HMMT 二月 2024 · 冲刺赛 · 第 30 题

HMMT February 2024 — Guts Round — Problem 30

[16] Let ABC be an equilateral triangle with side length 1 . Points D , E , F lie inside triangle ABC such that A , E , F are collinear, B , F , D are collinear, C , D , E are collinear, and triangle DEF is equilateral. Suppose that there exists a unique equilateral triangle XY Z with X on side BC , Y on side AB , and Z on side AC such that D lies on side XZ , E lies on side Y Z , and F lies on side XY . Compute AZ .

解析

[16] Let ABC be an equilateral triangle with side length 1 . Points D , E , F lie inside triangle ABC such that A , E , F are collinear, B , F , D are collinear, C , D , E are collinear, and triangle DEF is equilateral. Suppose that there exists a unique equilateral triangle XY Z with X on side BC , Y on side AB , and Z on side AC such that D lies on side XZ , E lies on side Y Z , and F lies on side XY . Compute AZ . Proposed by: Jaedon Whyte, Maxim Li 1 √ Answer: 3 1+ 2 Solution: A Z E Y O D F B C X First, note that point X can be constructed from intersection of ⊙ ( DOF ) and side BC . Thus, if there is a unique equilateral triangle, then we must have that ⊙ ( DOF ) is tangent to BC . Furthermore, ⊙ ( DOF ) is tangent to DE , so by equal tangents, we have CD = CX . We now compute the answer. Let x = AZ = CX = CD = BF . Then, by power of point, 2 (1 − x ) 2 BF · BD = BX = ⇒ BD = . x Thus, by law of cosine on △ BDC , we have that ( ) 2 2 2 (1 − x ) (1 − x ) 2 x + + x · = 1 x x 4 (1 − x ) 2 2 x + + (1 − x ) = 1 2 x 4 (1 − x ) = 2 x (1 − x ) 2 x √ 1 − x 3 = 2 x 1 x = √ . 3 1 + 2