2013 AMC 10A Problem 23

By power of a point from $C$, $CB\cdot CX$ $=AC^2-AB^2$ $=97^2-86^2$. This equals $(97-86)(97+86)$ $=11\cdot183$ $=2013$ $=3\cdot11\cdot61$. Both $CX$ and $BX$ are integers, so $BC=BX+CX$ is an integer factor paired with $CX$. Also $CX<BC<86+97=183$, so the only possible pair is $CX=33$, $BC=61$. Thus, D is the correct answer.