In $\triangle ABC$, $AB=40$, $AC=47$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at $B,X$, and $\overline{BX},\overline{CX}$ have integer lengths. What is $BC$?

1. By power of a point (theorem 39): $CX\cdot CB=AC^2-AB^2=47^2-40^2=3\cdot7\cdot29$.
2. Since $CX(CX+BX)=3\cdot7\cdot29$ with $CX,BX$ positive integers, matching $CX=3\cdot7=21$ and $CX+BX=29$ (the factorization consistent with $BC
3. $BC=\boxed{29}$.