# 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>. |

## Latest revision as of 20:22, 25 November 2017

## Problem

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

## Solution

Let be the center of the circle and be the point of tangency of the circle and and let be the point of intersection of lines and Because of the symmetry, feet. Let the length of be . The length of is . By Pythagorean Theorem, . Because , , and are radii of the same circle, . So, . Squaring both sides, we obtain . Subtracting from both sides and adding , our equation becomes , so our answer is .