, 2021

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 $\frac{3 n}{2}$. 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 $72 \cdot 36 = 2592$.