler lines Bogdan Suceav˘a and Paul Yiu Abstract. . A M ONTHLY problem Consider a triangle ABC with incenter I. Similarly. the point E is the foot of the altitude on DZ. namely.b Forum Geometricorum Volume 6 (2006) 191–197. E . F . ∠AZE = ∠CDE = ∠CED = ∠AEZ. is the nine-point circle of the triangle. It follows that AY = AF = AE = AZ. E. AB at D. F are on the same circle. I. E. By studying a generalization. the point of tangency of the incircle and the nine-point circle. Communicating Editor: Jean-Pierre Ehrmann. The triangle Ta and its orthocenter Here is an alternative solution. Therefore AZ = AE. Similarly. Let Y (respectively Z) be the intersection of DF (respectively DE) and the line through A parallel to BC. Now. the incircle touching the sides BC. 1. CA. and A is the midpoint of Y Z. 2006. The circle through A. The circle in question is indeed the nine-point circle of triangle DY Z. Given a triangle. See [3]. we construct three triangles associated its incircle whose Euler lines intersect on the Feuerbach point. F Publication Date: June 4. This is Problem 10710 of the American Mathematical Monthly with slightly different notations. Denote by the Euler line M Ha of triangle DY Z. The Euler line of triangle DY Z contains the Feuerbach point of triangle ABC. and these two points are on the same nine-point circle. is also on the nine-point circle. F . the point of tangency of the incircle and the nine-point circle of the latter triangle. Since ∠HaED = ∠Ha F D are right angles. We show that the parallel through N to the line IHa intersects at a point N such that N N = R2 . Suceav˘a and P. and N be respectively the circumcenter. where R is the circumradius of triangle ABC. Ha lies on the circle containing D. being the midpoint of the segment DHa . and ninepoint center of triangle ABC. It is well known that N is the midpoint of OH. 2. The intersection Ha = EY ∩ F Z is the orthocenter of triangle DY Z. The Feuerbach point on an Euler line The center of the nine-point circle of DY Z is the midpoint M of IA. Let O. The line M Ha is therefore the Euler line of triangle DY Z. The Euler line of Ta C . which is the incircle of triangle ABC. At the same time. and has DHa as a diameter. Proof. H. E. orthocenter. Z A Y Ha F H N E M E I F O N H A B I O D D J Ia Figure 2.192 B. Theorem 1. note that Ha is the antipodal point of the D on the incircle of triangle ABC. Yiu is the foot of the altitude on DY . It follows that I. Since N N and IHa are directly parallel. the line HA is parallel to IHa . which is their external center of similitude. Consider the excircle (Ia ) on the side BC. See Figure 3. Since DHa is a diameter of the incircle. This means that N is a point on the nine-point circle of triangle ABC. AH intersects at a point H such that AH = Ha I = r. Z A Y N Ha Fe F F E E I N B D A C Figure 3. Consider also the reflection I of I in O. and O that of II . It is well known that I Ia passes through the point of tangency D of (Ia ) and BC. We first show that JO = ra : 2ra JM Ia A JO = · IHa = ·r = · r = ra . and the excircle (Ia ). See Figure 2.The Feuerbach point and Euler lines 193 Clearly. The Euler line of Ta passes through the Feuerbach point Remark. it is well known that AH = 2 · OA . It is well known that the two circles are tangent internally at the Feuerbach point Fe . The midpoint of IIa is also the midpoint J of the arc BC of the circumcircle (not containing the vertex A). . with radius ra . Let the line through O parallel to IHa intersect at O . the inradius of triangle ABC. we have 2N N =HH + OO =(HA − H A) + (JO − R) =2 · A O − r + ra − R =DI + D I + ra − (R + r) =r + (2R − ra ) + ra − (R + r) =R. IM IA 2r Since N is the midpoint of OH. the lines N Ha and N I intersect at the external center of similitude of the nine-point circle and the incircle. Since M is the midpoint of IA. If A is the midpoint of BC. the Feuerbach point Fe is indeed the pedal of D on the Euler line of triangle DY Z. we can also construct the triangles Tb and Tc (containing respectively E with a side parallel to CA and F with a side parallel to AB). Tb . F respectively. and the parallel through A to BC. More precisely. in the construction of Ta .194 B. Theorem 1 also applies to these triangles. and Tc . we obtain another triangle Ta . The excircle case If. Suceav˘a and P. Analogous to Ta . D F . The Euler line of Ta passes through Sc = X442 C E . we replace the incircle by the A-excircle (Ia ). Ta is the triangle DY Z bounded by the lines D E . Yiu Denote the triangle DY Z by Ta . and CA. 3. The method in §2 leads to the following conclusions. if the excircle (Ia ) touches BC at D . The Feuerbach point is the common point of the Euler lines of the three triangles Ta . Z A F Y N N Sc E O A B Fa D Ma F Ia Ha Figure 4. AB at E . Corollary 2. F = (s − b : s − a : 0). The lines At D. 0. and IAB intersect at a point on the Euler line. Pt = b+c−a . Their Euler lines are the images of the Euler lines of ICA and IAB under the same homothety. Ha . F .The Feuerbach point and Euler lines 195 (1) The nine-point circle of Ta contains the excenter Ia and the points E . From this we conclude that the Euler lines of Ta . E = (s − c : 0 : s − a). 4. − 12 . See Figure 4. is a point on the Euler line of triangle ABC. and shown that it is the image of the Euler line of triangle IBC under the homothety h := h G. This. It appears in [1] as the triangle center X442 . every homothetic image of ABC in I is perspective with Ha Hb Hc . Tb . A generalization The concurrency of the Euler lines of Ta . which is the triangle center X21 in [1]. Tb . the vertices of the intouch triangle are D = (0 : s − c : s − b). t) has vertices At =(a + t(b + c) : (1 − t)b : (1 − t)c). F on the incircle. Therefore. Bt =((1 − t)a : b + t(c + a) : (1 − t)c). a also contains the internal center of similitude of the nine-point circle (N ) and the excircle (Ia ). On the other hand. E. (2) The orthocenter Ha of Ta is the antipode of D on the excircle (Ia ). Every homothetic image of ABC in I is perspective with the intouch triangle DEF . K. These three lines intersect at the point (a + t(b + c))(b + c − a + 2at) : ··· : ··· . Tc . Ct =((1 − t)a : (1 − t)b : c + t(a + b)). Tc concur at the image of Sc under the homothety h. We work with homogeneous barycentric coordinates. Tb . Bt E. L. (3) The Euler line a of Ta contains the point N . 0. Here. Nguyen [2] has recently studied the line containing Fa and Ma . Proof. its center is the midpoint Ma of the segment AIa . ICA. Tc respectively. Theorem 3. Hb . which is the point of tangency Fa of these two circles. More generally. Hc are the orthocenters of Ta . Recall that the Euler lines of triangles IBC. and Ct F have equations (1 − t)(b − c)(s − a)x −(s − a)(b + (c + a)t)x (s − a)(c + (a + b)t)x + + − (s − b)(a + (b + c)t)y (1 − t)(c − a)(s − b)y (s − b)(c + (a + b)t)y − + + (s − c)(a + (b + c)t)z (s − c)(b + (c + a)t)z (1 − t)(a − b)(s − c)z = = = 0. the Schiffler point Sc . The image of ABC under the homothety h(I. The same is true for the two analogous triangles Tb and Tc . again. This is clearly equivalent to the following theorem. can be paraphrased as the perspectivity of the “midway triangle” of I with the triangle Ha Hb Hc . They are the antipodes of D. Γ is indeed the Jerabek hyperbola of the intouch triangle. Its antipode on the incircle is therefore the Euler reflection point of the intouch triangle. Bt E and Ct F are perpendicular to F D and DE respectively. Since it contains also the circumcenter I of DEF . every homothetic image of ABC in P = (u : v : w) is perspective with the cevian triangle of the isotomic conjugate 1 1 1 of the superior of P . Q= b+c−a A E Fe Ct F Bt I Ge Q At B Pt D C Figure 5. namely. and is therefore perpendicular to EF . Therefore. The conic Γ clearly contains I and the Gergonne point. and similarly. Note also that D = Pt a for t = − b+c or − s−a a . . Its center is the point a(a2 (b + c) − 2a(b2 + c2 ) + (b3 + c3 )) : ··· : ··· . of the infinite point of its Euler line. we get the cevian triangle of the Gergonne point which is the intouch triangle. with respect to the intouch triangle. More generally. Γ contains D. corresponding respectively to t = 0 and t = 1. the line At D is parallel to the bisector of angle A. Note that the fourth intersection of Γ with the incircle is the isogonal conjugate. It is clear that the perspector Pt traverses a conic Γ as t varies. With P = I. E and F . It follows that Γ is a rectangular hyperbola. Proposition 4. Simiarly. It is a cirumconic of the intouch triangle DEF . which is the triangle center X65 in [1].196 B. Yiu Remark. Suceav˘a and P. the point v+w−u : w+u−v : u+v−w . since its coordinates are quadratic functions of t. as t = ∞. Now. The perspector of At Bt Ct and Ha Hb Hc is the reflection of P−t in the incenter. for an arbitrary point P . The Jerabak hyperbola of the intouch triangle The reflection of Γ in the incenter is the conic Γ which is the locus of the perspectors of Ha Hb Hc and homothetic images of ABC in I. The perspector P∞ is therefore the orthocenter of triangle DEF . The Feuerbach point is the Euler reflection point of the intouch triangle. L.edu Paul Yiu: Department of Mathematical Sciences. Florida 33431-0991. References [1] C. USA E-mail address: bsuceava@fullerton. Monthly. Encyclopedia of Triangle Centers.edu . and − 2(R−r) respectively. is the triangle cente X1317 . Boca Raton.html. (1) The fourth intersection of Γ with the incircle. Florida Atlantic University. Suceav˘a and A.edu/ck6/encyclopedia/ETC. Amer. Theorem 5. CA 928346850. [2] K. Math. and 1537. available at http://faculty. 5 (2005) 149–164. These are the perspector for the homothetic images with ratio +1 and −1 respectively. 106 (1999) 68. It must be the Feuerbach point on Γ . solution. Fullerton. Sinefakopoulos. being the antipode of the Feuerbach point. A E Fe F B I D O C Figure 6. On the complement of the Schiffler point. Kimberling. of ABC with ratios t = −1. Bogdan Suceav˘a: Department of Mathematics. These are the perspectors for the homothetic images R r . Forum Geom. Problem 19710. Nguyen. 107 (2000) 572–573. (Note: X145 is the reflection of the Nagel point in the incenter). [3] B. This means that the reflections of OI (the Euler line of the intouch triangle) concur at F . − R+r (2) The hyperbola Γ contains the following triangle centers apart from I and Fe : X8 and X390 (which is the reflection of the Gergonne point in the incenter).evansville. The conic Γ also contains Xn for the following values of n: 145. 224. USA E-mail address:
[email protected] Feuerbach point and Euler lines 197 This must also be the perspector of Ha Hb Hc (the antipode of DEF in the incircle) and a homothetic image of ABC. The Feuerbach point as the Euler reflection point of DEF Remarks. California State University. . Triangle A∗ B ∗ C ∗ is the anticevian triangle of the infinite point Q = (u(v − w) : v(w − u) : w(u − v)) of the trilinear polar of P . which depends on P only. . A A A∗ Z Y B∗ P τP B B X C C C∗ Figure 1 Proposition 1. We construct a simple perspectivity that is invariant under isotomic conjugation. We write τ (P ) := P/Q = (u(v − w)2 : v(w − u)2 : w(u − v)2 ). similarly define B and C . C ∗ = AA ∩ BB . b b FORUM GEOM ISSN 1534-1178 A Simple Perspectivity Eric Danneels Abstract. construct the parallels through B to XY and through C to XZ to intersect at A . Introduction In this note we consider a simple transformation of the plane of a given reference triangle ABC. Given a point P with cevian triangle XY Z. 1. A∗ B ∗ C ∗ is perspective with every cevian triangle. As an anticevian triangle. The author thanks Paul Yiu for his help in the preparation of this paper. Communicating Editor: Paul Yiu. Publication Date: June 12. 2006. We shall prove Proposition 1 in §2 below.b Forum Geometricorum Volume 6 (2006) 199–203. Construct A∗ = BB ∩ CC . B ∗ = CC ∩ AA . In particular. it is perspective with XY Z at the cevian quotient P/Q. x y z It has perspector τ (P ). the hyperbola passes through A. τ (P ) is (1) the center of the circumconic through P and its isotomic conjugate P• . X523 X519 . Proposition 2. X312 X94 . X253 X57 . X1088 X524 . (1) The circumconic through P and P • has equation u(v 2 − w2 ) v(w2 − u2 ) w(u2 − v 2 ) + + = 0. P. Proof. X8 X30 . More precisely. X88 X98 . k(x + y + z)2 + (ux + vy + cz) u v w For k = −1. then the other is ux + yv + wz = 0. (1) G/P . X536 • X694 . X671 • X538 . Here is a list of triangle centers with their images under τ . Its center is the cevian quotient (2) The pencil of hyperbolas with asymptotes the trilinears polars of P and P• has equation x y z + + = 0. X75 X4 .200 E. X694 • X1026 . The labeling of triangle centers follows Kimberling [1]. X264 X7 . Remark. X1022 X2394 . Wilson Stothers [2] has found that one asymptote of a circum-hyperbola determines the other. Danneels Let P • denote the isotomic conjugate of P . X2400 τ (P ) X2972 X11 X1650 X2087 X868 X2310 X1648 X1645 X1635 X1637 X676 . X2396 τ (P ) X244 X125 X122 X2170 X2088 X1649 X1647 X1646 X2086 X2254 X2491 P. P • X1 . X69 X20 . Proposition 3. X2407 X2398 . τ is invariant under isotomic conjugation: τ (P• ) = τ (P ). This gives a stronger result than (2) above. This is τ (P ). B. (2) the perspector of the circum-hyperbola with asymptotes the trilinear polars of P and P • . X538 • X1022 . X323 X99 . X325 X200 . X903 • X536 . and this circum-hyperbola has equation u(v − w)2 v(w − u)2 w(u − v)2 + + = 0. (and center P given in (1) above). X1026 X2395 . if ux + vy + wz = 0 is an asymptote of a circum-hyperbola. C. X1494 • X88 . P • X3 . x y z with perspector P = (u(v 2 − w2 ) : w(w2 − u2 ) : w(u2 − v 2 )). Similarly. B1 . C =(u(v − w) : v(u − w) : w(u + v)). Now the lines AA . C ∗ are the midpoints of the segments AA1 . BB1 . C to the trilinear polar L. Then. B ∗C ∗ : v(w − u) w(u − v) z x + = 0. CC intersect at the points A∗ =BB ∩ CC = (−u(v − w) : v(w − u) : w(u − v)). This is clearly the anticevian triangle of the point 1 1 1 1 1 1 − : − : − . A∗ . and infinite point (−u(v+w) : v(w + u) : w(u − v)). B. AB at A1 . CC1 . (1) Here is an easy alternative construction of A∗ B ∗ C ∗ . Proof of Proposition 1 The line XY has equation vwx+wuy−uvz = 0. A∗ B ∗ : u(v − w) v(w − u) Proposition 4. C ∗ A∗ : u(v − w) w(u − v) y x + = 0. The two analogously defined points are B =(u(w − v) : v(w + u) : w(u − v)). C ∗ =AA ∩ BB = (u(v − w) : v(w − u) : −w(u − v)). B ∗ . The parallel through B to XY is the line w(u − v)x + u(v + w)z = 0. These two lines intersect at A = (u(v + w) : v(w − u) : w(v − u)). Construct the parallels through A. This completes the proof of Proposition 1. Q = (u(v − w) : v(w − u) : w(u − v)) = v w w u u v which is the infinite point of the trilinear polar L. .A simple perspectivity 201 2. (2) The equations of the sidelines of triangle A∗ B ∗ C ∗ are z y + = 0. BB . C1 respectively. See Figure 2. Remarks. intersecting the sidelines BC. CA. the parallel through C to XZ is the line v(u − w)x + u(v + w)y = 0. The trilinear polar of τ (P ) with respect to the cevian triangle of P passes through P . B ∗ =CC ∩ AA = (u(v − w) : −v(w − u) : w(u − v)). C ∗ A∗ ∩ ZX =(−u(v − w) : v(w + u − 2v) : w(u − v)). 3. The trilinear polar of τ (P ) with respect to XY Z is the perspectrix of the triangles XY Z and A∗ B ∗ C ∗ . Now. Generalization Since the construction in §1 is purely perspective we can replace the line at infinity by an arbitrary line : px + qy + rz = 0. The parallel through B to XY becomes the line joining B to the intersection and XY . u v w This clearly contains the point P = (u : v : w). . the sidelines of these triangle intersect at the points B ∗ C ∗ ∩ Y Z =(u(v + w − 2u) : v(w − u) : −w(u − v)). A∗ B ∗ ∩ XY =(u(v − w) : −v(w − u) : w(u + v − 2w)). The line through these three points has equation w−u u−v v−w x+ y+ z = 0. Danneels A B1 A∗ Z B∗ Y P τP B A1 X C C∗ C1 Figure 2 Proof. etc. The perspector becomes τ (P ) = (u(qv − rw)2 : v(rw − pu)2 : w(pu − qv)2 ).202 E. Eric Danneels: Hubert d’Ydewallestraat 26.be . available at http://faculty. Kimberling.evansville.html. where L is the trilinear polar of P .A simple perspectivity 203 : r21w . References [1] C. Hyacinthos message 8476. 8730 Beernem. October 30. and the following remain Then τ (P ) = τ (P ) where P = p21u : q21v valid: (1) A∗ B ∗ C ∗ is the anticevian triangle of Q = L ∩ . Stothers.danneels@pandora. 2003. Encyclopedia of Triangle Centers. [2] W. (2) The perspectrix of XY Z and A∗ B ∗ C ∗ contains the point P .edu/ck6/encyclopedia/ETC. Belgium E-mail address: eric. . with circumcenter O. define Bc . beginning with the orthogonal projections of the vertices on the circumradii OA. 2006. A Ac Hb Bc Cb O Ab Ba Hc B H C Ha Ca Figure 1 Theorem 1. OC. Publication Date: June 19. let Ab and Ac be the pedals (orthogonal projections) of the vertex A on the lines OB and OC respectively.b Forum Geometricorum Volume 6 (2006) 205–212. Communicating Editor: Paul Yiu. The triangles AAb Ac . Ba BBc and Ca Cb C are congruent to the orthic triangle Ha Hb Hc . In this paper we prove some interesting results on triangles associated with these pedals. . OB. and other members of the Hyacinthos group for help and advice during the preparation of this paper. Similarly. we construct two triangles each with circumcircle tangent to the nine-point circle at the center of the Jerabek hyperbola. See Figure 1. b b FORUM GEOM ISSN 1534-1178 Pedals on Circumradii and the Jerabek Center Quang Tuan Bui Abstract. Ca and Cb . Ba . Peter Moses. The author thanks Paul Yiu. 1. Introduction Given a triangle ABC. Given a triangle ABC. (SC − SA ). Ac =(SA (SB + SC ) : SB (SC + SA ) : SC (SB − SA )). 0) =SC (−(SB + SC ). Theorem 3. Bc Ba and Ca Cb bound a triangle T2 whose circumcircle is tangent to the nine-point circle at the Jerabek center. Since (SA (SB + SC ). We verify that the point P = (SA (SB + SC ) : SC (SC − SA ) : SC (SA + SB )) is the pedal Ab of A on the line OB. SC (SA + SB )) =(SA (SB + SC ). −(SBC + 2SCA + SAB ). The Jerabek center J is the triangle center X125 in Kimberling’s Encyclopedia of Triangle Centers [1]. Ab =(SA (SB + SC ) : SC (SC − SA ) : SC (SA + SB )). SB (SC + SA ). this is a point on the line OB. The lines Bc Cb . 0. Ca =(SA (SB − SC ) : SB (SC + SA ) : SC (SA + SB )). Proposition 4. the infinite point of the line AP is (SA (SB + SC ). Ba =(SA (SC − SB ) : SB (SC + SA ) : SC (SA + SB )). The circumcircle of T1 is tangent to the nine-point circle of ABC at the Jerabek center. SC (SA + SB )) + (0. Recall that the Jerabek center J is the center of the circum-hyperbola through the circumcenter O.206 Q. This hyperbola is the isogonal conjugate of the Euler line. (1) The labeling of triangle centers. except for the common ones. 2(SBC + SCA + SAB ). The coordinate sum of P being (SB +SC )(SC +SA ). follows [1]. (SA + SB )). 0) =(SA (SB + SC ). SC (SA + SB )) − (0. the circumcircles of T1 and T2 are also tangent to each other at J. SC (SC − SA ). Basic results can be found in [2]. Cb =SA ((SB + SC ) : SB (SA − SC ) : SC (SA + SB )). SC (SC − SA ) − SB (SC + SA ). SB (SC + SA ). SC (SC − SA ). Hence. The Jerabek center J. SC (SA + SB )). Ca Ac and Ab Ba bound a triangle T1 homothetic to ABC. Proof. for example. . Bui Theorem 2. The lines Ab Ac . Bc =(SA (SB + SC ) : SB (SC + SA ) : SA (SA − SB )). In this paper we work with homogeneous barycentric coordinates and adopt standard notations of triangle geometry. has coordinates (SA (SB − SC )2 : SB (SC − SA )2 : SC (SA − SB )2 ). The homogeneous barycentric coordinates of the pedals of the vertices of triangle ABC on the circumradii are as follows. 0). The infinite point of OB is (SA (SB + SC ). T. SC (SA + SB )) − ((SC + SA )(SA + SB ). §4.5]. 2. Proof of Theorem 1 Note that the points Ab and Ac lie on the circle with diameter OA. Since these four triangles have equal circumradii R 2 . Therefore the angles in triangles AAb Ac and Ha Hb Hc are the same. A Ac Ka Hb Gb Bc Gc Cb O Ab Hc B H Ha Kb N Ba Kc C Ga Ca Figure 2 ∠Ac AAb =π − ∠Ab OAc = π − ∠BOC = π − 2A = ∠Hc Ha Hb . so do the midpoints of AC and AB.Pedals on circumradii and the Jerabek center 207 By the theorem in [2. they are congruent. Therefore. similarly for triangles Ba BBc and Ca Cb C. ∠AAb Ac =∠AOAc = π − ∠COA = π − 2B = ∠HaHb Hc . . This completes the proof of Theorem 1. the two lines AP and OB are perpendicular since −SA · (SB + SC ) · SA (SB + SC ) −SB · (SC − SA ) · (SBC + 2SCA + SAB ) +SC · (SA + SB ) · SC (SA + SB ) =0. ∠Ab Ac A =∠Ab OA = π − ∠AOB = π − 2C = ∠Hb Hc Ha . b cos B. Proof. . and Ab Ba . (Ka ) and (Ka ). Gb . See Figure 2. C1 . Bui Remarks. B1 =(SA (SB + SC )2 : SB (SC − SA )2 : SC (SA + SB )2 ). With reference to Figure 1. The triangle T1 bounded by the lines Bc Cb .208 Q. Ca Ac . Ca Ac . Kc are the midpoints of the circumradii OA. SB (SC + SA )x − (SBB + SCA )y + SB (SC + SA )z =0. The lines Bc Cb . It follows that the quadrilateral Bc Cb CB is an isosceles trapezoid. Proposition 6. SC (SA + SB )x + SC (SA + SB )y − (SCC + SAB )z =0. and Ab Ba have equations −(SAA + SBC )x + SA (SB + SC )y + SA (SB + SC )z =0. Similarly. and c cos C respectively. OC. T. (ii) Bc B = Cb C. Ca Ac . we have (i) ∠Bc BC = π2 − ∠OCB = π2 − ∠OBC = ∠Cb CB. From the coordinates of A1 . The triangle T1 We now consider the triangle T1 bounded by the lines Bc Cb . C1 =(SA (SB + SC )2 : SB (SC + SA )2 : SC (SA − SB )2 ). and also to the medial triangle Ga Gb Gc . (Kb ). triangle Ka Kb Kc is homothetic to (i) ABC at O. Therefore. we obtain the coordinates of the vertices of T1 : A1 =(SA (SB − SC )2 : SB (SC + SA )2 : SC (SA + SB )2 ). OB. Triangle T1 is homothetic to (i) ABC at the procircumcenter (a4 SA : b4 SB : c4 SC ). it is clear that the homothetic center of triangles A1 B1 C1 and ABC is the point (SA (SB + SC )2 : SB (SC + SA )2 : SC (SA + SB )2 ) = (a4 SA : b4 SB : c4 SC ). 1This is the triangle center X 184 in [1]. the lines Ca Ac and CA are parallel. The midpoints Ga . the nine-point center of the medial triangle. 3. (3) The circles (Kb ) and (Kc ) intersect at the circumcenter O and the midpoint Ga of BC. 1 (ii) the medial triangle Ga Gb Gc at the Jerabek center J. the lines Bc Cb and BC are parallel. Ab Ba is homothetic to triangle ABC. The quadrilateral Bc Cb CB is an isosceles trapezoid. with ratio of homothety −1. similarly for the other two pairs (Kc ). Gc lie on the nine-point circle (N ) of triangle ABC. and (ii) the medial triangle Ga Gb Gc at X140 . (1) The side lengths of these triangles are a cos A. Proof. with ratio of homothety 12 . as are Ab Ba and AB. Kb . Lemma 5. (2) If Ka . B1 . From these. it must also lie on the circumcircle of the other. It is routine to check that this contains the Jerabek center J whose coordinates are given in (1). 4. Gb B1 . Since the homothetic center J lies on the circumcircle of the medial triangle.Pedals on circumradii and the Jerabek center 209 For (ii). Gc C1 are respectively (SAA − SBC )x − SA (SB − SC )y + SA (SB − SC )z =0. Proof of Theorem 2 Theorem 2 is now an immediate consequence of Proposition 6(ii). SB (SC − SA )x + (SBB − SCA )y − SB (SC − SA )z =0. A Ac C1 J Bc B1 Cb Gc Gb X184 N O Ab Ba A1 B C Ga Ca Figure 3 . −SC (SA − SB )x + SC (SA − SB )y + (SCC − SAB )z =0. and the two circumcircles are tangent at J. the equations of the lines Ga A1 . T. From these.210 Q. the vertices of triangle T2 are the points A2 =((SB − SC )2 : 3SAB + SBC + SCA − SCC : 3SCA + SAB + SBC − SBB ). Triangles ABC and A2 B2 C2 are perspective at Q= 1 1 1 : : a2 b2 + b2 c2 + c2 a2 − b4 − c4 a2 b2 + b2 c2 + c2 a2 − c4 − a4 a2 b2 + b2 c2 + c2 a2 − a4 − b4 . and Ca Cb : −2SBC x + (S 2 − SBB )y + (S 2 − SCC )z =0. A Ac C2 Bc Cb O Ab Ba Q B2 A2 B C Ca Figure 4 Proposition 7. C2 =(3SCA + SAB + SBC − SCC : 3SBC + SCA + SAB − SAA : (SA − SB )2 ). . we obtain the equations of the lines Ab Ac . Bui 5. (S 2 − SAA )x + (S 2 − SBB )y − 2SAB z =0. (S 2 − SAA )x − 2SCA y + (S 2 − SCC )z =0. The triangle T2 From the coordinates of the pedals. B2 =(3SAB + SBC + SCA − SCC : (SC − SA )2 : 3SBC + SCA + SAB − SAA ). Bc Ba . it is enough to consider the pedal of the circumcenter O on the radical axis (SA + SB )(SA + SC )(SAB + SCA − 2SBC )2 x = 0. Remark. C2 whose coordinates are obtained by cyclic permutations of SA . Q= 3SBC + SCA + SAB − SAA These are equivalent to those given above in terms of a. T2 is perspective with ABC at 1 : ··· : ··· . Q is the image of the midpoint of HQ. which is the center of the Jerabek hyperbola. +(x + y + z) cyclic To verify that this circle is tangent to the circumcircle (SB + SC )yz + (SC + SA )zx + (SA + SB )xy = 0. This completes the proof of Theorem 3.Pedals on circumradii and the Jerabek center 211 Proof. B2 . 2). SC . Proof of Theorem 3 It is easier to work with the image of triangle T2 under the homothety h(H. c. b. The triangle center Q does not appear in [1]. and also on the Jerabek hyperbola SA (SBB − SCC ) SB (SCC − SAA ) SC (SAA − SBB ) + + = 0. and B2 . 2). 2 Under the homothety h(H. C2 given above. SB . x y z This shows that the circle A2 B2 C2 is tangent to the circumcircle at Q . The images of the vertices are A2 =(SA (SB + SC )(SBB − 4SBC + SCC ) + SBC (SB − SC )2 : (SC + SA )(SA (SB + SC )(3SB − SC ) + SBC (SB − SC )) : (SA + SB )(SA (SB + SC )(SB − 3SC ) + SBC (SB − SC ))). cyclic This is the point SC + SA SA + SB SB + SC : : . 2Q is the triangle center X 74 . From the coordinates of A2 . 6. the circumcircle of T2 is tangent to the nine-point circle at J. The circumcircle of A2 B2 C2 has equation 8S 2 · SABC ((SB + SC )yz + (SC + SA )zx + (SA + SB )xy) (SA + SB )(SA + SC )(SAB + SCA − 2SBC )2 x = 0. Q = SCA + SAB − 2SBC SAB + SBC − 2SCA SBC + SCA − 2SAB which is clearly on the circumcircle. Under the inverse homothety. Hanoi. Encyclopedia of Triangle Centers. 296/86 by-street.212 Q. Kimberling. Florida Atlantic University lecture notes. Vietnam E-mail address: bqtuan1962@yahoo. T. Introduction to the Geometry of the Triangle. Quang Tuan Bui: 45B. Bui C2 A Ac Q C2 J Hb Bc B2 Cb O N Ab Ba B2 Hc H B A2 A2 C Ha Ca Figure 5 References [1] C. Minh Khai Street. 2001.edu/ck6/encyclopedia/ETC.com .html. [2] P.evansville. Yiu. available at http://faculty. In both cases. Ωi (P ) =(u(v + w) : v(w + u) : w(u + v)) respectively. the perspector of this conic is one of its foci. The given trilinear equation is : π π π + β sin( B + + γ sin C + = 0. According to Vigari´e. In this paper. Denote by L(P ) its trilinear polar. b b FORUM GEOM ISSN 1534-1178 Simmons Conics Bernard Gibert Abstract. curiously. P is said to be the perspector of the conic and L(P ) its perspectrix. Ωc (P ) is also the perspector of the medial triangle and the anticevian triangle AP BP CP of P . C. we find a definition of a conic called “ellipse de Simmons” with a reference to E. Simmons [7]. The contacts of this “ellipse” with the sidelines of reference triangle ABC are the vertices of the cevian triangle of X13 . The centers of Γc (P ) and Γi (P ) are Ωc (P ) =(u(v + w − u) : v(w + u − v) : w(u + v − w)). We study the conics introduced by T. p. The locus of the trilinear pole of a line passing through P is a circumconic denoted by Γc (P ) and the envelope of trilinear polar of points of L(P ) is an inscribed conic Γi (P ). working with barycentric coordinates. this “ellipse” was introduced by T. Ωi (P ) is the complement of the isotomic conjugate of P . Tome 3. Circumconics and inscribed conics Let P = (u : v : w) be any point in the plane of triangle ABC which does not lie on one sideline of ABC. the corresponding conic with the other Fermat and isodynamic points is not mentioned in [1]. Vigari´e series of papers [8] in 1887-1889. . Introduction In [1. Simmons and generalize some of their properties. In other words. α sin A + 3 3 3 It appears that this conic is not always an ellipse and. we generalize the study of inscribed conics and circumconics whose perspector is one focus.227]. Note that L(P ) is the polar line of P in both conics. Publication Date: June 26. 2. Communicating Editor: Paul Yiu. 1. C. 2006. and has foci the first isogonic center or Fermat point (X13 in [5]) and the first isodynamic point (X15 in [5]).b Forum Geometricorum Volume 6 (2006) 213–224. This center must be the isotomic conjugate of the anticomplement of P .2. Let X be the fourth intersection of the conic and the circumcircle (X is the trilinear pole of the line KP ).214 B. The directrix associated to the perspector/focus in both Simmons conics is also the trilinear polar of this same perspector/focus. Elements of the conics S13 S14 perspector and focus X13 X14 other real focus X15 X16 center X396 X395 focal axis parallel to the Euler line idem non-focal axis L(X14 ) L(X13 ) directrix L(X13 ) L(X14 ) other directrix L(X18 ) L(X17 ) Remark. A similar construction in the anticevian triangle of P gives the foci of Γc (P ). The line BC and its perpendicular at Pa meet one axis at two points. A similar construction in the cevian triangle Pa Pb Pc of P gives the axes of Γi (P ). Proof.17]. Hence. There are two and only two non-degenerate inscribed conics whose perspector P is one focus : they are obtained when P is one of the isogonic centers. These strophoids are orthopivotal cubics as seen in [4. The circle with center Ωi (P ) which is orthogonal to the circle having diameter these two points meets the axis at the requested foci. the strophoids intersect at the isogonic centers. Construction of the axes of Γc (P ) and Γi (P ). 3. A computation shows that P must lie on three circum-strophoids with singularity at one vertex of ABC. The axes of Γc (P ) are the parallels at Ωc (P ) to the bisectors of the lines BC and AX. These conics will be called the (inscribed) Simmons conics denoted by S13 = Γi (X13 ) and S14 = Γi (X14 ). all the conics of the pencil have axes with the same directions (parallel and perpendicular to the Euler line) and are centered on the rectangular hyperbola . the infinite point of the perpendiculars to the Euler line. This will be generalized below. Construction of the foci of Γc (P ) and Γi (P ). Gibert 2. The four (not always real) base points of the pencil form a quadrilateral inscribed in the nine point circle and whose diagonal triangle is the anticevian triangle of X523 . They are the isogonal transforms of the three Apollonian circles which intersect at the two isodynamic points. If P is one focus of Γi (P ). 2. p. Inscribed conics with focus at the perspector Theorem 1. the other focus is the isogonal conjugate P∗ of P and the center is the midpoint of P P∗ .1. Hence. In Figure 1 we have four real base points on the nine point circle and on two parabolas P1 and P2 . The two (inscribed) Simmons conics generate a pencil of conics which contains the nine-point circle. Theorem 2. X523 . This hyperbola passes through the in/excenters. X396 . X523 ) (See [3]). X30 . more generally. Another approach is the following. The two (inscribed) Simmons conics generate a tangential pencil of conics which contains the Steiner inellipse. the intersection is X523 at infinity (the perspector of the Kiepert hyperbola) hence the common tangent must be the trilinear polar of this point. The fourth common tangent to two inconics is the trilinear polar of the intersection of the trilinear polars of the two perspectors. their centers X396 and X395 lie on the line GK.Simmons conics 215 P2 X16 NPC X13 A G B C X15 P1 X14 S14 S13 Figure 1. Indeed. See Figure 2. those of the diagonal triangle above. Note that the polar lines of any of its points in both Simmons inconics are parallel. any inconic with perspector on the . These conics must be tangent to the trilinear polar of the root X523 which is the line through the centers X115 and X125 of the Kiepert and Jerabek hyperbolas. cyclic It must also contain the vertices of the anticevian triangle of any of its points and. The locus of foci of all inconics with center on this line is the (second) Brocard cubic K018 which is nK0 (K. X1749 and is centered at X476 (Tixier point). Theorem 3. Simmons ponctual pencil of conics which is the polar conic of X30 (point at infinity of the Euler line) in the Neuberg cubic. In fact. in particular. X5 . In the case of the Simmons inconics. This is the diagonal conic with equation : 2 (b2 − c2 )(4SA − b2 c2 )x2 = 0. X395 . (3) The focal axes meet the non-focal axes at the vertices of a rectangle with center X230 on the orthic axis and on the line GK. See Figures 2 and 3. (4) The orthic axis is the mediator of the non-focal axes. since G lies on the Kiepert hyperbola. the singular focus of the cubic. (2) This line X115 X125 meets the sidelines of ABC at three points on K018. In particular. X395 and two other points P1 . P2 on the cubic K018 and collinear with X111 . This is also the case of the inconic with center K.216 B. Gibert L(X17) X16 P1 K018 X395 S14 A NPC X230 X13 K X396 X111 G X15 X125 X14 13 X115 P2 C S B L(X18) Polar conic of X30 L(X13) L(X14) Figure 2. These vertices are X396 . perspector H sometimes called K-ellipse (see [1]) although it is not always an ellipse. The two Simmons inconics S13 and S14 Kiepert hyperbola must be tangent to this same line (the perpector of each conic must lie on the Kiepert hyperbola since it is the isotomic conjugate of the anticomplement of the center of the conic). . the Steiner inellipse must also be tangent to this line. (1) The contacts of this common tangent with S13 and S14 lie on the lines through G and the corresponding perspector. Remarks. Simmons conics 217 P A S13 X1648 S14 X13 G B Steiner inellipse X15 X115 X111 X14 C Figure 3. Circumconics with focus at the perspector A circumconic with perspector P is inscribed in the anticevian triangle Pa Pb Pc of P .e. 1X 1648 is the tripolar centroid of X523 i. the isobarycenter of the traces of the line X115 X125 . it must then be a Fermat point of ABC. This is the in-parabola with perspector X671 (on the Steiner ellipse). Simmons tangential pencil of conics (5) The pencil contains one and only one parabola P we will call the Simmons parabola. The fourth common point of these conics is X476 (Tixier point) on the circumcircle. There are two and only two non-degenerate circumconics whose perspector P is one focus : they are obtained when P is one of the isogonic centers.1 4. . It lies on the line GK. In other words. Theorem 4. focus X111 (Parry point). P is a focus of the circumconic if and only if it is a Fermat point of Pa Pb Pc . Hence. The centers and other real foci are not mentioned in the current edition of [6] and their coordinates are rather complicated. The focal axes are those of the Simmons inconics. it is the inconic with perspector P in Pa Pb Pc . Thus. According to a known result 2. See Figure 4. 2The angular coordinates of a Fermat point of P P P are the same when they are taken either a b c with respect to Pa Pb Pc or with respect to ABC. They will be called the Simmons circumconics denoted by Σ13 = Γc (X13 ) and Σ14 = Γc (X14 ). touching the line X115 X125 at X1648 . Gibert X16 L(X13) L(X14) X395 A X13 X396 X15 Y14 X476 B C Σ14 X14 Σ13 Y13 Figure 4. This shows that one can find six other circumconics with focus at a Fermat point but. The two Simmons circumconics Σ13 and Σ14 A digression: there are in general four circumconics with given focus F . B. C. CC the circles passing through F with centers A.1. this focus is not the perspector. One of these lines is the trilinear polar L(Q) of the interior point 1 1 1 : BF : CF and the remaining three are the sidelines of the cevian triangle Q = AF of Q. 5. We have seen that these Simmons inconics are quite remarkable in the sense that the directrix corresponding to the perspector/focus F (which is the polar line of F in the conic) is also the trilinear polar of F . Some related loci We now generalize some of the particularities of the Simmons inconics and present several higher degree curves which all contain the Fermat points. Directrices and trilinear polars. See Figure 5. 5. These four lines are the directrices of the sought circumconics and their construction is therefore easy to realize. The generalization gives the following . Let CA .218 B. These circles have two by two six centers of homothety and these centers are three by three collinear on four lines. in this case. CB . Q003 is a very remarkable curve with equation a2 (SB y − SC z)y 2 z 2 = 0 cyclic which (at the time this paper is written) contains 70 points of the triangle plane. Directrices of circumconics with given focus Theorem 5. See [3] and [4]. The locus of the focus F of the inconic such that the corresponding directrix is parallel to the trilinear polar of F is the Euler-Morley quintic Q003. The Stothers quintic Q012 has equation a2 (y − z)(x2 − yz)yz = 0. we have the inconic with focus F at one of the extraversions of X1156 (on the Euler-Morley quintic). Theorem 6. or equivalently such that the pencil of conics generated by these two conics contains a circle. The Simmons inconics (or circumconics) have their perspectors at a focus hence on an axis. cyclic Q012 is also the locus of point M such that the circumconic and inconic with same perspector M have parallel axes. More generally. See [3]. The locus of the perspector P of the inconic (or circumconic) such that P lies on one of its axis is the Stothers quintic Q012.Simmons conics 219 CA A Q B L(Q) F C CC CB Figure 5.2. In Figure 6. . 5. Perspector lying on one axis. The focus F . An inconic with directrix parallel to the trilinear polar of the focus The center of the inconic in Theorem 6 must lie on the complement of the isotomic conjugate of Q012. .220 B. Q039 is also the locus of point P whose pedal triangle has a Brocard line passing through P . The locus of the focus F of the inconic such that F . Theorem 7. 5. Gibert ctr ix Q003 dir e A F B C L(F) Figure 6. The locus of P such that the polar lines of P and its isogonal conjugate P ∗ in one of the Simmons inconics are parallel are the two isogonal pivotal cubics K129a and K129b. More generally.3. the center Ω (midpoint of F F∗ ) and the perspector P (the isotomic conjugate of the anticomplement of Ω) of the inconic may be seen as a special case of collinear points. we have the inconic with perspector X673 (on the Stothers quintic) and center X3008 . its isogonal conjugate F∗ (the other focus). F∗ and P are collinear is the bicircular isogonal sextic Q039. The center of the circumconic in Theorem 6 must lie on a septic which is the G−Ceva conjugate of Q012. Remark. See [3]. Perspector lying on the focal axis. cyclic In Figure 7. another quartic with equation a2 (y + z − x)(y − z)(y 2 + z 2 − xy − xz) = 0. An inconic with perspector on one axis Ic X16 A F1 B X13 G I F2 X15 X1379 Ib asymptote X14 C X325 X1380 Ia asymptote Kiepert hyperbola Figure 8.Simmons conics 221 A X3008 X673 G C B Q012 Figure 7. The bicircular isogonal sextic Q039 . the homology must send the directrix to the line at infinity and. Appendices 6. with the conic S13 we obtain K129b = pK(K. The contacts lie on the line GK and on the circle with center G passing through X115 . the center of the Kiepert hyperbola. see remark 3 at the end of §3) into the infinite point X30 of the Euler line or the line X13 X15 . Furthermore. If we take one of these lines as an axis of homology. The axis of such homology must be parallel to the directrix and its center must be the common focus. The following table gives these eccentricities. must transform the point P1 (or P2 . ∆1 and ∆2 meet the Euler line at two points lying on the circle with center G passing through X125 . the two Simmons circumconics Σ13 and Σ14 are transformed into two circles Γ13 and Γ14 having the same . it is possible to find infinitely many homologies (perspectivities) transforming these two conics into concentric circles with center X13 (or X14 ) and the radius of the first circle is twice that of the second circle. X395 ). This is also true for Σ14 and S14 . In his paper [7]. for example. See [3]. conic eccentricity OH 1 S13 √ × √∆ 2(cot ω + 3) OH ×√ ∆ 2(cot ω − 3) OH 2 √ × √∆ 2(cot ω + 3) S14 Σ13 Σ14 1 2 2(cot ω − √ √ OH ×√ ∆ 3) where ω is the Brocard angle. cyclic ∆1 and ∆2 are the tangents to the Steiner inellipse which are perpendicular to the Euler line. T. 6. Simmons has shown that the eccentricity of Σ13 is twice that of S13 .2. C. 6. the center of the Jerabek hyperbola. Gibert More precisely. X396 ) and with the conic S14 we obtain K129a = pK(K. Since Σ13 and S13 (or Σ14 and S14 ) have the same focus and the same directrix.1. Let ∆1 and ∆2 be the two lines with equations (b2 + c2 − 2a2 + a4 + b4 + c4 − b2 c2 − c2 a2 − a2 b2 ) x = 0 cyclic and (b2 + c2 − 2a2 − a4 + b4 + c4 − b2 c2 − c2 a2 − a2 b2 ) x = 0.222 B. ∆ the area of ABC and OH the distance between O and H. Kimberling. Congressus Numerantium. available at http://faculty. Forum Geom. Obviously.-P. Librairie Albert Blanchard. third edition.Simmons conics 223 radius. Encyclopedia of Triangle Centers.html. Orthocorrespondence and orthopivotal cubics. Courbes G´eom´etriques Remarquables. [6] C.gibert/ [4] B. the two Simmons inconics are also transformed into two circles having the same radius. Lemoyne. the line M P1 meets ∆1 at m. Gibert. P1 Γ13 L(X 13) A X13 X476 m B M’ Γ14 P2 X14 C L(X 14) M ∆1 Σ13 Σ14 Figure 9. 129 (1998) 1–285. 1967.evansville. Kimberling. See Figure 9. Cubics in the Triangle Plane.fr/bernard. Ehrmann and B.gibert/ [3] B. available at http://perso. 3 (2003) 1–27. Gibert. Homologies and circles For any point M on Σ13 . . Special Isocubics in the Triangle Plane.wanadoo. A similar construction with M on Σ14 and P2 instead of P1 will give Γ14 . Triangle centers and central triangles. [2] J.fr/bernard. Brocard and T. [5] C. available at http://perso.edu/ck6/encyclopedia/ETC. Gibert. Paris. The parallel to the Euler line at m meets the line M X13 at M on Γ13 . References [1] H.wanadoo. 199–202. Etude bibliographique et terminologique. 57–61. London. G´eom´etrie du triangle. 242– 244. 1888. 199–203.St Etienne. Vigari´e. Gibert [7] T. 276–279. 102–04. 58–62.fr . 127–131. 3 (1889) 18–19. 27–30. France E-mail address: bg42@orange. C. 83–86. 42100 . Journal de Math´ematiques Sp´eciales. 1 (1887) 34–43. 3e s´erie. 154–157. Part V of John J.224 B. 182–185. 75–177. ´ [8] E. Milne’s Companion to the Weekly Problem Papers. 2 (1888) 9–13. Simmons. 55–59. 77–82. Recent Geometry. McMillan. Bernard Gibert: 10 rue Cussinel. 217–219. 127–132. 248–250. We present a synthetic proof here. Bellavitis’ theorem Eisso J. If ABCD is not a kite. Theorem 1 (Bellavitis. for the convex case [1]. then w1 + w2 + w3 + w4 = 180◦ A w1 C F B w2 F w4 w3 B D C A Figure 1 Proof. AB CB = CD . If the side lengths of a convex quadrilateral ABCD satisfy AB · CD = BC · DA. Similarly let the angles inside the quadrilateral that the other diagonal BD forms with the remaining pair of opposite sides be w2 . If AB = AD then BC = CD and AC is the perpendicular bisector of BD. In this note we give a synthetic proof of Bellavitis’ theorem and then generalizing this theorem. by trigonometry. we have AD Hence. for not only convex quadrilaterals.b Forum Geometricorum Volume 6 (2006) 225–227. b b FORUM GEOM ISSN 1534-1178 A Synthetic Proof and Generalization of Bellavitis’ Theorem Nikolaos Dergiades Abstract. w4 . 2006. then from AB · CD = BC · DA. Hence ABCD is a kite. we give a synthetic geometric proof for both theorems direct and converse. See Figure 1. . Communicating Editor: Paul Yiu. This Publication Date: September 18. let the diagonal AC form with one pair of opposite sides angles w1 . and it is obvious that w1 + w2 + w3 + w4 = 180◦ . 1854). C lies on the A-Apollonius circle of triangle ABD. Inside a convex quadrilateral ABCD. w3 . as Eisso Atzema proved. 1. From this approach evolves clearly the connection between hypothesis and conclusion. Atzema has recently given a trigonometric proof of Bellavitis’ theorem [1]. The reflection of AC in AF meets the Apollonius circle at C . Hence the lines AC . We make use of oriented angles and arcs. w2 = θ(BC. we consider noncyclic quadrilaterals below. We continue to use the notation w1 = θ(AB. the point C is the reflection of C in BD. where AF and AF are the internal and external bisectors of angle BAD and CF is the bisector of angle BCD.226 N. Since arc CF = arc F C . Similarly the reflection of AC in CF meets the Apollonius circle at A that is the reflection of A in BD. Theorem 2. XZ) the oriented angle from XY to XZ. A w1 A B w2 B w3 C Figure 2 C w4 D . So we have w2 + w3 =w2 + ∠BCB = ∠CB D = ∠AB D w1 + w4 =∠B AD + w4 = ∠BB A. Dergiades circle has diameter F F . DB). A generalization There is actually no need for ABCD to be a convex quadrilateral. (1) (2) From (1) and (2) we get w1 + w2 + w3 + w4 = ∠BB A + ∠AB D = 180◦ . Since it is clear that w1 + w2 + w3 + w4 = 180◦ for a cyclic quadrilateral. In an arbitrary noncyclic quadrilateral ABCD. AC). 2. w4 = θ(DA. w3 = θ(CD. the side lengths satisfy the equality AB · CD = BC · DA if and only if w1 + w2 + w3 + w4 = ±180◦ . Denote by θ(XY. CA). BD). CA are reflections of each other in BD and are met at a point B on BD. 2w3 =arc C A. Kimberling. Greece E-mail address: ndergiades@yahoo. 2w2 =arc CC + arc C B . Triangle centers and central triangles. is equivalent to the fact that D lies on the B-Apollonius circle of ABC because for a pedal triangle we know that B A = DC · sin C = B C = DA · sin A or sin A BC DC = = . i. This means that the circumcevian triangle of D is isosceles. C .e.gr . From these. Zanna 27. Since arc BC +arc CC +arc C A+arc AB = ±360◦ . A theorem by Giusto Bellavitis on a class of quadrilaterals. in turn.. This.. Note that 2w1 =arc BC. Thessaloniki 54643. Forum Geom. Congressus Numerantium. 129 (1998) 1 – 285. §7. B . 6 (2006) 181–185.18] The condition (3) is equivalent to B A = B C . J. It is well known that A B C is similar to with the pedal triangle A B C of D. w1 + w2 + w3 + w4 = ±180◦ if and only if (arc BC + arc CC + arc C A + arc AB) + arc C B + arc A B = ±360◦ . 2w4 =arc AB + arcA B . See [2. (3) B A = B C . The triangle A B C is the circumcevian triangle of D relative to ABC. Since ABCD is not a cyclic quadrilateral the lines DA. References [1] E. Atzema. DC meet the circumcircle of triangle ABC at the distinct points A . [2] C. Nikolaos Dergiades: I. DB. the above condition holds if and only if arc C B = arc B A . DA sin C BA From this we have AB · CD = BC · DA.A synthetic proof and generalization of Bellavitis’ theorem 227 Proof. . This topic was introduced. b b FORUM GEOM ISSN 1534-1178 On Some Theorems of Poncelet and Carnot Huub P. Introduction The scope of Euclidean Geometry was substantially extended during the seventeenth century by the introduction of the discipline of Projective Geometry. The further development of Projective Geometry was about two hundred years later. we extend a theorem by Carnot [1] from a triangle to a complete quadrilateral (Theorem 3). Communicating Editor: Floor van Lamoen. mainly by a French group of mathematicians (Poncelet. Some relations in a complete quadrilateral are derived. After more than thirteen centuries it was continued by two Frenchmen. Publication Date: September 25.b Forum Geometricorum Volume 6 (2006) 229–234. Meanwhile. no angles. The latter one published in 1640 his well-known Essay pour les coniques. Chasles. The author thanks Floor van Lamoen and the referee for their useful suggestions during the preparation of this paper.M. Carnot. the properties are unaltered when the geometric figure is subjected to a projection. Desargues and his famous pupil Pascal. 1. In connection with these relations some special conics related to the angular points and sides of the quadrilateral are discussed. Brianchon and others). In this article we present an (almost forgotten) result of Poncelet [4]. obtained in an alternative way and we derive some associated relations (Theorem 1). This short study contains the well-known hexagrammum mysticum. Until then geometers were mainly concentrating on the metric (or Euclidean) properties in which the measure of distances and angles is emphasized. An important tool in Projective Geometry is a semi-algebraic instrument. Furthermore. called the cross ratio. no circles and no parallelism but concentrates on the descriptive (or projective) properties. A theorem of Carnot valid for a triangle is extended to a quadrilateral. by M¨obius (1827) and Chasles (1829). the related subject of perspective had been studied by architects and artists (Leonardo da Vinci). nowadays known as Pascal’s Theorem. These properties have to do with the relative positional connection of the geometric elements in relation to each other. 2006. Projective Geometry has no distances. Projective Geometry was started by the Grecian mathematician Pappus of Alexandria . van Kempen Abstract. . independently of each other. It will be shown that this approach leads to surprising results derived along unexpected lines. C. A) we find CG DH AP · · = 1. A. G. (5) GD HA P C . van Kempen We take as starting-point the theorems of Ceva and Pappus-Pascal. F ) and find that in triangle ABC the lines AF . (2) P C GD P B EA BP DH AP CF · · · =1. (4) EB F C P A Similarly with the triples (Q. Let the intersections of PQ with AD and BC be H and F respectively and those of PR with CD and AB be G and E respectively (Figure 1). M. Q and R. Proof and extension of a Theorem by Poncelet Theorem 1. 2. E) and (R. (1) EB F C GD HA AP CG DP BE · · · =1. P. BP and CE are concurrent so that by Ceva’s theorem AE BF CP · · = 1. Let the diagonal points of a complete quadrilateral ABCD be P. (3) P D HA P C F B R C G F D P H Q A E B Figure 1 Proof.230 H. The first one is a close companion of the theorem of the Grecian mathematician Menelaus. C) and (R. Then AE BF CG DH · · · =1. In the analysis we will follow as much as possible the purist/synthetic approach. H. We apply the Pappus-Pascal theorem to the triples (Q. B. Let E . We will examine this problem by using Ceva’s theorem. Clearly a relation similar to (1) holds: AE BF CG DH · · · = 1. taking a convex quadrilateral ABCD in which AB + CD = BC + DA.On some theorems of Poncelet and Carnot 231 Relation (1) immediately follows from (4) and (5). (7) E B F C G D H A R C G G F D F H P M H Q A E E B Figure 2 It is well known [5] that the point of intersection of E G and F H is P . Q) and (H. We now consider a special case of Theorem 1. and (3) follows from (4) and (6). BC. F . CD and DA respectively. Again in the same way with triples (E. . This raises the questions whether or not a relation similar to (1) will hold. G and H be the points of tangency of the incircle with AB. p. See for instance [2. R) we find that AE BP DH · · = 1. Poncelet [4] has derived relation (1) by using cross ratios. (6) EB P D HA (2) follows from (5) and (6). D.49]. so that it is circumscriptable (Figure 2). This can be seen for a general quadrilateral with an inscribed conic from subsequent application of Brianchon’s theorem to hexagons AE BCG D and BF CDH A. Proof. By Ceva’s theorem we have in triangle ABC AE BF1 CP · · = 1. CE) in Figure 3. CF1 . M. CG. Theorem 2. Further Analysis We start with a given quadrilateral ABCD where E and G are arbitrary points on the lines AB and CD respectively. Now consider the two pencils (AH1 . (8) EB F1 C P A Now if S = AG ∩ DB and H1 = AD ∩ CS. BC. We then construct points F1 and H1 on BC and AD respectively such that (1) holds. then the points A. AE) and (CH1 . If in the quadrilateral ABCD the points E. van Kempen 3. C. Let T = CE ∩ BD and F1 = BC ∩ AT . We will use the cross ratio of pencils in relation to the cross-ratio of ranges. We have the cross-ratio equality between ranges and pencils: . These concepts are extensively described by Eves [3].232 H. E. F1 . AF1 . AG. then Ceva’s theorem applied to triangle ADC gives CG DH1 AP · · = 1. G and H1 lie on a conic and K = EG∩F1 H1 lies on BD. P. G and H1 lie on AB. G C D S P H1 K F1 T A E B Figure 3 First we consider the triangles ABC and ADC. We can do so by the following construction (Figure 3). F1 . (9) GD H1 A P C By multiplication of (8) and (9) we find the desired equivalence of (1). Here we have to switch to the field of Projective Geometry. CD and DA respectively such that S = AG ∩ CH1 and T = AF1 ∩ CE lie on BD. Therefore we consider the hexagon GGEHHF . Assume that relation (1) holds. F1 . S. G and H which is tangent to CD at G. consider the unique conic Γ through E. BC. K and T are collinear. (10) From this equality we see that A. The lines GE. Theorem 3. which touches its sides in the points E. G C D L F2 H2 P A E B Figure 4 With the help of Theorem 2 we prove an extension of Carnot’s theorem in [1] for a triangle to a complete quadrilateral. F . First we apply Desargues’ theorem to triangle GF C and triangle EHV (Figure 5). This means that the intersection points of the corresponding sides are collinear. T. H2 and E lie on a conic. F2 . F . By using triangle ABD and triangle CBD instead of triangle ABC and triangle ADC as above. Applying Pascal’s theorem to the hexagon AF1 H1 CEG we find that the diagonal BD is the Pascal line and consequently the points S. we can also construct F2 and H2 such that relation (1) holds (Figure 4). Now we apply Theorem 2. F1 . If in the quadrilateral ABCD the points E. F1 . Using Pascal’s theorem for the hexagon BGEDF2 H2 we find that L = EG ∩ F2 H2 lies on AC. G and H. E) = (D. F H and CV concur in P . So U = GF ∩ EH. E. G. G and H1 lie on a conic. T ) = C(H1 . CD and DA respectively and EG and F H concur in P = AC ∩ BD. G and H lie on AB. G. Let V = DE ∩ BH and W = DF ∩ BG. D.On some theorems of Poncelet and Carnot 233 A(H1 . We find that . B = F C ∩ HV and D = GC ∩ EV are collinear. G. C. By Theorem 2 we know that BF GDHE and AEF CGH lie on two conics. B) = (S. B. F . We examine the direction of the tangent to Γ at the point H. Next. E). D. finding that B. Proof. then (1) is satisfied if and only if there is a conic inscribed in quadrilateral ABCD. V. Carnot. Applying Theorem 3 to the results of Theorem 2 we find Corollary 4. This means that the tangents to Γ at G and H intersect on BD. Since both P and U are collinear with B and D. This leads to the conclusion that H and H ∗ are in fact the same point. Brussels 1945. 2582 NC Den Haag. De Boeck. Paris 1822. If in the quadrilateral ABCD of Theorem 2 the lines EG and F1 H1 concur in P . Boston 1972. F . F . available at http://www. H. J. Now assume that a conic is tangent to the sides of quadrilateral ABCD at the points E. and H∗ . as stated earlier. M. Yiu. With fixed E. Essai sur la th´eorie des transversals.fau. Compl´ements de G´eom´etrie plane. Euclidean Geometry.math. Poncelet. F and G there is exactly one point H∗ on AD such that the equivalent version of relation (1) holds. which proves the sufficiency part. R. G. G and H. Eves. where P = AC ∩ BD.M. The Netherlands E-mail address: hvkempen@versatel. Allun & Bacon. which implies that AD is the tangent to Γ at H.html. (1998). P. they must be the same conic. Trait´e des propri´et´es projectives des figures. A survey of Geometry. As these two conics have three double points in common. Note that of course EG and F H intersect in P . M.nl . then F1 H1 of Figure 3 and F2 H2 of Figure 4 coincide. Huub P. By the sufficiency part this leads to a conic tangent to the sides at E. N. Bachelier.234 H. This proves the necessity part. In the same way we prove that the lines AB and BC are tangent to Γ at E and F respectively. the line BD is the Pascal line. P. van Kempen C G D W H P =K F V A E B Figure 5 GE ∩ HF = P and GF ∩ EH = U . Paris 1806.edu/yiu/Geometry. van Kempen: Prins Mauritsplein 17. References [1] [2] [3] [4] [5] L. Deaux. not on a sideline of A1 A2 A3 . which says that any two orthogonal lines through H intersect the sides of the triangle in segments with collinear midpoints. and the line l at infinity. a3 = A1 A2 of A1 A2 A3 . 2006. and from this it follows that the line connecting the midpoints of the segments determined by r and r on a1 and a2 .b Forum Geometricorum Volume 6 (2006) 235–240. this involution becomes the orthogonal involution (where orthogonal lines correspond) and we find the well-known Droz-Farny Theorem. the orthocenter of A1 A2 A3 . b b FORUM GEOM ISSN 1534-1178 The Droz-Farny Theorem and Related Topics Charles Thas Abstract. line through P parallel to ai ). through the midpoint of the segment determined on a3 by r and r . At each point P of the Euclidean plane Π. see [1]. [3]). From this it follows that in the case where P = H. a2 = A3 A1 . . In this paper we investigate two closely related loci that have a strong connection with the Droz-Farny Theorem. not on the sidelines of a triangle A1 A2 A3 of Π. Communicating Editor: J. the pencil B of parabola’s with tangent lines the sides a1 = A2 A3 . The proof given in [3] (and [5]) probably is one of the shortest: Consider. i. and thus are conjugate pairs in the involution I. and consider the tangent lines r and r through P to a non-degenerate parabola P of this pencil B. there exists an involution in the pencil of lines through P . the orthogonal Droz-Farny lines through P . in the Euclidean plane Π. the tangent lines through P to a variable parabola of the pencil B are conjugate lines in an involution I of the pencil of lines through P . Next. Let P be any point of Π. The Droz-Farny Theorem Many proofs can be found for the original Droz-Farny Theorem (for some recent proofs. the orthocenter of A1 A2 A3 .e. 3 are the tangent lines through P of the degenerate parabola’s of the pencil B. by the Sturm-Desargues Theorem. Publication Date: October 2. and this involution I contains in general just one orthogonal conjugate pair. and we find the Droz-Farny Theorem. If P = H. A variable tangent line of P intersects r and r in corresponding points of a projectivity (an affinity. such that each pair of conjugate lines intersect the sides of A1 A2 A3 in segments with collinear midpoints. In the following we call these orthogonal lines through P . and not on l. 2. Chris Fisher. i = 1. the points at infinity of r and r correspond). is a tangent line of P. this involution becomes the orthogonal involution in the pencil of lines through H. Among examples of these loci we find the circumcirle of the anticomplementary triangle and the Steiner ellipse of that triangle. 1. Remark that (P Ai . in the Euclidean plane. cos1A3 ). according as X lies on the same or opposite side of ai as Ai ). A2 . i = 1. 4F 2 ( (1) we see that the set is given by the equation l1 2 l2 2 l3 2 x + x + x ) − s(l1 x1 + l2 x2 + l3 x3 )2 = 0. the incenter I is (1. their vertices are six points of a conic. A3 . d2 . Next. since. with respect to a triangle A1 A2 A with side-lenghts l1 . then. and the orthogonal DrozFarny lines through any point P are the orthogonal tangent lines through P of the parabola. cos A3 ). 1. a2 . and (d1 . if F is the xi .e. for which the vertices A1 . and the Lemoine (or symmedian) point K is (l1 . 2. di is positive or negative. δ1 1 δ2 2 δ3 3 (2) . l13 ).236 Two other characterizations of the orthogonal Droz-Farny lines through P are obtained as follows: Let X and Y be the points at infinity of the orthogonal DrozFarny lines through P . l12 . If X has trilinear coordinates (x1 . δ3 ) with respect to A1 A2 A3 . and P (and also through H. 1). the directrix of any parabola of the pencil B passes through H. if P = H. and if di is the ”‘signed”’ distance from X to the side ai (i. since any rectangular hyperbola through A1 . 3. x3 ) to ai are connected by the equation l1 2 l2 2 l3 2 d + d + d = s. x2 . cos A2 . It follows that the orthogonal Droz-Farny lines through P are the lines through P which are parallel to the (orthogonal) asymptotes of this rectangular hyperbola through A1 . and not at infinity. the involution I is the orthogonal involution. and H. tangent to a1 . and A3 . x2 . 2. we have di = l1 x1 +l2F 2 x2 +l3 x3 actual trilinear coordinates of X with respect to A1 A2 A3 . The first locus Let us recall some basic properties of trilinear (or normal) coordinates (see for instance [4]). Our first locus is defined as follows ([5]): Consider a fixed point P . passes through H). x3 ). Remark that l1 d1 + l2 d2 + l3 d3 = 2F . of any point P of the Euclidean plane. A2 . namely the rectangular hyperbola through A1 . x3 ) with respect to A1 A2 A3 . cos1A2 . and with directrix P H. l2 . x2 . not on a sideline of A1 A2 A3 . A2 . and suppose that s is a given real number and the set of points of the plane for which the distances di from (x1 . δ1 1 δ2 2 δ3 3 Using di = 2Fi xi l1 x1 +l2 x2 +l3 x3 . the circumcenter O is (cos A1 . l2 . with actual trilinear coordinates (δ1 . a3 . are homogeneous projective coordinates. A2 . The line at infinity has in trilinear coordinates the equation l1 x1 + l2 x2 + l3 x3 = 0. P . d3 ) are the area of A1 A2 A3 . Since the two triangles A1 A2 A3 and P XY are circumscribed triangles about a conic (a parabola of the pencil B). Trilinear coordinates (x1 . The centroid G of A1 A2 A3 has trilinear coordinates (l11 . l3 . A3 . δ2 . the orthocenter H is ( cos1A1 . l3 ). A3 are the basepoints and the incenter I of the triangle the unit point. and (l2 l3 . and also by (3). the medial triangle of ∆ = A1 A2 A3 is the triangle whose vertices are −1 −1 the midpoints of the sides of ∆. We look for the locus of the points Q of Π. a3 of ∆ = A1 A2 A3 are connected by 1 1 1 l1 2 l2 2 l3 2 d1 + d2 + d3 = 4F 2 ( + + ) = S. s): it is the conic determined by (1) and (2). s) belong to a pencil. s = S = 4F 2 ( l1 δ1 l2 δ2 l3 δ3 3. i = 1. δ2 . where qi is the line through Q. x2 . and a straightforward calculation shows that all conics of this pencil have center P . l12 .237 or. Assume that P (p1 . (3) p1 p2 p3 We denote this conic by K(P. d3 to the sides a1 . which means that they have the same asymptotes and the same axes. and circumscribed about the anticomplementary triangle −1 −1 A−1 1 A2 A3 of ∆. we find after an easy calculation that Q1 has coordinates (0. For P and ∆ fixed and s allowed to vary. not at infinity and not on a sideline of ∆. Proof. Substituting the coordinates (−l11 . where (δ1 . l3 l1 . we find immediately that 1 1 1 + + ). The locus K(P. x3 ). parallel to P Ai . ∆. or ( l11 . and A3 . and A3 of this anticomplementary triangle are (−l2 l3 . the conics K(P. 3. respectively. p2 . where (p1 . l13 ). and the anticomplementary triangle A−1 1 A2 A3 of ∆ is the triangle whose medial triangle is ∆. is the conic with center P . −l1 p3 ). −l3 l1 . Lemma 1. and by the value of s. 2. and if we give Q the coordinates (x1 . in (2). This locus was the subject of the paper [2]. δ1 δ2 δ3 l1 δ1 l2 δ2 l3 δ3 where (δ1 . ∆. S) of the points for which the distances d1 . δ2 . d2 . are collinear. with trilinear coordinates with respect to ∆ = A1 A2 A3 . and have the same points at infinity. Since l1 x1 + l2 x2 + l3 x3 = 0 is the equation of the line at infinity. l3 l1 . p3 ) of P : l2 l3 l1 2F ( x21 + x22 + x23 )(l1 p1 + l2 p2 + l3 p3 ) − s(l1 x1 + l2 x2 + l3 x3 )2 = 0. according as the product δ1 δ2 δ3 > 0 or < 0. or ( l11 . The conics K(P. δ3 ) are the actual trilinear coordinates of P with regard to ∆ = A1 A2 A3 . l1 p2 x1 + . p3 ) is a point of Π. − l12 . An easy calculation shows that −1 −1 the trilinear coordinates of the vertices A−1 1 . δ3 ) are the actual trilinear coordinates of a given point P . ∆. l12 . ∆. such that the points Qi = qi ∩ ai . Next. −l1 l2 ). a2 . p2 . l1 l2 ). if we use general trilinear coordinates (p1 . (l2 l3 . p2 . s) can be (homothetic) ellipses or hyperbola’s: this depends on the location of P with regard to ∆. A2 . l13 ). p3 ) are any triple of trilinear coordinates of P with regard to ∆. and again a straightforward calculation shows that we find ellipses or hyperbola’s. the point at infinity of P A1 has coordinates (l2 p2 + l3 p3 . −l1 p2 . l1 l2 ). − l13 ) −1 −1 of A−1 1 . A2 . The second locus We work again in the Euclidean plane Π. Q2 . and of Q3 : (x1 (l1 p1 + l3 p3 ) + p1 x2 l2 . l1 l2 (l1 p1 + l2 p2 )). The connection between the Droz-Farny -lines and the conics Recall from §2 that S = 4F 2 ( 1 1 1 1 1 1 + + ) = 2F (l1 p1 + l2 p2 + l3 p3 )( + + ). 0. = −GP Moreover. ∆) has trilinear coordinates (l2 l3 (l2 p2 + l3 p3 ). the centroid of ∆. p2 . with equation l1 x1 + l2 x2 + l3 x3 = 0. x2 . t2 ) = (−3. the condition that Q1 . x2 (l1 p1 + l2 p2 ) + x3 p2 l3 . l1 p1 + l2 p2 + l3 p3 ) and the projective coordinates (t1 . by C(P. 0). after a rather long calculation. The center M of C(P. l1 p1 + l2 p2 + l3 p3 ). = −GP . t2 ) = (−1. and deleting the singular part l1 x1 + l2 x2 + l3 x3 = 0.238 x2 (l2 p2 + l3 p3 ). gives us the following equation for the locus of the point Q: p3 (l1 p1 + l2 p2 )x1 x2 + p1 (l2 p2 + l3 p3 )x2 x3 + p2 (l3 p3 + l1 p1 )x3 x1 = 0. and give P∞ coordinates (t1 . l3 l1 (l3 p3 + l1 p1 ). in other words: 2GM Proof. 0) and G(0. (4) This is our second locus. and Q3 are collinear. Next. p3 x2 l2 + x3 (l1 p1 + l3 p3 )). and we denote this conic. if P∞ is the point at infinity of the line P G. where f is the homothety with center G. An easy calculation shows that the polar point of this point M with regard to the conic (4) is indeed the line at infinity. and where P is the point with trilinear coordinates (p1 . x3 ) for a point with respect to A1 A2 A3 and trilinears (x1 . l1 δ1 l2 δ2 l3 δ3 l1 p1 l2 p2 l3 p3 . and (x1 (l1 p1 + l2 p2 ) + x3 p1 l3 . t2 ) of M follow from t t t t2 1 2 1 0 1 −3 l1 p1 + l2 p2 + l3 p3 1 : =− . x1 l1 p3 + x3 (l2 p2 + l3 p3 )). we find for the coordinates of Q2 . (M P GP∞ ) = 2 0 1 0 1 0 1 −3 l1 p1 + l2 p2 + l3 p3 which gives (t1 . 4. Lemma 2.207]): x1 = l2 l3 (l2 x2 + l3 x3 ) x2 = l3 l1 (l3 x3 + l1 x1 ) x3 = l1 l2 (l1 x1 + l2 x2 ). p. ∆). x2 . It is the image f (P ). or. then (t1 . the equation 2GM 1 is equivalent with the equality of the cross-ratio (M P GP∞ ) to − 2 . circumscribed about A1 A2 A3 . respectively. and homothetic ratio − 12 . p3 ) with regard to ∆. where ∆ = A1 A2 A3 . x3 ) for the same point with respect to the medial triangle of A1 A2 A3 (see [4. In the same way. Next. Remark that the second part of the proof also follows from the connection between trilinears (x1 . 1). t2 ) (thus P∞ = t1 P + t2 G). choose on the line P G homogeneous projective coordinates with basepoints P (1. l12 (l22 + l32 − l12 )d21 + l22 (l32 + l12 − l22 )d22 + l32 (l12 + l22 − l32 )d23 = 4F 2 (l12 + l22 + l32 ).239 where F is the area of ∆ = A1 A2 A3 . ∆−1 ) is also the circumcirle of ∆−1 . the orthocenter of ∆−1 . If P = K(l1 . l2 . If P = H. with regard to ∆ = A1 A2 A3 .2. ∆. where S = 2F ( cosl1A1 + cosl2A2 + cosl3A3 )( cosl1A1 + cosl2A2 + cosl3A3 ). ∆. s). with 1 1 1 S =2F ( 2 + 2 + 2 )(l12 + l22 + l32 ) l1 l2 l3 =2F 2 (l22 l32 + l32 l12 + l12 l22 )(l12 + l22 + l32 ) . which is also the circumcenter of its anticomplementary triangle ∆−1 . line through P . it is clear that (P Ai . in the foregoing section. C(f −1 (H). and where (δ1 . Proof. (p1 . parallel to ai ). since any two orthogonal lines through H are axes of these conics. 2. K(H. s ∈ R. is the circumcircle of ∆−1 and it is the locus of the points for which the distances d1 . ∆. p2 . (2) The common axes of the conics K(P. the Lemoine point of ∆ = A1 A2 A3 . d2 . is the de Longchamps point X(20) of ∆. 5. S) and C(f −1 (P ). f is the homothety with center G and homothetic ratio − 12 . s) are circles with center H. both conics have center P and are circumscribed about the complementary triangle f−1 (∆) of A1 A2 A3 . ∆. l3 ). Remark that f −1 (H).1. Examples 5. Moreover. f −1 (∆)) are the orthogonal Droz-Farny -lines through P . S). d3 to the sides of ∆ are related by (l1 cos A1 )d21 + (l2 cos A2 )d22 + (l3 cos A3 )d23 cos A1 cos A2 cos A3 =4F 2 ( + + ) l1 l2 l3 l2 + l22 + l32 . Remark that f −1 (∆) is the anticomplementary triangle ∆−1 of ∆. (2) For the conic with center P . δ2 . δ3 ) are the actual trilinear coordinates of P with respect to this triangle. are conjugate diameters. In particular. p3 ) are trilinear coordinates of P with regard to ∆A1 A2 A3 . S). ∆. l12 l22 l32 . (1) The conics K(P. circumscribed about the anticomplementary triangle of A1 A2 A3 . and of the conic C(f −1 (P ). the conics K(H. f −1 (∆)) coincide. then K(K. which is easily seen from the fact that this circumcircle is the locus of the points for which the feet of the perpendiculars to the sides of ∆−1 are collinear. We have: Theorem 3. And the result follows from section 1. =2F 2 1 l1 l2 l3 or equivalently. the orthocenter of ∆ = A1 A2 A3 . Furthermore. 3. (1) Because of Lemma 1 and 2. i = 1. 5. The axes of this ellipse are the orthogonal Droz-Farny lines through I with respect to ∆. Moreover. Bradley. On ellipses with center the Lemoine point and generalizations. the centroid of ∆ = A1 A2 A3 . 80 (1996) no. since G is also the centroid of ∆−1 . Countless Simson line configurations.M. 5. ∆−1 ) is also this Steiner ellipse and its axes are the orthogonal Droz-Farny lines through G with respect to ∆. 1. is the locus of the points for which the distances d1 . 1). circumscribed about ∆−1 . where f −1 (I) is the incenter of ∆−1 (which is center X(8) of ∆. and it is the ellipse with center I.-L.. Congressus Numerantium. Triangle centers and central triangles. 129 (1998) 1–285. d2 . Math. l3 +ll12 −l2 .J. l1 +ll23 −l3 )) is the same ellipse. Moreover the locus C(f −1 (I). 1. 314–321. is the locus of the points for which the distances d1 . Kimberling. and it is the ellipse with center K. 11 (1993).4. Gazette. i. circumscribed about ∆−1 . where f −1 (K) is the Lemoine point of ∆−1 (or X(69) with coordinates (l2 l3 (l22 + l32 − l12 ). Ehrmann and F.T.240 is the locus of the points for which the distances d1 .. d3 to the sides of ∆ are related by l12 d21 + l22 d22 + l32 d23 = 12F 2 . with S = 18F . The axes of this ellipse are the orthogonal Droz-Farny lines through K with respect to ∆. Forum Geom. d2 . l13 ). A projective generalization of the Droz-Farny line theorem. Bradley and J.be . S22. Charles Thas: Department of Pure Mathematics and Computer Algebra. van Lamoen. B-9000 Gent. nr. d3 to the sides of ∆ are related by l1 d21 + l2 d22 + l3 d23 = 4F 2 ( l11 + l12 + l13 ). Thas.. the Nagel point with coordinates (l2 +ll31 −l1 . and it is the ellipse with center G.e. ∆−1 ).4. [5] C. the locus C(f −1 (K). ∆. 1 2 3 circumscribed about ∆−1 . Krijgslaan 281. ∆. l3 l1 (l32 + l12 − l22 ).-P. Ayme. ser.3. Forum Geom. then K(G. S). is the same ellipse. Nieuw Archief voor Wiskunde.thas@UGent. then K(I. 4 (2004). [4] C. If P = I(1. it is the Steiner ellipse of ∆−1 . 1–7. [3] J. [2] C. References [1] J. 225–227. S). d2 . ∆−1 ). the incenter of ∆ = A1 A2 A3 . Belgium E-mail address: charles. l12 . If P = G( l11 . 219–224. A purely synthetic proof of the Droz-Farny line theorem. 488. 5. d3 to the sides of ∆ are related by d21 + d22 + d23 = 4F 2 ( l12 + l12 + l12 ). 4 (2004). with S = 2F (l1 + l2 + l3 )( l11 + l12 + l13 ). l1 l2 (l12 + l22 − l32 )). The locus C(G. and DA respectively. We explore a configuration consisting of two quadrilaterals which share the intersection of diagonals. The extended butterfly theorem for quadrilaterals In this note we consider some properties of pairs of quadrilaterals ABCD and A B C D with A . 2006. known as the Butterfly Theorem for Quadrilaterals. We prove results analogous to the Sidney Kung’s Butterfly Theorem for Quadrilaterals in [2]. D C C X V B I D U A A B Z Figure 1 The segments AC. CD. |U I| |V C| |IC| Publication Date: October 9. B . BC. in [2] the following equality. was established: |AI| |AU | |IV | · = . i. (1) . when ABCD and A B C D share the same intersection of diagonals. 1. A C and B D are analogous to the three chords from the classical butterfly theorem (see [1] for an extensive overview of its many proofs and generalizations). When the intersection of the lines A C and B D is the intersection I of the lines AC and BD.b Forum Geometricorum Volume 6 (2006) 241–246. C and D on lines AB.. Communicating Editor: Paul Yiu.e. b b FORUM GEOM ISSN 1534-1178 On Butterflies Inscribed in a Quadrilateral ˇ Zvonko Cerin Abstract. and E = A C ∩ B C . B(f. Theorem 1. we have similar relations |XI| |XA| |IC| · = . The points A u+1 . (2) |U E| |V C| |IC| Proof. fu gu f +v g divide segments AB g. q) forsome real numbers f . We shall use analytic geometry of the plane. 0) and D(p. Their coordinates are a bit more complicated. U = AC ∩ D A . I = AC ∩ BD. which are real numbers different from −1. Then the vertices C and D are intersections of the lines CD and DA with the lines A E and B E. u+1 and B v+1 . If E lies on the line AC. For the intersections X = AC ∩ A B and Z = AC ∩ C D . Cerin 242 where U and V are intersections of the line AC with the lines D A and B C (see Figure 1). 0).ˇ Z. Let A B C D be an inscribed quadrilateral of ABCD. Next. g). C(1. Let the rectangular coordinates of the point E be (h. |U I| |V Z| |IZ| D C tA = ABD tC = BCD t4 t3 C V I D t1 A U t2 B E A B Figure 2 Our first result is the observation that (1) holds when the diagonals of A B C D intersect at a point E on the diagonal AC of ABCD. then |AI| |AU | |EV | · = . |AI| |CZ| |IZ| and |XU | |IV | |XI| · = . V = AC ∩ B C . p and q. k). in this situation. v+1 and BC in ratios u and v. we determine the points U and . respectively. Without loss of generality we can assume that A(0. not necessarily the intersection I = AC ∩ BD (see Figure 2). However. t4 = CC B . k) and Q4 (h. The difference D in this case is the quotient −(f q − gp)2 · k · P1 (h. Moreover. 2. then the point E lies on the line AC. P2 (h. k) · P2 (h. 1 : 1 or 3 : 1. C. k) · (g + f q − q − gp)2 · (guh + (u + 1 − f u)k − gu)2 D := where P5 (h. the extended Butterfly Theorem for Quadrilaterals holds not only for points E on the line AC but also when the point E is on a line through D (with the equation P1 (h. Indeed. t2 = D A E. in this case we can easily prove the following converse of Theorem 1. C and D (with the equation P2 (h. In other words. (qh − pk)2 · (g + f q − q − gp)2 · (guh + (u + 1 − f u)k − gu)2 where P1 (h. k) = 0) and on a conic through A. . the difference D is zero so that the extended Butterfly Theorem for Quadrilaterals holds. Q4 (h. k) = .e.. Both are somewhat impractical to write down explicitly. when the point E is on the line AC. Note that these are linear and quadratic polynomials in h and k. i. t3 = EB C . tA = ABD and tC = CDB (see Figure 2). k) =(u(q − g) + q(1 + uv))h + (uv(1 − p) + u(f − p) − p)k − u(qv − gp + f q). If the relation (2) holds when the points A and B divide segments AB and BC both in the ratio 1 : 3. There is a version of the above theorem where the points U and V are intersections U = AE ∩ A D and V = CE ∩ B C . k) are polynomials of degrees 5 and 4 respectively in variables h and k. since k is a factor in the numerator we see that when k = 0. if we substitute for u = v = 13 . A relation involving areas Let us introduce shorter notation for six triangles in this configuration: t1 = D AA . k) =2qguh2 + (gu − 2ugp − 2f qu + uq − vuq + q) hk− (p + uv + up + uf − pvu − 2puf ) k2 − 2guqh + u (qv + gp + f q) k. k) . 1 or 3 both into P1 and P2 we get three equations whose only common solutions in h and k are coordinates of points A. Remark. k) = 0). and D (which are definitely excluded as possible solutions).On butterflies inscribed in a quadrilateral 243 V as intersections of the line AC with the lines A D and B C and with a small help from Maple V find that the difference |AI|2 |AU |2 |EV |2 · − |U E|2 |V C|2 |IC|2 (f q − gp)2 · k · P5 (h. Our next result shows that the above relationship also holds for areas of these triangles. and ε(A. tA ) and h(C. The following theorem shows that the radii of circumcircles of the six triangles satisfy the same pattern without any restrictions on the point E. b) and (z. The only solution of the system E1 = 0. Let us keep again the same assumptions and notation from the proof of Theof side lengths . t2 ). h(E. Since ε(A. 4(x(b − c) + y(c − a) + z(a − b))2 . k) . If we keep the same assumptions and notation from the proof of Theorem 1.. we see that the square of the circumradius orem 1. h) = (p. t1 ). E = D. ε(C. However. R(t1 ) R(t2 ) · R(t3 ) R(t4 ) = R(tA ) R(tC ) . when u = 12 and v = 12 then E1 = 4 · P1 (h. k) = (2g − 7q) h + (7p − 2f − 4) k + 2f q + 4q − 2gp. (a) If E lies on the line AC. ε(E. for this solution the triangle t2 degenerates to a segment so that its area is zero which is unacceptable. Other relations Note that the above theorem holds also for (lengths of) the altitudes h(A. (y. k) = (2g − 7q) h + (7p − 2f − 1) k + q − 2gp + 2f q.ˇ Z. for (b).e. h(C. tC ). ε(A. area(t2 ) area(t4 ) area(tC ) (3) (b) If the relation (3) holds when the points A and B divide AB and BC both in the ratio 1 : 2 or 2 : 1. (a) is clearly true because k = 0 when the point E is on the line AC. tA ) and ε(C. t2 ). Proof. t4 ). t1 ) · |D A |. then area(t1 ) area(t3 ) area(tA ) · − area(t2 ) area(t4 ) area(tC ) (f q − gp) · k · P1 (h. c) is given by [(y − z)2 + (b − c)2 ] · [(z − x)2 + (c − a)2 ] · [(x − y)2 + (a − b)2 ] . tC ) because (for example) area(t1 ) = 12 h(A. Theorem 3. Cerin 244 Theorem 2. i. Proof. = (qh − pk) · (g + f q − q − gp) · (guh + (u + 1 − f u)k − gu) Hence. then the point E lies on the line AC. Let G(t) denote the centroid of the triangle t = ABC. a). For a triangle t let R(t) denote the radius of its circumcircle. q). On the other hand. E2 = 0 is (h. h(E. t) the distance of G(t) from the side BC opposite to the vertex A. while for u = 2 and v = 2 then E2 = P1 (h. then area(t1 ) area(t3 ) area(tA ) · = . 3. t4 ). t3 ). t1 ). ε(E. t3 ). Since R(t) = product4area(t) of a triangle with vertices in the points (x. t) = 2area(t) 3|BC| there is a version of the above theorem for the distances ε(A. h(A. In the next result we look at the distances of a particular vertex of the six triangles from its orthocenter. Kung. t3 ). A Butterfly Theorem for Quadrilaterals. |AH(t1 )| |EH(t2 )| · |EH(t3 )| |CH(t4 )| = |AH(tA )| |CH(tC )| . Since |AH(t)| = 2δ(A. |CH(t4 )|2 . Theorem 4.hr . a). t1 ). R(tA )2 and R(tC )2 . and |CH(tC )|2 .On butterflies inscribed in a quadrilateral 245 Applying this formula we find R(t1 )2 . R(t2 )2 . |EH(t2 )|2 . R(t3 )2 . (y. Mag. 2 R(t )2 R(tA )2 1) · 3 − R(t In a few seconds Maple V verifies that the difference R(t 2 is equal R(t2 )2 R(t4 )2 C) to zero. c) is given as [(y − z)2 + (b − c)2 ] · [x2 + a2 + yz − x(y + z) + bc − a(b + c)]2 . Bankoff. δ(C. |AH(tA )|2 . 60 (1987) 195–210. Again the pattern is independent from the position of the point E. Hrvatska. Let δ(A. δ(E. 78 (2005) 314–316. Mag. δ(E. R(t4 )2 . References [1] L. tC ). |EH(t3 )|2 . t) there is a version of the above theorem for the distances δ(A. This time one can see that |AH(t)|2 for the triangle t = ABC with the vertices in the points (x. δ(A. Europa E-mail address: cerin@math. Proof. The Metamorphosis of the Butterfly Problem. 10020 Zagreb. (x(b − c) + y(c − a) + z(a − b))2 Applying this formula we find |AH(t1 )|2 . Math.. Kopernikova 7. [2] S. t) denote the distance of O(t) from the side BC opposite to the vertex A. b) and (z. Math. t2 ). t4 ). For a triangle t let H(t) denote its orthocenter.. tA ) and δ(C. In a few seconds Maple V verifies that the difference |AH(t1 )|2 |EH(t3 )|2 |AH(tA )|2 · − |CH(t 2 is equal to zero. |EH(t2 )|2 |CH(t4 )|2 C )| Let O(t) be the circumcenter of the triangle t = ABC. . 2006. (2) Publication Date: October 16. In each case. By an I-triangle we mean a triangle U V W whose vertices U . the condition for perspectivity is F (u. V . Introduction Let ABC be a given triangle with incenter I. v. Communicating Editor: Paul Yiu.b Forum Geometricorum Volume 6 (2006) 247–253. w) := (b− c)vw + (c− a)wu+ (a− b)uv + (a− b)(b− c)(c− a) = 0. An I-triangle is perspective with the medial triangle if and only if it is perspective with the intouch triangle. W are on the angle bisectors AI. V = (a : v : c). BI. b b FORUM GEOM ISSN 1534-1178 On Triangles with Vertices on the Angle Bisectors Eric Danneels Abstract. A P V W I U B C Q Figure 1. CI respectively. W = (a : b : w) (1) for some u. The homogeneous barycentric coordinates of the vertices of an I-triangle can be taken as U = (u : b : c). Theorem 1. w. . The author thanks Paul Yiu for his help in the preparation of this paper. We present several interesting examples with new triangle centers. We study interesting properties of triangles whose vertices are on the three angle bisectors of a given triangle. 1. Such triangles are clearly perspective with ABC at the incenter I. We show that such a triangle is perspective with the medial triangle if and only if it is perspective with the intouch triangle. v. Proof. Danneels Let D. if an I-triangle is perspective with the intouch triangle at Q = (x : y : z). W are as given in (1). E. w1 = 2(s − c)w − c(a + b) + (a − b)2 u1 = From these. then U is the intersection of the line DP with the bisector IA. V . V . . W1 are the inversive images of U . v. the coordinates of V and W can be determined. (ii) the intouch triangle at (a((a(b + c) − (b − c)2 )x − (b + c − a)(b − c)(y − z)) : · · · : · · · ). V1 = (a : v1 : c). The triangle U V W is perspective with the intouch triangle at x(y + z − x) y(z + x − y) z(x + y − z) : : . Proof. then U1 V1 W1 is perspective with (i) the medial triangle at ((y + z − x)((a(b + c) − (b − c)2 )x − (b + c − a)(b − c)(y − z)) : · · · : · · · ). It has coordinates ((b − c)x : b(y − z) : c(y − z)). then U1 V1 W1 is also an I-triangle perspective with the medial and intouch triangles. If an I-triangle U V W is perspective with the medial triangle at (x : y : z). then it is perspective with the medial triangle at P = ((s − a)x((s − b)y + (s − c)z − (s − a)x) : · · · : · · · ). w1 ) = 64abc(s − a)(s − b)(s − c) · F (u. If the coordinates of U . If P = (x : y : z) is the perspector of an I-triangle U V W with the medial triangle. Let U V W be an I-triangle perspective with the medial and the intouch triangles. W in the incircle. v1 = 2(s − b)v − b(c + a) + (c − a)2 (c(a + b) − (a − b)2 )w − 2(s − c)(a − b)2 . F be the midpoints of the sides BC. v1 . W1 = (a : b : w1 ). 2(s − a)u − a(b + c) + (b − c)2 (b(c + a) − (c − a)2 )v − 2(s − b)(c − a)2 . where (a(b + c) − (b − c)2 )u − 2(s − a)(b − c)2 . then U1 = (u1 : b : c). AB of triangle ABC. w) = 0. U1 . If U1 . CA. 2 cyclic (2(s − a)u − a(b + c) + (b − c) ) It follows from (2) that U1 V1 W1 is perspective to both the medial and the intouch triangles. Theorem 2. Similarly. Q= s−a s−b s−c Conversely. F (u1 .248 E. (ii) the intouch triangle at a ((a(b + c) + (b − c)2 )x + 2s(b − c)(y − z)) : · · · : · · · . Let Xa . Therefore. We have IXa = IIa and IIa = AIa − AI = AI · s−a s−a . Let U V W be an I-triangle perspective with the medial and the intouch triangles. (ii) the intouch triangle at Qx = (a(b + c − a)(a2 (b + c) + (b + c − 2a)(b − c)2 ) : · · · : · · · ). Xb . s−a 2. If U2 (respectively V2 and W2 ) is the inversive image of U (respectively V and W ) in the A. 0).and C-) excircle. Xc be the inversive images of the excenters Ia . 1. c) + (b + c − a)(c + a − b)(a + b − c)(0. .Ic in the r2 s − 1 = a·AI incircle.On triangles with vertices on the angle bisectors 249 Theorem 3. 0. Some interesting examples We present some interesting examples of I-triangles perspective with both the medial and intouch triangles. b. then U2 V2 W2 is also an I-triangle perspective with the medial and intouch triangles.1. If an I-triangle U V W is perspective with the medial triangle at (x : y : z).(respectively B. abc a 4R a · AI 2 AI = = = . Xb =(a2 + b2 + c2 − 2ab − 2bc − 2ca)(a. Xa =(a2 + b2 + c2 − 2ab − 2bc − 2ca)(a. Hence. 0). c) + (b + c − a)(c + a − b)(a + b − c)(0. 2. b. triangles ABC and a c b Xa Xb Xc are homothetic with ratio 4R : −r. 1). Proposition 4. The perspectors in these examples are new triangle centers not in the current edition of [1]. Ib . Xa Xb Xc is an I-triangle perspective with (i) the medial triangle at Px = (a2 (b + c) + (b + c − 2a)(b − c)2 : · · · : · · · ). = 2 A 2 IXa r (s − a) (s − a)(s − b)(s − c) r sin ( 2 ) AI BI CI = IX = IX = 4R and by symmetry IX r . 0. Xc =(a2 + b2 + c2 − 2ab − 2bc − 2ca)(a. c) + (b + c − a)(c + a − b)(a + b − c)(1. then U2 V2 W2 is perspective with (i) the medial triangle at (s − a)(y + z − x)((a(b + c) + (b − c)2 )x + 2s(b − c)(y − z)) : · · · : · · · . b. 2. Yb =(a2 + b2 + c2 + 2bc − 2ca + 2ab)(a. 2. Ya =(a2 + b2 + c2 − 2bc + 2ca + 2ab)(−a. c) + (a + b + c)(a + b − c)(b + c − a)(0. Qy = b+c−a . Yc =(a2 + b2 + c2 + 2bc + 2ca − 2ab)(a.250 E. −b. (ii) the intouch triangle at a(a2 (b + c) + (2a + b + c)(b − c)2 ) : ··· : ··· . B-. Danneels A Cc Xc Px Xb Qx I Bb Xa B Aa C Ba Figure 2. b. c) + (a + b + c)(c + a − b)(a + b − c)(1. C-excircles. Proposition 5. Ya Yb Yc is an I-triangle perspective with (i) the medial triangle at Py = ((b + c − a)2 (a2 (b + c) + (2a + b + c)(b − c)2 ) : · · · : · · · ). Yb . 0). 0). 0. b. 0. 0). −c) + (a + b + c)(b + c − a)(c + a − b)(1. Let Ya . Yc be the inversive images of the incenter I with respect to the A-. 1. On triangles with vertices on the angle bisectors 251 Cb Ib Bc A Yb Ic Cc Yc Qy B Ac I Py Bb Aa C Ya Ab Ba Ca Ia Figure 3.4. Wc be the inversive images of Ya . Vc =(3c2 + (a − b)2 )(a. b. 0). Wb . Va =(3a2 + (b − c)2 )(−a. 2. Let Va . Vb . Yc with respect to the incircle. Proposition 6. 0. b. 0). −b. Let Wa . 2. −c) + 2c(b + c − a)(c + a − b)(0. Xc with respect to the A-. Vc be the inversive images of Xa . Xb . 0. c) + 2a(c + a − b)(a + b − c)(1. Va Vb Vc is an I-triangle perspective with (i) the medial triangle at Pv = (a(b + c − a)3 (a2 + 3(b − c)2 ) : · · · : · · · ). 1. Vb =(3b2 + (c − a)2 )(a. . (ii) the intouch triangle at Qv = a(a2 + 3(b − c)2 ) : ··· : ··· b+c−a . 1). Yb . c) + 2b(a + b − c)(b + c − a)(0.3. C-excircles. B-. Wa =(3a2 + (b − c)2 )(a. Wa Wb Wc is an I-triangle perspective with (i) the medial triangle at Pw = (a(a2 + 3(b − c)2 ) : · · · : · · · ). 1). b. 0). Danneels Cb Ib Bc A Vb Ic Cc Vc Qv Bb Pv I B Ac Aa C Va Ab Ba Ca Ia Figure 4. 1. c) − 2b(a + b − c)(b + c − a)(0. . Wc =(3c2 + (a − b)2 )(a. Proposition 7. 0. b. c) − 2a(c + a − b)(a + b − c)(1.252 E. b. 0). (ii) the intouch triangle at Qw = (a(b + c − a)2 (a2 + 3(b − c)2 ) : · · · : · · · ). 0. c) − 2c(b + c − a)(c + a − b)(0. Wb =(3b2 + (c − a)2 )(a. available at http://faculty. Encyclopedia of Triangle Centers.be . Belgium E-mail address: eric.danneels@telenet. Eric Danneels: Hubert d’Ydewallestraat 26. Reference [1] C.edu/ck6/encyclopedia/ETC.evansville.html. 8730 Beernem.On triangles with vertices on the angle bisectors 253 A Cc Wb Wc Pw Qw Bb I Wa B Aa C Ba Figure 5. Kimberling. . Motivation In [1]. it would make perfect sense to consider r0 = 0 and r−k = −rk not to mention rk for non-integer values of k as well. Pn−1 . define dk as the distance between P0 and Pk . Notice that for 1 ≤ k ≤ n − 1.n−k. b b FORUM GEOM ISSN 1534-1178 Formulas Among Diagonals in the Regular Polygon and the Catalan Numbers Matthew Hudelson Abstract. Specifically. . we derive formulas for all diagonals in terms of the shortest diagonals and other formulas in terms of the next-to-shortest diagonals. assuming unit side length.n−h} rh rk = r|k−h|+2i−1 i=1 and 1 = rk(2j−1) rk s j=1 where s = min{j > 0 : jk ≡ ±1 mod n}. Thus. .h. jπ dj sin n Defining sin kπ n . Also.b Forum Geometricorum Volume 6 (2006) 255–262. rk = sin πn the formulas given in [1] are min{k. Specifically. Publication Date: October 23. We also show that the formulas in terms of the shortest diagonals involve the famous Catalan numbers. Communicating Editor: Paul Yiu. Fontaine and Hurley develop formulas that relate the diagonal lengths of a regular n-gon. the only restriction on n in the definition of rn is that n not be zero or the reciprocal of an integer. We look at relationships among the lengths of diagonals in the regular polygon. rk = ddk1 . but there is no a priori restriction on k in the definition of rk . 2006. . Then the law of sines yields sin kπ dk n = . given a regular convex n-gon whose vertices are P0 . . These formulas are independent of the number of sides of the regular polygon. P1 . . 1. we have r2 r5 = r−2 + r0 + r2 + r4 + r6 . j=0 The third equality holds since the sum telescopes. s 1 = rk(2j−1) rk j=1 where s is any (not necessarily the smallest) positive integer such that ks ≡ ±1 mod n. The reciprocal formula in [1] is proven almost as easily: Proposition 2. Letting h and k be integers. Short proofs of rk formulas Using the identity sin α sin β = 12 (cos(α − β) − cos(α + β)). to those in [1]: Proposition 1. we can provide some short proofs of formulas equivalent. Given an integer k relatively prime to n. rh rk = k−1 rh−k+2j+1 . . we see that these two sums are in fact equal. Reversing the roles of h and k. and perhaps simpler. j=0 Proof. Hudelson 2.256 M. For integers h and k. we have kπ π hπ sin rh rk = (sin )−2 sin n n n (h − k)π (h + k)π π −2 1 cos − cos = (sin ) n 2 n n k−1 π (h − k + 2j)π (h − k + 2j + 2)π 1 = (sin )−2 cos − cos n 2 n n j=0 π = (sin )−2 n = k−1 k−1 j=0 π (h − k + 2j + 1)π sin sin n n rh−k+2j+1 . Note that we can switch the roles of h and k to arrive at the formula h−1 rk−h+2j+1 . consider the example when k = 2 and h = 5. rk rh = j=0 To illustrate that this is not contradictory. From Proposition 1. Recalling that r0 = 0 and r−j = rj . we have r5 r2 = r4 + r6 . Proposition 3. Proof. From powers of r2 to Catalan numbers We use the special case of Proposition 1 when h = 2. rk rk(2j−1) = sin kπ π sin n n sin Here. to develop formulas for powers of r2 . the third equality follows from the telescoping sum. We proceed by induction on m. When we have m = 0. For nonnegative integers m. and the fifth follows from the definition of s. This establishes the basis step. the formula reduces to r2 0 = r1 and both sides equal 1. we assume the result for m = n and begin with the sum on the right-hand side for m = n + 1. Starting at the right-hand side. 3. we have s j=1 −1 s kπ k(2j − 1)π sin n n j=1 s π kπ −1 1 k(2j − 2)π 2jkπ = sin sin cos − cos n n 2 n n j=1 −1 kπ 2skπ 1 π cos 0 − cos = sin sin n n 2 n −1 kπ skπ π sin2 = sin sin n n n −1 π kπ π = sin sin sin2 n n n −1 π kπ sin = sin n n 1 = . namely rk r2 = rk−1 + rk+1 . For the inductive step. .Formulas among diagonals in the regular polygon and the Catalan numbers 257 Proof. r2 m = m m i=0 i r1−m+2i . 258 M. if m − k is odd.. there is no contribution to αm. A second contribution occurs when −k = 1 − m + 2i.k = (m + k − 1) 2 (m − k − 1) 2 0. if m − k is even.k rk .e. we use Proposition 3 and the identity r−k = −rk to consider an expression for r2 m as a linear combination of rk ’s where k > 0. i i i−1 The third equality is by means of the change of index j = i − 1 and the fact that n = 0 if c < 0 or c > n.k . Notice that if m and k have the same parity. In fact. Piecing this information together. we find m m − 1 . we can determine αk directly. Notice that if m − k is odd. Hudelson n+1 i=0 n+1 n n n+1 r−n+2i = + r−n+2i i i i−1 i=0 n+1 n+1 n n = r r + i − 1 −n+2i i −n+2i i=0 i=0 n n n n + r r = j 2−n+2j i −n+2i j=0 n i=0 n (r−n+2i + r2−n+2i ) i i=0 n n = r r i 2 1−n+2i = i=0 = r2 r2 n = r2 n+1 which completes the induction. 1 2 (m − k + 1) 2 (m + k − 1) . or i = 1 2 (m − k − 1). The first equality uses the standard identity for n+1 n n binomial coefficients = + . we wish to determine the coefficients αk in the sum m+1 m αm. Now. r2 = k=1 From Proposition 2. or i = 1 2 (m + k − 1). 1 αm. the sum is known to end at k = m + 1. i. then m m = 1 . One contribution occurs when k = 1 − m + 2i. c The fifth equality is from Proposition 1 and the sixth is from the induction hypothesis. Intuitively. We now derive the inverse relationship. We have 2p 2p − α1 = p p−1 (2p)! (2p)! − p!p! (p − 1)!(p + 1)! p (2p)! 1− = p!p! p+1 1 2p = p+1 p = which is the closed form for the pth Catalan number.Formulas among diagonals in the regular polygon and the Catalan numbers Therefore.k array: 0 1 2 3 4 5 1 1 0 1 0 2 0 2 0 1 0 2 0 5 3 0 0 1 0 3 0 4 0 0 0 1 0 4 5 0 0 0 0 1 0 6 0 0 0 0 0 1 As a special case of this formula. if we subtract the column corresponding to s = −k from that corresponding to s = k. Inverse formulas. . if m + k is even. αm. We start with Proposition 1: r2 rk = rk−1 + rk+1 . then we obtain Pascal’s triangle (with an extra row corresponding to m = 0 attached to the top): 0 1 2 3 4 5 −2 −1 0 1 2 3 4 5 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 2 0 1 0 0 1 0 3 0 3 0 1 1 0 4 0 6 0 4 0 6 0 0 0 0 0 1 Next.k 259 m m − 1 . if m + k is odd. writing the rk values as linear combinations of powers of r2 . polynomials. indexed horizontally by k ∈ Z and vertically by m ∈ N. we obtain the αm. consider what happens when m = 2p and k = 1. 1 = (m − k + 1) 2 (m − k − 1) 2 0. if we arrange the coefficients of the original formula in a table. and binomial coefficients We have a formula for the powers of r2 as linear combinations of the rk values. 4. we have 0 1 2 3 4 5 6 1 1 0 0 0 0 0 0 1 0 0 0 0 0 2 0 1 0 0 0 0 3 −1 0 1 0 0 0 4 0 −2 0 0 −3 0 1 0 0 5 1 3 0 −4 0 1 0 6 0 6 0 −5 0 1 7 −1 0 This array displays the coefficient β(k. i i By inspection. m)r2m . As a result. Hudelson We demonstrate that for natural numbers k. k − 1 − i Pk (t) = (−1)i tk−1−2i . Now assume the existence of Pk−1 (t) and Pk (t). We note that P0 (t) = 0 and P1 (t) = 1. m) in the formula β(k. The third equality is obtained by replacing j with i − 1. we show for k ≥ 0. this holds for k = 0 as the binomial coefficients are all zero in the sum. rk = m . rk = k − 1 − i i i (−1)i r2 k−1−2i . Also. Therefore. we obtain the desired formula for rk in terms of powers of r2 : Proposition 4. Armed with this. Pk+1 (t) = tPk (t) − Pk−1 (t) k − 2 − j k − 1 − i i k−1−2i − (−1) t (−1)j tk−2−2j =t i j i j k−1−i k − 1 − i i k−2i = − (−1) t (−1)i−1 tk−2i i i−1 i i k − i = (−1)i tk−2i i i as desired. Assembling these coefficients into an array similar to the αm. we use the recurrence to establish the induction.k array in the previous section.260 M. Given k ≥ 1. when k = 1. the i = 0 term is the only nonzero contributor to the sum. Now. it is immediate that this formula holds for k = 0. there exist polynomials Pk (t) such that rk = Pk (r2 ). 1. Then from the identity rk+1 = r2 rk − rk−1 we have Pk+1 (t) = tPk (t) − Pk−1 (t) which establishes a second-order recurrence for the polynomials Pk (t). For all natural numbers k. Qk (t) such that rk = Qk (r3 ). . Q1 (t) = 1. this leads to the conclusion that the columns of the original α(m. we use r3 rk = rk+2 + rk + rk−2 which leads to rk+2 = (r3 − 1)rk + rk−2 . We√ and so Q2 (t) = t + 1. Therefore. With this. We now claim Proposition 5.Formulas among diagonals in the regular polygon and the Catalan numbers 261 where k is the column number and m is the row number. Q1 . k) array are inverses in the sense of ”array multiplication”: α(m. From the above identity. and Q3 (t) = t so all that remains is √ have r3 = r2 2 − 1 from Proposition 4. the k(th) column of the β(k. m = p . 2x2 This game is similar but slightly more complicated in the case of r3 . we show that for k ≥ 0. the function that generates the m(th) column of the α(m. Using machinery in §5. m) array can be generated using the generating function xk (1 + x2 )−1−k . Here. Q2k (t) = √ t+1 k − 1 − i i Q2k+1 (t) = k − i i i i (−1)i (t − 1)k−1−2i (−1) (t − 1) i k−2i + k − 1 − i i i (−1)i (t − 1)k−1−2i . Q0 (t) = 0. An interesting observation is that this array and the α(m. By to deterinspection. there are functions. otherwise. p) = i 1. k) array is m √ 2 1 − 4x 1 − xm−1 . i)β(i. we have Qk+2 (t) = (t − 1)Qk (t) − Qk−2 (t). but not necessarily polynomials.1 of [2]. This establishes a fourth-order recurrence relation for the functions Qk (t) so determining the four functions Q0 . in this case. k) array can be generated using an inverse function. 0. r2 = r3 + 1 mine Q2 (t). and Q3 will establish the recurrence. Q2 . Also. Pullman. Fontaine and S. k − 1 − i √ (−1)i (r3 − 1)k−1−2i r2k = r3 + 1 i i k − i k − 1 − i r2k+1 = (−1)i (r3 − 1)k−2i + (−1)i (r3 − 1)k−1−2i . Hudelson Proof. and Q3 for k = 0. The proof for Q2k+1 is similar. These are easily checked to match the functions Q0 .edu . 1. i i i i References [1] A. Proof by picture: Products and reciprocals of diagonal length ratios in the regular polygon. [2] H. Forum Geom. We proceed by induction on k. The third equality is obtained by replacing j with i − 1. Academic Press. 6 (2006) 97–101. For k ≥ 2.262 M. 1994. Q2 . Hurley. WA 99164-3113 E-mail address: hudelson@math. Wilf. Department of Mathematics. Generatingfunctionology. and is omitted. Washington State University. Q1 . we have Q2k (t) = (t − 1)Q2(k−1) (t) − Q2(k−2) (t) k − 2 − i √ (−1)i (t − 1)k−2−2i = t + 1 (t − 1) i i k − 3 − j − (−1)j (t − 1)k−3−2j j j k − 2 − i k − 2 − i √ + (−1)i (t − 1)k−1−2i = t+1 i i−1 i k − 1 − i √ i k−1−2i = t+1 (−1) (t − 1) i i as desired.. As a corollary. we have Corollary 6. treating each sum separately.wsu. Let P be an interior point of a given triangle ABC. . Theorem 2. lies on the Euler line of triangle ABC. Paul Yiu has subsequently discovered the following generalization. the barycentric square root of H. Given an acute √ triangle ABC with orthocenter H. A √ H H O H B C Figure 1. the present author proposed the following problem. b b FORUM GEOM ISSN 1534-1178 A Note on the Barycentric Square Roots of Kiepert Perspectors Khoa Lu Nguyen Abstract. 1. Communicating Editor: Paul Yiu. We prove that the orthocenter of the cevian triangle of the barycentric square root of P lies on the Euler line of ABC if and only if P lies on the Kiepert hyperbola. the orthocenter H of the cevian triangle of H. Theorem 1. Introduction In a recent Mathlinks message. The author is grateful to Professor Yiu for his generalization of the problem and his help in the preparation of this paper. The locus of point√P for which the orthocenter of the cevian triangle of the barycentric square root P lies on the Euler line is the part of the Kiepert hyperbola which lies inside triangle ABC. 2006.b Forum Geometricorum Volume 6 (2006) 263–268. Publication Date: October 30. (1) . The orthic axis is perpendicular to the Euler line. P and H are parallel. If P has homogeneous barycentric coordinates (u : v : w).. C A and CA. In particular. A B and AB. §§5.264 K. The trilinear polar P and the orthic axis H intersect at the point (u(SB v − SC w) : v(SC w − SA u) : w(SA u − SB v)).4. It is enough to prove Theorem 2. By Desargues’ theorem. A C B A1 P B A C B1 C1 Figure 2. for example. for example. P : u v w For the orthocenter H = (SBC : SCA : SAB ). is also called the orthic axis. Proposition 3. (SB − SC )vw + (SC − SA )wu + (SA − SB )uv = 0. Equivalently. from the fact that the orthic axis H is the radical axis of the circumcircle and the nine-point circle. i.5]. the trilinear polar of P . their intersection is a point at infinity if and only if u(SB v − SC w) + v(SC w − SA u) + w(SA u − SB v) = 0. the trilinear polar H : SA x + SB y + SC z = 0. Nguyen The barycentric square root is defined only for interior points. This is the reason why we restrict to acute angled triangles in Theorem 1 and to the interior points on the Kiepert hyperola in Theorem 2. B1 . L. It follows easily. C1 lie on a line P . This proposition is very well known. and A1 . 2. B1 . the three points A1 . C1 be respectively the intersections of B C and BC. then the trilinear polar is the line z x y + + = 0. Trilinear polars Let A B C be the cevian triangle of P . See.e. [2. Proposition 5. We may the barycentric square √ assume u. Similar constructions on the other two sides Then X is the trace of P √ P on CA and AB respectively. Proposition 4. Let M is the midpoint of BC. Proof.A note on the barycentric square roots of Kiepert perspectors 265 Note that this equation defines the Kiepert hyperbola. then A1 X 2 = A1 B · A1 C. Thus. Points on the Kiepert hyperbola are called Kiepert perspectors. Since A1 . A X P A1 B Figure 3. A1 X 2 = A1 B · A1 C. v. C harmonically. M X2 = M A1 · M A . root of P to be the point P with barycentric coordinates √ Paul Yiu [2] has given the following construction of P . 3. w > 0. CZ. we have M B 2 = M C 2 = M A1 · M A (Newton’s theorem). The trilinear polar P is parallel to the orthic axis if and only if P is a Kiepert perspector. A X M C . Hence. and ∠M X A1 = ∠M A X = 90◦ . (3) Construct the bisector √ of angle BX C to intersect BC at X. (2) Construct the perpendicular to BC at the trace A of P to intersect CA at X . on BC. and define √ √ √ ( u : v : w). (1) Construct the circle CA with BC as diameter. If the trilinear polar P intersects BC at A1 . A divide B. It follows that triangles M X A1 and M A X are similar. BY . with homogeneous barycentric coordinates (u : v : w). This means that A1 X is tangent at X to the circle with diameter BC. The barycentric give the traces Y and Z of √ square root P is the common point of AX. The barycentric square root of a point Let P be a point inside triangle ABC. Proof. If H is the orthocenter of XY Z. Y Y1 . with diameters XX1 . E. then the line OH is perpendicular to the trilinear polar P . This follows easly from ∠A1 X X =∠A1 X B + ∠BX X =∠X CB + ∠XX C =∠X XA1 . By Proposition 5. then A1 is the midpoint of XX1 . and XY Z the cevian triangle of its barycentric square root P . It is therefore on the radical axis of the three circles. If X1 is the intersection of Y Z and BC. This completes the proof of Theorem 2. this is the case precisely when P lies on the Kiepert hyperbola. and hence parallel to the orthic axis by Proposition 3. These three √ circles are coaxial. As shown in the previous section. ξC . Now. Therefore. Indeed. H is the radical center of the three circles. they are the generalized Apollonian circles of the point P . L. since D. and is perpendicular to the line P which contains their centers. We show that the circumcenter O of triangle ABC also has the same power with respect to these circles.266 K. C1 on the trilinear polar P . EF Y Z. ξB . Proposition 7.. F lie on the circles ξA . It follows that the line OH is the radical axis of the three circles. and with center on BC is orthogonal to the circle CA . i. triangle A1 XX is isosceles. Corollary 6. Since DEXY . ξC respectively. their centers are the points A1 . See [3]. ZZ1 . F Z intersect at H . X1 divide B. The circle through X. EY . Nguyen To complete the proof it is enough to show that A1 X = A1 X . See Figure 4. Proof of Theorem 2 Let P be an interior point √ of triangle ABC. (2) Consider the circles ξA . which is therefore the midpoint of XX1 . If X1 is the intersection of Y Z and BC. B1 . The orthocenter H of XY Z lies on the Euler line of triangle ABC if and only if the trilinear polar P is parallel to the Euler line. ξB .e. and the common chords DX. this has center A1 . and F DZX are cyclic. C harmonically. 4. Proof. Consider the orthic triangle DEF of XY Z. O also lies on the radical axis of the three circles. it follows from (2) that H has equal powers with respect to the three circles. X1 . the power of O with respect to the circle ξA is A1 O2 − A1 X 2 = OA21 − R2 − A1 X 2 + R2 = A1 B · A1 C − A1 X 2 + R2 = R2 by Proposition 5. then X. and H D · H X = H E · H Y = H F · H Z. . The same is true for the circles ξB and ξC . By Proposition 4. The orthocenter of the cevian triangle of P lies on the Brocard axis if and only if P is an interior point on the Jerabek hyperbola. Coordinates In homogeneous barycentric coordinates. 5. Proof. √ Theorem 8.A note on the barycentric square roots of Kiepert perspectors 267 B1 A Y D Z P √ A1 B O H F E X1 P X C C1 Z1 Figure 4. the orthocenter of the cevian triangle of (u : v : w) is the point . The Brocard axis OK is orthogonal to the Lemoine axis. The locus of points whose trilinear polars are parallel to the Brocard axis is the Jerabek hyperbola. 02139. Florida Atlantic University lecture notes. [2] P. Johnson. Generalized Apollonian circles. we obtain a2 SA · SABC + SBC a2 SA : · · · : · · · . Khoa Lu Nguyen: Massachusetts Institute of Technology.com .268 K. which is the point More generally. Nguyen 1 1 SB + w u : ··· : ··· . Introduction to the Geometry of the Triangle. MA. Yiu. Advanced Euclidean Geometry. student. 1925. USA E-mail address: treegoner@yahoo. A. SA + Sθ SB + Sθ SC + Sθ √ the orthocenter of the cevian triangle of P is the point a2 SA (SA + Sθ )(SB + Sθ )(SC + Sθ ) +SBC cyclic a2 SA + Sθ + a2 Sθ SA SA + Sθ : · · · : · · · . 8 (2004) 225–230. [3] P. cyclic References [1] R. L. if P is the Kiepert perspector 1 1 1 K(θ) = : : . 77 Massachusetts Avenue. + SC 1 1 + u v −SA 1 1 + v w 2 + SB 1 1 − 2 u2 w + SC 1 1 − 2 u2 v 2 2 2 to the square root of the orthocenter. Dover reprint. with (u : v : w ) = Applying this 1 1 1 SA : SB : SC . 2001. Cambridge. cyclic H in Theorem 1. Journal of Geometry and Graphics. Yiu. y1 = e1 + δtv. 2006. Then the point U = u : v : w is given by u = bn − cm. b. F are distinct points. E = d2 : e2 : f2 . are defined) by homogeneous trilinear coordinates. B. CF concur. and L. 1. where l : m : n is a point. so that the “plane” of ABC is infinite dimensional. F are points and L is a line other than the line at infinity. B. E. by an equation lα + mβ + nγ = 0.) Suppose that D. b. c of triangle ABC are variables or indeterminates. C. c. ϕ = ad3 + be3 + cf3 . (1) Write the vertices of DEF as D = d1 : e1 : f1 . This work examines cases in which a translation D E F of DEF in the direction of L is perspective to ABC. The line L∞ at infinity is given by aα + bβ + cγ = 0. BE . F = d3 : e3 : f3 . such that the set {A. One of these cases is the limiting case that D = L ∩ L∞ . in the sense that the lines AD . v = cl − an. and points are defined as functions of a. . and note that D = E = F = U.b Forum Geometricorum Volume 6 (2006) 269–284. and let δ = ad1 + be1 + cf1 . call this point U. b b FORUM GEOM ISSN 1534-1178 Translated Triangles Perspective to a Reference Triangle Clark Kimberling Abstract. E. D. Points and lines will be given (indeed. D. F } consists of at least five distinct points. CF concur. = ad2 + be2 + cf2 . Communicating Editor: Paul Yiu. (To say “transfigured plane” means that the sidelengths a. C. F is on L∞ implies that none of δ. let L∞ be the line at infinity and L a line other than L∞ . w = am − bl. none on L∞ . BE . E. The hypothesis that none of D. Suppose A. ϕ is 0. The line L is given parametrically as the locus of point D = Dt = x1 : y1 : z1 by x1 = d1 + δtu. in the sense that the lines AD . Publication Date: November 6. Introduction In the transfigured plane of a triangle ABC. We wish to translate triangle DEF in the direction of L and to discuss cases in which the translated triangle D E F is perspective to ABC. . z1 = f1 + δtw. E. they are zero not only for Euclidean triangles. . then Dt Et Ft is homothetic to ABC and hence perspective to ABC. As Dt traverses the line DU. Two basic theorems Theorem 1. y2 = e2 + tv. the concurrence determinant. The lines ADt . p0 are all zero. respectively. E. Proof. c are positive real numbers satisfying (a > b + c. b ≥ c + a. does not imply the “same” theorem in the more general setting of triangle algebra. for which a. (6) Thus. The degree of homogeneity of t is that of (x1 − d1 )/(δu). c > a + b) or (a ≥ b + c. or else P (t) is zero for at most two values of t. so that F = Ft = x3 : y3 : z3 is given by x3 = d3 + ϕtu y3 = e3 + ϕtv z3 = f3 + ϕtw. for every t. C. the triangle Dt Et Ft of translation of DEF in the direction of L is either perspective to ABC for all t or else perspective to ABC for at most two values of t. z2 α − x2 γ = 0. D. CFt are given by the equations −z1 β + y1 γ = 0. Thus. The geometric theorem. BEt . (3) where p0 =d3 e1 f2 − d2 e3 f1 . z2 = f2 + tw. perspectivity and parallelism (hence homothety) are defined by zero determinants. however. B. (5) p2 =δvw(d3 − ϕd2 ) + wu(ϕe1 − δe3 ) + ϕuv(δf2 − f1 ). This is well known in geometry. Suppose L is a line and U = L ∩ L∞ . either p0 . F } consists of at least five distinct points. in which case Dt Et Ft is perspective to ABC for all t. In these parameterizations. t represents a homogeneous function of a. 0 −z1 y1 z2 0 −x2 −y3 x3 0 − y3 α + x3 β = 0. formally of degree 2 in t: P (t) = p0 + p1 t + p2 t2 . (4) p1 =u(ϕe1 f2 − e3 f1 ) + v(δd3 f2 − ϕd2 f1 ) + w(e1 d3 − δe3 d2 ). b. (2) is a polynomial P . Kimberling The point E traverses the line through E parallel to L. If triangle DEF is homothetic to ABC. b. 2.270 C. so that E = Et = x2 : y2 : z2 is given by x2 = d2 + tu. b > c + a. When such determinants are “symbolically zero”. Suppose ABC and DEF are triangles such that {A. p1 . in which the objects are defined in terms of variables or indeterminants. c. Specifically. F The point traverses the line through F parallel to L. It is of interest to express the coefficients p0 . as are DE and AB. To that end. 29). we have ∆ = 0. The zero determinants yield bF1 = cE1 . Therefore. as defined by a zero determinant (e. then L1 L3 . For example. E. Then Dt Et Ft is perspective to ABC for all t. so that D. as defined by the identity −1 D1 D2 D3 d1 e1 f1 d2 e2 f2 = 1 E1 E2 E3 .. cD2 = aF2 . the lines F D and CA are parallel.Translated triangles perspective to a reference triangle 271 c ≥ a + b). (7) These equations can be used to give a direct but somewhat tedious proof of Theorem 2. p2 more directly in terms of a. but also for a. b. which implies that Dt Et Ft is perspective to ABC. If T1 is homothetic to T2 and T2 is homothetic to T3 . so that p0 = 0. w in (5) and (6). ABC and DEF are homothetic. likewise. E. Therefore Dt Et Ft is homothetic to ABC. It is understood that A. Suppose ABC and DEF in Theorem 1 are homothetic. F are not collinear. D1 = e2 f3 − f2 e3 . ∆ d3 e3 f3 F1 F2 F3 where d1 e1 f1 ∆ = d2 e2 f2 = d1 D1 + e1 E1 + f1 F1 . p1 . then T1 is homothetic to T3 . and DEF and Dt Et Ft are homothetic. substitute from (1) for u. in the case that ABC and DEF are homothetic. we digress to prove only that p0 = 0. C are not collinear. Ψ = 0. line EF is parallel to line BC. getting p1 = lp1l + mp1m + np1n and p2 = mnp2l + nlp2m + lm2n . p. F is the determinant equation ∆ = 0. where Ψ = e3 d2 f1 − e1 d3 f2 . and abcΨ∆ = 0 by (7). [1]. . Proof. b. B. Among geometric theorems that readily generalize to algebraic theorems are these: If L1 L2 and L2 L3 . Then L − R factors as abcΨ∆. v. F.g. Theorem 2. Next. etc. Let L and R denote the products of the left-hand sides and the right-hand sides in (7). c and the coordinates of D. As the defining equation for collinearity of D. then T1 is perspective to T2 . d3 e3 f3 that is. c as indeterminates. E. If T1 is homothetic to T2 . we shall use cofactors. aE3 = bD3 . and p2l =2a2 bc(e1 d2 f3 − e2 d3 f1 ) − ab2 e3 F3 + ac2 f2 E2 − bc2 f1 D1 + ba2 d3 F3 − ca2 d2 E2 + cb2 e1 D1 . Let At = BEt ∩ CFt . where the matrix p1 q 1 p2 q 2 M= p3 q 3 is nonsingular with adjoint (cofactor) matrix P1 Q1 # P2 Q2 M = P3 Q3 r1 r2 r3 R1 R2 . see also [2]. p2 more directly in terms of a. BEt . 3. p1n =a2 d3 E3 + b2 e3 D3 − caf1 E1 − bcf2 D2 .272 C. Theorem 4. Ct is a conic. Suppose a point P = p : q : r is given parametrically by p =p1 t2 + q1 t + r1 . Lemma 3. p2m =2b2 ca(f2 e3 d1 − f3 e1 d2 ) − bc2 f1 D1 + ba2 d3 F3 − ca2 d2 E2 + cb2 e1 D3 − ab2 e3 F3 + ac2 f2 E2 . Bt = CFt ∩ ADt . p1m =c2 f2 D2 + a2 d2 F2 − bce3 D3 − abe1 F1 . Ct = ADt ∩ BEt . the line L is not parallel to a sideline of triangle ABC. b. R3 Then P lies on this conic: (Q1 α + Q2 β + Q3 γ)2 = (P1 α + P2 β + P3 γ)(R1 α + R2 β + R3 γ). Kimberling where p1l =b2 e1 F1 + c2 f1 E1 − abd2 F2 − cad3 E3 . c and the coordinates of D. p1 . r =p3 t2 + q3 t + r3 . F is now completed. CFt is also the point of concurrence of three conics. The task of expressing the coefficients p0 . If. (8) In Theorem 4. q =p2 t2 + q2 t + r2 . E. we shall show that the point of concurrence of the lines ADt . . Intersecting conics We begin with a lemma proved in [7]. p2n =2c2 ab(e3 d1 e2 − d1 f2 e3 ) − ca2 d2 E2 + cb2 e1 D1 − ab2 e3 F3 + ac2 f2 E1 − bc2 f1 D1 + ba2 d3 F3 . in Theorem 1. Bt . then the locus of each of the points At . (11) By Lemma 3. Note that the conic {At } passes through B and C. by factoring |M | . Consider the case u = 0. Except for special cases. z traverse conics as in Theorem 4. as indicated by Figure 1. contrary to the hypothesis. . 1 y A D' x C B H E' F' z Figure 1. or w must be zero. determine 1Figure 1 can be viewed dynamically using The Geometer’s Sketchpad. v. Likewise. at the perspectors described in Theorem 1. see [6] for access. bt = x2 y3 . As the lemma applies only to nonsingular M .) Point H on line AD is an independent point. E = A. but this is the point in which line BC meets L∞ . then u : v : w = 0 : c : −b. y. we can. E. Intersecting conics Lemma 3 shows how to write out equations of conics starting with a matrix M . The choice of triangle DEF is given by the equations D = C. ct = z2 x3 . and the conics meet twice (with x = y = z). as D traverses line AH. F are not shown. so that at = (d2 + tu) (d3 + ϕtu) = ϕu2 t2 + (ϕd2 u + d3 u)t + d2 d3 . and triangle D E F is a translation of DEF in the direction of line AH. F = B. points x. the loci {Bt } and {Ct } are conics. The point At = at : bt : ct is given by at = x2 x3 . in which case u. the locus {At } is a conic unless ϕuvw = 0. (10) 2 ct =(f2 + tw) (d3 + ϕtu) = ϕuwt + (ϕf2 u + d3 w)t + f2 d3 . (The labels D. (9) bt = (d2 + tu) (e3 + ϕtv) = ϕuvt2 + (ϕd2 v + e3 u)t + d2 e3 .Translated triangles perspective to a reference triangle 273 Proof. they are. vq. P ) = pu(rv + qw) : qv(pw + ru) : rw(qu + pv). P ) = rv − qw : pw − ru : qu − pv. ϕu2 ϕd2 u + d3 u d2 d3 |M | = ϕuv ϕd2 v + e3 u d2 e3 ϕwu ϕf2 u + d3 w f2 d3 =ϕu (uf2 − wd2 ) (vd3 − ue3 ) (be3 d2 − be2 d3 + cd2 f3 − cd3 f2 ) . wr is not zero. as it is assumed that E = U and F = U. and DE is not parallel to AB. is the isogonal conjugate of P. Geometric definitions of isogonal and isotomic conjugates are given at MathWorld [8]. that DEF is a cevian triangle of a point. and F D is not parallel to CA. We shall also employ these terms and notations: crossdifference of U and P = CD(U. For various choices of DEF and L. In conclusion. Unless otherwise noted. the factor be3 d2 − be2 d3 + cd2 f3 − cd3 f2 is 0 if and only if line EF is parallel to line BC. The quotient U/P is defined by U/P = U · P −1 . the perspectivities indicated by Theorem 1 are of particular interest. the points U = u : v : w and P = p : q : r are arbitrary. In order to describe the configurations. ϕu = 0. briefly. Geometric interpretations of these “cross operations” are given at [4]. By hypothesis. it will be helpful to adopt certain terms and notations. Also. Kimberling criteria for nonsingularity. or an anticevian triangle. Finally. 4. If at least one of the products up. The multiplicative inverse of P. In connection with {At } and (9)-(11). crosspoint of U and P = CP (U. (uf2 − wd2 ) (vd3 − ue3 ) = 0. given by P −1 = p−1 : q −1 : r −1 . or a rotation of triangle ABC about its circumcenter. . crosssum of U and P = CS(U. P ) = rv + qw : pw + ru : qu + pv. Terminology and notation The main theorem in this paper is Theorem 1. defined if pqr = 0. Such choices are considered in Sections 3-6.274 C. if EF is not parallel to BC. the product U · P is defined by the equation U · P = up : vq : wr. The isotomic conjugate of P is defined if pqr = 0 by the trilinears a−2 p−1 : b−2 q −1 : c−2 r −1 . then the three loci are conics and Theorem 3 applies. and a parametric representation is given by x : y : z = x(s) : y(s) : z(s). we shall use abbreviations: a1 = (b2 +c2 −a2 )/(2bc). More generally. where x(s) =a1 + s(a1 − 2b1 c1 ). Suppose DEF is a triangle in the transfigured plane of ABC. symmedian point =X6 = a : b : c. trilinears for any point are homogeneneous functions of a. In this paper. X3 = a1 : b1 : c1 . Thus. When working with lines algebraically. b1 = (c2 +a2 −b2 )/(2ca). at most two U -ppt’s. such that D E F is a U -translation of DEF and D E F is perspective to ABC (in the sense that the lines AD . c. a point on L∞ ). b. That one . its point of intersection with L∞ will be called the direction of the line. each DEF has. c. CF concur) will be called a U -ppt of DEF. such as incenter =X1 = 1 : 1 : 1 = the multiplicative identity. as when DEF is a cevian triangle or an anticevian tiangle. by L(wy − vz)x2 + M (uz − wx)y 2 + N (vx − uy)z 2 = 0. In view of Theorem 1. P ). all having the same degree of homogeneity. and the line X3 X6 is the Brocard axis. centroid =X2 = 1/a : 1/b : 1/c. except for special cases. b. The cubic Z(U.e. thus. It will be helpful to introduce some related terminology. if DEF is perspective to ABC.Translated triangles perspective to a reference triangle 275 Notation of the form Xi as in [3] will be used for certain special points. we shall sometimes use trilinears for X3 expressed directly in terms of a. BE . for any line. and U is a direction (i. it is sometimes helpful to do so with reference to a parameter and the point in which the line meets L∞ . there is “usually” just one ppt. circumcenter =X3 = cos A : cos B : cos C. y(s) =b1 + s(b1 − 2c1 a1 ).. For details on these and other families of cubics. The parameter s is not necessarily a numerical variable. As cos A = (b2 + c2 − a2 )/(2bc). Two families of cubics will occur in the sequel. The line X2 X3 is the Euler line. thus in a parametric expression of the form p + su. The point X30 will be called the direction of the Euler line. see [5]. c. c1 = (a2 +b2 −c2 )/(2ab). this point is X30 = a1 − 2b1 c1 : b1 − 2c1 a1 : c1 − 2a1 b1. the degree of homogeneity of s is that of p minus that of u. The designation “ppt” means “proper perspective translation”. rather. P ) is given by (vqy − wrz)px2 + (wrz − upx)qy 2 + (upx − vqy)rz 2 = 0. In the case of the Euler line. and the cubic ZC(U. Instead of the trigonometric trilinears for X3 . it is a function of a. A triangle D E F . b. other than DEF itself. z(s) =c1 + s(c1 − 2a1 b1 ). for every U. The remainder of this article is mostly about special translations. .4 and Case 6. U ) satisfies ∆0 = 0 and ∆1 = 0. none of which lies on the cubic ∆1 = 0. X3 (the circumcenter). especially in Case 5.4. let U = X511 = cos(A + ω) : cos(B + ω) : cos(C + ω). where the cubic ∆1 = 0 is classified as ZC(511. In this case. X297 . X31 ). Z(X2 . and X2010 .276 C. Translated cevian triangles In this section. X694 . ∆1 = u v w . For given U. Further results in Case 1 are given in Theorem 5. ∆0 = u x x y z y z In particular. In this case.3: ∆0 = 0 and ∆1 = 0. d3 e3 f3 x y 0 and the perspectivity determinant (2) is given by −t(∆0 + t∆1 ). The cevian triangle DEF of X is not homothetic to ABC. Now for any given U.1: ∆0 = 0 and ∆1 = 0. Then the cubic ∆0 = 0 passes through the following points. there is no U -ppt. Kimberling ppt is of primary interest in the next three sections. Case 5. For quite a different example. specifically. DEF is the cevian triangle of a point X = x : y : z. Also. where ω denotes the Brocard angle. these being X1113 and X1114 . Then the cubic ∆1 = 0 passes through the two points in which the Euler line meets the circumcircle. so that ∆1 = 0. for any U. where 2 2 2 2 2 2 ax2 by 2 cz 2 a ux b vy c wz v w . take U = X523 . X325 . ∆0 =bcvw(b2 v 2 − c2 w2 ) + cawu(c2 w2 − a2 u2 ) + abuv(a2 u2 − b2 v 2 ) = − (bv − cw)(cw − au)(au − bv)(au + bv + cw) =0. Clearly this holds for X = X2 . the equations ∆0 = 0 and ∆1 = 0 represent cubics in x.2: ∆0 = 0 and ∆1 = 0. Rows 1 and 3 of the determinant ∆1 are equal. and these points do not also lie on the cubic ∆0 = 0. Another such example is obtained by simply taking X to be U. Other points on the cubic ∆0 = 0 are given at [5]. thus DEF is given as a matrix by 0 y z d1 e1 f1 d2 e2 f2 = x 0 z . z. Case 5. y. 5. For example. yet Dt Et Ft is perspective to ABC for every t. We shall consider four cases: Case 5. X6 (the symmedian point). X2009 . the point X = CP (X2 . 511)). for all U. let X be the isotomic conjugate of U . L(30. X6 · U −1 ) and Z(X75 · U −1 . respectively. Dt Et Ft is perspective to ABC for every t. pairs X and X are shown here: X 4 7 54 68 69 99 183 190 385 401 668 670 671 903 X 3 256 52 52 6 690 262 900 325 297 691 888 690 900 Returning to Case 5.. The perspector P traverses a circumconic. 2 An equation for the circumconic described in Theorem 5 is found from (8): b2 c2 (b2 v 2 − c2 w2 )βγ + c2 a2 (c2 w2 − a2 u2 )γα + a2 b2 (a2 u2 − b2 v 2 )αβ = 0. Dt Et Ft is perspective to ABC for t = −∆0 /∆1 . Let X be the isotomic conjugate of U . . AB. AB. The locus of the perspector Pt of triangles Dt Et Ft and ABC is a conic that passes through A. CA. Theorem 5. (14) z = (au − bv)xy + cz(uy − vx) Note that if X and U are triangle centers for which X is a point. and the point X 2 = b4 c4 v 2 w2 : c4 a4 w2 u2 : a4 b4 u2 v 2 . bv/(cawu). (13) (cw − au)zx + by(wx − uz) ax − by . cw/(abuv). Substituting and simplifying give Pt =b3 c3 vw(bv − acuwt)(cw − abuvt) : c3 a3 wu(cw − bavut)(au − bcvwt) : a3 b3 uv(au − cbwvt)(bv − cawut). the locus of Pt is a conic.4 is invariant of the point U. (15) Proof. the cevian triangle of the isotomic conjugate of U is DEF.4: ∆0 = 0 and ∆1 = 0. then X is a triangle center. C for t = au/(bcvw). C. Triangle D E F is thus a movable translation of DEF in the direction of U. 2Figure 2 can be viewed dynamically using The Geometer’s Sketchpad. The perspector is the point Pt = x2 x3 : x2 y3 : z2 x3 . B. In this case. In fact. See Figure 2. An arbitrary point U on L∞ is given by U = Au ∩ L∞ . and D E F stays perspective to ABC. (12) x = (bv − cw)yz + ax(vz − wy) cz − ax y = . B. respectively. Suppose X is the isotomic conjugate of a point U1 on L∞ but not on a sideline BC. and X is on L∞ . X = CD(X6 . the user can vary u freely. in the subcase that X is the isotomic conjugate of U. Then the perspector X in Case 5. Clearly. after cancellation of common factors. The cevian triangle of U is def.Translated triangles perspective to a reference triangle 277 Case 5. and P0 is the point given by (15). it is natural to ask about the perspectors. By Theorem 3. Pt passes through A.e. Point D is movable on line DU. U1−1 ). and to find the following theorem. is the point X = x : y : z given by by − cz .1. i. For the special case U = X511 . where u is an independent point. which. see [6] for access. The perspector is the point x2 x3 : x2 y3 : z2 x3 . CA. Suppose U is any point on L∞ but not on a sideline BC. Theorem 6. = Λ(−(u + v + w) + s(a2 u + b2 v + c2 w)) x = Coordinates y and z are found in the same manner. by − cz (bv − cw)yz + ax(vz − wy) (by − cz)/a . Kimberling F' F u E' A P E D' C B D f Figure 2. w1 = (a − b)(1 + abs). Proof. Thus (bv − cw)yz + ax(vz − wy) = Λ(−a(u + v + w) + as(a2 u + b2 v + c2 w)). and by (12). . U1−1 ). (16) so that (bv−cw)yz+ax(vz−wy) = Λ bv − av − aw + cw + as(b2 v − abv + c2 w − acw ). and multiplying through by the common denominator gives x : y : z =(by − cz)/a : (cz − ax)/b : (ax − by)/c =u1 (bv1 − cw1 ) : v1 (cw1 − au1 ) : w1 (au1 − bv1 ) =CD(X6 . which is to say that the perspector X is on L∞ . ax + by + cz = 0. Clearly. Represent U1 parametrically by u1 = (b − c)(1 + bcs). v1 = (c − a)(1 + cas). Write X = x : y : z = b2 c2 v1 w1 : c2 a2 w1 u1 : a2 b2 u1 v1 where the point U1 = u1 : v1 : w1 is on L∞ and u1 v1 w1 = 0. where Λ = −a3 b3 c3 (b − c)(c − a)(a − b)(1 + bcs)(1 + cas)(1 + abs). Cevian triangle and circumconic as in Theorem 5.278 C. In this case. where ax2 by 2 cz 2 ax2 by 2 cz 2 v w .2 in the cevian case are also points for which ∆0 = 0 and ∆2 = 0. given as a matrix by d1 e1 f1 −x y z d2 e2 f2 = x −y z . The cubic ∆0 = 0 is already discussed in Section 3. In this case. The Steiner circumellipse is given by bcβγ + caγα + abαβ = 0.3: ∆0 = 0 and ∆2 = 0. for which the perspector is X = X900 = CD(X6 . for any U. then the points listed for Case 5. these points do not also lie on the cubic ∆0 = 0. Case 6. the point X = U is on both cubics. The equation ∆2 = 0 represents the cubic Z(X2 /CS(X6 . (X670 . Proof. Then the cubic ∆2 = 0 passes through the points in which the Brocard axis X3 X6 meets the circumcircle these being X1312 and X1313 . X888 ). Case 6. (X903 . Dt Et Ft is perspective to ABC for every t. For given U. the isotomic conjugate X of U satisfies ∆0 = 0 and ∆1 = 0. U −1 )). As X traverses the Steiner circumellipse. X900 ). there is no ppt. take U = X523 . X101 ). 6. Translated anticevian triangles In this section. X690 ). It is easy to prove that the point CP (X2 . d3 e3 f3 x y −z The perspectivity determinant (2) is given by −2t(∆0 + t∆2 ). We consider four cases as in Section 3. Other examples (X. (X668 . (X671 . Case 6. Theorem 7 is exemplified by taking X = X190 . DEF is the anticevian triangle of a point X = x : y : z. For a diffferent example. X690 ).Translated triangles perspective to a reference triangle 279 Corollary 7. the perspector X traverses the line at infinity. U ) also lies on both cubics. to which Theorem 6 applies. These examples show that the mapping X → X is not one-to-one. for all U. Clearly this holds for X = X2 . the isotomic conjugate of X is then X514 . ∆2 = (bv + cw)u (cw + au)v (au + bv)w ∆0 = u x y z x y z .1: ∆0 = 0 and ∆2 = 0.2: ∆0 = 0 and ∆2 = 0. Now for any given U. The corollary follows from the easy-to-verify fact that the isotomic conjugacy map ping carries the Steiner circumellipse to L∞ . . X ) are these: (X99 . let U = X511 . For example. X891 ). As X traverses the Steiner inellipse. U1−1 ). = Λ(−(u + v + w) + s(a2 u + b2 v + c2 w)) x = Coordinates y and z are found in the same manner. The perspector is the point x2 x3 : x2 y3 : z2 x3 . and multiplying through by the common denominator gives x : y : z = CD(X6 . U1−1 ). which on cancellation of common factors. U1 ) satisfies (21). Now.4: ∆0 = 0 and ∆2 = 0. we have bwy 2 − cvz 2 + (bv − cw)yz + ax(vz − wy) =Λ(bv − av − aw + cw + as(b2 v − abv + c2 w − acw). In this case. (18) 2 2 cuz − awx + (cw − au)zx + by(wx − uz) ax − by . AB. the same point as at the end of the proof of Theorem 6. In fact. Dt Et Ft is perspective to ABC for t = −∆0 /∆2 . is the point X = x : y : z given by by − cz . it is easy to show that if U1 is on point X = x : y : z = CP (X2 . Proof. The Steiner inellipse is given by a2 α2 + b2 β 2 + c2 γ 2 − 2bcβγ − 2caγα − 2abαβ = 0. using (16). U1 ). Corollary 9. (20) Proof. (21) L∞ . where U1 is a point on L∞ but not on a sideline BC. bwy 2 −cvz 2 +(bv−cw)yz+ax(vz−wy) = Λ(−a(u+v+w)+as(a and by (17). we note that. Suppose X = CP (X2 . Following the steps of the proof of Theorem 6. Then the perspector X in Case 6. then the First. the perspector X traverses the circumconic (20). where = 2abc(b − c)(c − a)(a − b)(1 + bcs)(1 + cas)(1 + abs). by − cz bwy 2 − cvz 2 + (bv − cw)yz + ax(vz − wy) (by − cz)/a . It is easy to check that this point satisfies (20). CA. Λ Thus 2 u+b2 v+c2 w)). (17) x = 2 2 bwy − cvz + (bv − cw)yz + ax(vz − wy) cz − ax y = . the mapping U1 → . Kimberling Case 6. (19) z = 2 2 avx − buy + (au − bv)xy + cz(uy − vx) Theorem 8.4 is invariant of the point U. Let X = x : y : z = bu1 v1 + cu1 w1 : cv1 w1 + av1 u1 : aw1 u1 + bw1 v1 . X = CD(X6 .280 C. and X lies on the circumconic given by au21 (bv1 − cw1 )βγ + bv12 (cw1 − au1 )γα + cw12 (au1 − bv1 )αβ = 0. X690 ). X514 ). of which exactly two involve s. specifically. P ) is given by x = p(bq + cr). Proof. q = b1 + sv. y. b1 − 2c1 a1 . a1 : b1 : c1 = cos A : cos B : cos C and U = X30 = u : v : w = a1 − 2b1 c1 : b1 − 2c1 a1 : c1 − 2a1 b1 . Theorem 11. the perspectivity problem for both families. This equals 0 for s = −1/3. Note that the mapping X → X is not one-to-one. After canceling all common factors that do not contain s. Proof. Then the perspector X . and the isotomic conjugate X = x : y : z by a−2 (a1 + su)−1 : b−2 (b1 + sv)−1 : c−2 (c1 + sw)−1 . which is CP (X2 . y = q(cr + ap). for given X = x : y : z on the Steiner inellipse. Let X = CP (X2 . the perspector is then X = X900 = CD(X6 . the latter being the point in which the Euler line meets L∞ . cevian and anticevian triangles. Write p = a1 + su. P ). Suppose P is on the Euler line and P = X2 . so that the point X = CP (X2 . (X1084 . The same holds for the results of substituting in (13) and (14). v. z in (12) gives a product of several factors. U1 ) = X is invertible. X888 ). X ) are these: (X115 . . is discussed for translations in a single direction. w) = (a1 − 2b1 c1 . As X = X2 . for which a1 + su : b1 + sv : c1 + sw = X2 . the remaining coordinates for X have a common factor 3s + 1. Other examples (X. X891 ). and Theorem 8 applies. namely the direction of the Euler line. Substituting for x. X101 ). An arbitrary point X on the Euler line is given parametrically by a1 + su : b1 + sv : c1 + sw. in the case of the anticevian triangle of X. Corollary 9 is exemplified by taking X = X1086 . Translation along the Euler line In this section. z = r(ap + bq). in the case of the cevian triangle of X as given by (12)-(14). Two points on the Euler line are the circumcenter. is the point P. then the perspector. The remaining coordinates are equivalent to those given just above for X . the point U1 = u1 : v1 : w1 given by ca ab bc : : u1 : v1 : w1 = by + cz − ax cz + ax − by ax + by − cz is on L∞ . r = c1 + sw. where (u.Translated triangles perspective to a reference triangle 281 CP (X2 . as given by (17)-(19). we can and do cancel 3s + 1. (X1015 . (X2482 . If X is the isotomic conjugate of a point X on the Euler line other than X2 . 7. X690 ). is X . c1 − 2a1 b1 ). Theorem 10. we wish to translate DEF in the direction of line DU. seeking translations D E F that are perspective to ABC. If P is on the circumcircle and X = CS(X6 . let X = X619 . Canceling those. so that by Theorem 1. X13 . leaves trilinears for P. Translated rotated reference triangle Let DEF be the rotation of ABC about the circumcenter of ABC. E = − csc(C + θ) : csc θ : csc(A − θ). is the point CD(X6 . there are at most two perspective translations. P ). (p. θ2 = cos θ. P ).282 C. ). p. Then the vertices D. First. Substituting into (17)-(19) and factoring gives expressions with several common factors. q. in the case of the anticevian triangle of X as given by (17)-(19). F are given by the rows of the matrices (rcθ2 − c1 θ1 )−1 (rbθ2 + b1 θ1 )−1 θ1−1 d1 e1 f1 d2 e2 f2 = (rcθ2 + c1 θ1 )−1 θ1−1 (raθ2 − a1 θ1 )−1 . F = csc(B − θ) : − csc(A + θ) : csc θ. X14 . then the perspector X . Except for rotations of 0 and π. The perspector in the case of the anticevian triangle of X is the point X13 . including the factor 3s + 1 which corresponds to the disallowed X2 . We conclude this section with a pair of examples. d3 e3 f3 (rbθ2 − b1 θ1 )−1 (raθ2 + a1 θ1 )−1 θ1−1 . 8. y = cp + ar. .39]) is used to express the rotation DEF of ABC counterclockwise with angle 2θ as follows: D = csc θ : csc(C − θ) : − csc(B + θ). Yff’s parameterization of the circumcircle ([1. r) = ( (b − c)(bc + s) (c − a)(ca + s) (a − b)(ab + s) Then the point X = CS(X6 . the complement of the Fermat point (or 1st isogonic center). let X = X618 . P ) is given by x = br + cq. triangle DEF is not perspective to ABC. θ1 = sin θ. the complement of the 2nd isogonic center. Kimberling Substituting into (17)-(19) and factoring give expressions with several common factors. Finally. Proof. Canceling those leaves trilinears for CD(X6 . Theorem 12. In this section. Represent an arbitrary point P = p : q : r on the circumcircle parametrically by 1 1 1 . E. Let r =((a + b + c)(b + c − a)(c + a − b)(a + b − c))1/2 /(2abc). Let U = u : v : w be a point on L∞ . z = aq + bp. The perspector in this case is the point X14 . P ). Triangle DEF is a variable rotation of ABC about its circumcenter O. Translated rotation of ABC. Three conics as in Theorem 4 meet in two points. Only one of the factors involves t. the loci of as θ varies from 0 to π. Conjecture. which according to the Conjecture are a pair of antipodes on the circumcircle. 3 It would perhaps be of interest to study. The perspectivity determinant (2) is factored using a computer. Triangle D E F is thus a movable translation of DEF in the direction of DH. Independent point H determines line DH. F E' A x E z H F' y D' F O C B D Figure 3. p2 = (a + b − c) (a − b + c) (b − a + c) (a + b + c) (avw + buw + cuv) . see [6] for access. D . for fixed H. See Figure 3. The perspectors given by (22) are a pair of antipodes on the circumcircle. E . hence roots ±(θ1 /r)(−abcs)−1/2 . and it is a polynomial P (t) as in (3). (c2 + a2 − b2 )/(2ca). with coefficients p0 =4abcθ12 . cos C) =((b2 + c2 − a2 )/(2bc). b1 . (22) where s = avw + buw + cuv. p1 =4θ12 abc (au + bv + cw) = 0. . c1 ) =(cos A. 3Figure 3 can be viewed dynamically using The Geometer’s Sketchpad. Point D is movable on line DH. cos B. (a2 + b2 − c2 )/(2ab)).Translated triangles perspective to a reference triangle 283 where (a1 . 284 C. Kimberling References [1] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1–285. [2] C. Kimberling, Conics associated with a cevian nest, Forum Geom., 1 (2001) 141–150. [3] C. Kimberling, Encyclopedia of Triangle Centers, 2000-, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html. [4] C. Kimberling, Glossary, 2002-, http://faculty.evansville.edu/ck6/encyclopedia/glossary.html. [5] C. Kimberling, Points on Cubics, 2003-, http://faculty.evansville.edu/ck6/encyclopedia/Intro&Zcubics.html . [6] C. Kimberling, Translated Triangles, 2005, http://faculty.evansville.edu/ck6/encyclopedia/TransTri.html . [7] E. A. Maxwell, General Homogeneous Coordinates, Cambridge University Press, Cambridge, 1957. [8] E. Weisstein, MathWorld, http://mathworld.wolfram.com/ . Clark Kimberling: Department of Mathematics, University of Evansville, 1800 Lincoln Avenue, Evansville, Indiana 47722, USA E-mail address:
[email protected] b Forum Geometricorum Volume 6 (2006) 285–288. b b FORUM GEOM ISSN 1534-1178 On the Derivative of a Vertex Polynomial James L. Parish Abstract. A geometric characterization of the critical points of a complex polynomial f is given, in the special case where the zeros of f are the vertices of a polygon affine-equivalent to a regular polygon. 1. Steiner Polygons The relationship between the locations of the zeros of a complex polynomial f and those of its derivative has been extensively studied. The best-known theorem in this area is the Gauss-Lucas Theorem, that the zeros of f lie in the convex hull of the zeros of f . The following theorem [1, p.93], due to Linfield, is also of interest: Theorem 1. Let λj ∈ R\{0}, j = 1, . . . , k, and let zj , j = 1, . . . , k be distinct λ complex numbers. Then the zeros of the rational function R(z) := kj=1 z−zj j are the foci of the curve of class k − 1 which touches each of the k(k − 1)/2 line segments zµ , zν in a point dividing that line segment in the ratio λµ : λν . 1 , where the zj are the zeros of f , Linfield’s Theorem Since f = f · kj=1 z−z j can be used to locate the zeros of f which are not zeros of f . In this paper, we will consider the case of a polynomial whose zeros form the vertices of a polygon which is affine-equivalent to a regular polygon; the zeros of the derivative can be geometrically characterized in a manner resembling Linfield’s Theorem. First, let ζ be a primitive nth root of unity, for some n ≥ 3. Define G(ζ) to be the n-gon whose vertices are ζ0 , ζ 1 , . . . , ζ n−1 . Proposition 2. Let n ≥ 3, and let G be an n-gon with vertices v0 , . . . , vn−1 , no three of which are collinear. The following are equivalent. (1) There is an ellipse which is tangent to the edges of G at their midpoints. (2) G is affine-equivalent to G (ζ) for some primitive nth root of unity ζ. (3) There is a primitive nth root of unity ζ and complex constants g, u, v such that |u| = |v| and, for k = 0, . . . , n − 1, vk = g + uζ k + vζ −k . Proof. 1)=⇒2): Applying an affine transformation if necessary, we may assume that the ellipse is a circle centered at 0 and that v0 = 1. Let m0 be the midpoint of the edge v0 v1 . v0 v1 is then perpendicular to 0m0 , and v0 , v1 are equidistant from m0 ; it follows that the right triangles 0m0 v0 and 0m0 v1 are congruent, and in Publication Date: November 13, 2006. Communicating Editor: Paul Yiu. 286 J. L. Parish particular that v1 also lies on the unit circle. Now let m1 be the midpoint of v1 v2 ; since m0 and m1 are equidistant from 0 and the triangles 0m0 v1 , 0m1 v1 are right, they are congruent, and m0 , m1 are equidistant from v1 . It follows that the edges v0 v1 and v1 v2 have the same length. Furthermore, the triangles 0v0 v1 and 0v1 v2 are congruent, whence v2 = v12 . Similarly we obtain vk = v1k for all k, and in particular that v1n = v0 = 1. ζ = v1 is a primitive nth root of unity since none of v0 , . . . , vn−1 coincide, and G = G(ζ). 2)=⇒1): G(ζ) has an ellipse – indeed, a circle – tangent to its edges at their midpoints; an affine transformation preserves this. 2)⇐⇒3): Any real-linear transformation of C can be put in the form z → uz + vz for some choice of u, v, and conversely; the transformation is invertible iff |u| = |v|. We will refer to an n-gon satisfying these conditions as a Steiner n-gon; when needed, we will say it has root ζ. The ellipse is its Steiner inellipse. (This is a generalization of the case n = 3; every triangle is a Steiner triangle.) The parameters g, u, v are its Fourier coordinates. Note that a Steiner n-gon is regular iff either u or v vanishes. 2. The Foci of the Steiner Inellipse Now, let Sζ be the set of Steiner n-gons with root ζ for which the constant g, above, is 0. We may use the Fourier coordinates u, v to identify it with an open subset of C2 . Let Φ be the map taking the n-gon with vertices v0 , v1 , . . . , vn−1 to the n-gon with vertices v1 , . . . , vn−1 , v0 . If f is a complex-valued function whose domain is a subset of Sζ which is closed under Φ, write ϕf for f ◦ Φ. Note that ϕu = ζu and ϕv = ζ −1 v; this will prove useful. Note also that special points associated with n-gons may be identified with complex-valued functions on appropriate subsets of Sζ . We define several useful fields associated with Sζ . First, let F = C(u, v, u, v), where u, v are as in 3) of the above proposition. ϕ is an automorphism of F . Let K = C(x, y, x, y) be an extension field of F satisfying x2 = u, y 2 = v, x2 = u, y 2 = v. Let θ be a fixed square root of ζ; we extend ϕ to K by setting ϕx = θx, ϕy = θ −1 y, ϕx = θ −1 x, ϕy = θy. Let K0 be the fixed field of ϕ and K1 the fixed field of ϕn . Elements of F may be regarded as complex-valued functions defined on dense open subsets of Sζ . Functions corresponding to elements of K may only be defined locally; however, given G ∈ Sζ such that uv = 0 and f ∈ K1 defined at G, one may choose a small neighborhood U0 of G which is disjoint from Φk (U0 ), k = 1, . . . , n − 1 and on which neither u nor v vanish; f may then be k defined on U = n−1 k=0 Φ (U0 ). For the remainder of this section, G is a fixed Steiner n-gon with root ζ. The vertices of G are v0 , . . . , vn−1 . We have the following. Proposition 3. The foci of the Steiner inellipse of G are located at f± = g ± (θ + θ −1 )xy. On the derivative of a vertex polynomial 287 Proof. Translating if necessary, we may assume that g = 0, i.e., G ∈ Sζ . Note first that f± ∈ K0 . (This is to be expected, since the Steiner inellipse and its foci do not depend on the choice of initial vertex.) For k = 0, . . . , n − 1, let mk = (vk + vk+1 )/2, the midpoint of the edge vk vk+1 . Let d± be the distance from f± to m0 ; we will first show that d+ + d− is invariant under ϕ. (This will imply that the sum of the distances from f± to mk is the same for all k.) Now, m0 = (v0 +v1 )/2 = ((1+ζ)u+(1+ζ −1 )v)/2 = (θ+θ −1 )(θx2 +θ −1 y 2 )/2. Thus, m0 −f+ = (θ+θ −1 )(θx2 −2xy +θ −1y 2 )/2 = (ζ +1)(x−θ −1 y)2 /2. Hence d+ = |m0 − f+ | = |ζ + 1|(x − θ −1 y)(x − θy)/2. Similarly, d− = |ζ + 1|(x + θ −1 y)(x + θy)/2, and so d+ + d− = |ζ + 1|(xx + yy), which is invariant under ϕ as claimed. This shows that there is an ellipse with foci f± passing through the midpoints of the edges of G. If n ≥ 5, this is already enough to show that this ellipse is the Steiner inellipse; however, for n = 3, 4 it remains to show that this ellipse is tangent to the sides, or, equivalently, that the side vk vk+1 is the external bisector of the angle ∠f+ mk f− . It suffices to show that Ak = (mk − vk )(mk − vk+1 ) is a positive multiple of Bk = (mk − f+ )(mk − f− ). Now A0 = −(ζ − 1)2 (u − ζ −1 v)2 /4, and B0 = (ζ + 1)2 (x − θ −1 y)2 (x + θ −1 y)2 /4 = (ζ + 1)2 (u − ζ −1 v)2 /4; thus, A0 /B0 = −(ζ − 1)2 /(ζ + 1)2 = −(θ − θ −1 )2 /(θ + θ −1 )2 , which is evidently positive. This quantity is invariant under ϕ; hence Ak /Bk is also positive for all k. Corollary 4. The Steiner inellipse of G is a circle iff G is similar to G(ζ). Proof. f+ = f− iff xy = 0, i.e., iff one of u and v is zero. (Note that θ + θ−1 = 0.) n−1 Define the vertex polynomial fG (z) of G to be k=0 (z − vk ). We have the following. Proposition 5. The foci of the Steiner inellipse of G are critical points of fG . n−1 −1 Proof. Again, we may assume G ∈ Sζ . Since fG /fG = k=0 (z − vk ) , it suffices to show that at f± . Now f+ is invariant under ϕ, and this sum vanishes n−1 k −1 = −1 −1 = (f − v ) vk = ϕk v0 ; hence n−1 k k=0 + k=0 ϕ (f+ − v0 ) . (f+ − v0 ) 2 2 −θ/((θy − x)(y − θx)) Now let g = θ /((θ − 1)x(θy − x)). Note that g ∈ K1 ; that is, ϕn g = g. A straightforward calculation shows that (f+ − v0 )−1 = g − ϕg; n−1 k k −1 = k+1 g) = g − ϕn g = 0, as therefore, n−1 k=0 ϕ (f+ − v0 ) k=0 (ϕ g − ϕ desired. The proof that f− is a critical point of fG is similar. 3. Holomorphs Again, we let G be a Steiner n-gon with root ζ and vertices v0 , . . . , vn−1 . For any integer m, we set vm = vl where l = 0, . . . , n − 1 is congruent to m mod n. The following lemma is trivial. Lemma 6. Let k = 1, . . . , n/2. Then: k (1) If k is relatively prime to n, let Gk be the n-gon with vertices v0k , . . . , vn−1 k k k given by vj = vjk . Then G is a Steiner n-gon with root ζ , and its Fourier coordinates are g, u, v. 288 J. L. Parish (2) If d = gcd(k, n) is greater than 1 and less than n/2, set m = n/d. Then, k,l for l = 0, . . . , d − 1, let Gk,l be the m-gon with vertices v0k,l , . . . , vm−1 given by vjk,l = vkj+l . Then, for each l, Gk,l is a Steiner m-gon with root ζ k , and the Fourier coordinates of Gk,l are g, ζ l u, ζ −l v. The Gk,l all have the same Steiner inellipse. (3) If k = n/2, the line segments vj vj+k all have midpoint g. In the three given cases, we will say k-holomorph of G to refer to Gk , the union of the m-gons Gk,l , or the union of the line segments vj vj+k . We extend the definition of Steiner inellipse to the k-holomorphs in Cases 2 and 3, meaning the common Steiner inellipse of the Gk,l or the point g, respectively. The propositions of Section II clearly extend to Case 2; since the foci are critical points of the vertex polynomials of each of the Gk,l , they are also critical points of their product. In Case 3, taking g as a degenerate ellipse – indeed, circle – with focus at g, the propositions likewise extend; in this case, θ = ±i, so θ + θ−1 = 0, and the sole critical point of (z − vj )(z − vj+k ) is (vj + vj+k )/2 = g. In Cases 1 and 2, it should be noted that the Steiner inellipse is a circle iff the Steiner inellipse of G itself is a circle – i.e., G is similar to G(ζ). It should also be noted that the vertex polynomials of the holomorphs of G are equal to fG itself; hence they have the same critical points. Suppose that G is not similar to G(ζ). If n is odd, G has (n − 1)/2 holomorphs, each with a noncircular Steiner inellipse and hence two distinct Steiner foci; these account for the n − 1 critical points of fG . If n is even, G has (n − 2)/2 holomorphs in Cases 1 and 2, each with two distinct Steiner foci, and in addition the Case 3 holomorph, providing one more Steiner focus; again, these account for n − 1 critical points of fG . On the other hand, if G is similar to G(ζ), then fG = (z − g)n − r n for some real r; the Steiner foci of the holomorphs of G collapse together, and fG has an (n − 1)-fold critical point at g. We have proven the following. Theorem 7. If G is a Steiner n-gon, the critical points of fG are the foci of the Steiner inellipses of the holomorphs of G, counted with multiplicities if G is regular. They are collinear, lying at the points g + (2 cos kπ/n)xy, as k ranges from 0 to n − 1. (For the last statement, note that cos(n − k)π/n = − cos kπ/n.) Reference [1] Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, Clarendon Press, Oxford, 2002. James L. Parish:Department of Mathematics and Statistics, Southern Illinois University, Edwardsville, Edwardsville, IL USA 62026 E-mail address:
[email protected] b Forum Geometricorum Volume 6 (2006) 289–295. b b FORUM GEOM ISSN 1534-1178 On Two Remarkable Lines Related to a Quadrilateral Alexei Myakishev Abstract. We study the Euler line of an arbitrary quadrilateral and the Nagel line of a circumscriptible quadrilateral. 1. Introduction Among the various lines related to a triangle the most popular are Euler and Nagel lines. Recall that the Euler line contains the orthocenter H, the centroid G, the circumcenter O and the nine-point center E, so that HE : EG : GO = 3 : 1 : 2. On the other hand, the Nagel line contains the Nagel point N , the centroid M , the incenter I and Spieker point S (which is the centroid of the perimeter of the triangle) so that N S : SG : GI = 3 : 1 : 2. The aim of this paper is to find some analogies of these lines for quadrilaterals. It is well known that in a triangle, the following two notions of centroids coincide: (i) the barycenter of the system of unit masses at the vertices, (ii) the center of mass of the boundary and interior of the triangle. D C Gb Ga G Gc Gd A B Figure 1. But for quadrilaterals these are not necessarily the same. We shall show in this note, that to get some fruitful analogies for quadrilateral it is useful to consider the centroid G of quadrilateral as a whole figure. For a quadrilateral ABCD, this centroid G can be determined as follows. Let Ga , Gb , Gc , Gd be the centroids of triangles BCD, ACD, ABD, ABC respectively. The centroid G is the intersection of the lines Ga Gc and Gb Gd : G = Ga Gc ∩ Gb Gd . See Figure 1. Publication Date: November 20, 2006. Communicating Editor: Paul Yiu. Hb for triangle ACD. Oa Ob Oc Od . if the quadrilateral is cyclic. Figure 2 shows the three associated quadrilaterals Ga Gb Gc Gd . Oc . then O is the center of its circumcircle. denote by Oa and Ha the circumcenter and the orthocenter respectively of triangle BCD. Therefore. Let O =Oa Oc ∩ Ob Od . Ob . .290 A. the quasicircumcenter O is the intersection of perpendicular bisectors of the diagonals of ABCD. H =Ha Hc ∩ Hb Hd . and similarly. and Od . Hd for triangle ABC. Clearly. Myakishev 2. Hc for triangle ABD. Ha D Hb C Gb Ga Ob Oc O H G Od Oa Gc Gd Hc Hd A B Figure 2 We shall call O the quasicircumcenter and H the quasiorthocenter of the quadrilateral ABCD. The Euler line of a quadrilateral Given a quadrilateral ABCD. and Ha Hb Hc Hd . B. OH : HG = 3 : −2. fO (D) = Od . Theorem 1. and the idea of the proof was due to Franc¸ois Rideau [3]. for Q = A. and similarly for Hb and Hc . (i) Note that fG (D) =fG (xA + yB + zC) =xGa + yGb + zGc 1 = (x(B + C + D) + y(A + C + D) + z(A + B + D)) 3 1 = ((y + z)A + (z + x)B + (x + y)C + (x + y + z)D) 3 1 = ((y + z)A + (z + x)B + (x + y)C + (xA + yB + zC)) 3 1 = (x + y + z)(A + B + C) 3 =Gd . (ii) It is obvious that triangles ABC and Oa Ob Oc are orthologic with centers D and Od . and the quasicircumcenter O are collinear. Consider three affine maps fG . fO and fH transforming the triangle ABC onto triangle Ga Gb Gc . the centroid G.On two remarkable lines related to a quadrilateral 291 The following theorem was discovered by Jaroslav Ganin. From Theorem 1 of [1]. D C Ob Oc Od Oa A B Figure 3 (iii) Since Ha divides Oa Ga in the ratio Oa Ha : Ha Ga = 3 : −2. Proof. the point fH (Q) divides the segment fO (Q)fG (Q) into the ratio 3 : −2. Furthermore. C. and Ha Hb Hc respectively. See Figure 3. (see [2]). In any arbitrary quadrilateral the quasiorthocenter H. write D = xA + yB + zC with x + y + z = 1. Oa Ob Oc . In the affine plane. It follows that for every point Q in the plane . We shall refer to the point N := N1 N3 ∩ N2 N4 as the Nagel point of the circumscriptible quadrilateral. From this we conclude that H divides OG in the ratio 3 : −2. Od Gd . ABD. This is clearly Hd . T4 be the points of tangency of the incircle with the sides AB. This line contains also the quasininepoint center E defined as follows. Ed be the nine-point centers of the triangles BCD. Applying the affine maps we have fG (Q) =Ga Gc ∩ Gb Gd = G. ACD. Let Ea . Similarly define N2 . T2 . Let T1 . CD and DA respectively. (iv) Let Q = AC ∩ BD. Let ABCD be a circumscriptible quadrilateral with incenter I. fH (D) divides fO (D)fG (D). Ec . 3. and the quasiorthocenter. Theorem 1 enables one to define the Euler line of a quadrilateral ABCD as the line containing the centroid. in the ratio 3 : −2. namely. The Nagel line of a circumscriptible quadrilateral A quadrilateral is circumscriptible if it has an incircle. fO (Q) =Oa Oc ∩ Ob Od = O. fH (Q) divides fO (Q)fG (Q) in the same ratio. Eb . the quasicircumcenter. In particular. T3 . Note that both lines divide the perimeter of the quadrilateral into two equal parts. C . We have shown that fH (D) = Hd .292 A. We define the quasininepoint center to be the point E = Ea Ec ∩ Eb Ed . The following theorem can be proved in a way similar to Theorem 1 above. E is the midpoint of OH. N3 . Theorem 2. Let N1 be the isotomic conjugate of T1 with respect to the segment AB. ABC respectively. N4 in the same way. B T1 N1 A T2 T4 I N2 N N4 D T3 N3 Figure 4. Myakishev of ABC. fH (Q) =Ha Hc ∩ Hb Hd = H. BC. (p + r)D). 0 · D). BT2 = BT1 = q. kz. we have established the first of the following three lemmas. r · C. I = ((−a+b+p+t)A. In Figure 4. E a b q p A B q T1 p T2 T4 I r t D t T3 r C Figure 5. it is also on the line N2 N4 since (iii) N2 = (0 · A. I = ((q + t)A. q·B. z. r · C. and DT4 = DT3 = t. CT3 = CT2 = r. Since b + p = a + q. 0 · C. and w at D. (If not. (−b+a+q +r)B. Therefore. we see that (i) N1 = (p · A. t·D) is on the line N1 N3 . −a · C. ky. N = (p · A. (−a+b+p+t)C. z at C. Clearly. y · B. the result follows. p = q = r = s. which intersect at E. Proof. w · D) to mean that P is the barycenter of a system of masses x at A. q · B. so that the system (p + q + r + t)E is equivalent to the system ((a + q + r)B. 0 · B. I = ((p + q)E. assume that AT1 = AT4 = p. (i) As the incenter of triangle EDC. t · D). then ABCD is a rhombus. and t at D. b+p+t EC = a+q+r and ED Note that EB a EA = b . Suppose the circumscriptible quadrilateral ABCD has a pair of non-parallel sides AD and BC. (b + p + t)A. 0 · C. r at C. t · D). kw for nonzero k without changing the point P . (p + r)B. w can be replaced by kx. z · C. 0 · D). the result is trivial). r·C. Then by putting masses p at A. Let a = EB and b = EA. and (iv) N4 = (p · A. and I = G.On two remarkable lines related to a quadrilateral 293 In Theorem 6 below we shall show that N lies on the line joining I and G. . q at B. q · B. Similarly. Therefore. r·C. y. Lemma 3. I = ((t + r)E. In what follows we shall write P = (x · A. (q + t)C. (−b+a+q +r)D). 0·B. t·D). x. −a · A. Lemma 4. (b + p + t)C). −b · D). so that the barycenter N = (p·A. (a + q + r)D. (ii) N3 = (0·A. (ii) As an excenter of triangle ABE. −b · B). y at B. q · B. 1 · C. It follows that G = (r · Ga .156–157]. Actually. 1 · D) and ratio − 13 . P = 1q · B. 1r · C . See Figure 7. the lines T1 T3 and T2 T4 also pass through P . q · Gd ) = ((q + t)A. t · D). The following theorem follows easily from Lemmas 3. by which we mean the center of mass S of the perimeter of the quadrilateral. (p + q + r)B. centroid G and incenter I are collinear. G Gc G = CP = r and Gd G = q BP DP = t . To conclude the proof. 5. G = ((p + q + t)A. with homothetic center p Gb G AP aG M = (1 · A. · D . without assuming an incircle. · C. Hence. Myakishev Lemma 5. Note that CP = pr q BP and DP = t . (r + p)B. it is enough to add up the corresponding masses. (q + t)C. Denote the point of intersection of the diagonals by P . The quadrilateral Ga Gb Gc Gd is homothetic to ABCD. B p q T1 q A p T2 P T4 I r t D t T3 r C Figure 6. (p + r + t)D). P = p q r t Consequently. . This line also contains the Spieker point of the quadrilateral. r · C. 1 1 1 1 · A. see [4. the Nagel point N . p · Gc ) = (p · A. (q + r + t)C. N G : GI = 2 : 1. pp. 4. For another proof. For a circumscriptible quadrilateral.294 A. Furthermore. q · B. Thus. 1t · D and also P = p1 · A. AP Proof. · B. according to one corollary of Brianchon’s theorem. Theorem 6 enables us to define the Nagel line of a circumscriptible quadrilateral. Theorem 6. (r + p)D) and G = (t · Gb . 1 · B. [4] P. [2] A. Hyacinthos message 12400. (p + r + 2t)D). With reference to Figure 6.geom@mtu-net.. Proof. (p + r)B. March 16. Alexei Myakishev: Smolnaia 61-2. Thus. Hyacinthos message 12402. [3] F. 2rC. we have N =(2pA. Theorem 7. S = ((2p + q + t)A. Myakishev. 2006. Florida Atlantic University Lecture Notes. 138. 2tD). 4 (2004) 135– 141. References [1] E. Yiu. 1998. Splitting into two systems of equal total masses. ((p + r)D). 2qB. Russia. From this the result is clear. each side of the circumscriptible quadrilateral is equivalent to a mass equal to its length located at each of its two vertices. (q + t)C. Euclidean Geometry. the Spieker point is the midpoint of the incenter and the Nagel point.ru . 2006. Forum Geom. Moscow. Dergiades.On two remarkable lines related to a quadrilateral 295 B T1 N1 A T2 T4 Gd Gc I G Gb N2 N N4 D Ga T3 N3 C Figure 7. A theorem on orthology centers. March 16. (p + 2q + r)B. For a circumscriptible quadrilateral. Rideau. 125445 E-mail address: alex. I =((q + t)A. Danneels and N. (q + 2r + t)C. . AB be a. b b FORUM GEOM ISSN 1534-1178 Intersecting Circles and their Inner Tangent Circle Max M. then A − B = c and A. A. CB . B. C be represented by complex numbers of moduli R. We seek the radius of the circle C. CA intersects CB at an angle θ. (R + α)(R + β) Publication Date: November 27. We set up a coordinate system with the origin at the center of C. If the three intersecting cirlces were just touching instead. b.b Forum Geometricorum Volume 6 (2006) 297–300. R B. with 0 θ. tangent externally to each of the given circles. As a step toward this goal of a single equation. With these labels. γ respectively. We derive the general equation for the radius of the inner tangent circle that is associated with three pairwise intersecting circles. the centers of the circles CA . The points of tangency of the inner tangent circle form the vertices of an inscribed triangle. And CC intersects CA at an angle φ. Letting the lengths of the sides BC. the radius of C. Tran Abstract. consider three circles CA . See [1]. CB intersects CC at an angle ρ. CC are respectively R+α R A. the triangle ABC and the inscribed triangle is one and the same. β. The circles CA and CB intersect at angle θ if and only if R + α R + β 2 2 R A − R B = α + β + 2αβ cos θ. ρ. CA. See Figure 1. 2006. See Figure 1. It seems to the author that there should be one equation that gives the radius of the inner tangent circle inscribed in a triangular region bounded by either straight lines or circular arcs. Communicating Editor: Paul Yiu. φ π. CB and CC with radii α. the inner tangent circle would be the inner Soddy circle. the above can be shown to be equivalent to c2 = 4R2 αβ cos2 θ2 . C and C. 2 (1) Corresponding relations hold for the pairs B. . c respectively. B = R2 − c2 . By an application of (1) and the use of a half angle formula. R+γ R C. Let the points of tangency A. We then look at three special cases of the equation. With the above coordiR+β nate system. CC intersect each other at the given angles if and only if a2 = 4R2 βγ cos2 2ρ . (R + α)(R + β)(R + γ) (3) . θ. β. (R + α)(R + β) These equations are then used to solve for R in terms of α. CB . A.298 M. Tran φ CC C A θ ρ B CB CA Figure 1 Thus the three circles CA . b = (R + α)(R + γ) 2 c2 = (2) 4R2 αβ cos2 θ2 . we multiply the equations in (2) and take square root to obtain abc = 8αβγR3 cos θ2 cos φ2 cos ρ2 . (R + β)(R + γ) 4R2 αγ cos2 φ2 . In the first step of this process. φ and ρ. γ. This process involves substituting the value of a2 . c2 into Heron’s formula. . dividing by 2 . of the inscribed triangle ABC is given by = abc . . αβ . Now. and performing a lengthy algebraic manipulation to yield the equation: θ ρ φ φ ρ θ 1 4 cos2 cos2 cos2 + cos4 + cos4 + cos4 2 R 2 2 2 2 2 2 θ φ θ ρ φ 2 2 2 2 2 2 ρ − 2 cos cos − 2 cos cos −2 cos cos 2 2 2 2 2 2 2 ρ 2 cos θ φ ρ 1 2 (cos2 + cos2 − cos2 ) − R α 2 2 2 0= 2 cos2 β φ 2 θ ρ φ + cos2 − cos2 ) 2 2 2 θ 2 cos2 2 2 φ 2 ρ 2 θ + (cos + cos − cos ) γ 2 2 2 + + (cos2 cos4 ρ2 cos4 φ2 cos4 + + α2 β2 γ2 − 2 cos2 θ 2 cos2 βγ φ 2 − (7) θ 2 2 cos2 θ 2 cos2 αγ ρ 2 − 2 cos2 φ 2 cos2 ρ2 .Intersecting circles and their inner tangent circle 299 Using (3) and (2) we obtain. R+β 2Rb cos θ2 cos ρ2 (4) ab cos θ2 γ = . Using the above equation together with equations (6) will enable us to get an equation for R in terms of the parameters of the intersecting circles. 4R (5) Consequently. Heron’s formula for the triangle ABC can be written in the form 162 = 2a2 b2 + 2b2 c2 + 2a2 c2 − a4 − b4 − c4 . bc cos ρ2 α = . R+γ 2Rc cos ρ2 cos φ2 The area. (6) . b2 . equations (4) and (5) lead to a2 = b2 = c2 = (R + α) cos ρ2 α cos θ2 cos φ2 (R + β) cos φ2 β cos θ2 cos ρ2 (R + γ) cos θ2 γ cos ρ2 cos φ2 . R+α 2Ra cos θ2 cos φ2 ac cos φ2 β = . Introduction to Geometry.edu . When the three circles CA . it is rather unwieldy. we get a cone and equation (7) becomes 2 φ 2 2 φ 2 cos cos 2 φ φ 1 2 − 2 − cos2 + . Forum Geom. φ R α γ αγ cos2 2 When the circles CA and CC are lines that intersect at an angle φ > 0 and are both tangent to the circle CB .300 M. we have β = ∞ and θ = ρ = 0. Unfortunately. New york 1969. Max M. USA E-mail address: mtran@kingsborough. ρ and φ equals zero. CB and CC . Reyes. thus giving the equation: 1 1 1 1 2 2 2 1 1 1 2 0= 2 − + + + 2+ 2+ 2− − − . becomes lines. [2] W. 0 = 2 cos4 R 2 γR 2 γ After solving for 1/R. The Lucas Circles and the Descartes Formula. References [1] H. Tran: Mathematics and Computer Science Department. R R α β γ α β γ αβ βγ αγ Solving for R1 gives the standard Descartes formula for the Inner Soddy circle. Second Edition. Coxeter. Tran Although the equation can be formal solved in general. R γ 1 − sin φ 2 the same as obtained from working with the cone directly. equation (7) no longer gives any useful result when all three circles. gives the equation 1 1 1 1 1 = + +2 . Brooklyn. θ. M. − + + − 0 = 2 cos R 2 R α γ α γ Solving for 1/R. CB and CC are mutually tangent. 3 (2003) 95–100. John Wiley and Sons. using some trigonometric identities and the fact that R1 > γ1 . and equation (7) becomes 2 cos2 φ2 1 1 1 1 2 1 4 φ . we get the equation φ 2 2 1 + sin 2 1 = . A. The inner tangent circle in this case is just the inscribed circle in the triangle. Let us consider some special cases. See [2]. S. and using the fact that R1 > α1 and R1 > γ1 .. Kingsborough Community College. 2001 Oriental Boulevard. CA . When CC is a line tangent to CA and CB . NY 11235-2398. we generalize these situations to include slanted positions and provide integer solutions to these problems.b Forum Geometricorum Volume 6 (2006) 301–310. E. The other walking down the hill goes by land to the same town. We solve the following Brahmagupta problems from [5] in the context of rational cosines triangles. In [4] these problems have been given Pythagorean solutions. One of them being a wizard travels through the air. Volume 1. Also. 2006. One of them involves a broken tree and the other a mountain journey. The author thanks the referee for his suggestions. D. Introduction School textbooks on geometry and trigonometry contain problems about trees. poles. In this paper. b b FORUM GEOM ISSN 1534-1178 Two Brahmagupta Problems K. S. Sastry Abstract. However. it is every day experience that such objects need not be vertical. Problem 2. R. I desire to know the distance of the town from the hill and how high the wizard rose. Smith reproduces two problems from Brahmagupta’s work Kutakh¯adyaka (algebra) in his History of Mathematics. Communicating Editor: Paul Yiu. Springing from the summit of the mountain he ascends to a certain elevation and proceeds by an oblique descent diagonally to neighboring town. We omit the numerical data given in Problem 1 to extend it to an indeterminate one like the second so that an infinity of integer solutions can be found. trees grow not only vertically (and tall offering a magestic look) but also assume slanted positions (thereby offering an elegant look). The assumption is that such objects are vertical. Normally such objects are represented by vertical line segments. On the top of a certain hill live two ascetics. 1. buildings. Publication Date: December 4. In this paper we regard the angles formed in such situations as having rational cosines. Their journeys are equal. Problem 1. hills etc. A bamboo 18 cubits high was broken by the wind. a distant planar view of a mountain is more like a scalene triangle than a right one. for example the leaning tower of Pisa. . However. Tell the lengths of the segments of the bamboo. to be solved using the Pythagorean theorem or trigonometric ratios. Buildings too need not be vertical in structure. Its tip touched the ground 6 cubits from the root. λ is negative. m u > . then the sides are rational in proportion. (†) v n . we deal with triangles of integer sides. c) = (m(u2 + v 2 ) − 2nuv. by after multiplication by the lcm of the denominators. v) = 1. If both cos θ and sin θ are rational. Let ABC be a member triangle in which ∠BAC = θ. A θ c b φ B C a Figure 1 Applying the law of cosines to triangle ABC we have a2 = b2 + c2 − 2bcλ. When θ is obtuse. If the angles of a triangle are rational cosine (respectively Heron) angles. Sastry 2. S.0< We replace λ by a rational number m n triangles in the θ = arccos m family: n m < 1. Given a rational cosine (respectively Heron) angle θ. Integer triangle family containing a given rational cosine angle θ. 2u(nu − mv). 2.R. in effect. u v for a. and obtain a parametrization of (a. and we give the following description. Let cos θ = λ be a rational number. c in proportional values: 1> u2 c b a = 2 = . with gcd(u.302 K. Our discussion requires that 0 < θ < π2 so we must have 0 < λ < 1. and they can be rendered integers. b. Our discussion depends on such families of triangles. 2 − 2λuv + v 2(λu − v) u − v2 n . Thus. m(u2 − v 2 )). Background material An angle θ is a rational cosine angle if cos θ is rational. b. it is possible to determine the infinite family of integer triangles (respectively Heron triangles) in which each member triangle contains θ. By the triangle inequality c − a < b so that 2λc − b v c−a = = . We then solve the resulting simultaneous equations u v c + a = (2λc − b) c − a = a. then θ is called a Heron angle. or (c − a)(c + a) = b(2λc − b). Let ∠ABC = φ as shown in Figure 1. b c+a u say.1. 5) that is Pythagorean. 2u(3u − 5v). (i) It has an integer length AB = c. m = 2: φ = arccos (a.. (1) It is common practice to list primitive solutions except under special circumstances.Two Brahmagupta problems 303 It is routine to check that mc − nb . Restatement. u = 4.e. We may put u = 3. u2 − v 2 ). i. Throughout this discussion an integer tree is one with the following properties. u > 2v. c) = (61. Furthermore. v = 1 to obtain the specific Heron triangle (a. b. BD + DC > BC. 3. AB > BC.e. cos θ = pp2 −q +q 2 for integers p. c) = (5u2 − 6uv + 5v 2 . gcd(u. See Figure 2. c) = (4..1. i. when θ (and therefore . v) = 1. u(u − 2v). we have cos θ = 35 . A θ d b E D e d φ B a C Figure 2 We note from triangle BDC. 3. (iii) When the wind breaks it at a point D the broken part AD = d and the unbroken part BD = e both have integer lengths. 56. Now with n = 3. ma n independently of the parameters u. (2) This has area = 12 bc sin θ = 25 bc. m = 5. When AB = BC the entire tree falls to the ground. Here are two particular integer triangle families. (iv) The top A of the tree touches the ground at C at an integer length BC = a. 75) with area = 1680. 2 2 (2) When θ is a Heron angle. b. (†) describes a Heron triangle family. gcd(u. b. c) = (u2 − uv + v 2 . (1) The π3 integer family is given by (†) with n = 1. 5(u2 − v 2 )). (†) yields (a. Generalization of the first problem 3. v = 1 yields the non-Pythagorean triangle (a. (ii) It makes a rational cosine angle φ with the horizontal. q = 1. v of the family described and that cos A = m in (†) above. In (1) we have removed gcd(a. q) = 1. with p = 2. c) = 2. On the other hand. b. b. For example. All the triangles in the configuration so formed have integer sidelengths. q with gcd(p. v) = 1. AD = 2 cos it follows that > m. Sastry φ) is a Heron angle. these elements uniquely determine triangle ABC. (v) φ = the inclination of the tree with the ground = arccos m 2 . 2 2 −m ) . . Suppose AB = AC = and BC = m to begin with. If heavy winds break an integer tree AB at D so that the resulting configuration is an isosceles triangle with AB = AC. and m = 2(p2 − q 2 ) for ( 2 + 1)q > p > q. Then the breaking point D on AB can be located as the intersection of ED. a = 6 and φ is implicitly given (or assumed) to be π2 . In other words.R. i. From Figure 3. In the original Problem 1. the perpendicular bisector of AC. 2 2 m 3 . Before dealing with the general solution of Problem 1. √ In particular. an integer cube. then AB becomes a Heron tree broken by the wind. we consider some interesting examples. to obtain integer values we multiply each by BD = − 2 2 −m2 = ( 2 2 −m2 2 2 2 − m . In the notation of Figure 2. if = p2 + q 2 .2. Moreover.2. c = 18. (iii) e = the unbroken part = (2 − m2 ). S.e. gives us an option to use either φ or θ as the rational cosine angle to determine triangle ABC and hence the various integer lengths a.. These yield 3 . A θ b d E D e d φ B a C Figure 3 Proof. 3. then the length of the broken part is the cube of an integer. 3. We achieve this goal with the help of (†). . AB is a Heron tree and Figure 2 represents a Heron triangle configuration. Examples.1. cos θ = 2 2 −m 2 θ = 2 2 −m2 . . the solution is given by (i) c = the length of the tree = (22 − m2 ).304 K. (iv) a = the distance between the foot and top of the tree = m(22 − m2 ). . cos φ = 2 . the determination of the configuration of Figure 2. the present restatement of Problem 1. (ii) d = the broken part = 3 . e. b. d =(p2 + q 2 )3 . triangle ADC is equilateral and . The angle of inclination of the tree with the horizontal is 5 . φ = arccos 13 We leave the details of the following two examples to the reader as an exercise. 4. For simplicity we first consider a special case of (†) in which θ = π3 .Two Brahmagupta problems 305 c = b =2(p2 + q 2 )(2pq + p2 − q 2 )(2pq − p2 + q 2 ).2. and e = 897 to come down at a = 2380. The solution in this case is elegant.2. determine parametric expressions for the various elements of the configuration formed. q = 2. If the wind breaks an integer tree AB at D in such a way that AC = BC. 3. Then we simply state the general solution leaving the details to the reader. Since ∠DAC = d = b = DC. we have π 3. the general solution of Problem 1 involves the use of integral triangles given by (†). For a numerical example. a =4(p2 − q 2 )(2pq + p2 − q 2 )(2pq − p2 + q 2 ). we put p = 3. The broken part DA comes down so that the top A of the tree touches the ground at C. A π 3 d b E D c e 2 π 3 d φ B C a Figure 4 We refer to Figure 4. General solution of Problem 1 Ideally. From (1). then both the lengths of the tree and the broken part are perfect squares. If the wind breaks an integer tree AB at D in such a way that AD = DC = BC. A particular case of Problem 1. 3.1. 4. If ∠DAC = π3 .3. This gives a Heron tree of length 3094 broken by the wind into d = 2197 = 133 (an integer cube).2. An integral tree AB is broken by the wind at D. then the common length is a perfect square. e =(p2 + q 2 )(−p2 + 3q 2 )(3p2 − q 2 ). (ii) broken part d = mu(nu − mv). The solution in (3) yields the solution of the Heron tree problem broken 2 2 by the wind when θ is a Heron angle. If the angles of this integral triangle are Heron angles. Let cos θ = m trees have (i) length c = mn(u2 − v 2 ). u > 2v. we put u = 4. q = 1. (3) ma Remark. u(u − 2v). It is an indeterminate one in its original form. Then we find that c = 225.2.. ∠BDC = 23 π.e. (iii) unbroken part e = mv(mu − nv). d = b =u(u − 2v). v = 2. cos φ = 5. The corresponding integral 4. we take u = 5. The second problem Brahmagupta’s second problem does not need any restatement. Here is a numerical example. 2a 2(u2 − uv + v 2 ) For a specific numerical example. inclined at an angle φ = arccos 37 38 . then the plain view becomes a . e =c − d = v(2u − v). Then cos θ = 35 .. Suppose p = 2. c =u2 − v 2 .e. To break a specific Heron tree of this family. Hence the family of triangles (a. 16). broken into d = b = 5. i. d = 140. a Heron angle. No tree in the π 3 family can be Heron because sin π3 = √ 3 2 is irrational. Sastry a =u2 − uv + v 2 . For example. when cos θ = pp2 −q +q 2 . with u = 5. Note also that in Figure 4. e = 16 and a = 19. n . we have (a. a = 183. b. 5. u2 + 2uv − 2v 2 2c − b = arccos . n = 3. i. m = 5.306 K. (iv) distance between the foot and the top of the tree on the ground a = n(m(u2 + v 2 ) − 2nuv). and (v) the angle of inclination with the ground φ where (m2 − 2n2 )u2 + 2mnuv − m2 v 2 . and the angle of inclination φ = arccos 207 305 . General solution of Problem 1. S. and obtain a tree of length c = 21. φ = arccos Remark. v(2u − v)) contains the angle 23 π in each member. b. v = 2. c) = (19. v = 1. e) = (u2 − uv + v 2 . An integral mountain is one whose planar view is an integral triangle. cos A = − 12 and ∠A = 23 π. e = 85.R. Now. m = p + q in (†) to find the answers. and then reaches the town C2 . The hypothesis of the problem is B1 B2 + B2 C2 = B1 C1 + C1 C2 . it creates an amusing situation as we shall soon see – in a sense there is more wizardry! B2 c2 − c1 B1 a2 c1 a1 b2 − b1 b1 A1 = A2 C1 C2 Figure 5 Figure 5 shows an integral mountain A1 B1 C1 . i. (ii) ∠B1 A1 C1 = ∠B2 A2 C2 .. We may now put 2 2 2 2 2 2 cos θ = pp2 −q +q 2 . (iii) c2 + a2 − b2 = c1 + a1 − b1 . At B1 live two ascetics.. n = p − q . we use the following description of the family of Heron triangles. i. we need integral answers to the questions (i) the distance between the hill and the town C1 C2 = b2 − b1 . One interesting feature of the second problem is the pair of integral triangles that we are required to generate for its solution. (ii) the height the wizard rose. Instead. the solution would not be elegant. As it turns out. if c2 sin A1 is to be an integer. the altitude c2 sin A1 through B2 of triangle A2 B2 C2 .. This description has previously appeared in this journal [4] so we simply state the description.e.e. The other one walks along the path B1 C1 C2 . . each member triangle containing θ. The wizard of them flies to B2 along the direction of A1 B1 . Furthermore. i.e. with (i) A2 = A1 . In such a case we have a Heron mountain. Therefore the integral mountain must be a Heron mountain.Two Brahmagupta problems 307 heron triangle. (4) Hence the solution to Problem 2 lies in generating a pair of integral triangles A1 B1 C1 and A2 B2 C2 . together the two conditions above imply that triangles A1 B1 C1 and A2 B2 C2 have a common excircle opposite to the vertices C1 . As the referee pointed out. then sin A1 should necessarily be rational. c2 + a2 − b2 = c1 + a1 − b1 . C2 . Furthermore. (pu − qv)(qu + pv). b. 2uv). ci ) to (6) as we wish: we have just to take sufficiently large values for λ. v) = (p. i = 1.1. The answers to the questions are (i) the distance between the hill and th twon. numerical solutions. 161. It is easy to verify that ci + ai − bi = 14.R. Numerical examples. there are four variables u1 . we note that (i) p = q ⇒ A = π2 and (a. The solution of Problem 2. v2 = 1 is a solution. u2 + v 2 ) are respectively the Pythagorean triangle family and the isoceles Heron triangle family. ci ) = 2. (p2 + q 2 )u2 v2 ). This gives (a1 . 15) and (a2 . (2) Suppose λ = 2 × 3 × 5 × 7 = 210. 5. Sastry Let cos A = A is given by p2 −q 2 p2 +q 2 . (ii) The wizard rose to a height c2 sin A1 = 30 × 45 = 24. q. v2 . v1 = 2 in (6). c2 ) = (145. 2. Note that gcd(ai bi . c) = (u2 + v 2 . The indexing of the six solutions below is unconventional in the interest of Figure 6. c2 + a2 − b2 = c1 + a1 − b1 simplifies to v2 (qu2 + pv2 ) = v1 (qu1 + pv1 ) = λ. c) =(pq(u2 + v 2 ). 30). (p2 + q 2 )uv). u2 . v1 . b2 − b1 = 147. (p2 + q 2 )u1 v1 ). S. q) = (u. i = 1. b. Next. has been divided out. Since ∠B1 A1 C1 = ∠B2 A2 C2 . v) ⇒ (a. 2(u2 − v 2 ). q) = 1. q = 1. 14.2. b2 . 5. c2 ) =(pq(u22 + v22 ). u2 − v 2 . The Heron triangle family determining the common angle (a. and (ii) (p. 2. b. bi . We continue to refer to Figure 5. It is easy to verify that u2 = 12. (pu1 − qv1 )(qu1 + pv1 ).308 K. Then (6) becomes vi (ui + 2vi ) = 210. (u. In fact it is possible to give as many solutions (ai . a constant. p and q remain the same in (5). u1 = 3. (pu2 − qv2 )(qu2 + pv2 ). (5) In particular. c) = (u2 + v 2 . This gives v2 (u2 +2v2 ) = 14 = λ. This creates an amusing situation as we see below. (a2 . and they generate an infinity of solutions satisfying the equation (6). p ≥ 1 and pu > qv. c1 ) = (13. . c1 ) =(pq(u21 + v12 ). b1 . b2 . (6) For given p. (1) We put p = 2. b1 . We now obtain two particular. Hence we have (a1 . i. j > i. 6. i = 1. . . 2. . . B4 . C5 . In other words.Two Brahmagupta problems 309 i vi ui ai bi ci 6 1 208 43265 43575 520 5 2 101 10205 10500 505 4 3 64 4105 4375 480 3 5 32 1049 1239 400 2 6 23 565 700 345 1 7 16 305 375 280 It is easy to check that for all six Heron triangles ci + ai − bi = 210. B2 . 4. See Figure 6. (520)B6 (505)B5 (480)B4 (400)B3 (345)B2 43265 (280)B1 1049 565 4105 10205 305 A1 C1 C2 C3 C4 C5 C6 (375) (700) (1239) (4375) (10500) (43575) Figure 6 Another famous problem. C6 . j = 1. has been solved in the context of Heron triangles in [2]. . the two ascetics may choose to live at any of the places B1 . B3 . 3. 2. C4 . ladders leaning against vertical walls. Then they may choose to reach any next town C2 . From this we deduce that (i) Bi Ci + Ci Ci+1 = Bi Bi+1 + Bi+1 Ci+1 and (ii) Bi Ci + Ci Cj = Bi Bj + Bj Cj . C3 . 5. B5 . R. 34 (2001-2002) 61–64. 38 (2005-2006) 68–73. S. [3] K. India. [4] K. Sastry: Jeevan Sandhya. Math.. Dover. 5 (2005)119–126. S. E. [2] K. Pythaogrean solutions. New York. Sastry. volume II. Smith. Spectrum. [5] D. History of the Theory of Numbers. K. Spectrum. Sastry. S.165– 224.R. R. 1971. S. E. Dickson. Raghuvana Halli. R. 1958 pp. R. Brahmagupta’s problems. . 560 062.152–160. History of Mathematics. Sastry. Chelsea. volume 1. S. Sastry References [1] L. Bangalore. Math. Construction of Brahmagupta n-gons. and Heron triangles. Heron ladders.310 K. DoddaKalsandra Post. Forum Geom. pp. H4.2 H6.n H2.2 H3.2 H3. We may repeat this operation to find a sequence of hexagons Hn = H1.2 .3 H3.1 H2.1 H4.2 H5. we say we have made the first square wreath around H1 .1 and H5. creating the second square wreath.2 H3.1 with counterclockwise orientation.2 H5. Publication Date: December 11. 1.1 H3.2 H4.2 H6.1 H2.3 H1. and abuse notations by identifying a point with its affix.2 and H4.1 H4.3 H4.2 H2.1 H6.1 H6.1 . b b FORUM GEOM ISSN 1534-1178 Square Wreaths Around Hexagons Floor van Lamoen Abstract. to get a third hexagon H3 . See Figure 1.1 H2.3 H5.1 . H3.1 H3.n H6. We attach squares externally on the sides H1. H2. Then we attach externally squares to the sides H6.n H3.2 H1. [8]. The author wishes to thank Paul Yiu for his assistance in the preparation of this paper. .1 H5. We investigate the figures that arise when squares are attached to a triple of non-adjacent sides of a hexagon. Communicating Editor: Paul Yiu.1 H6.n H5.2 H4.b Forum Geometricorum Volume 6 (2006) 311–325. Square wreaths around hexagons Consider a hexagon H1 = H1.1 H1. to form a new hexagon H2 = H1.1 H5.2 H5.2 H6.3 H2. and this procedure is repeated with alternating choice of the non-adjacent sides.3 Figure 1 We introduce complex number coordinates.2 H1. Thus.2 . As a special case we investigate the figure that starts with a triangle.n as a complex number. we shall also regard Hm.n H4.n and square wreaths.2 . 2006.1 H2. the first subscript m taken modulo 6. Following Nottrot. n−2 + (1 − i)H2.n = − (1 − i)H2.n−1 − H2k. 2. The triangle H1.n H5.n =H2k.n−2 − H5.n−2 have the same centroid.n−1 . we easily determine the vertices of the hexagons in the above iterations.n−1H5. H5.n−2 .n−2 . This means.n .n−1 .n H6.n has centroid 1 1 (H1.n−1 .n−2 ) . 3. then H2.n−1. we have H1.n H6. (3) H2k+1.n−1) =(1 + i)H2k. 2. H3. H3.n−1 H6.n−2 + 2H5.n H3. H6. If n is even. H3.n−1 H3.312 F.n−1 − i(H2k+1.n−2 . H4. H2k+1.n−1H4.n−1 respectively.n are the opposite sides of the squares erected on H2.n =(1 + (−1)n i)H2k. (1) H2k.n H5.n−1 + (1 − i)H2k+1.n−1 − H2k+1.n−1 .n =H2k−1.n ) = (H1.n−2 − H3. 4) twice. 3.n−1) =i · H2k−1.n H3. 3 3 which is the centroid of triangle H1. H2k−1. Hm.n−2 .n−1 .n and H5.n−2H3. so do triangles H2.n−2 H4.n−1) =(1 + i)H2k−1. (2) If n is odd.n H4.n =H2k+1.n−2 + H5.n are the opposite sides of the squares erected on H1. Triangles H1.n = (1 + (−1)n i)Hm.n−1 − H2k−1.n−1 + (−1)m+n i · Hm+(−1)m+n+1 . reading first subscripts modulo 6. for k = 1.n and H2.n−1 − i · H2k+1.n−1 respectively. H4.n =(1 + (−1)n i)H2k+1.n−1.n−2H6.n−1 .n−1 H1. This means.n .n−2 + H3.n−1 + i(H2k. van Lamoen Assuming a standard orientation of the given hexagon H1 in the complex plane. Proposition 1.n−1) =i · H2k.n−1 − i(H2k.n−2H5. 2. (4) The above recurrence relations (1.n = − (1 − i)H4.n−2 .n H2.n−2 H5.n−1 + (−1)n+1 i · H2k+1+(−1)n .n−2 + 2H1.n + H5.n−1 + i(H2k−1.n−2 + (1 − i)H6.n + H3.n H5. The proof for the other pair is similar.n and H1. 3.n = − (1 − i)H6.n−2 + (1 − i)H4.n−1 and H5.n H3. for k = 1. or even more succintly.n H4.n−1 + (1 − i)H2k. then H1. 4) may be combined into H2k.n−1 + (−1)n i · H2k+(−1)n+1 .n−2 H3. M.n−1 − i · H2k.n−1.n−1 − H2k.n−2 − H1. we have H2k. . Proof.n−1 and H6. Applying the relations (1.n H1.n .n =H2k. 3.n−1 H2. 2.n−2 + 2H3. 1 + 37H2.7 =37H1. (5) Proof. the sequence of vertices Hm. 5.1 + 13(1 − i)H2.1 − (1 − i)H6.7 = 6H1.1 − 6(1 + i)H5. H6.1 − 6H5.1 and H2. we have the recurrence relation (5) for m = 1.1 + 8H2.n satisfies the recurrence relation Hm.1 .7 = − 24(1 − i)H1.1 − 30H4.3 + H1. 2.1 + (1 − i)H2. 3.1 − 6H6.3 H4. from these equations gives H1.1 − 18iH2.n = 6Hm.1 . m = 2.5 − 6H1. 3.6 =18iH1. For each m = 1. 4.1 H3. 4. 3.4 =3(1 + i)H1.1 + (1 − i)H3.1 − 6(1 − i)H6.1 − (1 + i)H2. H2.1 + 5(1 + i)H6. H2.n−4 + Hm.1 .1 − H5.1 + 24(1 − i)H3.n−2 − 6Hm.3 =2H1.7 − 6H2.1 .1 .1 − 6(1 + i)H3.1 − 6iH5. H1. The same relations hold if we simultaneously increase each first subscript by 2.3 H2. 6. Midpoint triangles Let M1 . Thus. 6.1 H5. H1.3 H6. H1.1 + (1 + i)H6. H3.1 − iH5. Elimination of Hm.1 .1 + (1 − i)H2. 5.1 .2 =(1 + i)H1. 5.1 + 11iH6. H2. 4.1 H1.1 − 6H4.3 .1 + 3(1 − i)H2.1 − H4.3 respectively. A similar reasoning shows that (5) holds for m = 2.1 − 11iH3.1 − H4.1 + 2iH6. m = 1.1 − 24(1 + i)H2.3 + H2.1 . 6. By using the recurrence relations (1.1 − (1 + i)H5.1 .1 − 30H5.1 + iH4.1 + (1 − i)H3.3 = − (1 − i)H1. We have 1 M1 = (H1.1 − 6H3. M3 be the midpoints of H4. H2.1 .1 − 6(1 − i)H4.1 − H3.1 ) 2 = − M1 + (1 + i)M2 + (1 − i)M3 .1 − 2iH3.1 and M1 .n−6 . M2 .1 − iH2.1 .1 .5 + 6H2. 6. M2 . Elimination of Hm.1 − 4iH2.3 − H2. Similarly.Square wreaths around hexagons 313 Theorem 2. or each second subscript by 1.6 =13(1 + i)H1. 2.1 .3 and H5. H1. M3 the midpoints of H1.1 − H6. H1. 3.5 =8H1.1 + 24(1 + i)H6.5 = − 5(1 − i)H1.1 + 2H2.1 − (1 − i)H4.1 − (1 + i)H3. 2. 3. 4).1 − 5(1 + i)H2. from these equations gives H2.1 + 6iH4. H2. H2.2 =iH1.3 ) 2 1 = ((1 + i)H1.4 =4iH1.1 .1 − H5.1 + 5(1 − i)H3.1 + (1 + i)H6. we have H1. 5.1 = 0.1 . . 4.1 . 2 1 N2 = ((1 + i)M3 + (1 − i)M1 ) . M1 − M1 =i(M2 − M3 ). For a permutation (j. the segments Mj Mk and Mk Mj are perpendicular to each other and equal in length. 3. From this we deduce the following corollaries. The midpoints of the segments Mj Mj . we have M2 − M3 =(1 − i)M1 − M2 + i · M3 . we conclude that M1 M1 is parallel to an angle bisector of M2 M3 and √ M3 M2 . while √ M M is parallel to an angle bisector of Mj Mk and Mk Mj . and is 2 times as long as each of these segments. j = 1. = 2 2 2 N1 = Thus. 3. 2. 2 1 N3 = ((1 + i)M1 + (1 − i)M2 ) . The same results for (k. (7) =2M1 − (1 + i)M2 − (1 − i)M3 =(M2 − M3 ) + (M3 − M2 ). 2 Note that M2 − M3 M2 + M3 +i· . M2 =(1 − i)M1 − M2 + (1 + i)M3 . Proposition 3. Proof. 2 2 M − M3 M + M3 −i· 2 . . Similarly. From (8). 2. are the points 1 N1 = ((1 + i)M2 + (1 − i)M3 ) . M3 (6) − M2 =(1 + i)M1 − i · M2 − M3 . k. and is 2 times as long as each of these segments. van Lamoen Similarly. ) of the integers 1. See Figure 2. for N2 and N3 . and also the center of the square constructed internally on M1 M2 M3 . From the above expressions for Mj . 2) follow similarly. (1. 1). (8) From (6) and (7). N1 is the center of the square constructed externally on the side M2 M3 of triangle M1 M2 M3 . 3. j = 1. 2. =i ((1 − i)M1 − M2 + iM3 ) . M2 M3 and M3 M2 are perpendicular and have equal lengths. M. ) = (3.314 F. M3 =(1 + i)M1 + (1 − i)M2 − M3 . Since the bisector angle M2 M1 M3 is parallel to the line N1 M1 . 2. The same is true for Mj N and Mk N . the lines Mj Mk and Mk Mj and the line through N perpendicular to M M are concurrent (at M ).1 H5. The segments Mj N and Mk N are equal in length and are perpendicular.1 H3.Square wreaths around hexagons 315 Corollary 4. The perspector is the outer Vecten point of M1 M2 M3 and the inner Vecten point of M1 M2 M3 .1 H2. The triangles M1 M2 M3 and M1 M2 M3 are perspective. and M1 lies on the circle with diameter M2 M3 .3 Figure 2 Let M1 M2 M3 be the desmic mate of M1 M2 M3 and M1 M2 M3 .2 M3 N3 H2. M1 N1 is perpendicular to this latter line. (M M N ) and (M M N ) are coaxial. If (j. By Proposition 3. k.2 N2 M1 H4.e. The circles (Mj Mk N ). 1 Corollary 5. ∠M2 M1 M3 is a right angle.3 M2 H3. .3 H4. Proposition 6.3 H3.1 M3 H6.1 H5.3 N1 H2.. 1The outer (respectively inner) Vecten point is the point X 485 (respectively X486 ) of [4]. i. See Figure 3. H4. so the midpoints of Mj Mk .2 M1 H6. M1 = M2 M3 ∩ M3 M2 etc.2 H1. (Mj Mk N ). Corollary 7.2 H5. the line being parallel to M M .1 H6. 3.3 H1.2 M2 H1. ) is a permutation of 1. M M and M M are collinear. Mj Mk . 1 H6.3 Figure 3 3. M2 and M3 .316 F.1 M3 N3 M3 M3 H4. van Lamoen Since the lines Mj Mj . This case has been studied before by Haight and Nottrot.3 M1 H2.1 = B.1 H5.3 M2 H3. Starting with a triangle An interesting special case occurs when the initial hexagon H1 degenerates into a triangle with H1.2 H5. 3. N2 .1 N1 H2.3 H3.1 = C. the intersection of the lines is the inferior (complement) of the outer Vecten point. which also passes through M1 . it is the center of the circle through N1 N2 N3 (see [4]).1 = H6.3 H4. H4. 2. Corollary 8.1 = H3.1 M2 H1. M. concur at the outer Vecten point of triangle M1 M2 M3 .2 H2. H2.2 M1 H6. M2 M2 . H4. .2 H1. N3 . Under this assumption.3 H1. The three lines joining the midpoints of M1 M1 .1 M2 H6. 2 As such. M3 M3 are concurrent at the center of the circle through N1 .1 = A.2 M1 H3.2 N2 H5. who examined especially the side lengths and areas of the squares in each wreath.1 = H5. j = 1. 2X 641 in [4]. 1 + (Hm.2j − Hm. a3 (1) = 1.2 ).2 ) =Hm. Lemma 10. (2) The sums a3 (k) = kj=1 a2 (j) satisfy the recurrence relation a3 (k) = 6a3 (k − 1) − 6a3 (k − 2) + a3 (k − 3). 4 It follows from (9) and (10) that Hm. (10) Proposition 9. Each trapezoid in the wreath bordered by Hn and Hn+1 has area a2 (n) · ABC.2 ). c. mb . 4Note that sequences a and a follow this third order recurrence relation as well.3 − Hm. the squares have sidelengths 2ma .2j−1 ) + (Hm. This is sequence A004254 in [9].2k−1 = a1 (k)(Hm. The sides of the squares are parallel and perpendicular to the sides or to the medians of triangle ABC according as n is odd or even.2 − Hm. b.2j+1 − Hm. The squares of the first wreath are attached to the triangle sides outwardly.1 + a2 (k)(Hm.2k k k−1 =Hm.1 + j=1 j=1 =Hm. 4 can be found if these initial squares are constructed inwardly. a2 (1) = 1. 2mb . a3 (2) = 6. The sequence a3 (k) is essentially sequence A089817 in [9]. We shall assume the sidelengths of triangle ABC to be a. c multiplied by a1 (k).2k = a2 (k)(Hm.1 ) + a2 (j) (Hm. 1 2 . If n = 2k − 1. (1) kj=1 a1 (j) = a2 (k).1 ).3 − Hm. and the median lengths ma . where a1 (k) = 5a1 (k − 1) − a1 (k − 2). 3 Haight [2] has computed the ratios of the sidelengths of the squares. b.2k − Hm. a2 (2) = 5. The trapzoids between H1 and H2 degenerate into triangles.2 − Hm.2 − Hm. 3Similar results as those in §§3. This also means that Hm. the squares have sidelengths a. (9) If n = 2k.2j ) j=1 j=1 k k−1 a1 (j) (Hm.Square wreaths around hexagons 317 between two consecutive hexagons Hn and Hn+1 are three squares and three alternating trapezoids of equal areas.3 − Hm.2k+1 − Hm. mc respectively. This is sequence A004253 in Sloane’s Online Encyclopedia of Integer sequences [9]. This also means that Hm. where a2 (k) = 5a2 (k − 1) − a2 (k − 2). a1 (2) = 4.1 ) + a3 (k − 1)(Hm. 2mc multiplied by a2 (k). a1 (1) = 1. a3 (3) = 30. 2 − Hm.1 + a2 (k)(Hm.n H6.n .2k =Hm.3 − Hm.3 − Hm. 2. 0) S1 (SB . SA .2j − Hm. −c ) (2. 2.1 Hm.n H3.n H4. These can be combined into a single relation Hm.2 ) Here is a table of the absolute barycentric coordinates (with respect to triangle ABC) of the initial values in the above recurrence relations.1 + k k (Hm.n .n H5.n B1. −1) 2 (0.3 − Hm.2 − Hm. m Hm. −1) 3 (0. −1.2k+1 =Hm.n H4. SC . van Lamoen Also. −1.2 ). SB ) (−1.n C2. −b2 . 1) S1 (−a2 .2 ).2j+1 − Hm. −1) 4 (0. 1. M. Hm. 0.2j−1) + j=1 =Hm. −c2 ) (−1.n midpoint C1. −1.2 − Hm.1 + a2 (n )(Hm.1 ) + a3 (k)(Hm. 0) S1 (SC .n = Hm.n H5.n A1.318 F.n H6.2 ). m 1 2 3 4 5 6 x (2a3 (n ) + 1)S + a2 (n )SB −a3 (n )S + a2 (n )SB −a3 (n )S − a2 (n )a2 −a3 (n )S − a2 (n )a2 −a3 (n )S + a2 (n )SC (2a3 (n ) + 1)S + a2 (n )SC y −a3 (n )S + a2 (n )SA (2a3 (n ) + 1)S + a2 (n )SA (2a3 (n ) + 1)S + a2 (n )SC −a3 (n )S + a2 (n )SC −a3 (n )S − a2 (n )b2 −a3 (n )S − a2 (n )b2 z −a3 (n )S − a2 (n )c2 −a3 (n )S − a2 (n )c2 −a3 (n )S + a2 (n )SB (2a3 (n ) + 1)S + a2 (n )SB (2a3 (n ) + 1)S + a2 (n )SA −a3 (n )S + a2 (n )SA Note that the coordinate sum of each of the points in the above is equal to S. SA ) (−1. 2) 5 (0. 0.3 − Hm.1 Hm.1 ) + a3 (k)(Hm. SC . 1) S1 (SC . −1. 0) S1 (−a2 .1 + k (Hm.1 + a2 (k)(Hm.2 − Hm. −b2 . 2) 6 (1. −1) From these we have the homogeneous barycentric coordinates Hm.2 − Hm.1 ) + a3 (k − 1)(Hm.1 ) + a2 (j) (Hm.2 1 2 1 (1. Here are the coordinates of the points Hm.3 − Hm.2k+1 =Hm. 0.2 ) j=1 j=1 =Hm. SA ) (2.n H3.n H2. SA .n A2. Hm. SB ) (−1.n H1.n B2.1 ) + a3 (n )(Hm. Consider the midpoints of the following segments segment H1.2j ) j=1 k a1 (j) (Hm. 0.2 − Hm. in which n = n2 and n = n−1 2 . 1. 0) S (SB .1 + a2 (k)(Hm.n H2.3 − Hm. n ) where a3 (n ) + 1 . Note that Tj. This perspector is the centroid if and only if (g − q) + 3(gp − f q) = 0. . The GP : GQ =((g − q) + 3(gp − f q))((g − q)S − (f − p)Sω ) : (3(f − p)2 + (g − q)2 )S + 2(f − p)(g − q)Sω . A1. Theorem 12.n are perspective. Clearly the triangle given above is perspective with ABC at the point 1 1 1 : : .n and ABC lie on the Kiepert hyperbola.n −2a3 (n )S − 2a2 (n )a2 (a3 (n ) + 1)S + 2a2 (n )SC (a3 (n ) + 1)S + 2a2 (n )SB −2(2a3 (n ) + 1)S − a2 (n )a2 2a3 (n )S + a2 (n )SC 2a3 (n )S + a2 (n )SB (a3 (n ) + 1)S + 2a2 (n )SC −2a3 (n )S − 2a2 (n )b2 (a3 (n ) + 1)S + 2a2 (n )SA 2a3 (n )S + a2 (n )SC −2(2a3 (n ) + 1)S − a2 (n )b2 2a3 (n )S + a2 (n )SA (a3 (n ) + 1)S + 2a2 (n )SB (a3 (n ) + 1)S + 2a2 (n )SA −2a3 (n )S − 2a2 (n )c2 2a3 (n )S + a2 (n )SB 2a3 (n )S + a2 (n )SA −2(2a3 (n ) + 1)S − a2 (n )c2 Proposition 11. Any two such triangles are perspective.n .n B2. The perspectors of Tj.n Cj. 2} and the Kiepert triangles Kφ 5 are of this form.Square wreaths around hexagons 319 For j = 1. Proof.n A2.n and T2. It is the Kiepert perspector K(φj. Otherwise.n C1. Also the medial triangle of a triangle of this form is again of the same form. which is the isogonal conjugate of the Kiepert perspector K − cot−1 perspector P in question divides GQ in the ratio f −p g−q . is perspective with the reference triangle. We simply give a description of this perspector P .n = 2a2 (n ) 5See for instance [6]. cot φ1. with f and g replaced by p and q respectively. the line joining this perspector to the centroid G intersects the Brocard axis at the point Q = (a2 ((g − q)SA − (f − p)S) : b2 ((g − q)SB − (f − p)S) : c2 ((g − q)SC − (f − p)S)).n B1.n Bj. (f + 1)S + gSA (f + 1)S + gSB (f + 1)S + gSC which is the Kiepert perspector K(φ) for φ = cot−1 f +1 g .n C2. Every triangle of the form : (f + 1)S + gSC : (f + 1)S + gSB −2f S − ga2 (f + 1)S + gSC : −2f S − gb2 : (f + 1)S + gSA (f + 1)S + gSB : (f + 1)S + gSA : −2f S − gc2 where f and g represent real numbers. The triangles T1.n the triangle with vertices Aj. Consider a second triangle of the same form. 2.n for j ∈ {1. denote by Tj. This is a special case of the following general result. even = cot−1 21−3 12 T1.n B1. 2 2 Proof. where wreaths.even = cot−1 21−3 √ 3 φ2.n+1 = − H2. Consider the case of A1.n+1 = − H5. a4 (1) = 1.n+1 by the vector H2.n A2.2k+1 21+3 √12 φ2.4 .3 is X591 . a4 (2) = 3.2k+1 = cot−1 φ1. 2 2 −−−−−−−→ 1 −−−−−−−−→ 1 −−−−−−−−→ C2. Pairs of congruent triangles Lemma 13.n+1 = − H5. on the other hand. a4 (n) = 4a4 (n − 1) + 4a4 (n − 2) − a4 (n − 3).n H5.2k+1 a3 (k)+1 2a2 (k) 2a3 (k−1) −1 φ2.n+1 . (1) T2.n+1 .odd = cot−1 21+3 3 T2.n C1.n+1 and .n H3.2 is the medial triangle of T1.n H5.n B2.n H1.n+1 H3.3 and T2. (3) The perspector of T2.n H2.n+1 H4.n+1 = − H3. then −−−−−−−→ 1 −−−−−−−−→ 1 −−−−−−−−→ A2.2k = cot a2 (k) 3 (k) φ2. 2 2 − −−−−−−− → 1 −−−−−−−−→ 1 −−−−−−−−→ B2.n+1 = − H6.n H3.n H4. Nottrot [8].n = triangle perspector Kφ with T1. Translate the trapezoid −−−−−−−−−→ H5.n+1 H3.n+1 and the trapezoid −−−−−−−−−→ H2.2 and T2.n+1 . a4 (3) = 16.n H1. (a) If n ≥ 2 is even.n A1. This is sequence A005386 in [9].n H4.2k T2. Together with the trapezoid H3. van Lamoen and 2a3 (n ) . 2 2 (b) If n ≥ 3 is odd.n+1 H6. these images form two triangles XH3.n H6.n A1.n H4.4 is the common circumcenter of T1.n H4.n H6. M.n H1.n+1 .320 F.n+1 = − H6.2k+1 = cot−1 2a a2 (k) φ1.n+1 .n+1 = − H4.n+1 H6.2 and T2.n+1 .n H5.n+1 = − H1. 2 2 − −−−−−−− → 1 −−−−−−−−→ 1 −−−−−−−−→ B1.n by the vector H5.n+1 H4.n+1 . then 1 −−−−−−−−→ 1 −−−−−−−−→ −−−−−−−→ A1.n+1 = − H3.2k φ1.3 . 2 2 −−−−−−−→ 1 −−−−−−−−→ 1 −−−−−−−−→ C1.n H1.n H2. a2 (n ) In particular the perspectors tend to limits when n tends to infinity. The perspectors and limits are given by cot φ2.odd = cot−1 √ Remarks.n+1 . has found that the sum of the areas of the squares between Hn and Hn+1 as a4 (n)(a2 + b2 + c2 ).n+1 = − H2.n+1 H2. (2) The perspector of T2. 4.n+1 = − H4. Note that a1 (n) = a4 (n) + a4 (n − 1) for n ≥ 2.n+1 = − H1.n+1 for even n.n C2.2k = cot−1 a3 (k−1)+1 2a2 (k) limit for k → ∞ √ φ1. 2 H1. Furthermore.n is perpendicular to and has the same length as B1.n−1 and H5.n−1 .n+1 .n−1 B1.n .n−1 . Consider the triangles H2.n+1 B2.n B2.n A1.n+1 is parallel to H5.n C2. A1. It is clear that the points X. Proof. See [5].n−1 A1.n−1 and H6.n H5.n .2 B1.n+1 .3 H1. In the even case it is thus clear that the homothetic center is the centroid.n homothetic at their common centroid G.n A1. (v) H3.n+1 and H1.n+1 is perpendicular to and has the same length as B1.n B2.n B2.n A2. We show that H2.n H5.n B2.n+1 = 12 XY = − 12 H1.2 C1.n+1 C1. By Lemma 13. Proposition 14. (iv) H2.n−1 .n+1 A2.2 A1. Let n ≥ 3 be an odd number.3 Figure 4.n .n+1 A1.n+1 . the following pairs of triangles are congruent.n C2. .n+1 .n−1 and H5. (2) If n ≥ 2 is even. the sides of the squares are parallel to and perpendicular to the sides of ABC or the medians of ABC according as n is odd or even.n C1.n C1. In the odd case it can be seen as the lines perpendicular to the sides of ABC are parallel to the medians of the triangles in the first wreath (the flank triangles).n B2.n B1.n+1 . and the segments H2.n B2.n B2. We consider the first of these cases.2 H5.n+1 and is half its length.n+1 and H6. the following pairs of triangles are congruent. (vi) H4.n H4.n+1 6As noted in the beginning of §3. Therefore. A1.n B1. 6 See Figure 4.n C2.n−1 and H1. Consider the triangles H2.3 H2.n H5.n+1 B1.2 H3. The other cases follow similarly.n B1. Similarly.n A2. and the same for H2.n−1 are perpendicular to each other and equal in length.n H1.n+1 all lie on a line through the centroid G.n B1. (1) If n ≥ 3 is odd.n+1 C2.n+1 and B1.2 A H5. (ii) H3.2 H4. Y H3.n−1 B1.n−1 B2. B2. (i) H2. the triangles H2.n H2.n+1 C1. −−→ − −−−−−−− → −−−−−−−→ A1.n H5.3 H4.n A1.n B2.n+1 .n−1 C1.n B2.n−1 and H5.n C1.3 B C Y G H3. Y .n B2.n−1 . B2.n+1 and H5.2 X H2.3 H6.Square wreaths around hexagons 321 H6.n+1 and B1. (iii) H4.n−1 C2.n are congruent.n−1 C2.n B1. It follows that B2. The centroid and the orthocenter befriend each other.n B2.n−1 A2.n+1 and H5.3 A1.n−1 B1. 4 H5.2 and H2.2 C2.4 H6.2 H1.n−1 and H5.3 H1.n+1 are perpendicular and equal in length.3 H4.3 H3. the triangles H2. A pair of Kiepert hyperbolas The triangles A1.322 F.n C2.2 B1.4 Figure 5.3 are congruent to ABC.n−1 and B1. Remark.2 C1. Therefore.n−1 C2.3 A H5. and B1. van Lamoen H6.n B1.4 H4.n are perpendicular and equal in length.4 H1.3 H6.2 .2 H3. M.2 B1.2 B1.n−1 H5. The other cases can be proved similarly.n+1 C1.4 C1. The same reasoning shows that the segments H2.3 H2.n C1. The fact that the segments B2.n+1 are perpendicular and equal in length has been proved in Proposition 3 for square wreaths arising from an arbitrary hexagon.n B2.3 H2.3 B2.n−1 and H5. H5. The counterparts of a point P in these triangles are the points with the barycentric coordinates relative to these three triangles as P relative to ABC.3 B C A1.n−1 C2.n+1 C1. .4 H3.4 H4. See Figure 5.3 B2.n+1 are congruent. 5.4 A2. = 0.2 C1. −((S + SA )(x + y + z) − Sz)X + ((S + SC )(x + y + z) − Sx)Z = 0. B.Square wreaths around hexagons 323 Theorem 15.3 .2 H3. Proof. ((S + SA )(x + y + z) − Sy)X − ((S + SB )(x + y + z) − Sx)Y = 0. (11) cyclic The locus of the perspector is the union of the line at infinity and the Kiepert hyperbola of triangle ABC. Here are some examples of points on the locus (11) with the corresponding perspectors on the Kiepert hyperbola (see [1.2 H1. The locus of point P in ABC whose counterparts in the triangles A1. They are concurrent if and only if (x + y + z)((SB − SC )yz + (SC − SA )zx + (SA − SB )xy) = 0. we see that it is homothetic to the Kiepert hyperbola. This means the perspector lies on the union of the line at infinity and the Kiepert hyperbola.3 P and C1. The counterparts of P = (x : y : z) form a triangle perspective with ABC if and only if the parallels through A. For a point P on the locus (11). These parallels have equations ((S + SB )(x + y + z) − Sz)Y − ((S + SC )(x + y + z) − Sy)Z = 0. let Q = (x : y : z) be the corresponding perspector.3 form a triangle A B C perspective to ABC is the union of the line at infinity and the rectangular hyperbola S (SB − SC )yz + (x + y + z) cyclic (SB − SC )(SA + S)x = 0. CQ are concurrent. B1. = 0.3 P .3 to AQ.3 H4. . B1.3 H6.2 and H2. They are concurrent if and only if (x + y + z) (b2 − c2 )((SA + S)x2 − SA yz) = 0. Rearranging the equation of the conic in the form (11).2 B1.3 P are concurrent. This means that the parallels through A1.2 .3 . These parallels have equations ((S + SB )y − (S + SC )z)X + ((S + SB )y − SC z)Y − ((S + SC )z − SB y)Z −((S + SA )x − SC z)X + ((S + SC )z − (S + SA )x)Y + ((S + SC )z − SA x)Z ((S + SA )x − SB y)X − ((S + SB )y − SA x)Y + ((S + SA )x − (S + SB )y)Z = 0. C to A1. H5. cyclic The locus therefore consists of the line at infinity and a conic. BQ. C1. 7]). 324 F. M. van Lamoen Q on Kiepert . See Figure 6.3 B1.2 B = C1. A0 is the intersection of the Kiepert hyperbola with the parallel through A to BC.3 C1.3 H3.2 ∩ A1.2 ∩ C1.3 A B A0 G G B C A1. Since the triangle A1. .3 C1.3 H5.2 A1.3 H2.hyperbola 3 K arctan 2 orthocenter centroid outer Vecten point A B C A0 B0 C0 P on locus centroid de Longchamps point G = (−2SA + a2 + S : · · · : · · · ) outer Vecten point A = B1.2 C = A1. Here. similarly for B0 and C0 .3 H4.2 ∩ B1.3 = (S + SB : S + SA : −(S + SA + SB )) B0 C0 C B1. the rectangular hyperbola (11) is the Kiepert hyperbola of triangle A1.3 = (S + SC : −(S + SC + SA ) : S + SA ) C1.3 H6.3 A Figure 6.3 .3 B1.3 H1.3 has centroid G.3 C1. We show that it is also the Kiepert hyperbola of triangle A B C .3 = (−(S + SB + SC ) : S + SC : S + SB ) B1. com/˜njas/sequences/index. Gibert. B . the centroid of triangle A B C is the point G in the above table. Forum Geom. J.html Floor van Lamoen: St.Square wreaths around hexagons 325 This follows from the fact that A . Hyacinthos message 13996. Kimberling. Hyacinthos message 13995. [3] C. C. 15 Aug 2006.nl . [4] C.. Moses. J. Online Encyclopedia of Integer Sequences.evansville. [9] N. [5] F. 129 (1998) 1–285. 14 (1975-1976) 77–81. On a generalization of Pythagoras’ theorem. Congressus Numerantium. [2] F. C have coordinates −S − 2SA : S + SA : S + SA A B S + SB : −S − 2SB : S + SB C S + SC : S + SC : −S − 2SC From these. 2006. available at http://www. [6] F. C. available at http://faculty. Pythagoras. Fruitlaan 3. [7] P. Butcher (ed.att. Auckland University Press (1971) 73–77. van Lamoen and P.). in J. Forum Geom. Friendship of triangle centers. van Lamoen.edu/ck6/encyclopedia/ETC. August 15.research.html. Yiu. 1 (2001) 125–132. Kimberling. Encyclopedia of Triangle Centers. M. Sloane. The rectangular hyperbola (11) is therefore the Kiepert hyperbola of triangle A B C .C. M. Triangle centers and central triangles. Vierkantenkransen rond een driehoek. Haight. 1 (2001) 1–6.. The Netherlands E-mail address: fvanlamoen@planet. A. G. A Spectrum of Mathematics. Willibrordcollege. [8] J. A. References [1] B. 4462 EP Goes. Nottrot. The Kiepert pencil of Kiepert hyperbolas. . Analysis. by (b). and does not lie on the sidelines of the triangle. Proposition 1 (Lalesco). (b) for any point M ∈ Γ. the orthocenter. b b FORUM GEOM ISSN 1534-1178 Some Geometric Constructions Jean-Pierre Ehrmann Abstract.2. then (a) P is the midpoint of HH . It follows that T maps the asymptotes of h to the asymptotes of h . as. If the hyperbola h intersects Γ again at U . C . it follows that h = T (h). The center W of h is the midpoint of H U . P = H. Simson lines through a given point 1. D be the endpoints of the diameter of Γ perpendicular to the Simson line S(U ). 1. Given a point P = H and not on any of the sidelines of triangle ABC. the asymptotes of h are S (D) and S (D ) and. 1. Hence. U . B. . let the rectangular hyperbola h through A. Communicating Editor: Floor van Lamoen. where T is the translation by the −−→ vector HP . C. and (iii) a problem of congruent incircles whose analysis leads to some remarkable properties of the internal Soddy center. If the Simson lines of A . where H is the orthocenter of A B C . B . C concur at P . T maps P ∈ h to H ∈ h . they are parallel to the asymptotes of h. We solve some problems of geometric construction. Let D. the center W of h is the midpoint of HU . S(U ) and S (U ) are parallel. C. by (a) above. P intersect the circumcircle Γ again at U .b Forum Geometricorum Volume 6 (2006) 327–334. Some of them cannot be solved with ruler and compass only and require the drawing of a rectangular hyperbola: (i) construction of the Simson lines passing through a given point. 2006. the Simson lines S(M ) and S (M ) of M with respect to ABC and A B C are parallel. The asymptotes of h are S(D) and S(D ). Construction 1. As a rectangular hyperbola is determined by a point and the asymptotes. by (b). We make use of the following results of Trajan Lalesco [3]. A triangle ABC and a point P are given. Problem. Moreover.1. Let h be the image of h under the translation T by the vector Publication Date: December 18. Let h be the rectangular hyperbola through A. B . (ii) construction of the lines with a given orthopole. Let h be the rectangular hyperbola through A . We want to construct the points of the circumcircle Γ of ABC whose Simson lines pass through P . W = T (W ). B. P . U A A W W O H P H C B C B Figure 1. 1.-P. in any case and in order to avoid imaginary lines. The above construction leads to a construction of the lines whose orthopole is P . let Q be the isogonal conjugate of the infinite point of the direction orthogonal to HP . The other intersections of h with Γ are the points whose Simson lines pass through P . Moreover. See Figure 1. pp. the orthopole of the line M N is the common point of the Simson lines of M and N (see [1]).241–242]). Orthopole. It is well known that. The line through Q parallel to the Simson line of M intersects the line HP at R. . if M and N lie on the circumcircle. Thus. Then h passes through U . Ehrmann −−→ HP . B . if we have three real points A .328 J. we can proceed this way: for each point M whose Simson line passes through P . Thus. Then P is the orthopole of the perpendicular bisector of QR.3. the orthopole of a line L lies on the directrix of the insribed parabola touching L (see [1. the lines with orthopole P are the sidelines of A B C . C whose Simson lines pass through P . Construct a point P inside ABC such that if B and C are the traces of P on AC and AB respectively. As 2BDa = AB+BC−AC and 2CDa = AC+BC−AB. (a) ⇐⇒ (b).2. Let ha be the hyperbola through A with foci B and C. Let P be a point inside ABC and Qa the incenter of P BC. Analysis. (b) P lies on the open arc ADa of ha . A B Wa C P I Qa B C Da Figure 2. (b)=⇒ (c). (a) P B − P C = AB − AC. AI and P Qa are the lines tangent to ha respectively at A and P .Some geometric constructions 329 2. Proposition 2. BWa is a bisector of ∠ABP . . Proof.1. Two congruent incircles 2. and Da the projection of the incenter I of triangle ABC upon the side BC. The following statements are equivalent. Problem. Wa is equidistant from the four sides of the quadrilateral. If Wa is their common point. the quadrilateral AB P C has an incircle congruent to the incircle of P BC. (c) The quadrilateral AB P C has an incircle. Hence. 2. (d) IQa ⊥ BC. (e) The incircles of P AB and P AC touch each other. we have BDa −CDa = AB−AC and Da is the vertex of the branch of ha through A. P C . Let’s notice that. P B − P C = AB − AC ⇐⇒ Sb = Sc . P B − P C = AB − AC ⇐⇒ Sa = Da ⇐⇒ IQa ⊥ BC. When the conditions of Proposition 2 are satisfied. (c) Wa Qa and ADa are parallel. V . So (c) ⇐⇒ (d) because Wa Qa is the tangent to ha at P . Proof. If the incircle of AB P C touches P B . Ma I is the conjugate diameter of the direction of ADa with respect to ha . AB respectively at U . A I P B C Figure 3. we have P B−P C = BU −CV = BV −CU = AB−AC. Proposition 3. . let us recall the classical construction of an hyperbola knowing the foci and a vertex: For any point M on the circle with center Ma passing through Da . Now. (a) The incircles of P BC and AB P C are congruent. If the incircles of P AC and P AB touch the line AP respectively at Sb and Sc . U . See Figure 3. we have 2BSa = P B + BC − P C. (a) ⇐⇒ (b) is obvious. If Sa is the projection of Qa upon BC. AC. Ehrmann (c)=⇒ (b). (a) ⇐⇒ (d). As the line Ma I passes through the midpoint of ADa . Hence. Hence.330 J. V .-P. we have 2ASb = AC + P A − P C and 2ASc = AB + P A − P B. as I is the pole of ADa with respect to ha . the following statements are equivalent. (b) P is the midpoint of Wa Qa . (b’) ⇐⇒ (c). (a) ⇐⇒ (e). (d) P lies on the line Ma I where Ma is the midpoint of BC. (ii) The homogeneous barycentric coordinates of P . c) a . For one of them M . b + 2(s − c) Qa : a a √ Wa : (a + 2 as. A further investigation leads to the following √ results. b − s + as. b. s s . See Figure 4. (i) P B + P C = as where s is the semiperimeter of ABC. Qa . Construction 2. where ra is the (iii) The common radius of the two incircles is ra 1 − s radius of the A−excircle. and N the reflection of B in M . Wa are as follows. L touches ha at L ∩ CN . . The perpendicular from B to ADa and the circle with center Ma passing through Da have two common points. Remark. √ √ c − s + as) P : (a.Some geometric constructions 331 if L is the line perpendicular at M to BM . A B Wa C P I Qa B Da Ma C Figure 4. c + 2(s − b) a. We have already known that P B − P C = c − b. the perpendicular at M to BM will intersect Ma I at P and the lines Da I and AI respectively at Qa and Wa . the semiperimeter. Ehrmann 3. ρ= A P I B C Figure 5. s − b). . 2s + 4R + r See [2] for more details and references. Its center is X(176) in [2] with barycentric coordinates ∆ ∆ ∆ . The internal Soddy circle is the circle tangent externally to each of these three circles. and R be respectively the area. and the circumradius of triangle ABC. The internal Soddy center Let ∆. (B. Proposition 4. The inner Soddy center X(176) is the only point P inside ABC (a) for which the incircles of P BC. See Figure 6. P AB touch each other. P CA. s. See Figure 5.-P.332 J. BC P A . CA P B have an incircle. r. c+ a+ s−a s−b s−c and its radius is ∆ . the inradius. b+ . s − a). The three circles (A. s − c) touch each other. (b) with cevian triangle A B C for which each of the three quadrilaterals AB P C . (C. (C. c+ . a point P inside ABC verifying these conditions must lie on the open arc ADa of ha and on the open arc BDb of hb and these arcs cannot have more than a common point. P B = ρ + s − b. AB respectively are the same ones Da .Some geometric constructions 333 A P B C Figure 6. P C = ρ + s − c. CA. X(482) = a + s−a s−b s−c 1Thanks to Franc¸ois Rideau for this nice remark. s − a). Qb Qc . a s IQa AWa (4) The four common tangents of the incircles of BC P A and CA P B are BC. P C − P A = a − c. (1) It follows from Proposition 2(d) that the contact points of the incircles of P BC. Dc than the contact points of incircle of ABC. P AB touch each other at the points where the internal Soddy circle touches the circles (A. P CA. P A − P B = b − a. P AB with BC. Db . where ra is the radius of the A-excircle. s − c). (5) The lines AQa . and = . (3) If Qa is the incenter of P BC. and Wa the incenter of AB P C . BQb . AP and Da I. these conditions are satisfied for P = X(176). Remarks. (B. Proposition 2 shows that the conditions in (a) and (b) are both equivalent to P B − P C = c − b. Proof. Moreover. s − b). As P A = ρ + s − a. we have ra ra Qa Da Wa I = . b+ . 1 (2) The incircles of P BC. P CA. CQc concur at 2∆ 2∆ 2∆ . . fr . rue des Cailloux. Ehrmann References [1] J.334 J. 20 (1919)? 232–261. The complete quadrilateral.html. W.-P.evansville. 2. 92110 . La geometrie du Triangle. Encyclopedia of Triangle Centers. ser. [3] T. available at http://faculty. France E-mail address: Jean-Pierre. Jacques Gabay reprint 1987. Lalesco. [2] C. Annals of Mathematics. Paris Vuibert 1937. Jean-Pierre Ehrmann: 6.
[email protected]/ck6/encyclopedia/ETC. Clawson.Clichy. b. Its Converse and a Generalization Amy Bell Abstract. rc . b. Since AH = 2R cos A etc. one of the quantities AH. 1. Specifically. Hansen’s right triangle theorem In an interesting article in Mathematics Teacher. CH and 2R. Thanks are also due to the referee for suggestions leading to improvements on the paper. W. we find two quadruples of segments with equal sums and equal sums of squares. c = 2R. rb . CH is negative if the triangle contains an obtuse angle. b b FORUM GEO M8 ISSN 1534-117 Hansen’s Right Triangle Theorem. Hansen’s theorem relating the inradius and exradii of a right triangle and its sides to an arbitrary triangle. rc and r. Hansen [2] has found some remarkable identities associated with a right triangle. Let ABC be a triangle with a right angle at C. under the direction of Professor Paul Yiu. We seek to generalize Hansen’s theorem to an arbitrary triangle. (1) The sum of the four radii is equal to the perimeter of the triangle: ra + rb + rc + r = a + b + c. This suggests that a possible generalization of Hansen’s theorem is to replace the triple a. (2) The sum of the squares of the four radii is equal to the sum of the squares of the sides of the triangle: ra2 + rb2 + rc2 + r 2 = a2 + b2 + c2 . 2006. c by the quadruple AH. Now. given a triangle.b Forum Geometricorum Volume 6 (2006) 335–342. rb . by replacing a. Publication Date: December 20. Note that a = BH and b = AH. BH. sidelengths a.. b. which we generically denote by H. On the other hand. c. Communicating Editor: Floor van Lamoen. W. this latter vertex is the orthocenter of the triangle. BH. Note also that CH = 0 since C and H coincide. c by appropriate quantities whose sum and sum of squares are respectively equal to those of ra . for a right triangle ABC with right angle vertex C. and three excircles of radii ra . We generalize D. . D. Theorem 1 (Hansen). A strong converse of Hansen’s theorem is also established. It has an incircle of radius r. the hypotenuse being a diameter of the circumcircle. The paper is a revision of part of the author’s thesis for the degree of Master of Science in Teaching (Mathematics) at Florida Atlantic University. A characterization of right triangles in terms of inradius and exradii Proposition 3. Two quadruples with equal sums and equal sums of squares 2. (3) rb = s − a. (1) . Theorem 2. Proof. Ib A Ic I O H B C Ia Figure 1. By the formulas for the exradii and the Heron formula. Bell We shall establish the following theorem. (2) ra = s − b. (1) ra + rb + rc + r = AH + BH + CH + 2R. (2) ra2 + rb2 + rc2 + r 2 = AH 2 + BH 2 + CH 2 + (2R)2 . (5) C is a right angle. (1) rc = s. (4) is equivalent to the condition (s − a)(s − b) = s(s − c). (3).336 A. The following statements for a triangle ABC are equivalent. (4) r = s − c. each of (1). (2). Let ABC be a triangle with orthocenter H and circumradius R. in the nineteenth century work of John Casey [1]. §2. 1 We present a unified detailed proof of these propositions here. Let the bisector of angle A intersect the circumcircle at M . (4) implies (5). and can be found. We establish a basic result. we study the excircles in relation to the circumcircle and the incircle. Lemma 4 and the statement of Proposition 6 can be found in [3. simpler and more geometric than the trigonometric proofs outlined in [3]. The line BM clearly contains the incenter I and the excenter Ia .6. below.1]. A formula relating the radii of the various circles As a preparation for the proof of Theorem 2. See Figure 2. (2).185– 193]. (a + b − c)(a + b + c) = 2ab. Proposition 6. M is the midpoint of the arc BM C.1]. for example. its converse and a generalization 337 rc = s B ra = s − b r = s−c C A rb = s − a Figure 2. 1The referee has pointed out that these results had been known earlier. The converse is clear. . An outline proof of Proposition 5 can be found in [4. Propositions 5 and 6 can also be found in [5. we have s2 − (a + b)s + ab = s2 − cs. a2 + b2 = c2 . pp. Inradius and exradii of a right triangle Assuming (1). (3). 3. Consider triangle ABC with its circumcircle (O). (a + b − c)s = ab.Hansen’s right triangle theorem. Lemma 4. Clearly. (a + b)2 − c2 = 2ab. §4. This shows that each of (1).4. M B = M I = M Ia = M C. See Figure 4. and M is the midpoint of IIa . This follows by an easy calculation of angles. Also. I Ia = 2 · OM = 2R. Now. the reflection of I in O. O is the midpoint of II . but this means that N lies on the circumcircle (ABC) and thus coincides with M . Bell A O I B X X C M Ia Figure 3. ra + rb + rc = 4R + r Proof. I Ib = I Ic = 2R. (i) ∠IBIa = 90◦ since the two bisectors of angle B are perpendicular to each other. so N B = N I = N Ia .338 A. It follows that I is the reflection of I in O. It follows that M Ia = M B = M I. Remark. Similarly. Note that this latter line is parallel to OM . It is enough to prove that M B = M I. Since M is the midpoint of IIa . (iii) From the circle (IBIa ) we see ∠BN A = ∠BN I = 2∠BCI = ∠BCA. Proposition 5. let I be the intersection of the line IO and the perpendicular from Ia to BC. . See Figure 3. The same reasoning shows that M C = M I = M Ia as well. The circle through the three excenters has radius 2R and center I . Proposition 5 also follows from the fact that the circumcircle is the nine point circle of triangle Ia Ib Ic . (ii) The midpoint N of Ia I is the circumcenter of triangle IBIa . We summarize this in the following proposition. and I is the orthocenter of this triangle. I Ia = 2R Proposition 6. whose center is the midpoint N of Ib Ic . r and ra .Hansen’s right triangle theorem. BH = 2R + r − rb . Therefore N is the antipodal point of M on the circumcircle. The line Ia I intersects BC at the point X of tangency with the excircle. 2(R + OD) = rb + rc . (1) Since AH = 2·OD. From (2). 4. the four points Ib . Note that I X = 2R − ra . we have 2 · OD = r + (2R − ra ). C are on a circle. The center N must lie on the perpendicular bisector of BC. Since O is the midpoint of II . CH = 2R + r − rc . and we have 2N D = rb + rc . we have ra + rb + rc = 4R + r. . by (2) we express this in terms of R. ra + rb + rc = 4R + r. its converse and a generalization 339 A I O I B C D M Ia Figure 4. Proof. (2) Consider the excenters Ib and Ic . similarly for BH and CH: AH = 2R + r − ra . B. See Figure 5. Ic . Thus. From this. Proof of Theorem 2 We are now ready to prove Theorem 2. which is the line OM . we have IX + I X = 2 · OD. Since the angles Ib BIc and Ib CIc are both right angles. (2) This follows from simple calculation making use of Proposition 6. . AH + BH + CH + 2R =8R + 3r − (ra + rb + rc ) =2(4R + r) + r − (ra + rb + rc ) =2(ra + rb + rc ) + r − (ra + rb + rc ) =ra + rb + rc + r. Bell Ib N A Ic I O I B X D X C M Ia Figure 5.340 A. This completes the proof of Theorem 2. AH 2 + BH 2 + CH 2 + (2R)2 =(2R + r − ra )2 + (2R + r − rb )2 + (2R + r − rc )2 + 4R2 =3(2R + r)2 − 2(2R + r)(ra + rb + rc ) + ra2 + rb2 + rc2 + 4R2 =3(2R + r)2 − 2(2R + r)(4R + r) + 4R2 + ra2 + rb2 + rc2 =r 2 + ra2 + rb2 + rc2 . ra + rb + rc = 4R + r From these. and similar expressions for BH. 2 2 2 + + (s − a)(s − b) (s − b)(s − c) (s − c)(s − a) =s((s − c) + (s − a) + (s − b)) ra rb + rb rc + rc ra = =s2 . its converse and a generalization 341 5. (2) ra2 + rb2 + rc2 = (4R + r)2 − 2s2 . In the following lemma we collect some useful and well known results. (4) a2 + b2 + c2 = 2s2 − 2(4R + r)r. One of cos A.Hansen’s right triangle theorem. Converse of Hansen’s theorem We prove a strong converse of Hansen’s theorem (Theorem 10 below). Using AH = 2R cos A and a = 2R sin A. (1) ra rb + rb rc + rc ra = s2 . the condition (3) holds if and only if AH2 + BH 2 + CH 2 + (2R)2 = a2 + b2 + c2 . . Lemma 8. This means that triangle ABC contains a right angle. we have AH 2 + BH 2 + CH 2 + (2R)2 − (a2 + b2 + c2 ) =4R2 (cos2 A + cos2 B + cos2 C + 1 − sin2 A − sin2 B − sin2 C) =4R2 (2 cos2 A + cos 2B + cos 2C) =8R2 (cos2 A + cos(B + C) cos(B − C)) = − 8R2 cos A(cos(B + C) + cos(B − C)) = − 16R2 cos A cos B cos C. By Theorem 2(2). A triangle ABC satisfies ra2 + rb2 + rc2 + r 2 = a2 + b2 + c2 (3) if and only if it contains a right angle. Proposition 7. cos C must be zero from above. b. (3) ab + bc + ca = s2 + (4R + r)r. (1) follows from the formulas for the exradii and the Heron formula. ra2 + rb2 + rc2 =(ra + rb + rc )2 − 2(ra rb + rb rc + rc ra ) =(4R + r)2 − 2s2 . cos B. CH. They can be found more or less directly in [3]. Proof. and c. Proof. From this (2) easily follows. Florida Atlantic University Lecture Notes.net . Amy Bell: Department of Mathematics. (4) =⇒ (1): Theorem 1 (1). by Proposition 6. Advanced Euclidean Geometry. Coconut Creek. Proposition 9. (4) follows from (3) since a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca) = 2 4s − 2(s2 + (4R + r)r) = 2s2 − 2(4R + r)r. Johnson. (1) =⇒ (3): This follows easily from Proposition 6. The following statements for a triangle ABC are equivalent. (2) ra2 + rb2 + rc2 + r 2 = a2 + b2 + c2 . Theorem 10. Proof. r An easy rearrangement gives (3). 6th edition. Casey. (1) ra + rb + rc + r = a + b + c. circumscribed triangles. 1960. (2) ⇐⇒ (4): Proposition 7 above. Dover reprint. Yiu. [3] R. [4] P. 1000 Cocunut Creek Boulevard. Hansen. and the four-square. Bell Again. ra2 + rb2 + rc2 + r 2 = a2 + b2 + c2 if and only if 2R + r = s. Broward Community College. Yiu. (3) R + 2r = s. Florida Atlantic University Lecture Notes. [5] P. W. Euclidean Geometry. 4s2 = (4R + r)2 + 2(4R + r)r + r 2 = (4R + 2r)2 = 4(2R + r)2 . s = 2R + r. 1998. A. A Sequel to the First Six Books of the Elements of Euclid. On inscribed and escribed circles of right triangles. Proof. 4R + r =ra + rb + rc + + = s−a s−b s−c = ((s − b)(s − c) + (s − c)(s − a) + (s − a)(s − b)) (s − a)(s − b)(s − c) 1 = 3s2 − 2(a + b + c)s + (ab + bc + ca) r 1 = (ab + bc + ca) − s2 . ra2 + rb2 + rc2 + r 2 = a2 + b2 + c2 if and only if (4R + r)2 − 2s2 + r 2 = 2s2 − 2(4R + r)r. References [1] J. By Lemma 8(2) and (4). 2001.342 A. (3) ⇐⇒ (2): Proposition 9 above. 96 (2003) 358–364. [2] D. Mathematics Teacher. 1888. North Campus. USA E-mail address: abmath@earthlink. three-square problem. Introduction to the Geometry of Triangle. (4) One of the angles is a right angle. FL 33066. and are called special Kiepert inscribed triangles. 2006. y. 1. We shall work with homogeneous barycentric coordinates and make use of standard notations of triangle geometry as in [2]. we show that the family of special Kiepert inscribed triangles all with the same centroid G form part of a poristic family between K and an inscribed conic with center which is the inferior of the Kiepert center. we study several construction problems related to the triangles which have K as their own Kiepert hyperbolas. consisting of the Kiepert perspectors 1 1 1 : : . The Kiepert hyperbola has equation K(x. Given a reference triangle and its Kiepert hyperbola K. z) := (SB − SC )yz + (SC − SA )zx + (SA − SB )xy = 0 (1) in homogeneous barycentric coordinates. lies on the Steiner inellipse. our objects of interest are Kiepert inscribed triangles whose centroids are Kiepert perspectors. Publication Date: December 28. In this paper we shall mean by a Kiepert inscribed triangle one whose vertices are on the Kiepert hyperbola K. Basic results on triangle geometry can be found in [3]. Communicating Editor: Jean-Pierre Ehrmann. t ∈ R ∪ {∞}. the Kiepert center Ki = ((SB − SC )2 : (SC − SA )2 : (SA − SB )2 ). Among other results.b Forum Geometricorum Volume 6 (2006) 343–357. K(t) = SA + t SB + t SC + t we study triangles with vertices on K having K as their own Kiepert hyperbolas. We shall assume the vertices to be finite points on K. Special Kiepert inscribed triangles Given a triangle ABC and its Kiepert hyperbola K. Its center. it is called the Kiepert cevian triangle of its perspector. . We shall make frequent use of the following notations. If a Kiepert inscribed triangle is perspective with ABC. b b FORUM GEOM ISSN 1534-1178 Some Constructions Related to the Kiepert Hyperbola Paul Yiu Abstract. and call such triangles special Kiepert inscribed triangles. Since the Kiepert hyperbola of a triangle can be characterized as the rectangular circum-hyperbola containing the centroid. Such triangles necessarily have their vertices on K. Given a conic C and a point M . For k = 2. f4 with SA .e. Yiu P (t) Q(t) f2 f3 f4 g3 = ((SB − SC )(SA + t) : (SC − SA )(SB + t) : (SA − SB )(SC + t)) = ((SB − SC )2 (SA + t) : (SC − SA )2 (SB + t) : (SA − SB )2 (SC + t)) = SAA + SBB + SCC − SBC − SCA − SAB = SA (SB − SC )2 + SB (SC − SA )2 + SC (SA − SB )2 = (SAA − SBC )SBC + (SBB − SCA )SCA + (SCC − SAB )SAB = (SA − SB )(SB − SC )(SC − SA ) Here. i = 1. 3. draw (i) the polar of M with respect to C. f3 . t3 . 2. P (t) is a typical infinite point. A Kiepert inscribed triangle with vertices K(ti ). SB . f3 . i. 3. SB . if and only if S 2 f2 + (SA + SB + SC )f3 − 3f4 = 0. to construct the chord of C with M as midpoint. is g3 (t1 −t2 )(t2 −t3 )(t3 −t1 ) (S 2 +2(SA +SB +SC )ti +3t2 · ABC. .344 P. SC replaced by t1 . If the line in (ii) intersects C at the two real points P and Q. The area of a triangle with vertices K(ti ). 3. f4 are the functions f2 . i = 1. SC of degree k.. polar of M P M Q Figure 1. t2 . We shall make use of the following simple construction. is special. Proposition 1. is a symmetric function in SA . with centroid on the Kiepert hyperbola. Construction of chord of conic with given midpoint Construction 3. i Proposition 2. and Q(t) is a typical point on the tangent of the Steiner inellipse through Ki . the function fk . 4. (ii) the parallel through M to the line in (i). where f2 . then the midpoint of P Q is M . 2. Some constructions related to the Kiepert hyperbola 345 A K1 Ki polar of M Q G H B C K2 M K3 Figure 2. Kiepert inscribed triangle with centroid K(t) . (ii) the parallel through M to the tangent of K at K(t). K1 A Ki K(t) G H B K3 M C K2 Figure 3. construct (i) K1 on K and M such that the segment K1 M is trisected at Ki and K(t). See Figure 3. Then K1 K2 K3 is a special Kiepert inscribed triangle with centroid K(t). (iii) the intersections K2 and K3 of K with the line in (ii). See Figure 2. Construction 4. Here is an interesting family of Kiepert inscribed triangles with prescribed centroids on K. Construction of Kiepert inscribed triangle with prescribed centroid and one vertex A simple application of Construction 3 gives a Kiepert inscribed triangle with prescribed centroid Q and one vertex K1 : simply take M to be the point dividing K1 Q in the ratio K1 M : M Q = 3 : −1. Given a Kiepert perspector K(t). w) defines the line y z x + + = 0. v. tA = − v−w w−u u−v Clearly. Applying Proposition 2. w) defines two points on a sideline of triangle ABC. v. v. w). w) clearly define the line at infinity and the Kiepert hyperbola K respectively. v. w)2 L(u. we find that the centroid of AP BP CP lies on the Kiepert hyperbola if and only if (u + v + w)K(u. (SC − SA )u + (SB − SC )v These are Kiepert perspectors with parameters tA . tC given by SB v − SC w SC w − SA u SA u − SB v . v. v. w) in (1). L(u. SB − SC SC − SA SA − SB P (u. This result and many others in the present paper are obtained with the help of a computer algebra system. (2) SB − SC SC − SA SA − SB which is the tangent of the Steiner inellipse at Ki . The centroid of the Kiepert cevian triangle of P lies on the Kiepert hyperbola if and only if P is (i) an infinite point. Proof. Yiu It is interesting to note that the area of the Kiepert inscribed triangle is indepen√ −3 dent of t. This is similarly the case for the other two factors of P (u. the equation reduces to (SA − SB )v 2 − 2(SB − SC )vw + (SC − SA )w2 . Theorem 5. or (ii) on the tangent at Ki to the Steiner inellipse. w)P (u. the vertices of its Kiepert cevian triangle are −(SB − SC )vw AP = :v:w . BP = u : (SB − SC )w + (SA − SB )u −(SA − SB )uv CP = u : v : . tC = − . w) = The factors u + v + w and K(u. v. v. It is 3 2 3 |g3 |f2 2 times that of triangle ABC. tB = − .346 P. . v. tB . This shows that the two points on the line BC are the intercepts of lines through A parallel to the asymptotes of K. if P is on the Kiepert hyperbola. w) = ((SA − SB )v 2 − 2(SB − SC )vw + (SC − SA )w2 ). y. Let P = (u : v : w) in homogeneous barycentric coordinates. where v w u + + . v. (SA − SB )v + (SC − SA )w −(SC − SA )wu :w . and the corresponding Kiepert cevian triangles have vertices at infinite points. On the other hand. the Kiepert cevian triangle AP BP CP degenerates into the point P . If we set (x. 2. the factor L(u. Special Kiepert cevian triangles Given a point P = (u : v : w). z) = (−(v + w). w) = 0. Each factor of P (u. Kiepert cevian triangles of infinite points Consider a typical infinite point P (t) = ((SB − SC )(SA + t) : (SC − SA )(SB + t) : (SA − SB )(SC + t)) in homogeneous barycentric coordinates. Translation of Kiepert parabola 3. A Ki B H G C Figure 4. Altogether. − G(x. y. which is the Kiepert parabola. v. z) = SB − SC (SC − SA )(SA − SB ) Since g3 · G(x. this conic is a translation of the inscribed conic (SB − SC )2 x2 − 2(SC − SA )(SA − SB )yz = 0. A B C A . See Figure 4. y.Some constructions related to the Kiepert hyperbola 347 Remark. z) = − f2 (x + y + z)2 + (SB − SC )2 x2 − 2(SC − SA )(SA − SB )yz. w) above determine a conic with equation 2(SB − SC )yz x2 = 0. the six points defined by P (u. It can be easily verified that P (t) is the infinite point of perpendiculars to the line joining the Kiepert perspector K(t) to the orthocenter H. 1 The Kiepert cevian triangle of P (t) has vertices 1This is the line S (S − S )(S + t)x = 0. is a Kiepert perspector. SB + SC + 2t (SC − SA )(SC + t)(SA + t) B(t) = (SB − SC )(SA + t) : : (SA − SB )(SC + t) . SC + SA + 2t A(t) = C(t) B(t) A A(t) H G K(t) B C Figure 5. we say that the Kiepert cevian triangle of the infinite point P (t) is the same as the Kiepert parallelian triangle of the Kiepert perspector K(t). Thus. Two distinct Kiepert perspectors have parameters satisfying (3) if and only if the line joining them is parallel to the orthic axis. It is interesting to note that the area of triangle A(t)B(t)C(t) is equal to that of triangle ABC. 2 similarly for B(t) and C(t). SAB + SAC − 2SBC − (SB + SC − 2SA )t which. (3) Proposition 6. The Kiepert cevian triangle of P (t) is the same as the Kiepert parallelian triangle of K(t) It is also true that the line joining A(t) to K(t) is parallel to BC. See Figure 5. . the centroid of triangle A(t)B(t)C(t) is the point SB − SC : ··· : ··· . Yiu (SB − SC )(SB + t)(SC + t) : (SC − SA )(SB + t) : (SA − SB )(SC + t) . SC + SA + 2t (SA − SB )(SA + t)(SB + t) C(t) = (SB − SC )(SA + t) : (SC − SA )(SB + t) : . 2This is the line −(S A + t)(SB + SC + 2t)x + (SB + t)(SC + t)(y + z) = 0.348 P. Now. but the triangles have opposite orientations. It is K(s) where s is given by 2f2 · st + f3 · (s + t) − 2f4 = 0. by Theorem 5. The line joining K(s) and K(t) is parallel to the orthic axis if and only if 1 1 S 1+s S +s S +s A1 B C 1 1 = 0. SA +t SB +t SC +t S − S B C SC − SA SA − SB For s .Some constructions related to the Kiepert hyperbola 349 Proof. The orthic axis SA x + SB y + SC z = 0 has infinite point P (∞) = (SB − SC : SC − SA : SA − SB ). this is the same condition as (3). A(t)B(t)C(t) has centroid K(s). Then. draw (i) the parallel through K(s) to the orthic axis to intersect the Kiepert hyperbola again at K(t). This can be easily accomplished with the help . The Kiepert cevian triangle of P (t) has centroid K(s) Construction 7. B(t). See Figure 6. Given a Kiepert perspector K(s). Special Kiepert inscribed triangles with common centroid G We construct a family of Kiepert inscribed triangles with centroid G.= t. 4. This leads to the following construction. C(t) B(t) AK(s) A(t) H B G K(t) C Figure 6. to construct a Kiepert cevian triangle with centroid K(s). (ii) the parallels through K(t) to the sidelines of the triangle to intersect K again at A(t). the centroid of the reference triangle ABC. C(t) respectively. (4) SA + t SB + t SC + t As t varies. Beginning with a Kiepert perspector K1 = K(t) and Q = G. Yiu of Construction 3. and to the line (4) at the point ((SA + t)2 : (SB + t)2 : (SC + t)2 ). Poristic triangles with common centroid G 2 2 3The polar of M has equation (S − S )(S B C AA − S − 2(SB + SC )t − 2t )x = 0 and has infinite point ((SA + t)(SA (SB + SC − 2t) − (SB + SC )(SB − SC + t)) : · · · : · · · ). which is the inscribed ellipse E tangent to the sidelines of ABC at the traces of 1 1 1 : : . The line through M parallel to its own polar with respect to K 3 has equation SC − SA SA − SB SB − SC x+ y+ z = 0. we easily determine M = ((SA +t)(SB +SC +2t) : (SB +t)(SC +SA +2t) : (SC +t)(SA +SB +2t)). (SB − SC )2 (SC − SA )2 (SA − SB )2 and to the Kiepert hyperbola at G.350 P. P A Ki G H P3 B C M P2 Figure 7. See Figure 7. It has center ((SC −SA )2 +(SA −SB )2 : (SA −SB )2 +(SB −SC )2 : (SB −SC )2 +(SC −SA )2 ). . this line envelopes the conic (SB − SC )4 x2 + (SC − SA )4 y 2 + (SA − SB )4 z 2 − 2(SB − SC )2 (SC − SA )2 xy − 2(SC − SA )2 (SA − SB )2 yz − 2(SA − SB )2 (SB − SC )2 zx = 0. the inferior of the Kiepert center Ki . its points can be parametrized as Q(t) = ((SB − SC )2 (SA + t) : (SC − SA )2 (SB + t) : (SA − SB )2 (SC + t)). Q (t) = SC − SA SA − SB SB − SC and SB + t SC + t SA + t : : Q (t) = SA − SB SB − SC SC − SA respectively. The Kiepert cevian triangle of Q(t) has vertices (SC − SA )(SA − SB )(SB + t)(SC + t) : (SC − SA )2 (SB + t) : (SA − SB )2 (SC + t) .Some constructions related to the Kiepert hyperbola 351 Theorem 8. there is a family of special Kiepert cevian triangles with perspectors on the line (2) which is the tangent of the Steiner inellipse at Ki . The area of triangle A (t)B (t)C (t) is (f2 · t2 + f3 · t − f4 )3 (f2 · (SA + t)2 − (SC − SA )2 (SA − SB )2 ) Among these.1. 5. These two points are clearly on the line (2). According to Theorem 5. SA + t (SA − SB )(SB − SC )(SC + t)(SA + t) 2 B (t) = (SB − SC ) (SA + t) : : (SA − SB )2 (SC + t) . A poristic triangle completed from a point on the Kiepert hyperbola outside the inscribed ellipse E (with center the inferior of Ki ) has its center at G and therefore has K as its Kiepert hyperbola. 5. SC + t A (t) = Theorem 9. Special Kiepert cevian triangles with the same area as ABC. Since this line also contains the Jerabek center Je = (SA (SB − SC )2 : SB (SC − SA )2 : SC (SA − SB )2 ). It is an ellipse inscribed in the triangle in Construction 4. A family of special Kiepert cevian triangles 5. SB + t (SB − SC )(SC − SA )(SA + t)(SB + t) C (t) = (SB − SC )2 (SA + t) : (SC − SA )2 (SB + t) : . the envelope is a conic with center which divides Ki Kg in the ratio 3 : −1. at the points SA + t SB + t SC + t : : . The three perspectors are collinear on the tangent of the Steiner inellipse at Ki . . The Kiepert cevian triangle of Q(t) is triply perspective to ABC.2. The triangles B (t)C (t)A (t) and C (t)A (t)B (t) are each perspective to ABC. Proof. if we replace G by a Kiepert perspector Kg . four have the same area as the reference triangle. Triple perspectivity. More generally. Oppositely oriented triangle triply perspective with ABC at three points on tangent at Ki . SB +SC −2SA The points Q(t) =(−2(SB − SC ) : SC − SA : SA − SB ). Q (t) =(SB − SC : SC − SA : −2(SA − SB )).1.352 P. C1 B1 G1 A Q Q Ki Q G A 1 B C Figure 8.2. B1 =(2(SB − SC ) : −(SC − SA ) : 2(SA − SB )). C1 =(2(SB − SC ) : 2(SC − SA ) : −(SA − SB )). This has centroid f3 SC − SA SA − SB SB − SC K − : : = . See Figure 8. Q (t) =(SB − SC : −2(SC − SA ) : SA − SB ). give the Kiepert cevian triangle A1 =(−(SB − SC ) : 2(SC − SA ) : 2(SA − SB )). t = SA (SB +SC )−2SBC . 2f2 SB + SC − 2SA SC + SA − 2SB SA + SB − 2SC A (t)B (t)C (t) is also the Kiepert cevian triangle of the infinite point P (∞) (of the orthic axis). Yiu 5. C2 =((SB − SC )2 : (SC − SA )2 : (SC − SA )(SB − SC )). which has centroid SA + SB + SC 1 1 1 K − : : . Q (∞) = SC − SA SA − SB SB − SC with the parallels through B. The points Q (∞) and Q (∞) are the intersection 1 1 1 : : C to the line joining A to the Steiner point St = SB −S SC −SA SA −SB . B2 =((SB − SC )2 : (SA − SB )(SB − SC ) : (SA − SB )2 ). −1): A2 =((SC − SA )(SA − SB ) : (SC − SA )2 : (SA − SB )2 ).Some constructions related to the Kiepert hyperbola 353 5.2. See Figure 9. B2 C2 A G2 Q = Ki Q Q O G A2 B C St Figure 9. = 3 SB + SC − 2SA SC + SA − 2SB SA + SB − 2SC The points Q (t).2. Q (∞) = SA − SB SB − SC SC − SA 1 1 1 : : . C These points give the Kiepert cevian triangle which is the image of ABC under the homothety h(Ki . t = ∞. Oppositely congruent triangle triply perspective with ABC at three points on tangent at Ki . 1 1 1 : : . With the Kiepert center Ki = Q(∞). we have the points Q(∞) =((SB − SC )2 : (SC − SA )2 : (SA − SB )2 ). Q (t) and G2 are on the Steiner circum-ellipse. B3 = : : SA + SB − 2SC SC + SA − 2SB SB + SC − 2SA SB − SC SC − SA SA − SB C3 = : : . Q(t) = ((SB − SC )(SB + SC − 2SA ) : (SC − SA )(SC + SA − 2SB ) : (SA − SB )(SA + SB − 2SC )) . SC + SA − 2SB SB + SC − 2SA SA + SB − 2SC A3 = with centroid SB − SC : ··· : ··· (SB − SC )2 + 2(SC − SA )(SA − SB ) . These give the Kiepert cevian triangle SB − SC SC − SA SA − SB . Q(t) is the infinite point of the line (2). Yiu 5. Q (t) = ((SB − SC )(SC + SA − 2SB ) : (SC − SA )(SA + SB − 2SC ) : (SA − SB )(SB + SC − 2SA )) . See Figure 10. Q (t) = ((SB − SC )(SA + SB − 2SC ) : (SC − SA )(SB + SC − 2SA ) : (SA − SB )(SC + SA − 2SB )) . Triangle triply perspective with ABC (with the same orientation) at three points on tangent at Ki .2. t = −f3 2f2 .3. A A3 Ki Q Q Q O G G3 C3 B C B3 Figure 10.354 P. : : SB + SC − 2SA SA + SB − 2SC SC + SA − 2SB SB − SC SC − SA SA − SB . CA. K3 K3 A Ki G H B K1 C M K2 Figure 11.Some constructions related to the Kiepert hyperbola 355 5. BAC. Qb . If K3 is a real intersection of K with the line in (iii). The lines AQa . Qc is ABC oppositely oriented as ACB. Construction of special Kiepert inscribed triangles given two vertices K1 .2. (iii) the reflection of the line K1 K2 in the polar in (ii). Qb . Special Kiepert inscribed triangles with two given vertices Construction 10. See Figure 11. AB respectively. Qc of the line (2) with the sidelines BC. (ii) the polar of M with respect to K. construct (i) the midpoint M of K1 K2 . Given two points K1 and K2 on the Kiepert hyperbola K. The common Kiepert cevian triangle of Qa . CBA. t = −SA . Qb . For t = −SA .4. These points are the intercepts Qa . CQc are the tangents to K at the vertices. Qc respectively. K2 . triply perspective with ABC at Qa . 6. then the Kiepert inscribed triangle K1 K2 K3 has centroid on K. we have Q(t) =(0 : SC − SA : −(SA − SB )). Q (t) =(−(SB − SC ) : 0 : SA − SB ). BQb . Q (t) =(SB − SC : −(SC − SA ) : 0). z) +2(x + y + z) ((SB − SC )vw + (SC − SA )(3u + w)w + (SA − SB )(3u + v)v)x =0. If there are two such real intersections K3 and K3 . z) − 3g3 · (x + y + z)2 = 0. If M = (u : v : w) in homogeneous barycentric coordinates. K2 uniquely determining K3 . (5) The parallel through M is the line (3(SB − SC )vw + (SC − SA )(u − w)w + (SA − SB )(u − v)v)x = 0. This is the case when K1 K2 is a tangent to the hyperbola 4f2 · K(x. 2). which is the image of K under the homothety h(Ki . The polar of M in K is the line ((SA − SB )v + (SC − SA )w)x = 0. These two intersections coincide if the line in Construction 10 (iii) above is tangent to K. It is therefore an intersection of K with this homothetic image. K3 A Ki Kg G H B K2 M C K1 Figure 12. A point K3 for which triangle K1 K2 K3 has centroid on K clearly lies on the image of K under the homothety h(M. y. this homothetic conic has equation (u + v + w)2 K(x. then the two triangles K1 K2 K3 and K1 K2 K3 clearly have equal area. y.356 P. Yiu Proof. See Figure 12. (6) The reflection of (6) in (5) is the radical axis of K and its homothetic image above. 3). Family of special Kiepert inscribed triangles with K1 . available at http://faculty. Yiu. Florida Atlantic University. given in Construction 4. Boca Raton. 2001. Forum Geom. The Kiepert pencil of Kiepert hyperbolas. Introduction to the Geometry of the Triangle.Some constructions related to the Kiepert hyperbola 357 The resulting family of special Kiepert inscribed triangles is the same family with centroid K(t) and one vertex its antipode on K. Encyclopedia of Triangle Centers. Florida 33431-0991. 1 (2001) 125–132.evansville. M. [2] F. Kimberling. Yiu.edu/ck6/encyclopedia/ETC. [3] P.. Paul Yiu: Department of Mathematical Sciences. Florida Atlantic University lecture notes.html. USA E-mail address: yiu@fau. References [1] C. van Lamoen and P.edu . . M. 47 Hajja. b b FORUM GEOM ISSN 1534-1178 Author Index Atzema. S.: Proof by picture: Products and reciprocals of diagonals length ratios in the regular polygon. C. J. 327 Fontaine. 205 ˇ Cerin.: Translated triangle perspective to a reference triangle. 335 Boskoff. 179 Hudelson. L. C. T. 229 Kimberling. 187 Bradley.: Hansen’s right triangle theorem. Q.: A characterization of the centroid using June Lester’s shape function. 97 Gibert. P. 97 van Kempen. H. F. B.: A projectivity characterized by the Pythagorean relation. 247 Dergiades.: A simple perspectivity. G.: On butterflies inscribed in a quadrilateral.: Pedals on circumradii and the Jerabek center. M.: The orthic-of-intouch and intouch-of-orthic triangles. N. Z. M. 53 A very short and simple proof of “the most elementary theorem” of euclidean geometry.: On some theorems of Poncelet and Carnot. 17 Myakishev.: A conic associated with the Euler line.: Some geometric constructions. its converse and a generalization.: Isocubics with concurrent normals. J. K.: The locations of triangle centers. W. K. 139 Formulas among diagonals in the regular polygon and the Catalan numbers. 311 Manuel. 289 Nguyen.: A theorem by Giusto Bellavitis on a class of quadrilaterals.: On the mixtilinear incircles and excircles. E. 255 Hurley. A. 171 van Lamoen. J.: A 4-step construction of the golden ratio. S. 199 On triangles with vertices on the angle bisectors.b Forum Geometricorum Volume 6 (2006) 359–360. 57 The locations of the Brocard points. A. 1 A note on the barycentric square root of Kiepert perspectors.: A synthetic proof and generalization of Bellavitis’ theorem. 167 Hofstetter. 79 Square wreaths around hexagons. 181 Bell. 241 Danneels. A. 225 Ehrmann.: Proof by picture: Products and reciprocals of diagonals length ratios in the regular polygon.: On two remarkable lines related to a quadrilateral.: Archimedean adventures. 71 Bui. E. 263 .-P. 269 Kiss.: Concurrent medians of (2n + 1)-gons. 235 Tran. A. 53 Stothers. M. G. S. 149 Salazar. 191 Thas. M. 343 . M. S. 187 The Feuerbach point and Euler lines.: The locations of triangle centers. 71 Spirova. P. W. K. C. D. 107 Rodr´ıguez. L.360 Author Index Okumura.: A generalization of Power’s Arhimedean circles. A. 1 Sastry. 285 Rabinowitz. 29 Parish. C.: The Feuerbach point and Euler lines. P.: A projectivity characterized by the Pythagorean relation. 17 Ruoff.: A generalization of Power’s Arhimedean circles. J.: A characterization of the centroid using June Lester’s shape function. D. 297 Watanabe. J. 57 The locations of the Brocard points. C. J. 191 Some constructions related to the Kiepert hyperbola.: On the mixtilinear incircles and excircles. 117 Suceav˘a B. 25 The Droz-Farny Theorem and related topics.: Intersecting circles and their inner tangent circle.: The cyclic complex of a cyclic quadrilateral.: A note on the Droz-Farny theorem. 17 Smith.: A conic associated with the Euler line. R.: Some Grassmann cubics and desmic structures. 101 Pamfilos.: On the derivative of a vertex polynomial. P. 101 Yiu.: Pseudo-incircles. 301 Semi˜ao.: On the generating motions and the convexity of a well known curve in hyperbolic geometry.: Two Brahmagupta problems.: A conic associated with the Euler line. H.