HMMT 二月 2005 · 冲刺赛 · 第 31 题
HMMT February 2005 — Guts Round — Problem 31
题目详情
英文原题
- [10] The L shape made by adjoining three congruent squares can be subdivided intofour smaller L shapes.
Each of these can in turn be subdivided, and so forth. If we perform 2005 successive
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subdivisions, how many of the 4 L ’s left at the end will be in the same orientationas the original one?
解析
英文解析
- The L shape made by adjoining three congruent squares can be subdivided into foursmaller L shapes.
Each of these can in turn be subdivided, and so forth. If we perform 2005 successive
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subdivisions, how many of the 4 L ’s left at the end will be in the same orientationas the original one?
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Solution: 4 + 2
After n successive subdivisions, let a be the number of small L’s in the same orientationnas the original one; let b be the number of small L’s that have this orientation rotatedn
° °
counterclockwise 90 ; let c be the number of small L’s that are rotated 180 ; andn
°
let d be the number of small L’s that are rotated 270 . When an L is subdivided,
it produces two smaller L’s of the same orientation, one of each of the neighboringnorientations, and none of the opposite orientation. Therefore,
( a , b , c , d ) = ( d + 2 a + b , a + 2 b + c , b + 2 c + d , c + 2 d + a ) .
n +1 n +1 n +1 n +1 n n n n n n n n n n n n
It is now straightforward to show by induction thatn − 1 n − 1 n − 1 n − 1 n − 1 n − 1
( a , b , c , d ) = (4 + 2 , 4 , 4 − 2 , 4 )
n n n n
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for each n ≥ 1. In particular, our desired answer is a = 4 + 2 .
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