HMMT 二月 2022 · 几何 · 第 10 题
HMMT February 2022 — Geometry — Problem 10
题目详情
- Suppose ω is a circle centered at O with radius 8. Let AC and BD be perpendicular chords of ω . Let P be a point inside quadrilateral ABCD such that the circumcircles of triangles ABP and CDP are √ √ tangent, and the circumcircles of triangles ADP and BCP are tangent. If AC = 2 61 and BD = 6 7, √ √ then OP can be expressed as a − b for positive integers a and b . Compute 100 a + b .
解析
- Suppose ω is a circle centered at O with radius 8. Let AC and BD be perpendicular chords of ω . Let P be a point inside quadrilateral ABCD such that the circumcircles of triangles ABP and CDP are √ √ tangent, and the circumcircles of triangles ADP and BCP are tangent. If AC = 2 61 and BD = 6 7, √ √ then OP can be expressed as a − b for positive integers a and b . Compute 100 a + b . Proposed by: Daniel Xianzhe Hong Answer: 103360 Solution: Let X = AC ∩ BD , Q = AB ∩ CD and R = BC ∩ AD . Since QA · QB = QC · QD , Q is on 2 the radical axis of ( ABP ) and ( CDP ), so Q lies on the common tangent at P . Thus, QP = QA · QB . 2 Similarly, RA · RC = RP . Let M be the Miquel point of quadrilateral ABCD : in particular, M = OX ∩ QR is the foot from O to QR . By properties of the Miquel point, ABM R and ACM Q are cyclic. Thus, 2 QP = QA · QB 2 RP = RA · RC 2 2 2 QP + RP = QM · QR + RM · RQ = ( QR + RM ) QR = QR . ◦ As a result, ∠ QP R = 90 . ′ ′ Now, let P the inverse of P with respect to ω . Note that by properties of inversion, ( ABP ) and ′ ′ ′ ( CDP ) are tangent, and ( ACP ) and ( BDP ) are also tangent. But now, 2 ′ 2 QP = QP = QA · QB 2 ′ 2 RP = RP = RA · RC 2 2 ′ 2 ′ 2 2 QP + RP = QP + RP = QR . ′ ′ Thus, P QP R is a cyclic kite, so P and P are reflections of each other across QR . In particular, since ′ ′ O, P, P are collinear, then M lies on line OP P . 2 2 r 2 r We can now compute OP by using the fact that OP + = 2 OM = , where r = 8. Since OX OP OX 64 2 can be computed to equal 2 quite easily, then OP + = 64, or OP − 64 OP + 64 = 0. Solving this OP √ √ √ √ yields OP = 32 ± 8 15, and because P is inside the circle, OP = 32 − 8 15 = 1024 − 960.