Quadrilateral with maximum area inside a semicircle

Question

The quadrilateral $ABCD$ inside a semicircle with radius 1 has a maximum area. Calculate that maximum area. Clues to solution include the observation that the vertices of such quadrilateral must be on perimeter of semicircle.

Answer 1

First we note that all of the vertices of such quadrilateral must be on perimeter of semicircle:

For triangle $ABC$ having maximum area implies that B must be right in the middle of arc $AC$ (to have a maximum height).Similarly arcs $BC$ and $CD$ should be equal.
Suppose that the angle of arc $AB$ is $\alpha(0\le\alpha\le\frac{\pi}{3})$. Then

$$\begin{split} S(ABCD) &=S(AOB)+S(BOC)+S(COD)-S(AOD) \\ &=3S(AOB)-S(AOD)\\ &=\frac32\sin(\alpha)-\frac12\sin(3\alpha)\\ &=\frac32\sin(\alpha)-\frac12(3\sin(\alpha)-4\sin^3(\alpha))\\ &=2\sin^3(\alpha)\\ &\le 2 \sin^3(\frac{\pi}{3})\\ &=2(\frac{\sqrt 3}{2})^3\\ &=\frac{3\sqrt3}{4} \end{split}$$
So the maximum area of such quadrilateral inside a semicircle is $\frac{3\sqrt3}{4}$.