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

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

Revision as of 20:22, 25 November 2017

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

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