2004 AIME II Problems/Problem 12

Problem

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$ and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p.$

Solution

Let the radius of the center circle be $r$ and its center be denoted as $O$ . The center obviously has an x-coordinate of $3$ .

So line $AO$ passes through the point of tangency of circle $A$ and circle $O$ . Now let $y$ be height from the base of trapezoid to O. Now, from Pythagoreum Theorem, $3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}$ .