9. In a rectangular box $A B C D E F G H$ with edge lengths $A B=A D=6$ and $A E=49$, a plane slices through point $A$ and intersects edges $B F, F G, G H, H D$ at points $P, Q, R, S$ respectively. Given that $A P=A S$ and $P Q=Q R=R S$, find the area of pentagon $A P Q R S$.
Answer:
$$
\frac{141 \sqrt{11}}{2}
$$

Let $A D$ be the positive $x$-axis, $A B$ be the positive $y$-axis, and $A E$ be the positive $z$-axis, with $A$ the origin. The plane, which passes through the origin, has equation $k_{1} x+k_{2} y=z$ for some undetermined parameters $k_{1}, k_{2}$. Because $A P=A S$ and $A B=A D$, we get $P B=S D$, so $P$ and $S$ have the same $z$-coordinate. But $P\left(0,6,6 k_{2}\right)$ and $S\left(6,0,6 k_{1}\right)$, so $k_{1}=k_{2}=k$ for some $k$. Then $Q$ and $R$ both have $z$-coordinate 49 , so $Q\left(\frac{49}{k}-6,6,49\right)$ and $R\left(6, \frac{49}{k}-6,49\right)$. The equation $Q R^{2}=R S^{2}$ then gives
$$
\left(\frac{49}{k}-6\right)^{2}+(49-6 k)^{2}=2\left(12-\frac{49}{k}-12\right)^{2} \text {. }
$$

This is equivalent to
$$
(49-6 k)^{2}\left(k^{2}+1\right)=2(49-12 k)^{2},
$$
which factors as
$$
(k-7)\left(36 k^{3}-336 k^{2}-203 k+343\right)=0 .
$$

This gives $k=7$ as a root. Note that for $Q$ and $R$ to actually lie on $F G$ and $G H$ respectively, we must have $\frac{49}{6} \geq k \geq \frac{49}{12}$. Via some estimation, one can show that the cubic factor has no roots in this range (for example, it's easy to see that when $k=1$ and $k=\frac{336}{36}=\frac{28}{3}$, the cubic is negative, and it also remains negative between the two values), so we must have $k=7$.
Now consider projecting $A P Q R S$ onto plane $A B C D$. The projection is $A B C D$ save for a triangle $Q^{\prime} C R^{\prime}$ with side length $12-\frac{49}{k}=5$. Thus the projection has area $36-\frac{25}{2}=\frac{47}{2}$. Since the area of the projection equals $[A P Q R S] \cdot \cos \theta$, where $\theta$ is the (smaller) angle between planes $A P Q R S$ and $A B C D$, and since the planes have normal vectors $(k, k,-1)$ and $(0,0,1)$ respectively, we get $\cos \theta=\frac{(k, k,-1) \cdot(0,0,1)}{\sqrt{k^{2}+k^{2}+1}}=\frac{1}{\sqrt{2 k^{2}+1}}=\frac{1}{\sqrt{99}}$ and so
$$
[A P Q R S]=\frac{47 \sqrt{99}}{2}=\frac{141 \sqrt{11}}{2} .
$$