HMMT 二月 2006 · TEAM2 赛 · 第 7 题
HMMT February 2006 — TEAM2 Round — Problem 7
题目详情
英文原题
- [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 theoctagon? (No proof is necessary.)
解析
英文解析
- [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 manyregions 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 unitsquares in a cross arrangement, with four half-squares wedged at the corners. Thecenter 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 .