Find the largest area of a rectangle inside a circular segment of $\frac{2\pi}{3}$.
Solution 1:
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.}}$
Solution 2:
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 $A_y$.