Mathematical Excalibur

Volume 11, Number 4November 2006 – December 2006 Olympiad Corner The 9th China Hong Kong Math Olympiad was held on Dec. 2, 2006. The following were the problems. Problem 1. Let M be a subset of {1, 2, …, 2006} with the following property: For any three elements x, y and z (x < y < z) of M, x + y does not divide z. Determine the largest possible size of M. Justify your claim. Problem 2. For a positive integer k, let f1(k) be the square of the sum of the digits of k. (For example f1(123) = (1+2+3)2 = 36.) Let fn+1(k) = f1(fn(k)). Determine the value of f2007(22006). Justify your claim. Problem 3. A convex quadrilateral ABCD with AC ≠ BD is inscribed in a circle with center O. Let E be the intersection of diagonals AC and BD. If P is a point inside ABCD such that Pole and Polar Kin Y. Li Let C be a circle with center O and radius r. Recall the inversion with respect to C (see Mathematical Excalibur, vol. 9, no. 2, p.1) sends every point P≠O in the same plane as C to the image point P’ on the ray OP such that OP·OP’=r2. The polar of P is the line p perpendicular to the line OP at P’. Conversely, for any line p not passing through O, the pole of p is the point P whose polar is p. The function sending P to p is called the pole-polar transformation (or reciprocation) with respect to O and r (or with respect to C). (3) Let x,y,z be the polars of distinct points X,Y,Z respectively. Then Z = x∩y ⇔ z = XY. Proof. By La Hire’s theorem, Z on x∩y ⇔ X on z and Y on z ⇔ z = XY. (4) Let W, X, Y, Z be on C. The polar p of P = XY∩WZ is the line through Q = WX∩ZY and R = XZ∩YW. Proof. Let S, T be the poles of s = XY, t = WZ respectively. Then P = s∩t. By fact (3), S = x∩y, T = w∩z and p = ST. For hexagon WXXZYY, we have Q=WX∩ZY, S=XX∩YY, R=XZ∩YW, where XX denotes the tangent line at X. By Pascal’s theorem (see Mathematical Excalibur, vol. 10, no. 3, p.1), Q,S,R are collinear. Similarly, considering the hexagon XWWYZZ, we see Q,T,R are collinear. Therefore, p = ST = QR. Next we will present some examples using the pole-polar transformation. Example 1. Let UV be a diameter of a semicircle. P,Q are two points on the semicircle with UP<UQ. The tangents to the semicircle at P and Q meet at R. If S=UP∩VQ, then prove that RS⊥UV. X p O r P' Y Following are some useful facts: (1) If P is outside C, then recall P’ is found by drawing tangents from P to C, say tangent at X and Y. Then P’ = OP ∩XY, where ∩ denotes intersection. By symmetry, OP⊥XY. So the polar p of P is the line XY. Conversely, for distinct points X, Y on C, the pole of the line XY is the intersection of the tangents at X and Y. Also, it is the point P on the perpendicular bisector of XY such that O, X, P, Y are concyclic since ∠ OXP = 90°= ∠ OYP. (2) (La Hire’s Theorem) Let x and y be the polars of X and Y, respectively. Then X is on line y ⇔ Y is on line x. Proof. Let X’, Y’ be the images of X, Y for the inversion with respect to C. Then OX·OX’ = r2 = OY·OY’ implies X, X’, Y, Y’ are concyclic. Now X is on y ⇔ ∠ XY’Y = 90° ⇔ ∠ XX’Y = 90° ⇔ Y is on x. ∠PAB+ ∠PCB = ∠PBC + ∠PDC = 90o , prove that O, P and E are collinear. Problem 4. Let a1, a2, a3,… be a sequence of positive numbers. If there exists a positive number M such that for every n = 1, 2, 3, …, 2 2 2 a 12 + a 2 + ... + a n < M an +1 , P then prove that there exists a positive number M’ such that for every n = 1, 2, 3, …, a1 + a 2 + ... + a n < M ' a n +1 . Editors: ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK ଽ υ ࣻ (KO Tsz-Mei) గ ႀ ᄸ (LEUNG Tat-Wing) ‫ ؃‬୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST ֔ ᜢ ‫( ݰ‬NG Keng-Po Roger), ITC, HKPU Artist: S R P K U V Q ྆ ‫( ़ ؾ‬YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance. On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is January 25, 2007. For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed envelopes. Send all correspondence to: Dr. Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: [email protected] Solution (due to CHENG Kei Tsi). Let K=PQ∩UV. With respect to the circle, by fact (4), the polar of K passes through UP∩VQ=S. Since the tangents to the semicircle at P and Q meet at R, by fact (1), the polar of R is PQ. Since K is on line PQ, which is the polar of R, by La Hire’s theorem, R is on the polar of K. So the polar of K is the line RS. As K is on the diameter UV extended, by the definition of polar, we get RS⊥UV. Since OO1 is the perpendicular bisector of AC. Since AD⊥OX. Then ∠ PQD = ∠ PED = ∠ DBO. (1997 Chinese Math Olympiad) Let ABCD be a cyclic quadrilateral. Similarly.E. Since Γ1. (continued on page 4) We claim Q = G. No. by fact (1). Quadrilateral ABCD has an inscribed circle Γ with sides AB. C. A. Therefore. This implies E is the radical center. Then PE·PA = PD·PB = PG·PO. Since the tangents at A and D intersect at X. which implies Q. BC. ∆PAC. O are concyclic. we will have P = BD ∩ OG. Let AB∩CD =E. ∆PBD with centers O. see Mathematical Excalibur. By fact (3). Combining with OE⊥O1O2 proved above. By fact (4).) Among Γ. F lie on a circle in that order. A line through C cuts the circle with center O at D. BC. so PE is the radical axis of Γ1. the polars of E. P is on the polar of Q. Q = G since they are both the point of intersection (other than O) of the circumcircle of ∆BOD and the circle with diameter OC. Let Q = OP∩CH. p. P and E are collinear. 4. By fact (1). Since GK∩HL = P. E. If lines BC. HL respectively.G are concyclic. Γ2 be the circumcircles of quadrilateral ABCD. we get OP⊥EF. since E = AC∩BD. Γ2. Example 5. H. C is a point on AB extended. The tangents to the circle at the points A and D. Solution (due to WONG Chiu Wai). D are concyclic. 3. D. Solution. G are concyclic. AD∩BC = F and GK∩HL = P. If O is the center of Γ. AB is a diameter of a circle with center O. So OE⊥O1O2. which implies P. by fact (3). We first show that the polar of O1 with respect to Γ is line AC. If BC||EF. 06 – Dec. 4. Γ2 intersect at P. B. K. To show Q = G. Prove that the lines AD. E. EF intersect. Let AB∩CD = P and AD∩BC = Q. ∠ PDH and ∠ PEH are 90°. Example 3. then by symmetry. by fact (1). Let the tangents from Q meet the circumcircle of ABCD at E and F. the polar of E with respect to Γ is line O1O2. F are lines GK. AD. Prove that O. all we need to show is that Solution. For this. Γ2. the polar of P is line EF. By La Hire’s theorem. 2. F are collinear. note that ∠ PQH. E. two of the pairwise radical axes are lines AC and BD. Γ1. P K C O1 B A O2 E P D O C E G A O O1 D F B C Solution (due to WONG Chiu Wai). H. prove that O. EF are either parallel or concurrent. Line CF intersect the circumcircle again at G.A. we see O. the polar of P with respect to the circle having center O is the line through BA∩DE = C and AD∩EB = H. (1998 Austrian-Polish Math Olympiad) Distinct points A. the polar of P passes through AD∩BC = Q. P and E are collinear. is the line EF. Vol. A E B F C D Q P O H L D F ∠PAB+ ∠PCB= ∠PBC+ ∠PDC= 90o . then prove that OP⊥EF. B. the polar of X is line AD. E. lines BC and EF are perpendicular to line OX. Since P = AB∩CD. Let O be the center of the circle and X = AA∩DD∩BF∩CE. (2006 China Hong Kong Math Olympiad) A convex quadrilateral ABCD with AC ≠ BD is inscribed in a circle with center O. vol. Therefore. OF is a diameter of the circumcircle of ∆BOD with center O1. which by fact (1). by fact (4). the polar of X = CE∩BF passes through BC∩EF. L respectively. Let E be the intersection of diagonals AC and BD. BC and EF are concurrent in this case. (2006 China Western Math Olympiad) As in the figure below. note ∠ APC = 360°− ( ∠ PAB + ∠ PCB+ ∠ ABC) = 270°− ∠ ABC = 90° + ∠ ADC and so E Q H D G O1 F O B A C ∠ AO1C =2(180°− ∠ APC) =2(90°− ∠ ADC) =180°−2 ∠ ADC =180°− ∠ AOC. Page 2 Example 6. 11. CD. By the definition of polar. DA tangent to Γ at G. no. E. Consider the pole-polar transformation with respect to the inscribed circle. Then OP⊥CH.Mathematical Excalibur. Q. 06 Example 2. the lines BF and CE are concurrent. O2 respectively. Example 4. D. Let Γ. the polar of O2 with respect to Γ is line BD. Once this shown. then by fact (4). which implies O. P ∠ AOC + ∠ AO1C = 180°. Consider the pole-polar transformation with respect to the circumcircle of ABCD. Nov. O1. which implies PE⊥O1O2. If P is a point inside ABCD such that A G B E Solution. B A O F E D X C . Prove that P. Γ1. Let AE∩BD=P. we get BC||EF||AD. (Next we will consider radical axis and radical center. C)+f (B. Problem 263. (Due to Mihaiela Vizental and Alfred Eckstein. Beaverton. prove that ab p − ba p is divisible by 6p. Italy) and Fai YUNG. Prove that there is a unique positive integer A < n2 such that [n2/A] + 1 is divisible by n. …. ∏ sin k k =1 90 = 3 10 . where f ( x. we have 2i 2ieix 179 179 ∏ sin k = ∏ 2iω o k =1 k =1 90 90 ω −1 k k/2 179 . Greece. Prove that among any 13 consecutive positive integers. N respectively. The solutions should be preceded by the solver’s name. CHRYSSOSTOMOS (Larissa. Department of Mathematics. USA). Greece. Note f ( x. home (or email) address and school affiliation. Show that there is a rational number q such that sin3 sin6 sin9 Lsin87 430 9 3 = 30 sin30o sin60o ∏sin3mo sin( 60o − 3mo ) sin( 60o + 3mo ) 4 m=1 3 = 40 sin9o sin18o sin27o Lsin81o 4 3 = 40 sin9o cos9o sin18o cos18o sin27o cos27o sin36o cos36o sin45o 4 3 2 = 85 sin18o sin36o sin54o sin72o 2 3 2 = 85 sin18o cos18o sin36o cos36o 2 3 2 = 87 sin36o sin72o 2 = 3 2 10− 2 5 10+ 2 5 3 = 89 10. C are in the interval (0. Prove that OM = ON. Determine (with proof) the maximum of Let ω = e 2πi / 180 . AB at M.A. c > 0. The deadline for submitting solutions is January 25. 2007. we get 3 1 1 1⎞ (a + b + c ) 2 ⎛ ⎜ + + ⎟ ≥ 6. Problem 261. 4 4 287 2 3 o o o o Solution. there exists a cell with no ant. then f (A.A)+f (C. The Hong Kong University of Science & Technology. z ) = 4 sin x + 3 sin y + 2 sin z . This leads to ⎡ n2 ⎤ n2 ⎡ n2 ⎤ kn − 1 = ⎢ ⎥ ≤ < ⎢ ⎥ + 1 = kn. x2. For positive integers m.A. If n > 2. 11. 4 Using this. then 1 ≤ A < n2 = 4 and only A = 3 works. n =0 k =1 ix − ix 2 ix Using sin x = e − e = e − 1 . j =1 n where x1.B. Kipp JOHNSON (Valley Catholic School. Teacher. Clear Water Bay. Jeff CHEN (Virginia. Let the tangent at D to the circumcircle intersect line BC at P. Koyrtis G. where in each cell. Koyrtis G. Hong Kong. Problem 265. (Source: 1993 Jiangsu Math Contest) Solution. Romaina) Show that if A. consider a (2m+1)×(2n+1) table. Note sin 3θ = sin θ cos 2θ + cos θ sin 2θ = sin θ (cos 2 θ − sin 2 θ + 2 cos 2 θ ) 3 1 = 4 sin θ ( cos 2 θ − sin 2 θ ) 4 4 = 4 sin θ sin(60o − θ ) sin(60o + θ ). 06 – Dec. Italy) and D. teacher). ⎝a b c⎠ sin1 sin 2 Lsin 89 sin 90 = q 10. there is exactly one ant. Let n > 1 be an integer. USA).R. 20 Math Problem Group (Roma. Multiplying by 2 on both sides. Greece. ⎣ ⎦ ∏ sin k o = ∏ k =1 k =1 90 ω k −1 2 o = P(1) 2179 = 90 2178. Prove that after that moment. Let line PO intersect lines AC.Mathematical Excalibur. Nov. a = 1/(2r + 3s + 4t). Arad. b. Kowloon. sinθ sin(60o − θ ) sin(60o + θ ) = sin 3θ . USA).A. by the AM-HM inequality. n.B) ≥ 3. Problem 258. Li. 2 sin x + 3 sin y + 4 sin z So. assume [n2/A]+1=kn for some positive integer k. Let S be the left-handed side. Samuel Liló Abdalla (Brazil). we have 60o − no ) sin( 60o + no ) S = sin30o sin60o ∏sinno sin( n=1 29 = ∑ ( x 4j − x5j ) . Therefore.R. b. Koyrtis G. π/2). 3⎝a b c⎠ (*) Let r = sin A. Kin Y. Let O be the center of the circumcircle of ∆ABC and let AD be a diameter. teacher). Oregon. CHRYSSOSTOMOS (Larissa. xn are nonnegative real numbers whose sum is 1. Problem 264. No. 06 Page 3 We claim the unique number is A =n+1. Please send submissions to Dr. Jeff CHEN (Virginia. 2 sin x + 3 sin y + 4 sin z For a. 20 Math Problem Group (Roma. Jeff CHEN (Virginia. s = sin B. ⎡ n2 ⎤ n2 + 1 ≥ ⎢ ⎥ + 1 ≥ n. G.C. teacher) and Fai YUNG. Then ∏ sin k = ∏ sin k = ∏ 2iω o o k =1 k = 91 k = 91 2 179 90 179 ωk −1 k/2 A≤ . every ant moves to a horizontal or vertical neighboring cell. Then P( z ) = ∑ z n =∏ ( z − ω k ). G. 4. z ) + 1 = 6 sin x + 6 sin y + 6 sin z . t = sin C. Problem 262. n2 1 = n +1+ . At a certain moment. ***************** Solutions **************** Problem 256. This is a contradiction as kA is an integer and cannot be strictly between n and n + 1. where [x] denotes the greatest integer less than or equal to x. USA). b = 1/(2s + 3t + 4r) and . 289 The case A = n does not work because [n2/n] + 1 = n + 1 is not divisible by n when n >1. CHRYSSOSTOMOS (Larissa. B. n −1 n −1 The case A = n + 1 works because ⎡ n2 ⎤ ⎢ n + 1⎥ + 1 = (n − 1) + 1 = n. Problem 257. If n = 2. y . For a prime number p > 3 and arbitrary integers a. o o o o Solution 1. ⎣ A⎦ A ⎣ A⎦ Solution 2. then [n2/A]+1 divisible by n implies Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration. we have 1 1 1⎞ (a + b + c )⎛ ⎜ + + ⎟ ≥ 9. which implies n < kA ≤ (n2+A)/n < n+1. Vol. For 0 < A < n. So A ≤ n + 1. one of them has sum of its digits (in base 10) divisible by 7. y . A ⎣ A⎦ This leads to Also. Jeff CHEN (Virginia. By (1). let his friends be M6j.D. There are 15m pairs of friendship. Consider the pole-polar transformation with respect to the incircle. (1998 IMO) Let I be the incenter of triangle ABC. C are concyclic. which is the polar of B. m = 2k for some positive integer k.Mathematical Excalibur. Through D. Due to tangency. BC. M1k and M2k for all k ≠ j. = ( B' K + KC ) + ( B ' M + MA) = KC + MA. hence student j is good. It is easy to check 1 to 25 are good. 2. Observe that B is sent to B’ = IB∩MK under the inversion with respect to the incircle. which implies ∠ AB’C > 90° and hence ∠ RIS < 90°. By (1) and (2). Then Pole and Polar (continued from page 2) Using (*). k. For the other n − k good students. So ∠ EPM = ∠ DEM. Prove that the circumcircle of ∆PQR passes through the midpoint M of side BC. Then the polar of S is B’C.B. Problem 260. every one of them has at least k worse friends. R. Jeff CHEN (Virginia. For the other n − k good students. By the intersecting chord theorem. Since B’ is on line MK. USA) and Fai YUNG. Suppose each student has m friends and n is the maximum number of good students. Then ∠ RIS = 180° − ∠ AB’C. Then 2 B 'T = B ' C + B ' A For m even. n ≤ 25. CF be the altitudes of acute triangle ABC. To finish. By the definition of polar. 2 ⎠ ⎝ k +1 ∠ EBM = ∠ EPM + ∠ BEP ∠ BEM = ∠ DEM + ∠ BED. A. Q. USA). Since KC and MA are nonparallel. (2) ∠ AEB = 90° = ∠ ADB implies A. we get RD·QD=MD·PD. Problem 259. j =1 k Solving for n. S are collinear. number the students 1. E. Nov. C. M(i−1)k and M(i+1)k for all k ≠ j. Then B. Every student has the same number of friends in the class. we also get IR⊥B’A. Consider the 6×5 matrix M with Mij= 5(i − 1) + j. the polars of B. So ∠ ACB = ∠ ARQ. let his friends be M6j. Since MK||RS. ∠ BEP = ∠ BCF = 90° − ∠ ABC = ∠ BAD = ∠ BED. Call a student good if his ability is better than more than half of his friends. m = 2k − 1 for some positive integer k. R B B' I K S M A L T C A E F P B D R M Q C Observe that (1) ∠ BFC = 90° = ∠ BEC implies B. 2 2 2 Therefore. we get IS⊥B’C. we get f ( A.E concyclic. (Source: 1994 Hubei Math Contest) Solution. 2⎠ Solution. 2. C) + f ( B. hence student j is good. Then 2⎛ 1 1 1⎞ ⎜ + + ⎟ = 6r + 6 s + 6t. in the odd case. A) + f (C. where friendships are mutual. AB respectively. …. L and M respectively. Q. Therefore. Let P be the intersection of lines EF and CB. F. Determine the maximum possible number of good students in this class. B'T < KC + MA CL + AL AC = = . their polars concur at C. 11. student j has at least (2k−j) ≥ k > m/2 worse friends. By the converse of the intersecting chord theorem. C concyclic. j =1 k Solving for n. Let AD. 06 – Dec. we get ⎛ 15 n ≤ 30. The line through B parallel to MK meets the lines LM and LK at R and S respectively. For Mij with 1 < i < 6. In a class of 30 students. …. their polars concur at B’. we need to take k = 5 (so m = 9). student j has at least (2k + 1 − j) > k = m/2 worse friends. RD·QD=BD·CD (*) Since ∠ BEC = 90° and M is the midpoint of BC. K. draw a line parallel to line EF intersecting line AB at R and line AC at Q.S are collinear. CA and AB at K. L. by La Hire’s theorem. Let T be the midpoint of AC. Example 7. 4. we get MB = ME and ∠ EBM = ∠ BEM. Commended solvers: Koyrtis G. we have ∠ ACB = ∠ AFE. For M6j. we get ∠ AFE = ∠ ARQ.5 − 30 < 26. B. From EF||QR. For j = 1. Let the incircle of ABC touch the sides BC. Since R. Greece. For j = 1. B) + 3 6r + 6s + 6t 6r + 6s + 6t 6r + 6s + 6t = + + 2r + 3s + 4t 2s + 3t + 4r 2t + 3r + 4s ≥ 6. BE. R. CA. P. 3⎝a b c⎠ Page 4 Using (*). we get ⎛ 31 k + 1 ⎞ n ≤ 31. Vol. K. 2. We have ME/MP = MD/ME and so MB2=ME2=MD·MP=MD(MD+PD) =MD2+MD·PD. Then ∑ (2k + 1 − j ) + (n − k )(k + 1) ≤ 15 ⋅ 2k. For m odd. No. CHRYSSOSTOMOS (Larissa. B’ is inside the circle with diameter AC. Then MD·PD = MB2− MD2 = (MB−MD)(MB+MD) = BD(MC+MD) = BD·CD. M are concyclic. M are lines MK. For an example of n = 25. For M1j. every one of them has at least k + 1 worse friends. Next. teacher). . so the polar of B’ is line RS. 30 from best to worst ability (no two with the same ability). L intersect at C and since L. since the polars of K.5 − ⎜ + ⎝k k⎞ ⎟ ≤ 30.5 − ⎜ + ⎟ ≤ 31. The result follows. k. ….B. Then right triangles EPM and DEM are similar. By a similar reasoning. B is on the polar of B’. let his friends be Mij and M5k for all i < 6 and k ≠ j. 06 c = 1/(2t + 3r + 4s). (Source: 1998 Hubei Math Contest) Solution. Now ∑ (2k − j ) + (n − k )k ≤ 15(2k − 1).5 − 62 < 24. we will show B’ is inside the circle with diameter AC. Prove that angle RIS is acute.