HMMT 十一月 2024 · 冲刺赛 · 第 5 题

HMMT November 2024 — Guts Round — Problem 5

[6] Let ABCD be a trapezoid with AB ∥ CD , AB = 20 , CD = 24 , and area 880 . Compute the area of the triangle formed by the midpoints of AB , AC , and BD .

解析

[6] Let ABCD be a trapezoid with AB ∥ CD , AB = 20, CD = 24, and area 880. Compute the area of the triangle formed by the midpoints of AB , AC , and BD . Proposed by: Pitchayut Saengrungkongka Answer: 20 Solution: B M A N P D C We first compute the height of the trapezoid. If h is the height, then the area is 1 880 = h (20 + 24) , 2 so h = 40. Now, let M , N , P be the midpoints of AB , AC , and BD . Notice that P N is parallel to h AB . Thus, the altitude from M to N P has length = 20. 2 To compute N P , let X be the midpoint of BC . Since XN is a midsegment of △ CAB , we have AB CD XN = = 10. Since XP is a midsegment of △ BCD , we have XP = = 12. Hence, N P = 2 2 XP − XN = 2. 1 Thus, the area of triangle M N P is · 2 · 20 = 20 . 2