PUMaC 2019 · 几何（A 组） · 第 4 题

PUMaC 2019 — Geometry (Division A) — Problem 4

Let BC = 6 , BX = 3 , CX = 5, and let F be the midpoint of BC . Let AX ⊥ BC and √ √ √ AF = 247. If AC is of the form b and AB is of the form c where b and c are nonnegative integers, find 2 c + 3 b .

解析

Let BC = 6 , BX = 3 , CX = 5, and let F be the midpoint of BC . Let AX ⊥ BC and √ √ √ AF = 247. If AC is of the form b and AB is of the form c where b and c are nonnegative integers, find 2 c + 3 b . Proposed by Alan Chung. Solution by Aleksa Milojevi´ c. Answer: 1288 Solution: Consider the circles C with center B and radius BX = 3 and C with center C 1 2 and radius CX = 5. Their radical axis is a line through X perpendicular to BC , i.e. the line 1 AX . As A is on this radical axis, we get that its power w.r.t. circles C and C is the same: 1 2 2 2 2 2 AB − BX = AC − CX . 1 2 2 On the other hand, by the median formula in triangle ABC we have that AF = BA + 2 1 2 1 2 2 2 CA − BC . The two equations we have got are enough to find AB and AC . Namely: 2 4 1 2 2 2 2 AB + AC = 2( AF + BC ) = 2 · 256 = 512 4 2 2 2 2 AB − AC = BX − CX = − 16 2 2 By solving this system we get AB = c = 248 and AC = b = 264. Therefore, 2 c + 3 b = 1288.