HMMT 二月 2026 · 团队赛 · 第 6 题
HMMT February 2026 — Team Round — Problem 6
题目详情
- [40] Let ABCD be a cyclic quadrilateral with circumcenter O , and let M be the midpoint of AB . ◦ Suppose that ∠ CM D = 90 and AD ⊥ BC . Prove that the reflection of O over AB lies on line CD .
解析
- [40] Let ABCD be a cyclic quadrilateral with circumcenter O , and let M be the midpoint of AB . ◦ Suppose that ∠ CM D = 90 and AD ⊥ BC . Prove that the reflection of O over AB lies on line CD . Proposed by: Tiger Zhang Answer: Solution 1: E D P N C B A M O Let AD and BC intersect at E . If E lies on rays DA and CB , then M lies inside the circle with ◦ diameter CD , so ∠ CM D > 90 , a contradiction. Thus, E lies on rays AD and BC . Claim 1. Line CD is the perpendicular bisector of EM . ◦ Proof. Since ∠ CED = ∠ CM D = 90 , quadrilateral CEDM is cyclic. We have ◦ ∠ CM E = ∠ CDE = 180 − ∠ CDA = ∠ ABE = ∠ BEM, ∼ so CE = CM . Analogously, DE = DM , so △ CDE △ CDM , implying CD is the perpendicular = bisector of EM . Let N be the midpoint of CD ; notice that N is the circumcenter of CEDM . Let CD and M O intersect at P . It suffices to show M O = M P . Claim 2. M N = M P . Proof. Assume WLOG that D , N , P , and C are collinear in this order. We have ◦ ∠ N P M = 90 − ∠ EM P = ∠ AM E = 2 ∠ BEM = ∠ CN M, so M N = M P . ◦ Notice that ∠ ON P = 90 . Since the midpoint of OP is the unique point that lies on both OP and the perpendicular bisector of N P , we conclude M is this midpoint, so M O = M P , as desired. Solution 2: ©2026 HMMT X D P C O 1 B A M O 2 O Let E be the intersection of AD and BC . Similar to the first solution, E lies on ray AD and BC . Let ω , ω be the circumcircles of △ ADM , △ BCM with centers O , O respectively. 1 2 1 2 Claim 3. ω and ω are symmetric with respect to point M , and both are tangent to EM . 1 2 Proof. Because M DEC and ABCD are cyclic, we have ∠ EM C = ∠ EDC = ∠ EBA and ∠ EM D = ∠ ECD = ∠ EAM . Therefore, ω , ω are tangent to EM . 1 2 Note that the reflection of ω across M must be a circle passing through M and B and tangent to 1 EM . Therefore, the reflection of ω across M is ω . 1 2 Claim 4. O O CD is a parallelogram. 1 2 ◦ ◦ Proof. Notice that ∠ M O D + ∠ M O C = 2 ∠ M AD + 2 ∠ M BC = 2(180 − ∠ AEB ) = 180 . Thus, 1 2 DO ∥ CO . Combined with the fact that ω , ω have same radius because they are symmetric across 1 2 1 2 BC , we get that O O CD is a parallelogram. 1 2 ◦ Finally, let P be the reflection of O across AB . Since ∠ O OO = 90 , we have P M = OM = O D = 1 2 1 O C . Moreover, 2 ◦ ◦ ◦ ∠ P M O = 90 + ∠ BM O = 90 + (90 − ∠ XM B ) = 2 ∠ CBA = ∠ CO M. 2 2 2 Thus, P M O C is an isosceles trapezoid, implying that P C ∥ O O . Combined with the fact that 2 1 2 O O CD is a parallelogram, we have that points P , C , and D are collinear, as desired. 1 2