HMMT 二月 2025 · 团队赛 · 第 3 题
HMMT February 2025 — Team Round — Problem 3
题目详情
- [30] Let ω and ω be two circles intersecting at distinct points A and B . Point X varies along ω , 1 2 1 and point Y on ω is chosen such that AB bisects the angle ∠ XAY . Prove that as X varies along ω , 2 1 the circumcenter of △ AXY (if it exists) varies along a fixed line.
解析
- [30] Let ω and ω be two circles intersecting at distinct points A and B . Point X varies along ω , 1 2 1 and point Y on ω is chosen such that AB bisects the angle ∠ XAY . Prove that as X varies along ω , 2 1 the circumcenter of △ AXY (if it exists) varies along a fixed line. Proposed by: Pitchayut Saengrungkongka Solution 1: A O O 1 2 O B X Y Let O , O , and O be the centers of ω , ω , and the circumcircle of △ AXY , respectively. 1 2 1 2 We claim that triangle OO O is isosceles with OO = OO , and thus in particular O always lies on 1 2 1 2 the perpendicular bisector of O O . 1 2 To this end, observe that OO ⊥ AX and O O ⊥ AB , so ∠ OO O = ∠ XAB . Analogously, 1 1 2 1 2 ∠ OO O = ∠ Y AB . So indeed OO O is isosceles, and we are done. 2 1 1 2 Remark. One may also consider the antipodes of A on ω and ω for an equivalent but more natural 1 2 angle-chase. Solution 2: Q A ∗ P N A N ∗ ∗ M Y B ∗ X O X ∗ B Y ′ M A ′ ′ Let A be the A -antipode in circle ( AXY ). It suffices to show that A lies on a fixed line. We will show that this line is one that is parallel to AB . Let M be the second intersection of line AB with circle ( AXY ), and let N be the antipode of M on ′ this circle. Since AM A N is a rectangle with N lying on the line through A perpendicular to AB , it suffices to show that N is fixed (independent of X and Y ). ∗ To this end, take an inversion at A with arbitrary radius, denoting images with • 7 → • . ∗ ∗ ∗ ∗ Observe that X and Y lie on the fixed lines ℓ = ω and ℓ = ω . Let ℓ be the line through A 1 2 1 2 ∗ perpendicular to AB , and suppose that ℓ and ℓ intersect ℓ at P and Q , respectively. 1 2 ∗ ∗ ∗ ∗ Since ∠ XAB = ∠ BAY , we have ∠ X AB = ∠ B AY . Circles ω , ω , and ( AXY ) are mapped to 1 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ lines B X , B Y , and X Y . As AN ⊥ AB , it follows that N is the intersection of X Y and ℓ . ∗ B ∗ ∗ ∗ ∗ ∗ ∗ Finally, observe that ( N , A ; P, Q ) = ( N , M ; X, Y ) is a harmonic bundle, as AM bisects ∠ X AY ∗ ∗ ◦ ∗ and ∠ M AN = 90 . Since A , P , and Q are fixed, so is N . Thus N is fixed, and O lies on the perpendicular bisector of AN , which is also fixed. Remark. An alternative approach to the last paragraph is to recall Blanchet’s theorem , which states ∗ ∗ ∗ ∗ that P Y , QX , and AB are concurrent. By Ceva and Menelaus, we get that ( N , A ; P, Q ) = − 1. ′ Remark. If one projects the kite/harmonic quadrilateral N XM Y from A onto the line through the ′ antipodes defined in the previous remark, we obtain that A N (a line parallel to AB ) passes through the midpoint of the two antipodes, directly finishing.