Maximum perimeter for triangle inscribed in circle How to prove that isosceles triangle has maximum perimeter from all trangles inscribed in circle?
I found that from all isosceles trinagles - equilateral has maximum perimeter:
Maximum perimeter of an isosceles triangle inscribed in the unit circle?, but I wonder how to prove that a triangle with maximum perimeter should be isosceles.
 A: Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry
$$ \ell / 2 = \sin (\theta) $$
Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to 
$$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$
and we know that 
$$ \theta_1 + \theta_2 + \theta_3 = \pi $$
while
$$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$
Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to 
$$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$
by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$. 
A: Hint Fix two arbitrary vertices, and optimize the perimeter of the triangle as a function of the position of a third vertex.
