AMC10 2013 10A, Problem 23

In $\triangle ABC$, $AB = 86$, and $AC=97$.  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$.  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$?

Solution: Let $BX = q$, $CX = p$, and $AC$ meets the circle at $Y$ and $Z$, with $Y$ on $AC$.  Then $AZ = AY = 86$.  Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$.  We know that $p+q>p$, so $p$ is either $3$, $11$, or $33$.  We also know that $p>11$ by the triangle inequality on $\triangle ACX$.  Thus, $p$ is $33$ so we get that $BC = p+q = \fbox{61}$.

Answer: 61