Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.

My attempt:


We know that every triangle has a unique incircle; we also have the following 2 theorems:

Theorem 1: Among all triangles of given perimeter, the equilateral one has the
  largest area.
Theorem 2: The radius $r$ of the incircle for a triangle  $\triangle ABC$ is
  given by $ r = 2 \frac{Area(\triangle ABC) } {Perimeter(\triangle ABC)}$

Let $A > 0$ and restrict our focus to all triangles satisfying
$Area(\triangle ABC) = A$.
By Theorem 2, if we have any two such triangles, the one with the smaller perimeter will give us a larger incircle radius $r$. 
Let $\mathcal T$ be any triangle with area $A$ and suppose it is not equilateral. By Theorem 1, the equilateral triangle with the same perimeter has a greater area. We can scale this equilateral triangle down to a triangle $\mathcal E$ where $Area(\mathcal E) = A$ and $Perimeter(\mathcal E) < Perimeter(\mathcal T)$.
Since this equilateral triangle $\mathcal E$ is an isosceles triangle, we arrive at the answer:
$\alpha = 60 °$.
