HMMT 二月 2005 · 冲刺赛 · 第 13 题
HMMT February 2005 — Guts Round — Problem 13
题目详情
英文原题
- [7] Triangle ABC has AB = 1, BC = 7, and CA = 3. Let ℓ be the line through
A perpendicular to AB , ℓ the line through B perpendicular to AC , and P the point 1
of intersection of ℓ and ℓ . Find P C .2
1 2
解析
英文解析
- Triangle ABC has AB = 1, BC = 7, and CA = 3. Let ` be the line through Aperpendicular to AB , ` the line through B perpendicular to AC , and P the point of 1
intersection of ` and ` . Find P C .2
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 32
° ° ° °
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 .
QBAC P