HMMT 十一月 2025 · 冲刺赛 · 第 24 题
HMMT November 2025 — Guts Round — Problem 24
题目详情
- [12] Let ABCDE be a convex pentagon such that ABCD is a rectangle and ∠ AEB = ∠ CED = 30 . √ Given that AB = 14 and BC = 20 3, compute the area of triangle ADE . © 2025 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2025, November 08, 2025 — GUTS ROUND Organization Team Team ID#
解析
- [12] Let ABCDE be a convex pentagon such that ABCD is a rectangle and ∠ AEB = ∠ CED = 30 . √ Given that AB = 14 and BC = 20 3, compute the area of triangle ADE . Proposed by: Pitchayut Saengrungkongka √ Answer: 60 3 Solution 1: E M F A D B C ′ E ′ Let M be the midpoint of AD . Let ⊙ ( ABE ) intersect AD and EM again at F and E , respectively. ◦ Notice EM is the perpendicular bisector of AD , as ∠ AEB = ∠ CED . Then, ∠ AF B = ∠ AEB = 30 , √ √ √ √ so AF = AB · 3 = 14 3. Since M is the midpoint of AD , we have AM = 10 3 and M F = 4 3. √ √ ′ ′ ′ By Power of a Point, E M · M E = AM · M F = 4 3 · 10 3 = 120 . Furthermore, E E ∥ AB and ABE E ′ is cyclic, so it is an isosceles trapezoid. Hence, E M = AB + EM = 14+ EM and EM · ( EM +14) = 120, √ √ so EM = 6. The area of △ ADE is then 1 / 2 · 6 · 20 3 = 60 3 . Solution 2: E A D M O 2 O 1 B C Let O be the circumcenter of △ EAB , O be the circumcenter of △ EDC , and M be the center of 1 2 ◦ ABCD . We have that ∠ AO B = 2 ∠ AEB = 60 and O A = O B , so △ AO B is equilateral. 1 1 1 1 © 2025 HMMT Since M E is the perpendicular bisector of AD and O M is the perpendicular bisector of AB , △ O M E 1 1 is a right triangle with O E = O B = AB = 14 and 1 1 √ √ √ 20 3 14 3 O M = dist( M, AB ) − dist( O , AB ) = − = 3 3 . 1 1 2 2 q √ 2 14 2 Thus, EM = 14 − 3 3 = 13, so the distance from E to AD is 13 − = 6, and the area of 2 √ √ 1 △ ADE is · 6 · 20 3 = 60 3 . 2