Problem - 3671
Let $a$, $b$, and $c$ be the lengths of $\triangle{ABC}$'s three sides. Compute the area of $\triangle{ABC}$ if the following relations hold: $$\frac{2a^2}{1+a^2}=b,\qquad \frac{2b^2}{1+b^2}=c,\qquad \frac{2c^2}{1+c^2}=a$$
The given conditions are equivalent to $$1+\frac{1}{a^2}=\frac{2}{b},\qquad 1+\frac{1}{b^2}=\frac{2}{c},\qquad 1+\frac{1}{c^2}=\frac{2}{a}$$
Therefore
$$1+\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}=\frac{2}{b}+\frac{2}{c}+\frac{2}{a}$$
$$\Big(1-\frac{1}{a}\Big)^2+\Big(1-\frac{1}{b}\Big)^2+\Big(1-\frac{1}{c}\Big)^2=0$$
$$\therefore\quad a=b=c=1$$
It follows that the area of $\triangle{ABC}$ is $\boxed{\frac{\sqrt{3}}{4}}$.