Circle inscribed in triangle What's the radius and area of circle of max area that can be inscribed in a isoceles triangle with $2$ equal sides of length $1$? Radius formula is given, $r = \dfrac{2A}{P}$, where $A$ is area of triangle and $P$ is perimeter of triangle.
I have no idea how to do this.
I'm guessing the triangle might be equilateral because equilateral is also isosceles.

First assume that $\triangle ABC$ has $AB = AC = 1$, and denote $S$ = area of $\triangle ABC$, and $P$ = perimeter of the triangle, then:
$r = \dfrac{2S}{P} = \dfrac{2\cdot \dfrac{1}{2}\cdot 1\cdot 1\cdot \sin A}{1+1+2\sin A} = \dfrac{1}{2}\cdot \dfrac{\sin A}{1+\sin A} = f(A)$.
Taking derivative of $f$ w.r.t $A$ we have: $f'(A) = \dfrac{\cos A}{2(1+\sin A)^2} = 0 \iff \cos A = 0 \iff A = \dfrac{\pi}{2}$. This shows that $A = \dfrac{\pi}{2}$ is a maxima, and thus $r_{max} = f(\dfrac{\pi}{2}) = \dfrac{1}{4}$, and the circle with maximum area equal to: $\pi\cdot \left(\dfrac{1}{4}\right)^2 = \dfrac{\pi}{16}$.
