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AIME 1995 · 第 1 题

AIME 1995 — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Square S1S_{1} is 1×1.1\times 1. For i≥1,i\ge 1, the lengths of the sides of square Si+1S_{i+1} are half the lengths of the sides of square Si,S_{i}, two adjacent sides of square SiS_{i} are perpendicular bisectors of two adjacent sides of square Si+1,S_{i+1}, and the other two sides of square Si+1,S_{i+1}, are the perpendicular bisectors of two adjacent sides of square Si+2.S_{i+2}. The total area enclosed by at least one of S1,S2,S3,S4,S5S_{1}, S_{2}, S_{3}, S_{4}, S_{5} can be written in the form m/n,m/n, where mm and nn are relatively prime positive integers. Find m−n.m-n.

AIME diagram

解析

Solution

The sum of the areas of the squares if they were not interconnected is a geometric sequence:

12+(12)2+(14)2+(18)2+(116)21^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2

Then subtract the areas of the intersections, which is (14)2+…+(132)2\left(\frac{1}{4}\right)^2 + \ldots + \left(\frac{1}{32}\right)^2:

12+(12)2+(14)2+(18)2+(116)2−[(14)2+(18)2+(116)2+(132)2]1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 - \left[\left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2 + \left(\frac{1}{32}\right)^2\right] =1+122−1322= 1 + \frac{1}{2}^2 - \frac{1}{32}^2

The majority of the terms cancel, leaving 1+14−110241 + \frac{1}{4} - \frac{1}{1024}, which simplifies down to 1024+(256−1)1024\frac{1024 + \left(256 - 1\right)}{1024}. Thus, m−n=255m-n = \boxed{255}.

Alternatively, take the area of the first square and add  34\,\frac{3}{4} of the areas of the remaining squares. This results in 1+34[(12)2+…+(1162)]1+ \frac{3}{4}\left[\left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{16}^2\right)\right], which when simplified will produce the same answer.