Find the largest area of a rectangle inside a circular segment of $\frac{2\pi}{3}$. 
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. 

 A: I just wanted to say to the optimistic geometer out there that this is the value of the maximum area when $r=1$, as per Wolfram Alpha:
$$
2 \\ \times \\
\left(1-\frac{2}{\left(\frac{1}{4}+\frac{1}{4}\left(\sqrt{2\left(9-\sqrt{33}\right)}-\sqrt{33}\right)\right)^2+1}\right)
\\ \times \\
\left(\frac{2\left(\frac{1}{4}+\frac{1}{4}\left(\sqrt{2\left(9-\sqrt{33}\right)}-\sqrt{33}\right)\right)}{\left(\frac{1}{4}+\frac{1}{4}\left(\sqrt{2\left(9-\sqrt{33}\right)}-\sqrt{33}\right)\right)^2+1}+\frac{1}{2}\right)$$
What does this tell us? To start, it doesn't tell us that a purely geometric solution is impossible. $_{_\text{However, we do learn that $+100$ rep isn't worth the challenge.}}$
A: 
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 $A_y$.
A: What makes you assume that a geometric construction is possible at all?
Anyways trying to find maximizing area by differentiation of
$$ \frac{A}{r^2} = \cos \theta\cdot ( 2 \sin \theta -1) $$
which results in maximum rectangle area solution
$$  \sin \theta=\frac{\sqrt{33}+1} {8}   $$
A geometric re-construction by such prior knowledge already sounds like cheating.
The power of the big circle drawn is $33$
A geometric construction sketch is nevertheless given. Circle diameters are not drawn to proportion on purpose because narrow lines would not show detail ... the right angle $ TQP$ does not even look right angled at $P$. A Geogebra sketch can be made though.
Required $\theta$ is $\angle TQP$ of green/red triangle.

