PUMaC 2011 · 几何（B 组） · 第 8 题

PUMaC 2011 — Geometry (Division B) — Problem 8

[ 8 ] Let ω be a circle of radius 6, and let ω be a circle of radius 5 that passes through the 1 2 center O of ω . Let A and B be the points of intersection of the two circles, and let P be a 1 point on major arc AB of ω . Let M and N be the second intersections of P A and P B with 2 ω , respectively. Let S be the midpoint of M N . As P ranges over major arc AB of ω , the 1 2 minimum length of segment SA is a/b , where a and b are positive integers and gcd( a, b ) = 1. Find a + b . 1

解析

We claim that the length of arc M N is constant as P varies. We can see this by noting that ⌢ ⌢ ⌢ ⌢ 1 M LB − AN = ∠ AP B , which is constant, and that M LB + M A is constant. Subtracting 2 ⌢ ⌢ ⌢ these two constant quantities, we get that M N = M A + AN is constant. Since OS is the distance from O to the midpoint of a chord of constant length, OS is constant as well. Thus, the locus of all points S is a part of a circle centered at O . It follows that the minimum distance from this locus to point A is the difference between the radii of ω and of the locus 1 of S . Now, to find the radius of the locus of S , consider the location of S when P is at the midpoint C of major arc AB . Since ω passes through O , we have that CA and CB are 2 tangent to ω . Thus, the segment M N becomes AB , and S coincides with T , the midpoint 1 of AB . By the similarity of triangles T AO and ACO , we have that OT /OA = OA/OC , so 2 2 OT = OA /OC = 6 / 10 = 18 / 5. Thus, the radius of the locus of S is 18/5, and the difference between the two radii is 6 − 18 / 5 = 12 / 5, so the answer is 12 + 5 = 17 . 4 Figure 7: Problem 7 diagram. Figure 8: Problem 8 diagram. 5