3. A square and an equaliteral triangle together have the property that the area of each is the perimeter of the other. Find the square's area.
Answer: $12 \sqrt[3]{4}$
Solution: Let $s, t$ be the side lengths of the square and triangle, respectively. Then $4 s=\frac{\sqrt{3} t^{2}}{4}$, and so
$$
3 t=s^{2}=\left(\frac{\sqrt{3} t^{2}}{16}\right)^{2}=\frac{3 t^{4}}{2^{8}},
$$
so that $t^{3}=2^{8}$ and $s^{2}=3 t=3 \sqrt[3]{2^{8}}=12 \sqrt[3]{4}$