# 1957 AHSME Problems/Problem 31

## Problem

A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length:

$\textbf{(A)}\ \frac{2 + \sqrt{2}}{3} \qquad \textbf{(B)}\ \frac{2 - \sqrt{2}}{2}\qquad \textbf{(C)}\ \frac{1+\sqrt{2}}{2}\qquad\textbf{(D)}\ \frac{1+\sqrt{2}}{3}\qquad\textbf{(E)}\ \frac{2-\sqrt{2}}{3}$

## Solution

$[asy] real x = 8*(2-sqrt(2))/2; // Square draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); // Corners draw((0,x)--(x,0)); draw((8-x,0)--(8,x)); draw((8,8-x)--(8-x,8)); draw((x,8)--(0,8-x)); [/asy]$

Let the side length of the regular octagon be $s$, and let the length of the legs of the isosceles right triangles be $x$. The triangles are $45-45-90$ triangles, so $s=x\sqrt2$. Because each side of the square is length $1$ and is composed of two legs of the triangles and one side of the octagon, $2x+s=1$. Substituting $s=x\sqrt2$ into this equation, we can now solve for $x$ to get our desired answer: \begin{align*} 2x+s &= 1 \\ 2x+x\sqrt2 &= 1 \\ x(2+\sqrt2) &= 1 \\ x &= \frac{1}{2+\sqrt2} \cdot \frac{2-\sqrt2}{2-\sqrt2} \\ x &= \frac{2-\sqrt2}{4-2} = \frac{2-\sqrt2}{2} \end{align*} Thus, our answer is $\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}$.