A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $$m$$ and $$n$$, respectively, then $$m + n^2$$ is equal to

We need to find $$m + n^2$$ where $$m$$ and $$n$$ are the area and perimeter of a square inscribed in a circle, which itself is inscribed in an equilateral triangle of side 12.

In an equilateral triangle of side $$a$$, the inradius is given by $$r = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6}$$, so for $$a = 12$$ we have $$r = \frac{12\sqrt{3}}{6} = 2\sqrt{3}$$.

A square inscribed in a circle has its diagonal equal to the diameter, so the diagonal of the square is $$2r = 4\sqrt{3}$$. Since for a square with side $$s$$ the diagonal is $$s\sqrt{2}$$, it follows that $$s\sqrt{2} = 4\sqrt{3} \implies s = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$$.

The area of the square is $$m = s^2 = (2\sqrt{6})^2 = 4 \times 6 = 24$$, and its perimeter is $$n = 4s = 4 \times 2\sqrt{6} = 8\sqrt{6}$$.

Since $$n^2 = (8\sqrt{6})^2 = 64 \times 6 = 384$$, we obtain $$m + n^2 = 24 + 384 = 408$$.

The correct answer is Option 1: 408.