PUMaC 2016 · 几何（B 组） · 第 3 题

PUMaC 2016 — Geometry (Division B) — Problem 3

Let ABCD be a square with side length 8. Let M be the midpoint of BC and let ω be the circle passing through M , A , and D . Let O be the center of ω , X be the intersection point (besides A ) of ω with AB , and Y be the intersection point of OX and AM . If the length of m OY can be written in simplest form as , compute m + n . n

解析

We will reference the following picture (which we need when we use words like “horizontal” and “vertical”). Let R be the radius of ω . The distance from M to the midpoint of AD can be expressed as √ 2 2 2 2 2 R + R − 4 , and this is equal to 8. We thus have R − 4 = (8 − R ) , and solving for R gives R = 5. The distance from O to AB is 4, so the Pythagorean theorem gives us that the vertical distance from O to X is 3. Thus, if we use a coordinate system with the natural x and y axes − 3 and origin O , then the line OX is described by the equation y = x . Meanwhile, the line 4 AM has slope 2 and y -intercept 5 so it is described by the line y = 2 x + 5. The intersection − 3 − 20 15 point Y thus has x -coordinate satisfying 2 x + 5 = x , i.e. x = , and this gives y = . 4 11 11 Thus, we have √ ( ) ( ) 2 2 − 20 15 25 OY = + = . 11 11 11 1 Our answer is thus 25 + 11 = 36 . Problem written by Eric Neyman. ′