HMMT 二月 2006 · 几何 · 第 1 题

英文解析

1. 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 intersection of lines GH and AB ; Y that of AB and CD ; Z that of CD and EF ; and W thatof 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.