Circle within a Circle

Geometry Level 3

Equilateral triangle $ABC$ has a circumcircle $\Gamma$ with center $O$ and circumradius $10$ . Another circle $\Gamma_1$ is drawn inside $\Gamma$ such that it is tangential to radii $OC$ and $OB$ and circle $\Gamma$ . The radius of $\Gamma_1$ can be expressed in the form $a \sqrt{b} -c$ , where $a, b$ and $c$ are positive integers, and $b$ is not divisible by the square of any prime. What is the value of $a + b + c$ ?

The answer is 53.

1 solution

Calvin Lin Staff
May 13, 2014

Since $ABC$ is an equilateral triangle, we have $\angle BAC = 60^\circ$ , $\angle BOC = 120 ^\circ$ . By symmetry, $OO_1$ bisects $\angle COB$ , so $\angle COO_1 = 60^\circ$ .

Let the radius of $\Gamma_1$ be $r$ , then $OO_1 = 10 - r$ . Let $\Gamma_1$ be tangential to $OC$ at $M$ . We have $O_1 M = r$ . Since $M$ is the point of tangency, we have $\angle OMO_1 = 90^\circ$ . Considering right triangle $OMO_1$ , we get $\sin MOO_1 = \frac {r}{10-r}$ , or that $\frac { \sqrt{3} } {2} = \frac {r} {10-r}$ . Solving for $r$ , we get $r = \frac {10 \sqrt{3} } { 2 + \sqrt{3} } = 10 \sqrt{3} ( 2 - \sqrt{3} ) = 20 \sqrt{3} - 30$ .

Hence, $a + b + c = 20 + 3 + 30 = 53$ .

Circle within a Circle

The answer is 53.

1 solution

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