Problem:
Circle C with radius 2 has diameter AB. Circle D is internally tangent to circle C at A. Circle E is internally tangent to circle C, externally tangent to circle D, and tangent to AB. The radius of circle D is three times the radius of circle E and can be written in the form m​−n, where m and n are positive integers. Find m+n.
Solution:
Let circles C,D, and E have centers C,D, and E, respectively. Let circle E be tangent to AB at F. Let circle E have radius s, and circle D have radius b=3s. Then CE=2−s,DE=b+s=4s,EF=s, and DC=2−3s. The Pythagorean Theorem applied to △CEF gives CF=(2−s)2−s2​=4−4s​ and to △DEF gives DF=(4s)2−s2​=s15​. Because DF=DC+CF, it follows that s15​=(2−3s)+4−4s​. Squaring and simplifying twice reduces this equation to 9s2+84s−44=0, which has solutions s=3−14±240​​. Thus b=3s=240​−14, and the requested sum is 240+14=254​.