HMMT 二月 2006 · TEAM2 赛 · 第 7 题

HMMT February 2006 — TEAM2 Round — Problem 7

[25] Suppose we have an octagon with all angles of 135 , and consecutive sides of alternating length √ 1 and 2 . We draw all its sides and diagonals. Into how many regions do the segments divide the octagon? (No proof is necessary.)

解析

[25] Suppose we have an octagon with all angles of 135 , and consecutive sides of √ alternating length 1 and 2 . We draw all its sides and diagonals. Into how many regions do the segments divide the octagon? (No proof is necessary.) Answer: 84 Solution: The easiest way to see the answer is to view the octagon as five unit squares in a cross arrangement, with four half-squares wedged at the corners. The center square is divided into 8 regions. The other 4 squares are each divided into 15 regions. The 4 half-squares are each divided into 4 regions. The answer is thus 8 + 4 × 15 + 4 × 4 = 84 .