A triangle having coordinates $(a\cos\phi, a \sin\phi) , (a\cos\theta, a\sin\theta) , (a\cos\psi, a \sin\psi)$....

Question

A triangle having coordinates $(a\cos\phi, a \sin\phi) , (a\cos\theta, a\sin\theta) , (a\cos\psi, a \sin\psi)$ having its area $$ \Delta = 2a^2 \sin\frac{\theta - \phi}{2}\sin\frac{\phi -\psi}{2}\sin\frac{\psi -\theta}{2}$$

Please suggest when will be the area of this triangle is maximum. Thanks.

Answer 1

Setting $\theta=0$ without loss of generality then setting the partial derivatives to 0 and solving yields $\phi=2\pi/3$, $\psi=4\pi/3$.

Answer 2

Let two vertexes be fixed and taken as a base. Then the position of the third vertex that maximizes the area is the one that maximizes the altitude, i.e. on the mediatrix. So the triangle must be isosceles.

Repeating for the three sides, the triangle must be equilateral and the vertexes are spaced by 120°.