HMMT-Nov 2014 guts, Problem 10

Let $A B C$ be a triangle with $C A=C B=5$ and $A B=8$. A circle $\omega$ is drawn such that the interior of triangle $A B C$ is completely contained in the interior of $\omega$. Find the smallest possible area of $\omega$.

Solution: We need to contain the interior of $\overline{A B}$, so the diameter is at least 8 . This bound is sharp because the circle with diameter $\overline{A B}$ contains all of $A B C$. Hence the minimal area is $16 \pi$. $\fbox{16 \pi}$.

Answer: 16 \pi