HMMT 十一月 2024 · THM 赛 · 第 3 题

HMMT November 2024 — THM Round — Problem 3

Points K, A, L, C, I, T, E are such that triangles CAT and ELK are equilateral, share a center I , and points E, L, K lie on sides CA, AT , T C respectively. If the area of triangle CAT is double the area of triangle ELK and CI = 2 , compute the minimum possible value of CK .

解析

Points K, A, L, C, I, T, E are such that triangles CAT and ELK are equilateral, share a center I , and points E, L, K lie on sides CA, AT , T C respectively. If the area of triangle CAT is double the area of triangle ELK and CI = 2, compute the minimum possible value of CK . Proposed by: Albert Wang, Isaac Zhu √ Answer: 3 − 1 Solution 1: C K I E A T L √ √ First, compute that triangle CAT has side length 2 3 and area 3 3. Since triangle ELK has half the 1 area of CAT , the triangles CEK, AEL, T LK must each have the area of CAT . Now by sine area 6 formula, we have √ 1 1 ◦ · 3 3 = [ CEK ] = CE · CK · sin(60 ) = ⇒ CE · CK = 2 6 2 √ Also, note that CE + CK = CA = 2 3 by symmetry. So we have the equation √ √ CE · (2 3 − CE ) = 2 = ⇒ CE = 3 ± 1 . √ The answer is 3 − 1 . Solution 2: Let M be the midpoint of CT . Note that KIM is a right triangle. Using the given ratio √ 2 2 of areas, we know that CI = 2 · KI , implying that KI = 2. Pythagorean theorem on right triangle √ ◦ ◦ ◦ KIM tells us that KM = 1. Since CIM is a 30 − 60 − 90 triangle, we know that CM = 3. The √ √ smallest possible value of CK = 3 ± 1 is then 3 − 1 .