HMMT 十一月 2017 · GEN 赛 · 第 4 题

HMMT November 2017 — GEN Round — Problem 4

Triangle ABC has AB = 10, BC = 17, and CA = 21. Point P lies on the circle with diameter AB . What is the greatest possible area of AP C ? 3

解析

Triangle ABC has AB = 10, BC = 17, and CA = 21. Point P lies on the circle with diameter AB . What is the greatest possible area of AP C ? Proposed by: Michael Tang 189 Answer: 2 To maximize [ AP C ], point P should be the farthest point on the circle from AC . Let M be the 1 1 midpoint of AB and Q be the projection of M onto AC . Then P Q = P M + M Q = AB + h , B 2 2 where h is the length of the altitude from B to AC . By Heron’s formula, one finds that the area of B √ 2 · 84 1 ABC is 24 · 14 · 7 · 3 = 84, so h = = 8. Then P Q = (10 + 8) = 9, so the area of AP C is B AC 2 1 189 · 21 · 9 = . 2 2 3