The complete quadrilateral and its propertiesNikolaos Rapanos email: [email protected] HMS-Preparation Notes for I.M.O. 2009 July 7, 2009 1 The basics Definition 1.1. The figure determined by four lines, no three of which are concurrent, and their six points of intersection, form the so-called complete quadrilateral. Equivalently, one can define the complete quadrilateral as the shape that is formed by a convex quadrilateral and the two intersections of its opposite sides. The segments AB, AC, BE, DC are the sides, and the segments AF, DE, BC are the diagonals of the complete quadrilateral F DAECB. Figure 1: The complete quadrilateral. 1 (see Figure 2) 2 . DAC. Have in mind that O1 O2 ⊥ BM etc. The circumcenters of the four triangles mentioned above. Prove by angle chasing that the quadrilaterals O1 O2 M O3 and O2 M O3 O4 are inscribable. Proof. Proof. (Simson’s Line of the complete quadrilateral) The feet of the perpendiculars from M to the sides of the complete quadrilateral are collinear. (see Figure 1) Figure 2: Aubert’s and Simson’s Lines. Proof. Let M be the intersection of the circles CF DB and CF EC . Using the fact.BEA concur at Miquel’s point M . (see Figure 1) Theorem 1. and Miquel’s point are concyclic.DAC.F EC.3. F EC. it turns out that the Simson’s Lines of the four triangles coincide. The circumcircles of the four triangles DF B. Theorem 1.Theorem 1.1. Prove by angle chasing that the quadrilaterals M DAC and M BAE are inscribable. Use Simson’s Line for the circumcircles of each of the four triangles DF B. BEA and the point M .2. that the circumcircles of the four triangles concur at M . C 0 = C3 ∩ AD and D0 = C2 ∩ AC. C2 . C3 form a bundle. Therefore. Figure 3: The circles whose diameters are the diagonals of the complete quadrilateral.5. (Aubert’s Line) The orthocenters of the four triangles DF B. CC 0 . Aubert’s Line is the homothetic image of Simson’s Line under the homothety H(M. C3 be the circles with diameters the diagonals of the complete quadrilateral F DAECB. DD0 . Any inscribed angle which sees a semi-circle is right. 2) maps Simson’s Line (which according to the previous theorem. 3 . C2 . which is parallel to Simson’s Line of the complete quadrilateral. happens to be the same for the four triangles) to another line passing through the orthocenter of each triangle. The circles C1 . so the orthocenter HADC of the triangle ADC is the common point of the segments AA0 .4.Theorem 1. (see Figure 2) Theorem 1. Proof. and thus they are parallel to each other.DAC and BEA lie on the same line. 2). form a bundle. It is well known that the homothety H(M. We will prove that the common chord of the circles coincides with Aubert’s Line of the complete quadrilateral. Let C1 . Let A0 = C1 ∩ DC.F EC. Proof. We work separately for each of the four circumcircles. Therefore. Gauss’s Line is perpendicular to Simson’s and Aubert’s Line. Similarly. (Gauss’s Line) The midpoints of the diagonals of the complete quadrilateral are collinear. HF EC and HEBA are also radical centers of the same circles. C3 . C2 . thus Gauss’s Line is perpendicular to Aubert’s Line. i.6. that the midpoints of the diagonals are collinear. Furthermore. C2 . we conclude Figure 4: Gauss’s Line. Theorem 1. HADC is the radical center of the circles C1 . and so we conclude that the circles C1 . C3 form a bundle and their common chord is Aubert’s Line. Proof. CHADC ·C 0 HADC = DHADC ·D0 HADC (because the quadrilateral C 0 D0 CD is inscribable) and DHADC · D0 HADC = AHADC · A0 HADC (because the quadrilateral A0 D0 AD is also inscribable). Since the circles C1 . thus HADC has equal powers with respect to the three circles. we prove that HF DB . C3 form a bundle (according to the preivious theorem) and their centers are the midpoints of the diagonals of the complete quadrilateral. 4 . C2 .e. It is also well known that the line joining the centers of two circles is perpendicular to their common chord. Proof. Consider the circles CF DB . 2 2.Remark. If the quadrilateral DBCE is inscribable. then the Miquel’s point M lies on BC. therfore ∠F M B + ∠F M C = 180o . 5 . If the quadrilateral F DAE is inscribable. and we advise the reader to study carefully and understand this proof. Figure 5: Special cases for Miquel’s point.1 More properties of the complete quadrilateral Special Cases Theorem 2. This proof for Gauss’s Line shows how we can take advantage of a bundle of circles.2. then the Miquel’s point M lies on the line AF . Thus F M (the radical axis of the first two circles) passes through A. Proof. The points of intersection of the first two circles are F and Miquel’s point M . CF EC and CDECB . Because of the hypothesis.1. Theorem 2. Observe that the radical center of the three circles is A. we have ∠F M B = ∠F DA and ∠F M C = ∠F EA. This method can be used to pove collinearities in general. (OB) = m.2. D) = −1 comes from the concept of the cross ratio. B. then the points A and C are also harmonic conjugates of B and D. B. C and D four points which lie in this order on it. The notation (A. D) = −1 when the points B and D are harmonic conjugates of A and C. B. Proof. We also say that the points B and D are the harmonic conjugates of A and C. If A.3 we will present how harmonic division appears together with a complete quadrilateral. B. we will present some properties of the harmonic conjugate points and in Theorem 2. (OC) = l. and often we use the notation (A. The four point (ABCD) is called a harmonic division. C. If the points B and D are harmonic conjugates of A and C and M the midpoint of AC. C. (OD) = n. D) = −1. Property 1. Firstly. Property 2. or harmonic 4-tuple. Remark. B. C. It is known that (A. and O an arbitrary point of d such that (OA) = k. C and D form a harmonic 4-tuple on d.2 The complete quadrilateral and harmonic division Let d be a line and A. Property 3. if BA DA = BC DC Figure 6: Harmonic Division. If the points B and D are harmonic conjugates of A and C. The definition of cross ratio and its consequences as well as its generalization to concyclic points which leads to the concept of the harmonic quadrilateral will be discussed further in the lectures. so above figure) we have 6 BA BC = DA DC or equivalently (according to the . then 2(kl + mn) = (k + l)(m + n) (1) Figure 7: Illustration of Property 3. then the points B and D lie on the same ray of d with M as the vertex ray. The following lemma is a consequence of the Appolonius circle property1 . then the third is also true. then 1 1 2 = + AC AB AD Proof. Property 4. Let four points A. • The division (ABCD) is harmonic. Lemma 2. if the points B and D divide the line segment AC harmonically in ratio p : q. C and D. in this order.3. The endpoints of each diagonal. we get P B CE DA · · =1 (2) P C EA DB Applying Ceva’s Theorem for the triangle ABC and the cevians AF. if two of the following propositions are true. Property 5. In particular. also follows that the converse is also true under the restriction that B and D lie on the same ray of d with M (the midpoint of AC) as the vertex of the ray. • XB is the internal angle bisector of ∠AXC • XB ⊥ XD Theorem 2. C and D are harmonic conjugate. Use the metric relation (1) for O ≡ M .m−k l−m = n−k n−l ⇔ nm − ml − kn + kl = nl − nm − kl + k ⇔ 2(kl + mn) = (k + l)(m + n) From this proof. Use the metric relation (1) for O ≡ A. B. C and D are harmonic conjugate and M the midpoint of AC. we get QC DB EA · · =1 QB DA CE 1 (3) Appolonius circle is the locus of a point moving so that the ratio of its distances from two fixed points is fixed. form a harmonic 4-tuple. lying on d. The converse is also true under the restriction that B and D lie on the same ray of d with M as the vertex of the ray. (Descarte’s Identity) If the points A. Applying Menelaus’s Theorem for the triangle ABC and the line DE. B. Then. 7 . Let P = DE ∩ BC and Q = AF ∩ BC. then M B · M D = M A2 = M C 2 Proof. Proof. B. CF . DF. and the intersection points of the diagonal with the other two diagonals of the complete quadrilateral. (Newton’s Identity) If the points A. then the circle with diameter the segment BD is the Appolonius circle for the fixed points A and C.1. and the ratio p : q. 1 Harmonic bundle Suppose we have a line d and a harmonic 4-tuple (ABCD) on d. 2. Z = OC ∩ O0 C 0 . C. (Pappus) The intersection points between a line and the four radii of a harmonic bundle. we get QB PB = PC QC thus the points P.2. is divided by the other three radii to two equal segments. Theorem 2. B.Figure 8: Harmonic 4-tuple and complete quadrilateral (Theorem 2. XD .the radii of the bundle) is harmonic . XC. If O(ABCD) and O0 (AB 0 C 0 D0 ) are harmonic bundles and E = OB ∩ O0 B 0 . Theorem 2. D) = −1 and (A. D0 ) = −1 then the lines BB 0 . B 0 . If (A. C form a harmonic 4-tuple. 8 . then we say that the bundle X(ABCD) (which consists of four lines XA. XB.7. A bundle O(ABCD) is harmonic if and only if a line parallel to one of its radii. Theorem 2.5. Theorem 2.3). DD0 are concurrent. B.4. C 0 . then E.6. If X is a point not lying on d. CC 0 . form a harmonic 4-tuple. Multiplying by parts (2) and (3). Z and W are collinear. W = OD ∩ O0 D0 . Here we will present some well-known theorems about the harmonic bundles without proofs since their proofs can be found in [1] and in other classic books of Euclidean Geometry. Q. 2. I the incenter and Ib the excenter. then the bundle A(HO9 GO) is harmonic. O the circumcenter. form a harmonic bundle. In a triangle ABC consider three points X. form a harmonic bundle. AB respectively.2 Concurrency through harmonic division The following lemma is a consequence of Theorem 2.3 and shows how we can use harmonic division to derive concurrency results.4. If X 0 is the point of intersection of Y Z with the extended side BC.O the circumcenter. • Two of the sides. Lemma 2. Y. then the four-point (BXCX 0 ) forms a harmonic division if and only if the cevians AX. 2.the contained median and the line parallel to the third side.2. Common harmonic bundles in a triangle ABC are the following: • The sides of one of the three angles and the internal and external bisectors of it. CA. 9 . G its centroid and O9 the Euler’s circle center. Z on the sides BC. then the bundle A(HIOIb ) is harmonic. If the bundle O(ABCD) is harmonic and the radii of another bundle O0 (A0 B 0 C 0 D0 ) are one to one perpendicular to the radii of the first.Figure 9: Illustration of Theorem 2. • If H is the orthocenter. BY and CZ are concurrent.8. Theorem 2. • If H is the orthocenter. then the second bundle is also harmonic. and therefore CT ⊥ BO. If the quadrilateral F DAE is inscribed in a circle C(O. so it is enough to show that if M is the intersection of OT and BC. Furthermore the intersection of OT and BC is the Miquel’s point of the complete quadrilateral F DAECB. The first pair of the harmonic 4-tuples gives that KL must be the polar of B. 10 . We already know (Theorem 2. Proof. B1 = CO ∩ BT . Let K = CT ∩ F E. L = CT ∩ AD. (CEQA) are also harmonic. P = BT ∩ DF. Q = BT ∩ AE. But this is true since CM · CB = CB1 · CO = CO2 − r2 = CF · CD. Combining this result with Pappus theorem for harmonic bundles. From the complete quadrilateral KT AECF according to Theorem 2. then OT is perpendicular to BC. and therefore BT ⊥ CO.1) that the Miquel’s point of the complete quadrilateral F DAECB lies on BC. K. (CF P D). then M lies on the circumcircle of the triangle F DB. we have that the bundle A(DF KC) is harmonic.9 states that OT ⊥ BC and also that M = OT ∩BC is the Miquel’s point in this configuration. C1 = BO ∩ CT. So T must be the orthocenter of the triangle OBC. Similarly we can prove that the 4-tuples (BDLA).3 An interesting result Theorem 2.2. F.3. r). we conclude that the points B. E form a harmonic 4-tuple. The second pair of the harmonic 4-tuples gives that P Q must be the polar of C. (Brockard) Denote by T the intersection of DE and BC.9. and thus OT ⊥ BC. Figure 10: Theorem 2. Now having in mind Theorem 2. According to Theorem 11 .9 and that it suffices to show that OF ⊥ F X. Solution. and let D. and according to Lemma 2.2. CA and AB respectively. Prove that EF ZY is cyclic. Y and D respectively. H its orthocenter. and M be the midpoint of AC. BE and CF concur at Gergonne’s point of triangle ABC we deduce that (T. Let O be its circumcenter. the intersection of OM and F X by Y . Notice also that the lines XD. BY. and F the foot of the altitude from C. It is quite common in such configurations to consider the intersection T of the C-altitude and the circumcircle of ABC. and the intersection of OF and AC by Z.3 Applications Problem 1.(Nikolaos Rapanos-BMO 2008) Clearly it suffices to prove that OF is perpendicular to F X. D. B. Since AD. Figure 11: Extend EF to cut BC at T . Solution. E. we construct the complete quadrilateral T ACBED as in the Figure 13. Now observe that T D2 = T Z · T Y = T F · T E and we are done. IMO-SL 96) An acute-angled scalene triangle ABC is given with AC ≥ BC. CZ are concurrent at Gergonne’s point of triangle XBC. we have T ∈ Y Z. F be the points of tangency of the incircle of triangle ABC with the sides BC. Denote by T the intersection of BC and EF . XC and BC in Z. (IMO-SL 1995) Let ABC be a triangle. We denote the intersection of P H and BC by X. Problem 2. C) = −1. Prove that the points F. Z and Y are concyclic. M. Let X be in the interior of ABC such that the incircle of XBC touches XB. Let P be the point (other than A) on the line AB such that AF = P F . (BMO 2008-P1. is not enough. Working with angle chasing and elementary synthetic geometry. Q are harmonic conjugate2 and we are done. Alternatively one could give an elegant solution using Butterfly Theorem. = BP BA Therefore it suffices to prove that ⇔ BP FB = BA BQ ⇔ F A−F B FB = BA BQ ⇔ FA FB =1+ BA BQ ⇔ FA QA = FB QB which is true since the points A. Here we present a solution based on the theory of the complete quadrilateral. we have to show that BF BQ = BX BE . similarities or other more advanced techniques. 2. From the hypothesis we have AF = P F and also it is well known that T F = F H.Figure 12: Comment. 12 .9 OF is perpendicular to DE. Indeed the official solution is based on similar triangles. thus AT P H is a rhombus and AT //P X which yields BX BE = BF BQ BP BA . F.3 for the complete quadrilateral T ACBED. we need to introduce something stronger in our solution like ratios. In such cases. therefore it is enough to show that F X is parallel to DE. unfortunately. B. Equivalently. 2 We can derive this by applying Theorem 2. AC such that BB 0 = CC 0 . Given a triangle ABC denote by D. AB respectively. Let ABC a triangle and D. If P P1 ⊥ AB and P P2 ⊥ AC. Seniors) Let H be the orthocenter of the triangle ABC which is inscribed in a circle of center K and radius R = 1. In a triangle ABC. we consider points B 0 . AB. 13 . Problem 5. Problem 4. C 0 on AB. Problem 7. 4 Problems Problem 3. If S is the intersection of the lines HK and BC and it is also true that KS · KH = 1. K. ZP. CA. CA. Z the points at which the incircle touches the sides BC. If P D is a diameter of the incircle. and if AP. If L = BI ∩ EF . EP cut BC at M. Prove that the symmetric point of D with respect to the midpoint M of BC lies on the angle bisector of ∠A. E. A triangle ABC is given and let two arbitrary lines BD and CE which intersect at P such that ∠ACP = ∠ABP . F the points where the incircle touches BC. E. N respectively. (Greek National Olympiad 2008. Problem 6. then prove that LB ⊥ LC. Let D be the intersection of lines BC 0 and B 0 C.Figure 13: We form the complete quadrilateral and we take advantage of the parallel lines and the harmonic conjugate points created. then prove that the midperpendicular line of the segment P1 P2 passes through a fixed point. then prove that KM = M N . compute the area of ABHCA. (Chinese IMO TST 2002) Let ABCD be a convex quadrilateral. References [1] A. E the feet of the internal and external angle bisectors of angle ∠A on BC respectively.Problem 8. CZ concur at the orthocenter H. BC1 A1 . then prove that the common chord of circles C1 and C2 passes through the midpoint of BC. P = AC ∩ BD. Problem 10. Skiadas. Show that the lines RP. [3] www. cut BC at K. AB of a triangle ABC respectively.mathlinks. CA. Consider the midpoints R. C2 6= C). and denote by K and L the points of intersection between the parallel line from D to EZ with AB and AC respectively. Problem 9. The circumcircles of triangles AB1 C1 . CA. L respectively. Points A3 . (Nikolaos Rapanos) Let ABC be a triangle inscribed in circle Γ. Also denote by C1 the circle of diameter DE and by R the second point of intersection between C1 and Γ. Prove that ∠BOC = ∠AOD. [2] Cosmin Pohoata. If AR and the tangent to Γ at point A. (IMO-SL 1998. Let E = AB ∩ CD. AIMO 2007/TST 2-P2) Points A1 . BC and the circumference of triangle GKL concur. and let O be the foot of the perpendicular from P to the line EF .ro 14 . B1 . F = AD ∩ BC. B2 . Euclidean Geometry. Call D. Greek Mathematical Olympiad 1999) Let a triangle ABC and their heights AD. AB respectively. C1 are chosen on the sides BC. B3 . CA1 B1 intersect the circumcircle of triangle ABC again at points A2 . (IMO-SL 2006 G9. Call G the intersection point of BC and EF . where C2 is any circle passing through K and L. Harmonic Division and its Applications. BE. C2 respectively (A2 6= A. Prove that the triangles A2 B2 C2 and A3 B3 C3 are similar. B2 6= B. P of the segments AH. ZE. B1 . C1 with respect to the midpoints of the sides BC. C3 are symmetric to A1 . Problem 11. 1973.
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