3. Let $n$ be the answer to this problem. Hexagon $A B C D E F$ is inscribed in a circle of radius 90 . The area of $A B C D E F$ is $8 n, A B=B C=D E=E F$, and $C D=F A$. Find the area of triangle $A B C$.
Proposed by: Joseph Heerens
Answer:
2592
Solution:

Let $O$ be the center of the circle, and let $O B$ intersect $A C$ at point $M$; note $O B$ is the perpendicular bisector of $A C$. Since triangles $A B C$ and $D E F$ are congruent, $A C D F$ has area $6 n$, meaning that $A O C$ has area $3 n / 2$. It follows that $\frac{B M}{O M}=\frac{2}{3}$. Therefore $O M=54$ and $M B=36$, so by the Pythagorean theorem, $M A=\sqrt{90^{2}-54^{2}}=72$. Thus, $A B C$ has area $72 \cdot 36=2592$.