Question

Square $S_1$ is $1\times1.$ For $i\geq1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_i,$ two adjacent sides of square $S_i$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1}$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$ The total area enclosed by at least one of $S_1,$ $S_2,$ $S_3,$ $S_4,$ $S_5$ can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m-n.$

Answer

Answer: 255

Concepts:geometric sequence area inclusion-exclusion

Difficulty rating: 1900

Small Hint:

Each square has one fourth the area of the preceding square

Big Hint:

Adjacent squares overlap in one fourth of the smaller square, and nonadjacent interiors do not overlap

Solution:

The sum of the five square areas is $\begin{aligned}1+\frac14+\frac1{16}&+\frac1{64}+\frac1{256}\\&=\frac{1364}{1024}.\end{aligned}$ The perpendicular-bisector placement makes the overlap of each adjacent pair one fourth of the smaller square. These four overlaps are disjoint and have total area $\begin{aligned}\frac1{16}+\frac1{64}&+\frac1{256}+\frac1{1024}\\&=\frac{85}{1024}.\end{aligned}$ Thus the union has area $\frac{1364-85}{1024}=\frac{1279}{1024},$ and $m-n=1279-1024=255.$

1995 AIME Problem 1

Answer: 255

Difficulty rating: 1900

Small Hint:

Big Hint:

Solution:

Problem 1 in Other Years