HMMT 二月 2005 · 冲刺赛 · 第 13 题

HMMT February 2005 — Guts Round — Problem 13

[7] Triangle ABC has AB = 1, BC = 7, and CA = 3. Let ℓ be the line through 1 A perpendicular to AB , ℓ the line through B perpendicular to AC , and P the point 2 of intersection of ℓ and ℓ . Find P C . 1 2

解析

Triangle ABC has AB = 1, BC = 7, and CA = 3. Let be the line through A 1 perpendicular to AB , the line through B perpendicular to AC , and P the point of 2 intersection of ` and ` . Find P C . 1 2 Solution: 3 √ − 7 3 − 1 3+1 − 1 ◦ √ By the Law of Cosines, ∠ BAC = cos = cos ( − ) = 150 . If we let Q be the 2 2 3 ◦ ◦ ◦ ◦ intersection of ` and AC , we notice that ∠ QBA = 90 − ∠ QAB = 90 − 30 = 60 . 2 √ It follows that triangle ABP is a 30-60-90 triangle and thus P B = 2 and P A = 3 . ◦ ◦ ◦ ◦ Finally, we have ∠ P AC = 360 − (90 + 150 ) = 120 , and 2 2 ◦ 1 / 2 1 / 2 P C = ( P A + AC − 2 P A · AC cos 120 ) = (3 + 3 + 3) = 3 . C A B Q P