DMC 12A

Abuse the symmetry of the first two constraints and multiply them together to get $ab:ab{c}^2=1:36\Rightarrow 1:c^2=1:36\Rightarrow c=6$.

The symmetric setup motivates finding the inradius of the large triangle, which is $1+\sqrt{3}$.

We can show that the condition is satisfied iff $\{(n+1)\pi \}<\{\pi \}$, where $\{x\}$ denotes the fractional part of $x$. Using the fact that $\pi \approx \frac{22}{7}$ and also $\pi <\frac{22}{7}$, we get that when $n$ increases by $7$, $\{(n+1)\pi \}$ decreases by a little more than $0$. Since $n$ is small, this means our condition will hold for $n=7k$, or $14$ different values.

The possibilities for $P$ lie on the circle with diameter $AB$. Then, our ellipse constraint allows for at most $4$ possibilities of $P$, and these possibilities will be symmetric in each quadrant. Therefore, the area-maximizing shape is a square inscribed in the circle.