HMMT 十一月 2023 · 团队赛 · 第 7 题

HMMT November 2023 — Team Round — Problem 7

[45] Let ABCD be a convex trapezoid such that ∠ BAD = ∠ ADC = 90 , AB = 20 , AD = 21 , and ◦ CD = 28 . Point P ̸ = A is chosen on segment AC such that ∠ BP D = 90 . Compute AP .

解析

[45] Let ABCD be a convex trapezoid such that ∠ BAD = ∠ ADC = 90 , AB = 20, AD = 21, and ◦ CD = 28. Point P ̸ = A is chosen on segment AC such that ∠ BP D = 90 . Compute AP . Proposed by: Pitchayut Saengrungkongka 143 Answer: 5 A D P C B X Solution 1: Construct the rectangle ABXD . Note that ◦ ∠ BAD = ∠ BP D = ∠ BXD = 90 , so ABXP D is cyclic with diameter BD . By Power of a Point, we have CX · CD = CP · CA. Note √ 2 2 that CX = CD − XD = CD − AB = 8 and CA = AD + DC = 35. Therefore, CX · CD 8 · 28 32 CP = = = , CA 35 5 and so 32 143 AP = AC − CP = 35 − = . 5 5 Solution 2: Since ABP D is cyclic and ∠ BDP = ∠ BAP = ∠ ACD , it follows that △ BP D ∼ △ ABC . Thus, if BD = 5 x , BP = 4 x , and DP = 3 x , then by Ptolemy’s theorem, we have AP · (5 x ) = AD · (3 x ) + AB · (4 x ) 21 · 3 + 20 · 4 143 AP = = 5 5