HMMT 二月 2025 · 冲刺赛 · 第 29 题
HMMT February 2025 — Guts Round — Problem 29
题目详情
- [16] Points A and B lie on circle ω with center O . Let X be a point inside ω . Suppose that XO = 2 2, ◦ XA = 1, XB = 3, and ∠ AXB = 90 . Points Y and Z are on ω such that Y ̸ = A and triangles △ AXB and △ Y XZ are similar with the same orientation. Compute XY .
解析
- [16] Points A and B lie on circle ω with center O . Let X be a point inside ω . Suppose that XO = 2 2, ◦ XA = 1, XB = 3, and ∠ AXB = 90 . Points Y and Z are on ω such that Y ̸ = A and triangles △ AXB and △ Y XZ are similar with the same orientation. Compute XY . Proposed by: Ethan Liu 11 Answer: 5 1 ◦ Solution 1: Consider a rotation about X by 90 followed by a homothety with ratio that sends 3 1 ′ ′ B to A . This sends ω to ω with radius of the radius of ω and center O . Since A is the image of 3 B under this rotation, we know A lies on both circles; the same argument shows Y must lie on both ′ ′ ′ circles. Thus, Y is the reflection of A over OO . In particular, this means that XY = AX , where X ′ is the reflection of X over OO . ′ ′ Let M be the midpoint of AB . Note that because △ OXO ∼ △ BXA , we also have △ XOX ∼ ′ ′ △ XM A , as they are both isosceles and ∠ XOX = 2 ∠ XOO = 2 ∠ XBA = ∠ XM A . This implies that ′ ′ XA 2 √ △ XOM ∼ △ XX A . Thus, we know that AX = OM · = OM . It remains to compute OM ; XM 10 √ 2 noting that the distance between O and the foot from X to AB is 10, and that the altitude of 5 √ 3 △ AXY has length 10, we get that the distance from O to AB is 10 s 2 √ √ 2 √ √ 3 2 11 10 + 2 2 − 10 = 10 10 5 10 11 ′ by the Pythagorean theorem, which means that XY = AX = . 5 M B A ′ O Y X ′ X O Z Solution 2: P Y B M A ′ X X N O Z Let M be the midpoint of AB . We will find M O first. √ Let the internal bisector of ∠ AXB intersect ⊙ ( AXB ) at P . From Ptolemy, XP = 2 2 = XO. Let ′ ′ X be the foot of altitude from X to M O . Observe that O is the reflection of P across XX . By the 3 ′ √ area of △ AXB , we have X M = . Therefore, 10 √ 11 10 ′ ′ ′ ′ ′ M O = OX + X M = X P + X M = 2 X M + M P = . 10 ◦ By the spiral similarity △ XAB 7 → △ XY Z , we have that AY ⊥ BZ and ∠ AOB + ∠ Y OZ = 90 . √ 11 10 11 Therefore, △ XAM ∼ △ XY N where N is the midpoint of Y Z . Thus, Y Z = and XY = . 5 5 − ′ ′ Note: if △ XAB ∼ △ XY Z , just consider another △ XY Z caused by reflecting △ XY Z across XO .