# Difference between revisions of "1953 AHSME Problems/Problem 15"

## Problem 15

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:

$\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ \textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}$

## Solution

The maximum diameter of the circular piece is the same as the side length of the square piece, so the circle is tangent to the square on all four sides. The maximum size a square piece that you can cut from the circle now has 4 edges that are the same as the square created by connecting the four midpoints of the original square. If the side length of the original square is $s$, then the sides of the final square have length $\sqrt{\frac{s^2}{2}}$, so its area is $\frac{s^2}{2}$, $\frac{1}{2}$ the area of the original square. $\textbf{(B)}$

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