HMMT 二月 2011 · 冲刺赛 · 第 9 题

英文解析

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 .
   √
   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.