Let $ABCD$ be an [isosceles trapezoid](https://artofproblemsolving.com/wiki/index.php/Isosceles_trapezoid), whose dimensions are $AB = 6, BC=5=DA,$and $CD=4.$ Draw [circles](https://artofproblemsolving.com/wiki/index.php/Circle) of [radius](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Positive_integer), $n$ is not [ divisible](https://artofproblemsolving.com/wiki/index.php/Divisibility) by the [ square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) of any [ prime](https://artofproblemsolving.com/wiki/index.php/Prime_number), and $k$ and $p$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime). Find $k+m+n+p.$
Let the radius of the center circle be $r$ and its center be denoted as $O$.


Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$. Let $y$ be the height from the base of the trapezoid to $O$. From the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), 
\[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]
We use a similar argument with the line $DO$, and find the height from the top of the trapezoid to $O$, $z$, to be $z = \sqrt {r^2 + 4r}$.
Now $y + z$ is simply the height of the trapezoid. Let $D'$ be the foot of the [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) from $D$ to $AB$; then $AD' = 3 - 2 = 1$. By the Pythagorean Theorem, $(AD')^2 + (DD')^2 = (AD)^2 \Longrightarrow DD' = \sqrt{24}$ so we need to solve the equation $\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt {24}$. We can solve this by moving one radical to the other side, and squaring the equation twice to end with a [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation). 
Solving this, we get $r = \frac { - 60 + 48\sqrt {3}}{23}$, and the answer is $k + m + n + p = 60 + 48 + 3 + 23 = \boxed{134}$.