TITLE: Find the largest area of a rectangle inside a circular segment of $\frac{2\pi}{3}$.
QUESTION [15 upvotes]: What is the largest area of a rectangle inside a circular segment of $\frac{2\pi}{3}$ and radius $r$? 
  One side of the rectangle lies on the circle's chord. 
  We want a geometrical solution (as opposed to analytic geometry or trigonometry).

If $r$ is the radius, then the rectangle has an "elevation" of $\frac{r}{2}$ above the $x$ axis.
If $A$ is the upper right vertex of the rectangle and $O$ the center of the circle, also $\theta$ the angle of $OA (= r)$ with the $x$ axis, then the area is 
$$ 2r\cos θ (r\sin θ-\frac{r}{2}) $$
I can't think of any purely geometrical solution.

REPLY [3 votes]: Still cheating -
using the quadratic equation for $\sin\theta$,
\begin{align} 
\sin^2\theta-\tfrac14\sin\theta-\tfrac12&=0
\tag{1}\label{1}
,
\end{align}
but the roots are found from its coefficients 
by the old geometric method as follows.

Point $H$ is $r$ units to the right of the center $O$.
Point $U$ is $r/4$ units above the $H$.
Point $V$ is $r/2$ units to the right of $U$.
Point $W$ is the center of the circle, $|OW|=|WV|$
Intersections of that circle with the vertical line $UH$
gives two points, $X_+$ and $X_-$,
whose $y$-coordinates are the roots of \eqref{1} scaled by $r$,
and the $y$-coordinate of the point $X_+$ 
is the sought coordinate $A_y$.