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:


If you draw segments from the center of the circle to each vertex, you cut the triangle into 3 smaller triangles.  Since the radii are perpendicular to the sides, the heights of all 3 triangles is $r$.  Let the length of the sides adjacent to $\alpha$ be $b$.  Then 2 of the triangles have area $br/2$.  The side opposite $\alpha$ has length $2b\sin(\alpha/2)$ so the area of the 3rd triangle is $br\sin(\alpha/2)$.  The total area, then is
$$A=\frac{br}{2}+\frac{br}{2}+br\sin\left(\frac{\alpha}{2}\right)=br\left((1+\sin\left(\frac{\alpha}{2}\right)\right).$$
We need to express $b$ in terms of $r$.   Another espression for the area is $(1/2) b^2\sin(\alpha/2)\cos(\alpha/2).$  Solving for $b$ and plugging in the above expression and solving for $r$ gives
$$r=\frac{A}{\sqrt{\frac{2A}{\sin \alpha}}\left(1+\sin(\alpha/2)\right)}.$$
So now we "simply" maximize $r$.  This is a computational mess.  Take the derivative of $r$ with respect to $\alpha$.  Set the junk equal to zero and you get $\pi/3$.  But I did this last bit with Maple.  It looks do-able by hand by a very patient person.
