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 p−k+mn​​, 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 E be the midpoint of AB,F be the midpoint of CD,x be the radius of the inner circle, and G be the center of that circle. Then GE⊥AB. Because the point of tangency of the circles centered at A and G is on AG,AG=x+3. Use the Pythagorean Theorem in △AEG to obtain GE=x2+6x​. Similarly, find that FG=x2+4x​. Because the height of the trapezoid is 24​, conclude that
x2+6x​+x2+4x​x2+6x​x2+6x24(x2+4x)​​=24​, so =24​−x2+4x​=24+x2+4x−224​x2+4x​, and =12−x​
This yields 23x2+120x−144=0, whose positive root is x=23−60+483​​. Thus k+m+n+p=60+48+3+23=134​.