HMMT 二月 2005 · TEAM2 赛 · 第 2 题
HMMT February 2005 — TEAM2 Round — Problem 2
题目详情
- [10] What are the dimensions of the rectangle of smallest area that is ( a, b )-tileable?
解析
- [10] What are the dimensions of the rectangle of smallest area that is ( a, b )-tileable? Solution: If a = 0, then a 1 × 2 b rectangle is tileable in an obvious way. If a > 0, then a 2 a × 2 b rectangle is tileable by dividing the rectangle into quarters and pairing each square with a square from the diagonally opposite quarter. The answer is therefore max { 1 , 2 a } × 2 b . Minimality of area follows from the next question. b b a a 1