HMMT 二月 2026 · 冲刺赛 · 第 18 题

HMMT February 2026 — Guts Round — Problem 18

[11] Let ABCD be a trapezoid with side AB parallel to side CD . Let P be the intersection of diagonals AC and BD . Given that the distances from P to sides AB , BC , CD , and DA are 3 , 6 , 8 , and 8 , respectively, compute the perimeter of ABCD .

解析

[11] Let ABCD be a trapezoid with side AB parallel to side CD . Let P be the intersection of diagonals AC and BD . Given that the distances from P to sides AB , BC , CD , and DA are 3 , 6 , 8 , and 8 , respectively, compute the perimeter of ABCD . Proposed by: Jason Mao 165 Answer: = 82 . 5 2 Solution: B A P D C ′ D ′ ′ Let D be the point such that ABD D is a parallelogram, and let AB = 3 x . Then: 8 • Since ABCD is a trapezoid, △ P AB ∼ △ P CD , so similarity gives CD = AB = 8 x . Thus, 3 ′ ′ CD = CD − DD = 8 x − 3 x = 5 x. • Since P is equidistant from CD and DA , we have ∠ P DA = ∠ P DC = ∠ P BA , so ′ BD = AD = AB = 3 x. 1 • Since △ P AB ∼ △ P CD , we have P A · P D = P B · P C . Multiplying both sides by sin ∠ AP D = 2 1 sin ∠ BP C , we derive [ P AD ] = [ P BC ] . Using the distances from P to AD and BC , 2 1 1 · 8 · (3 x ) = · 6 · BC = ⇒ BC = 4 x. 2 2 (3 x )(4 x ) ′ 2 Thus, BCD has side lengths 3 x , 4 x , and 5 x , making it a right triangle of area = 6 x . The 2 2 2 · 6 x 12 x ′ height of the trapezoid is then the distance from B to CD , which is = . Hence, 5 x 5 2 1 12 x 1 12 x 48 x [ ADC ] = · · CD = · · 8 x = . 2 5 2 5 5 Using the distances from P to AD and CD , we also know 1 1 [ ADC ] = [ AP D ] + [ DP C ] = · 8 · 3 x + · 8 · 8 x = 44 x. 2 2 Thus, 2 48 x 44 · 5 55 = 44 x = ⇒ x = = , 5 48 12 and the perimeter of ABCD is 165 3 x + 4 x + 8 x + 3 x = 18 x = . 2 ©2026 HMMT