PUMaC 2019 · 几何（B 组） · 第 6 题

PUMaC 2019 — Geometry (Division B) — Problem 6

Let Γ be a circle with center A , radius 1 and diameter BX . Let Ω be a circle with center C , radius 1 and diameter DY , where X and Y are on the same side of AC . Γ meets Ω at two points, one of which is Z . The lines tangent to Γ and Ω that pass through Z cut out a sector ◦ of the plane containing no part of either circle and with angle 60 . If ∠ XY C = ∠ CAB and √ √ a + b ◦ ∠ XCD = 90 , then the length of XY can be written in the form for integers a, b, c c where gcd( a, b, c ) = 1. Find a + b + c .

解析

Let Γ be a circle with center A , radius 1 and diameter BX . Let Ω be a circle with center C , radius 1 and diameter DY , where X and Y are on the same side of AC . Γ meets Ω at two points, one of which is Z . The lines tangent to Γ and Ω that pass through Z cut out a sector ◦ of the plane containing no part of either circle and with angle 60 . If ∠ XY C = ∠ CAB and √ √ a + b ◦ ∠ XCD = 90 , then the length of XY can be written in the form for integers a, b, c c where gcd( a, b, c ) = 1. Find a + b + c . Proposed by Zachary Stier. Answer: 16 Solution: Let the circles have radii a, c and let the angle at Z be θ . We first compute AC . √ ◦ ◦ 2 2 ∠ AZC = θ + 2(90 − θ ) = 180 − θ so the Law of Cosines gives b = AC = a + c + 2 ab cos θ . The given angle conditions make AXY C cyclic, and the right angle makes XY the diameter (of 3 2 length d ) of the circumcircle. Using Ptolemy and Pythagoras, we get the equation d − d ( a + √ √ 2 2 3 3 b + c ) − 2 abc = 0; plugging in: d − 5 d − 2 3 = 0. Taking d = d 3 gives 3 d − 5 d − 2 = 0 which √ √ √ 3+ 11 has root − 1, so the original has root − 3; factoring gives the only positive value, d = 2 for a final answer 16 .