, 2020

Solution:

Let the two diameters split the circle into four sectors of areas $A$, $B$, $A$, and $B$, where $A + B = \frac{\pi}{2}$. Without loss of generality, let $A \leq B$.

If our radius cuts into a sector of area $A$, the area of the smallest sector will be of the form $\min(x, A - x)$. Note that $\min(A - x, x) \leq \frac{A}{2} \leq \frac{\pi}{8}$.

If our radius cuts into a sector of area $B$, then the area of the smallest sector will be of the form $\min(A, x, B - x) \leq \min\left(A, \frac{B}{2}\right) = \min\left(A, \frac{\pi}{4} - \frac{A}{2}\right)$. This equals $A$ if $A \leq \frac{\pi}{6}$ and it equals $\frac{\pi}{4} - \frac{A}{2}$ if $A \geq \frac{\pi}{6}$. This implies that the area of the smallest sector is maximized when $A = \frac{\pi}{6}$, and we get an area of $\frac{\pi}{6}$.