HMMT February

Solution:

Note that the condition for $\Gamma$ in the problem is equivalent to the following condition: if $\min (\angle APB, \angle BPC, \angle CPD, \angle DPA) > 60^{\circ}$, then $P$ is contained in the interior of $\Gamma$.

Let $X_1, X_2, X_3$, and $X_4$ be the four points in $ABCD$ such that $ABX_1$, $BCX_2$, $CDX_3$, and $DAX_4$ are all equilateral triangles. Now, let $\Omega_1, \Omega_2, \Omega_3$, and $\Omega_4$ be the respective circumcircles of these triangles, and let the centers of these circles be $O_1, O_2, O_3$, and $O_4$.

Note that the set of points $P$ such that $\angle APB, \angle BPC, \angle CPD, \angle DPA > 60^{\circ}$ is the intersection of $\Omega_1, \Omega_2, \Omega_3$, and $\Omega_4$. We want to find the area of the minimum circle containing this intersection.

Let $\Gamma_1$ and $\Gamma_2$ intersect at $B$ and $B'$. Define $C', D'$ and $A'$ similarly. It is not hard to see that the circumcircle of square $A'B'C'D'$ is the desired circle. Now observe that $\angle AB'D' = \angle AB'D = 60^{\circ}$. Similarly, $\angle AD'B' = 60^{\circ}$, so $AB'D'$ is equilateral. Its height is the distance from $A$ to $B'D'$, which is $\frac{1}{\sqrt{2}}$, so its side length is $\frac{\sqrt{6}}{3}$. This is also the diameter of the desired circle, so its area is $\frac{\pi}{4} \cdot \frac{6}{9} = \frac{\pi}{6}$.