7. Point $P$ is located inside a square $A B C D$ of side length 10 . Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the circumcenters of $P A B, P B C, P C D$, and $P D A$, respectively. Given that $P A+P B+P C+P D=23 \sqrt{2}$ and the area of $O_{1} O_{2} O_{3} O_{4}$ is 50 , the second largest of the lengths $O_{1} O_{2}, O_{2} O_{3}, O_{3} O_{4}, O_{4} O_{1}$ can be written as $\sqrt{\frac{a}{b}}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$.
Answer: 16902
Solution: Note that $\mathrm{O}_{1} \mathrm{O}_{3}$ and $\mathrm{O}_{2} \mathrm{O}_{4}$ are perpendicular and intersect at $O$, the center of square $A B C D$. Also note that $\mathrm{O}_{1} \mathrm{O}_{2}, \mathrm{O}_{2} \mathrm{O}_{3}, \mathrm{O}_{3} \mathrm{O}_{4}, \mathrm{O}_{4} \mathrm{O}_{1}$ are the perpendiculars of $\mathrm{PB}, \mathrm{PC}, \mathrm{PD}, \mathrm{PA}$, 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 that since the area of $O_{1} O_{2} O_{3} O_{4}=50$, we have that $\left(d_{1}+d_{3}\right)\left(d_{2}+d_{4}\right)=100$. Also note that the area of octagon $A O_{1} B O_{2} C O_{3} D O_{4}$ is twice the area of $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$, which is the same as the area of $\mathrm{ABCD}$. Note that the difference between the area of this octagon and $A B C D$ is $\frac{1}{2} \cdot 10\left[\left(d_{1}-5\right)+\left(d_{2}-5\right)+\left(d_{3}-5\right)+\left(d_{4}-5\right)\right]$. Since this must equal 0 , we have $d_{1}+d_{2}+d_{3}+d_{4}=20$. Combining this with the fact that $\left(d_{1}+d_{3}\right)\left(d_{2}+d_{4}\right)=100$ 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 $P^{\prime}$, then $P C P^{\prime} D$ is cyclic. In other words, we have $\angle A P B$ and $\angle C P D$ are supplementary.
Fix any $\alpha \in\left(0^{\circ}, 180^{\circ}\right)$. 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 $P^{\prime}$ denote the unique point on $A C$ such that $\angle A P^{\prime} B=\alpha$, 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 in we have $\angle C P^{\prime} D=\angle C P^{*} D=180^{\circ}-\alpha$. Thus, we have $P=P^{\prime}$ or $P=P^{*}$, so $P$ must lie on one of the diagonals. Without loss of generality, assume $P=P^{\prime}(P$ is on $A C)$. Note that $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is an isosceles trapezoid with bases $\mathrm{O}_{1} \mathrm{O}_{4}$ and $\mathrm{O}_{2} \mathrm{O}_{3}$. Additionally, the height of the trapezoid is $\frac{A C}{2}=5 \sqrt{2}$. Since the area of 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.