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:

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

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 $x=\boxed{C)5}$ .

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1962 AHSME Problems/Problem 25

Problem

Solution