HMMT 二月 2018 · 冲刺赛 · 第 12 题

HMMT February 2018 — Guts Round — Problem 12

[ 7 ] 4 P N R has side lengths P N = 20, N R = 18, and P R = 19. Consider a point A on P N . 4 N RA ′ ′ ′ ′ is rotated about R to 4 N RA so that R , N , and P lie on the same line and AA is perpendicular to P A P R . Find . AN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2018, February 10, 2018 — GUTS ROUND Organization Team Team ID#

解析

[ 7 ] 4 P N R has side lengths P N = 20, N R = 18, and P R = 19. Consider a point A on P N . 4 N RA ′ ′ ′ ′ is rotated about R to 4 N RA so that R , N , and P lie on the same line and AA is perpendicular to P A P R . Find . AN Proposed by: Henrik Boecken 19 Answer: 18 ′ ′ Denote the intersection of P R and AA be D . Note RA = RA , so D , being the altitude of an isosceles ′ triangle, is the midpoint of AA . Thus, ′ ∠ ARD = ∠ A RD = ∠ N RA P A P R 19 so RA is the angle bisector of P N R through R . By the angle bisector theorem, we have = = AN RN 18