HMMT November

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 $\frac{\pi}{6}$.