PUMaC 2019 · 几何（A 组） · 第 8 题

PUMaC 2019 — Geometry (Division A) — Problem 8

Let γ and Γ be two circles such that γ is internally tangent to Γ at a point X . Let P be a point on the common tangent of γ and Γ and Y be the point on γ other than X such that P Y is tangent to γ at Y . Let P Y intersect Γ at A and B , such that A is in between P and B and let the tangents to Γ at A and B intersect at C . CX intersects Γ again at Z and ZY √ p AX 1 intersects Γ again at Q . If AQ = 6, AB = 10 and = . The length of QZ = r , where XB 4 q p and q are coprime positive integers, and r is square free positive integer. Find p + q + r . 1

解析

Let γ and Γ be two circles such that γ is internally tangent to Γ at a point X . Let P be a point on the common tangent of γ and Γ and Y be the point on γ other than X such that P Y is tangent to γ at Y . Let P Y intersect Γ at A and B , such that A is in between P and B and let the tangents to Γ at A and B intersect at C . CX intersects Γ again at Z and ZY √ p AX 1 intersects Γ again at Q . If AQ = 6, AB = 10 and = . The length of QZ = r , where XB 4 q p and q are coprime positive integers, and r is square free positive integer. Find p + q + r . Proposed by Mel Shu. Solution by Igor Medvedev. Answer: 28 Solution: P lies on the polar of C so CZ must be the polar of P so P Z is a tangent to Γ. In ′ particular, P Z = P Y so there exists a circle γ tangent to P Y and P Z at Y and Z respectively. ′ Then the homothety centered at Z which takes γ to Γ takes Y to Q , so Q is the midpoint of arc AQB . It follows that AQ = BQ . Similarly XY is a bisector of angle ∠ AXB . AX BY Then = by the angle bisector theorem. Hence BY = 2, AY = 8. By Stewart’s XB Y A 2 2 theorem applied to ABQ we have QY = AQ − BY · Y A = 20. Then by power of a point, √ 16 18 √ Y Z = . Then QZ = 5. 5 20 3