# 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.

Hint Fix two arbitrary vertices, and optimize the perimeter of the triangle as a function of the position of a third vertex.

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 $\theta_1 = \theta_2 = \theta_3$.