HMMT 十一月 2025 · 团队赛 · 第 10 题
HMMT November 2025 — Team Round — Problem 10
题目详情
- [55] Let ABCD be an isosceles trapezoid with AB parallel to CD , and let P be a point in the interior of ABCD such that ∠ P BA = 3 ∠ P AB and ∠ P CD = 3 ∠ P DC. 2 Given that BP = 8, CP = 9, and cos ∠ AP D = , compute cos ∠ P AB . 3 © 2025 HMMT
解析
- [55] Let ABCD be an isosceles trapezoid with AB parallel to CD , and let P be a point in the interior of ABCD such that ∠ P BA = 3 ∠ P AB and ∠ P CD = 3 ∠ P DC. 2 Given that BP = 8, CP = 9, and cos ∠ AP D = , compute cos ∠ P AB . 3 Proposed by: Pitchayut Saengrungkongka √ 3 5 Answer: 7 Solution 1: A B X P P Y C D Let the perpendicular bisector of AB (which is also the perpendicular bisector of CD ) intersect AP and DP at X and Y , respectively. Observe that AX = BX , so ∠ P AB = ∠ XAB = ∠ XBA . Then, ∠ P XB = ∠ XAB + ∠ XBA = 2 ∠ P AB = ∠ P BA − ∠ P AB = ∠ P BA − ∠ XBA = ∠ P BX, so P X = BP = 8. Similarly, P Y = CP = 9. So Law of Cosines on △ P XY gives q p 2 2 2 2 2 XY = P X + P Y − 2 · P X · P Y · cos ∠ AP D = 8 + 9 − 2 · 8 · 9 · = 7 . 3 ◦ Finally, since XY ⊥ AB , we have 90 − ∠ P XY = ∠ P AB . So cos ∠ P AB = sin ∠ P XY . By Law of Sines on △ P XY , s √ 2 P Y 2 9 3 5 sin ∠ P XY = sin ∠ XP Y · = 1 − · = , XY 3 7 7 √ 3 5 so cos ∠ P AB = . 7 © 2025 HMMT Solution 2: A B ′ P P P C D ′ Let ∠ P AB = α and ∠ P DC = β . Let P be the reflection of P across the perpendicular bisector of ′ ′ ′ AB . Denote d by the length of the segment P P . Then, ∠ AP P = α and ∠ P AP = 2 α . ′ Notice that AP = BP = 8. Then, by the law of sines, sin 2 α sin α sin α = = . ′ ′ P P AP 8 √ 2 Thus, cos α = d/ 16. Similarly, cos β = d/ 18. Then, we get sin α = 256 − d / 16 and sin β = √ 2 324 − d / 18 accordingly. 2 2 The condition cos AP D = is equivalent to cos( α + β ) = . Expanding everything out, we have 3 3 p 2 2 2 d − (256 − d )(324 − d ) 2 = 3 288 p 2 2 2 d − 192 = (256 − d )(324 − d ) 2 196 d = 46080 √ 48 5 d = 7 √ 3 5 Therefore, cos P AB = cos α = d/ 16 = . 7 © 2025 HMMT