HMMT 二月 2024 · 团队赛 · 第 8 题
HMMT February 2024 — Team Round — Problem 8
题目详情
- [50] Let P be a point in the interior of quadrilateral ABCD such that the circumcircles of triangles P DA , P AB , and P BC are pairwise distinct but congruent. Let the lines AD and BC meet at X . If O is the circumcenter of triangle XCD , prove that OP ⊥ AB .
解析
- [50] Let P be a point in the interior of quadrilateral ABCD such that the circumcircles of triangles P DA , P AB , and P BC are pairwise distinct but congruent. Let the lines AD and BC meet at X . If O is the circumcenter of triangle XCD , prove that OP ⊥ AB . Proposed by: Pitchayut Saengrungkongka Solution 1: D P ′′ ′ C C C ′ A B B X Because the circles have equal radii, ∠ P DA = ∠ ABP , so if ( P DA ) intersects line AB again at a point ′ ′ ′ ′ B , then we have ∠ P B B = ∠ P BB , which means P B = P B , similarly for the second intersection of ′ ( P CB ) with AB , A ; thus, ( P DA ) and ( P CB ) are congruent mirror images across the P -altitude, as ′ ′ they are ( P AB ) and ( P BA ) , respectively. ′ ′ Consider C , the reflection of C across the P -altitude. We want to prove that C lies on ( XCD ) , as ′ then the circumcenter of ( XCD ) will lie on the perpendicular bisector of CC . Because of our earlier ′ observation, C , D, P, A are concyclic. We present two approaches to finishing the angle chase from here: ′′ ′ • Add point C , the intersection of CC with ( XDA ) . Because AB is parallel to the line between ′′ ′′ the centers, and so is C C , then ABCC is a parallelogram; thus, ′ ′ ′′ ′ ′ ∠ C DA = ∠ C C A = ∠ C CB = ∠ C CX. ′ ′ • Add point B , the reflection of B over the P -altitude. Note that B lies on ( XDA ) ; in particular, ′ ′ ′ ′ C CBB is an isosceles trapezoid, as C B is the reflection of CB over the P -altitude of △ P AB . Thus, ′ ′ ′ ′ ′ ′ ′ ∠ C DA = 180 − ∠ C B A = 180 − ∠ C B B = ∠ C CB = ∠ C CX. ′ Remark. It is possible to do the last angle chasing without adding any additional points (beyond C ). However, the details are much messier. Solution 2: ′ H ′′ D ′ D ′ C ′′ C P ′ X ′ ′ B A Invert about P . Because the circles ( P DA ) , ( P AB ) , ( P BC ) all have equal radii and pass through P , ′ ′ ′ ′ ′ ′ ′ the resulting lines D A , A B , B C are equal distances away from P ; letting H be the intersection of ′ ′ ′ ′ ′ ′ ′ lines D A and B C , it follows that P is an incenter or excenter of △ A B H . Also, in the original diagram, the P -altitude of △ P AB includes the second intersection of the circles ( P DA ) and ( P CB ) ′ ′ (as in the first solution); thus this P -altitude inverts to line P H . Finally, X is the intersection of ′ ′ ′ ′ ( P D A ) and ( P B C ) . ′ ′ ′ Note that the center of ( XCD ) lies on the P -altitude of P AB iff the inverse of the center of ( X C D ) ′ ′ ′ ′ ′′ does. Thus we want to show that the center of ( X C D ) lies on H P . Let D be the second intersection ′ ′ ′ ′ ′ of X C D with B H . Then ′ ′′ ′ ′ ′′ ′ ′′ ′ ∡ ( D D , H P ) = ∡ D D H + ∡ D H P ′ ′ ′ ′ ′ = ∡ D X C + ∡ B H P ′ ′ ′ ′ ′ ′ = ∡ D X P + ∡ P X C + ∡ P H A ′ ′ ′ ′ ′ ′ = ∡ D A P + ∡ P B C + ∡ P H A ′ ′ ′ ′ ′ ′ = ∡ P A B + ∡ P B H + ∡ P H A ◦ = 90 ′ ′ ′ (where the last step follows from the fact that P is an incenter or excenter of △ A B H ), and ′ ′ ′′ ◦ ′′ ′ ′ ′′ ∡ H D D = 180 − ∡ D HD − ∡ D D H ◦ ′′ ′ ′′ ′ ′ ′′ ′ = 180 − ∡ D H P − ( ∡ D H P + ∡ D D H ) ◦ ′′ ′ = 90 − ∡ D H P ′ ′′ ′ = ∡ D D H , ′ ′ ′′ ′ ′ ′′ so △ H D D is isosceles, and thus H P is the perpendicular bisector of D D . Thus the center of ′ ′ ′ ′′ ′ ( X C D D ) lies on H P , which means we’re done. Solution 3: D P C A B ′ B X ′ A ′ ′ Let A be the other intersection of line P A with ( P DX ) and B be the other intersection of P B with ′ ′ ( P CX ) . Consider circles ( P A B ) and ( XCD ) . Note that A and B have equal power with respect ′ ′ to both circles, because of ( P DA X ) and ( P CB X ) . Thus, AB is the radical axis of the two cricles. However, ′ ∡ P A X = ∡ P DX = ∡ P DA = ∡ ABP and ′ ∡ XB P = ∡ XCP = ∡ BCP = ∡ P AB, ′ ′ where the last step follows from the fact that the circles have equal radii. Because ∡ B P A = ∡ BP A , ′ ′ ′ ′ it follows that A , X, B are collinear, and in fact △ P AB ∼ △ P B A . In particular, this means ′ ′ that the P -altitude of △ P AB passes through the circumcenter of △ P B A , as the circumcenter and ′ ′ orthocenter are isogonal conjugates. Thus, as the circumcenter of P A B lies on the P -altitude, and ′ ′ the line between the centers of ( P A B ) and ( XCD ) must be perpendicular to their radical axis AB , then the circumcenter of ( XCD ) must lie on the P -altitude as well, completing the proof.