[Geometry]
Problem: The graphs of $y=x^{2}-|x|-12$ and $y=|x|-k$ intersect at distinct points $A, B, C$, and $D$, in order of increasing $x$-coordinates. If $A B=B C=C D$, compute $k$.

Solution: First, note that both graphs are symmetric about the $y$-axis, so $C$ and $D$ must be reflections of $B$ and $A$, respectively, across the $y$-axis. Thus $x_{C}=-x_{B}$ and $y_{C}=y_{B}$, so $B C=2 x_{C}$. For $x<0$, the equations become $y=x^{2}+x-12$ and $y=-x-k$; setting the $x$-expressions equal to each other yields the equation $x^{2}+2 x+(k-12)=0$, from which $x=-1 \pm \sqrt{13-k}$. Therefore $x_{B}=-1+\sqrt{13-k}$ and $B C=2-2 \sqrt{13-k}$. (Note that the existence of two distinct negative values of $-1 \pm \sqrt{13-k}$ forces $12<k \leq 13$.)

Thus the $x$-coordinates of the four points are

$$
\begin{aligned}
& x_{A}=-1-\sqrt{13-k} \\
& x_{B}=-1+\sqrt{13-k} \\
& x_{C}=1-\sqrt{13-k} \\
& x_{D}=1+\sqrt{13-k} .
\end{aligned}
$$

To compute $y_{A}$, use the second equation $y=|x|-k$ to obtain $y_{A}=1+\sqrt{13-k}-k=$ $(1-k)+\sqrt{13-k}$; similarly, $y_{B}=(1-k)-\sqrt{13-k}$. Therefore

$$
\begin{aligned}
A B & =\sqrt{\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}} \\
& =\sqrt{(2 \sqrt{13-k})^{2}+(-2 \sqrt{13-k})^{2}} \\
& =2 \sqrt{2(13-k)}
\end{aligned}
$$

Because $A B=B C, 2 \sqrt{2(13-k)}=2-2 \sqrt{13-k}$. Let $u=\sqrt{13-k}$; then $2 \sqrt{2} u=2-2 u$, from which $u=\frac{2}{2+2 \sqrt{2}}=\frac{1}{1+\sqrt{2}}$, which equals $\sqrt{2}-1$ by rationalizing the denominator. Thus

$$
13-k=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}, \text { so } k=\mathbf{1 0}+\mathbf{2} \sqrt{\mathbf{2}} \text {. }
$$

Because $10+2 \sqrt{2} \approx 12.8$, the value of $k$ determined algebraically satisfies the inequality $12<k \leq 13$ observed above.
Let $C=(a, b)$. Because $C$ and $D$ lie on a line with slope 1 , $D=(a+h, b+h)$ for some $h>0$. Because both graphs are symmetric about the $y$-axis, the other two points of intersection are $A=(-a-h, b+h)$ and $B=(-a, b)$, and $a>0$.

In terms of these coordinates, the distances are $A B=C D=\sqrt{2} h$ and $B C=2 a$. Thus the condition $A B=B C=C D$ holds if and only if $\sqrt{2} h=2 a$, or $h=\sqrt{2} a$.

The foregoing uses the condition that $C$ and $D$ lie on a line of slope 1 , so now use the remaining equation and subtract:

$$
\begin{aligned}
b & =a^{2}-a-12 \\
b+h & =(a+h)^{2}-(a+h)-12 \\
h & =2 a h+h^{2}-h
\end{aligned}
$$

Because the points are distinct, $h \neq 0$. Dividing by $h$ yields $2-2 a=h=\sqrt{2} a$. Thus $a=\frac{2}{2+\sqrt{2}}=2-\sqrt{2}$.

Finally, because $C$ lies on the two graphs, $b=a^{2}-a-12=-8-3 \sqrt{2}$ and $k=a-b=$ $10+2 \sqrt{2}$.

Final answer: $10+2 \sqrt{2}$