Problem (Geometry):
Let $r=1$ and $R=5$. A circle with radius $r$ is centered at $A$, and a circle with radius $R$ is centered at $B$. The two circles are internally tangent. Point $P$ lies on the smaller circle so that $\overline{B P}$ is tangent to the smaller circle. Compute $B P$.

Solution:
Draw radius $A P$ and note that $A P B$ is a right triangle with $\mathrm{m} \angle A P B=90^{\circ}$. Note that $A B=R-r$ and $A P=r$, so by the Pythagorean Theorem, $B P=\sqrt{(R-r)^{2}-r^{2}}=\sqrt{R^{2}-2 R r}$. With $r=1$ and $R=5$, it follows that $B P=\sqrt{\mathbf{1 5}}$.

Final answer: $\sqrt{15}$