HMMT 二月 2025 · 冲刺赛 · 第 15 题

HMMT February 2025 — Guts Round — Problem 15

[9] Right triangle △ DEF with ∠ D = 90 and ∠ F = 30 is inscribed in equilateral triangle △ ABC such that D , E , and F lie on segments BC , CA , and AB , respectively. Given that BD = 7 and DC = 4, compute DE .

解析

[9] Right triangle △ DEF with ∠ D = 90 and ∠ F = 30 is inscribed in equilateral triangle △ ABC such that D , E , and F lie on segments BC , CA , and AB , respectively. Given that BD = 7 and DC = 4, compute DE . Proposed by: Pitchayut Saengrungkongka √ Answer: 13 Solution 1: A F E B C D ◦ ◦ From ∠ E = 60 , we get that ∠ AEF = 120 − ∠ CED = ∠ CDE . Therefore, △ AEF ∼ △ CDE . Since EF : DE = 2 : 1, the ratio of similarity must be 2 : 1, so AE = 2 CD = 8. Recall ABC has side length √ √ 2 2 2 7 + 4 = 11, so EC = 11 − 8 = 3. Law of Cosines on △ CDE gives DE = 3 + 4 − 3 · 4 = 13 . Solution 2: A X F E B C D ◦ ◦ Let ⊙ ( DEF ) meet AC again at point X . Then, ∠ F XA = 180 − ∠ F XE = ∠ F DE = 90 and ◦ ◦ ◦ ◦ ∠ XDC = 180 − ∠ DCX − ∠ DXC = 120 − ∠ DXE = 120 − ∠ DF E = 90 . It follows that CX = 2 CD = 8, so AX = 11 − CX = 3, and AF = 2 AX = 6. Thus, Law of Cosines on △ AEF gives √ √ √ 2 2 EF = 8 + 6 − 8 · 6 = 2 13, implying that DE = 13 .