HMMT 二月 2006 · GEN2 赛 · 第 6 题

HMMT February 2006 — GEN2 Round — Problem 6

Octagon ABCDEF GH is equiangular. Given that AB = 1, BC = 2, CD = 3, DE = 4, and EF = F G = 2, compute the perimeter of the octagon. 2 2 2

解析

Octagon ABCDEF GH is equiangular. Given that AB = 1, BC = 2, CD = 3, DE = 4, and EF = F G = 2, compute the perimeter of the octagon. √ Answer: 20 + 2 Solution: Extend sides AB, CD, EF, GH to form a rectangle: let X be the intersec- tion of lines GH and AB ; Y that of AB and CD ; Z that of CD and EF ; and W that of EF and GH . √ √ As BC = 2, we have BY = Y C = 2. As DE = 4, we have DZ = ZE = 2 2. As √ F G = 2, we have F W = W G = 2. We can compute the dimensions of the rectangle: W X = Y Z = Y C + CD + DZ = √ √ 3 + 3 2, and XY = ZW = ZE + EF + F W = 2 + 3 2. Thus, HX = XA = XY − √ √ √ AB − BY = 1+2 2, and so AH = 2 HX = 4+ 2, and GH = W X − W G − HX = 2 . The perimeter of the octagon can now be computed by adding up all its sides. 2 2