, 2022

Solution:

Note that $O_{1} O_{3}$ and $O_{2} O_{4}$ are perpendicular and intersect at $O$, the center of square $A B C D$. Also note that $O_{1} O_{2}, O_{2} O_{3}, O_{3} O_{4}, O_{4} O_{1}$ are the perpendiculars of $P B, P C, P D, P A$, respectively. Let $d_{1} = O O_{1}$, $d_{2} = O O_{2}$, $d_{3} = O O_{3}$, and $d_{4} = O O_{4}$. Note that since the area of $O_{1} O_{2} O_{3} O_{4} = 50$, we have that $(d_{1} + d_{3})(d_{2} + d_{4}) = 100$. Also note that the area of octagon $A O_{1} B O_{2} C O_{3} D O_{4}$ is twice the area of $O_{1} O_{2} O_{3} O_{4}$, which is the same as the area of $A B C D$. Note that the difference between the area of this octagon and $A B C D$ is $\frac{1}{2} \cdot 10 \left[ (d_{1} - 5) + (d_{2} - 5) + (d_{3} - 5) + (d_{4} - 5) \right]$. Since this must equal $0$, we have $d_{1} + d_{2} + d_{3} + d_{4} = 20$. Combining this with the fact that $(d_{1} + d_{3})(d_{2} + d_{4}) = 100$ gives us $d_{1} + d_{3} = d_{2} + d_{4} = 10$, so $O_{1} O_{3} = O_{2} O_{4} = 10$.

Note that if we translate $A B$ by $10$ to coincide with $D C$, then $O_{1}$ would coincide with $O_{3}$, and thus if $P$ translates to $P'$, then $P C P' D$ is cyclic. In other words, we have $\angle A P B$ and $\angle C P D$ are supplementary.

Fix any $\alpha \in (0^{\circ}, 180^{\circ})$. There are at most two points $P$ in $A B C D$ such that $\angle A P B = \alpha$ and $\angle C P D = 180^{\circ} - \alpha$ (two circular arcs intersect at most twice). Let $P'$ denote the unique point on $A C$ such that $\angle A P' B = \alpha$, and let $P^*$ denote the unique point on $B D$ such that $\angle A P^* B = \alpha$. Note that it is not hard to see that we have $\angle C P' D = \angle C P^* D = 180^{\circ} - \alpha$. Thus, we have $P = P'$ or $P = P^*$, so $P$ must lie on one of the diagonals. Without loss of generality, assume $P = P'$ ($P$ is on $A C$).

Note that $O_{1} O_{2} O_{3} O_{4}$ is an isosceles trapezoid with bases $O_{1} O_{4}$ and $O_{2} O_{3}$. Additionally, the height of the trapezoid is $\frac{A C}{2} = 5 \sqrt{2}$. Since the area of the trapezoid is $O_{1} O_{2} O_{3} O_{4}$, we have the midlength of the trapezoid is $\frac{50}{5 \sqrt{2}} = 5 \sqrt{2}$.

Additionally, note that $\angle P O_{1} B = 2 \angle P A B = 90^{\circ}$. Similarly $\angle P O_{2} B = 90^{\circ}$. Combining this with the fact that $O_{1} O_{2}$ perpendicular bisects $P B$, we get that $P O_{1} B O_{2}$ is a square, so $O_{1} O_{2} = P B = \frac{23 \sqrt{2} - 10 \sqrt{2}}{2} = \frac{13 \sqrt{2}}{2} = \sqrt{\frac{169}{2}}$.

Since this is the second largest side of $O_{1} O_{2} O_{3} O_{4}$, we are done.