# Difference between revisions of "1962 AHSME Problems/Problem 25"

## Problem

Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$. The radius of the circle, in feet, is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$

## Solution

Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$. The length of $OE$ is $EF-OF=-x+8$. By Pythagorean Theorem, $OA=OD=\sqrt{x^2+4^2}=\sqrt{x^2+16}$. Because $OA$, $OD$, and $OE$ are radii of the same circle, $-x+8=OE=OA=AD=\sqrt{x^2+16}$. So, $\sqrt{x^2+16}=-x+8$. Squaring both sides, we obtain $x^2+16=x^2-16x+64$. Subtracting $x^2+16$ from both sides and adding $16x$, our equation becomes $16x=80$, so our answer is $x=\boxed{C)5}$.