HMMT 二月 2011 · 冲刺赛 · 第 9 题

HMMT February 2011 — Guts Round — Problem 9

[ 6 ] Segments AA , BB , and CC , each of length 2, all intersect at a point O . If ∠ AOC = ∠ BOA = ′ ◦ ′ ′ ∠ COB = 60 , find the maximum possible value of the sum of the areas of triangles AOC , BOA , ′ and COB .

解析

[ 6 ] Segments AA , BB , and CC , each of length 2, all intersect at a point O . If ∠ AOC = ∠ BOA = ′ ◦ ′ ′ ∠ COB = 60 , find the maximum possible value of the sum of the areas of triangles AOC , BOA , ′ and COB . √ Answer: 3 ′ ′ ′ Extend OA to D and OC to E such that AD = OA and C E = OC . Since OD = OE = 2 and ◦ ∠ DOE = 60 , we have ODE is an equilateral triangle. Let F be the point on DE such that DF = OB ′ ′ ′ ′ ∼ ∼ and EF = OB . Clearly we have 4 DF A 4 OBA and 4 EF C OB C . Thus the sum of the areas = = ′ ′ ′ ′ of triangles AOC , BOA , and COB is the same as the sum of the areas of triangle DF A , F EC , ′ and OAC , which is at most the area of triangle ODE . Since ODE is an equilateral triangle with side √ ′ length 2, its area is 3. Equality is achieved when OC = OA = 0.