6. Let $A B C D$ be a rectangle inscribed in circle $\Gamma$, and let $P$ be a point on minor arc $A B$ of $\Gamma$. Suppose that $P A \cdot P B=2, P C \cdot P D=18$, and $P B \cdot P C=9$. The area of rectangle $A B C D$ can be expressed as $\frac{a \sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers and $b$ is a squarefree positive integer. Compute $100 a+10 b+c$.
Answer: 21055

Solution: We have
$$
P D \cdot P A=\frac{(P A \cdot P B)(P D \cdot P C)}{(P B \cdot P C)}=\frac{2 \cdot 18}{9}=4 .
$$

Let $\alpha=\angle D P C=180^{\circ}-\angle A P B$ and $\beta=\angle A P D=\angle B P C$. Note that $\alpha+\beta=90^{\circ}$. We have, letting $x=A B=C D$ and $y=A D=B C$,
$$
2[P A D]+2[P B C]=y(d(P, A D)+d(P, B C))=y \cdot x=[A B C D] .
$$

Here $d(X, \ell)$ is used to denote the distance from $X$ to line $\ell$. By the trig area formula, the left-hand side is
$$
P A \cdot P D \cdot \sin \beta+P B \cdot P C \cdot \sin \beta=13 \sin \beta .
$$

Similarly, we have $[A B C D]=16 \sin \alpha$. Thus, letting $K=[A B C D]$,
$$
1=\sin ^{2} \alpha+\sin ^{2} \beta=\frac{K^{2}}{13^{2}}+\frac{K^{2}}{16^{2}}=\frac{425}{13^{2} \cdot 16^{2}} K^{2}
$$
giving $K=\frac{208}{\sqrt{425}}=\frac{208 \sqrt{17}}{85}$.