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 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
.