Two similar geometry problems Nguyen Van Linh Highshool for gifted student-Hanoi University of science Abstract In this note we give nice proofs of two similar geometry problems and their applications. The strong ties between them makes us suprised. Let begin with problem which appeared in Vietnam IMO training 2009: Problem 1: Given triangle ABC and its circumcircle (O), its orthocenter H. P is an arbitrary point inside triangle ABC. Let A 1 , B 1 , C 1 be the second intersections of AP, BP, CP and (O), A 2 , B 2 , C 2 be the reflections of A 1 , B 1 , C 1 across the midpoints of BC, CA, AB, respectively. Prove that H, A 2 , B 2 , C 2 are concylic. Solution: First we will express a lemma: Lemma 1 ∠C 2 B 2 A 2 = ∠AP C Proof: B' C' A' M A 2 X N B 2 C 2 Z Y C 1 B 1 A 1 O P A B C 1 Denote M, N the reflections of A 2 , C 2 across the midpoint of AC then ∆MB 1 N is the reflection of ∆A 2 B 2 C 2 across the midpoint of AC. Since AM//CA 2 , AM = CA 2 and BA 1 //CA 2 , BA 1 = CA 2 we get AMA 1 B is a parallelogram. Similarly, BC 1 NC is a parallelogram too. Let A , B , C be the midpoints of AA 1 , BB 1 , CC 1 then A , C are the midpoints of BM, BN, respectively. A homothety H 1 2 B : M → A , B 1 → B , N → C therefore ∆MB 1 N → ∆A B C . Thus ∠M B 1 N = ∠A B C . On the other side, ∠OA P = ∠OB P = ∠OC P = 90 o hence O, A , B , C , P are concyclic, which follows that ∠A 2 B 2 C 2 = ∠MB 1 N = ∠A B C = ∠A P C . We are done. Return to problem 1: B 3 A 3 C 3 B 4 C 4 A 4 H C 2 B 2 A 2 Z Y X C 1 B 1 A 1 O P A B C Construct three lines through A 1 , B 1 , C 1 and perpendicular to AA 1 , BB 1 , CC 1 , respectively. They inter- sect each other and make triangle A 3 B 3 C 3 , intersect (O) again at A 4 , B 4 , C 4 . Note that AA 4 , BB 4 , CC 4 are diameters of (O) therefore A 4 , C 4 are the reflections of H across the mid- points of BC, AB. We get HA 2 A 4 A 1 , HC 2 C 4 C 1 are parallelograms, which follows that HA 2 //A 1 A 4 , HC 2 //C 1 C 4 . But PC 1 B 3 A 1 is a cyclic quadrilateral and applying lemma 1 we have ∠C 2 HA 2 = ∠C 1 B 3 A 1 = ∠C 1 P A 1 = ∠CAP = ∠C 2 B 2 A 2 or H, A 2 , B 2 , C 2 are concyclic. Our proof is completed. Corollary 1: Let A 5 , B 5 , C 5 be the second intersections of AH, BH, CH and (A 2 B 2 C 2 ), respectively. Then ∆A 5 B 5 C 5 ∼ ∆ABC. Proof: Since A 5 , B 5 , C 5 , H are concyclic then ∠C 5 A 5 B 5 = ∠C 5 HB 5 = ∠BAC. Similarly we get the result. Note: Corollary 1 is only a special case of this problem: 2 Given triangle ABC and an arbitrary P inside. Let A 1 B 1 C 1 is the pedal triangle of triangle ABC wrt P. Q is another arbitrary point inside triangle ABC. Let A 2 , B 2 , C 2 be the projections of Q on P A 1 , P B 1 , P C 1 then ∆A 2 B 2 C 2 ∼ ∆ABC Corollary 2: A 2 A 5 , B 2 B 5 , C 2 C 5 are concurrent. Proof: A 4 C 4 B 4 S B 3 C 3 A 3 B 5 A 5 C 5 H C 2 B 2 A 2 C 1 B 1 A 1 O P A B C First we will show that A 3 P//AH. Since C 4 B 4 CB is a rectangle we get C 4 B 4 //BC. But A 3 C 1 P B 1 is cyclic quadrilateral thus ∠P A 3 B 1 = ∠P C 1 B 1 = 90 o − ∠A 3 C 1 B 1 = 90 o − ∠A 3 B 4 C 4 Therefore A 3 P ⊥ C 4 B 4 or A 3 P//AH. On the other side, H, A 2 , B 2 , C 2 , A 5 , B 5 , C 5 are concyclic then ∠C 2 A 2 A 5 = ∠C 2 HA 5 = ∠(C 1 C 4 , AH) = ∠(C 1 C 4 , A 3 P ) = ∠B 3 A 3 P Similarly, ∠A 5 A 2 B 2 = ∠C 3 A 3 P, ∠C 5 C 2 A 2 = ∠B3C 3 P, ∠C 5 C 2 B 2 = ∠A 3 C 3 P, ∠B 5 B 2 C 2 = ∠A 3 B 3 P, ∠B 5 B 2 A 2 = ∠C 3 B 3 P Applying Ceva-sine theorem we obtain: sin ∠C 2 A 2 A 5 sin ∠A 5 A 2 B 2 . sin ∠B 5 B 2 A 2 sin ∠B 5 B 2 C 2 . sin ∠C 5 C 2 B 2 sin ∠C 5 C 2 A 2 = sin ∠B 3 A 3 P sin ∠C 3 A 3 P . sin ∠C 3 B 3 P sin ∠A 3 B 3 P . sin ∠A 3 C 3 P sin ∠B3C 3 P = 1 Our proof is completed then. Next, we come to another problem : Problem 2: Let ABC be a triangle and (O) its circumcircle, P be an arbitrary point inside triangle ABC.AP, BP, CP intersects (O) again at A 1 , B 1 , C 1 . Let A 2 , B 2 , C 2 be the reflections of A 1 , B 1 , C 1 across the lines BC, CA, AB, respectively, H be the orthocenter of triangle ABC. Prove that H lies on the circumcircle of triangle A 2 B 2 C 2 . Solution: 3 First we will express three lemmas: Lemma 2: Given triangle ABC and its circumcircle (O). Let D be an arbitrary point on (O). A line d through D and d ⊥ BC.d ∩ (O) = {G} then AG is parallel to the Simson line of point D. Proof: G E F O A B C D Denote d ∩ BC = {E}, F the projection of D on AC then EF is the Simson line of D. Since DEF C is a cyclic quadrilateral we get ∠GAC = ∠GDC = ∠EF A. Therefore AG//EF . Lemma 3: Given triangle ABC and its circumcircle (O). Let E, F be arbitrary points on (O). Then the angle between the Simson lines of two points E and F is half the measure of the arc EF . Proof: LJ K G H I O A B C E F See on the figure, IJ, HG are the Simson lines of two points E and F , respectively. IJ ∩ HG = {K} Denote L the projection of K on BC. We have KL//F G and F HGC is a cyclic quadrilateral thus ∠GKL = ∠HGF = ∠ACF (1) Similarly, ∠LKJ = ∠EBA(2) 4 From (1) and (2) we are done. Lemma 4: Let ABC be a triangle and (O) its circumcircle, J be an arbitrary point inside triangle ABC.AJ, BJ, CJ intersects (O) again at D, E, F . Let D , E , F be the reflections of D, E, F across the lines BC, CA, AB, respectively. Then ∆DEF ∼ ∆D E F . Proof: L U S M T N R D' Q F' P E' F'' D'' E'' F E D O J A B C Let M, N be the reflections of F , D across the line AC. It is easy to see that EN = E D , EM = E F , MN = F D . So we only need to prove that ∆MEN ∼ ∆F ED. Now denote R = D N ∩ AC, Q = D D ∩ BC, U = RQ ∩ ED, S = BC ∩ ED, L = B E ∩ AC. We have RQ//ND hence ∠N DE = ∠RUE = ∠QDS − ∠D QR = 90 o − ∠BSD − ∠D CR = 90 o − ∠BSD − (∠ACB − ∠BCD) = 90 o − ∠ACB − EDC = 90 o − ∠ACB − ∠EBC = 90 o − ∠(AC, BE). Similarly, ∠MF E = 90 o − ∠(AC, BE) thus ∠N DE = ∠MF E(3) On the other side, ND MF = 2RQ 2T P = D C. sin C F A. sin A = DC. sin C F A. sin A = JC. sin C JA. sin A (4) Let D E F be the pedal triangle of triangle ABC wrt J. This result is well-known (it also appeared in Mathvn magazine No.3): ∆D E F ∼ ∆DEF . Therefore DE F E = D E F E = JC. sin C JA. sin A (5) From (3), (4) and (5) we get ∆DN E ∼ ∆FME. Then DE NE = F E ME or DE D E = F E F E Similarly we claim DE D E = F E F E = EF E F and the result follows. Return to our problem: 5 C 4 A 4 C 3 A 3 H B 2 A 2 C 2 B 1 C 1 A 1 O P A B C Denote A 3 , C 3 the reflections of H across the lines BC, AB, respectively then A 3 , C 3 lie on (O). Let A 1 A 2 ∩ (O) = {A 4 , A 1 }, C 1 C 2 ∩ (O) = {C 4 , C 1 } Since AA 3 //A 1 A 4 and A, A 3 , A 1 , A 4 are concyclic we get AA 3 A 1 A 4 is an isoceles trapezoid. But HA 3 A 1 A 2 is also an isoceles trapezoid too hence HA 2 //AA 4 . Similarly, HC 2 //CC 4 . Therefore ∠C 2 HA 2 = ∠(AA 4 , CC 4 ). Applying lemma 1, if we denote d a , d c the Simson lines of A 1 , C 1 then ∠(AA 4 , CC 4 ) = ∠(d 1 , d 2 ). Applying lemma 2, ∠(d 1 , d 2 ) = ∠C 1 B 1 A 1 . So ∠C 2 HA 2 = ∠C 1 B 1 A 1 . Applying lemma 3, ∆A 1 B 1 C 1 ∼ ∆A 2 B 2 C 2 then ∠C 1 B 1 A 1 = ∠C 2 B 2 A 2 . Therefore ∠C 2 HA 2 = ∠C 2 B 2 A 2 which follows that A 2 , B 2 , C 2 , H are concylic. We complete the proof. Note: Problem 2 is the generalization of the Fuhmann circle. Corollary 3: Let A 3 , B 3 , C 3 be the second intersections of AH, BH, CH and (A 2 B 2 C 2 ), respectively. Then A 2 A 3 , B 2 B 3 , C 2 C 3 concur at P . Proof: L A 3 B 3 C 3 H C 2 Z B 2 A 2 C 1 B 1 A 1 O P A B C 6 ∠C 2 A 2 A 3 = ∠A 3 HC 2 = ∠(AH, C 2 H) = 90 o − ∠(BC, C 2 H). Let L, Z be the projections of C 1 on BC, AB, respectively then LZ is the Simson line of C 1 . Applying lemma 2 we get HC 2 //LZ Therefore ∠(BC, C 2 H) = ∠BLZ = ∠BC 1 Z. Hence ∠C 2 A 2 A 3 = 90 o − ∠BC 1 Z = ∠C 1 BA = ∠C 1 CA. From corollary 1, we have ∆A 3 B 3 C 3 ∼ ∆ABC and from lemma 4, ∆A 2 B 2 C 2 ∼ ∆A 1 B 1 C 1 . Thus there exist a similarity transformation which takes A 3 C 2 B 3 A 2 C 3 to AC 1 BA 1 CB 1 and we have A 2 A 3 , B 2 B 3 , C 2 C 3 concur at P . Moreover, AP A 1 P = A 3 P A 2 P . But AA 3 //A 1 A 2 then applying Thales’s theorem we get P ≡ P. Our proof is completed then. Now, what does make problem 1 and 2 be similar? Let A 4 , B 4 , C 4 be the reflections of A 2 , B 2 , C 2 across the midpoints of BC, CA, AB. Since (BA 2 C) is the reflection of (ABC) across the midpoint of BC then A 4 ∈ (ABC) On the other side, BC is the midline and parallel to A 4 A 1 of triangle A 1 A 2 A 4 thus A 4 A 1 //BC, which follows that ∠CAA 1 = ∠BAA 4 . Similarly, ∠CBB 1 = ∠ABB 4 , ∠ACC 1 = ∠BCC 4 . Thus AA 4 , BB 4 , CC 4 concur at an isogonal conjugate point Q of P wrt triangle ABC. C 1 B 1 A 1 P B 3 A 3 C 3 H C 2 B 2 A 2 X C 4 B 4 A 4 O S A B C As that result above we came to the following conclusion: Problem 1 and 2 are equivalent. Therefore there are at least two solutions for each problem and its corollary! At the end we will prove another result about those nice problems: Corollary 4: Let I be the circumcenter of triangle A 2 B 2 C 2 . K is the reflection of H across I. AK, BK, CK cut (I) again at A 5 , B 5 , C 5 . Then A 3 A 5 , B 3 B 5 , C 3 C 5 are concurrent. Proof: 7 B 5 A 5 C 5 K B 3 C 3 A 3 H I C 2 A 2 B 2 O A B C Since KA 3 ⊥ AH we have ∠C 5 C 3 A 3 = ∠C 5 KA 3 = ∠KCB. Similarly, ∠C 5 C 3 B 3 = ∠KCA, ∠A 3 B 3 B 5 = ∠KBC, ∠B 5 B 3 C 3 = ∠ABK, ∠B 3 A 3 A 5 = ∠CAK, ∠C 3 A 3 A 5 = ∠BAK. Then applying Ceva-sine theorem for triangle A 3 B 3 C 3 we are done. References: [1] Ha Khuong Duy- Vietnam IMO training 2009, problem 142, 167. 8 . problems Nguyen Van Linh Highshool for gifted student-Hanoi University of science Abstract In this note we give nice proofs of two similar geometry problems and their applications. The strong ties between. A line d through D and d ⊥ BC.d ∩ (O) = {G} then AG is parallel to the Simson line of point D. Proof: G E F O A B C D Denote d ∩ BC = {E}, F the projection of D on AC then EF is the Simson line. across the lines BC, AB, respectively then A 3 , C 3 lie on (O). Let A 1 A 2 ∩ (O) = {A 4 , A 1 }, C 1 C 2 ∩ (O) = {C 4 , C 1 } Since AA 3 //A 1 A 4 and A, A 3 , A 1 , A 4 are concyclic we get