$A C$ and $B D$ are two perpendicular chords in a circle. The chords intersect at $E$, as shown, such that $B E=3, E C=2$, and $A E=6$. The exact perimeter of the quadrilateral $A B C D$ may be written in the form $m \sqrt{n}+p \sqrt{q}$, where $m, n, p$, and $q$ are positive integers, $q>n$, and neither $q$ nor $n$ has a divisor that is a perfect square greater than 1. Determine the value of $\sqrt{m n}+\sqrt{p+q}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d0cbae19e125c7a52478g-4.jpg?height=369&width=396&top_left_y=1553&top_left_x=1537)
If $A B$ and $D C$ are drawn, then $\angle A B D=\angle A C D$ since they are subtended by the same arc.

![](https://cdn.mathpix.com/cropped/2024_04_17_698c2a84b151b65f3cc1g-14.jpg?height=371&width=396&top_left_y=1511&top_left_x=927)

As well, we are given that $\angle A E B=\angle D E C=90^{\circ}$, so $\triangle A E B$ and $\triangle D E C$ are similar since they have two equal angles.

This means $\frac{C E}{B E}=\frac{D E}{A E}$ or $D E=\frac{A E \times C E}{B E}=\frac{6 \times 2}{3}=4$.

Since $\angle A E B=\angle B E C=\angle C E D=\angle D E A=90^{\circ}$, we can apply the Pythagorean theorem four times to compute the perimeter of $A B C D$ as

$$
\begin{aligned}
A B+B C+C D+D A & =\sqrt{A E^{2}+B E^{2}}+\sqrt{B E^{2}+C E^{2}}+\sqrt{C E^{2}+D E^{2}}+\sqrt{D E^{2}+A E^{2}} \\
& =\sqrt{6^{2}+3^{2}}+\sqrt{3^{2}+2^{2}}+\sqrt{2^{2}+4^{2}}+\sqrt{4^{2}+6^{2}} \\
& =\sqrt{45}+\sqrt{13}+\sqrt{20}+\sqrt{52} \\
& =3 \sqrt{5}+\sqrt{13}+2 \sqrt{5}+2 \sqrt{13} \\
& =5 \sqrt{5}+3 \sqrt{13}
\end{aligned}
$$

Neither 5 nor 13 has a perfect square factor other than 1 , and $13>5$, so we must have $n=5$ and $q=13$, from which it follows that $m=5$ and $p=3$.

Therefore, $\sqrt{m n}+\sqrt{p+q}=\sqrt{25}+\sqrt{16}=9$.

ANSWER: 9