Problem (Geometry):
The taxicab distance between points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ is defined as $d(A, B)=$ $\left|x_{A}-x_{B}\right|+\left|y_{A}-y_{B}\right|$. Given some $s>0$ and points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$, define the taxicab ellipse with foci $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ to be the set of points $\{Q \mid d(A, Q)+d(B, Q)=s\}$. Compute the area enclosed by the taxicab ellipse with foci $(0,5)$ and $(12,0)$, passing through $(1,-1)$.

Solution:
Let $A=(0,5)$ and $B=(12,0)$, and let $C=(1,-1)$. First compute the distance sum: $d(A, C)+d(B, C)=19$. Notice that if $P=(x, y)$ is on the segment from $(0,-1)$ to $(12,-1)$, then $d(A, P)+d(B, P)$ is constant. This is because if $0<x<12$,

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-x|+|5-(-1)|+|12-x|+|0-(-1)| \\
& =x+6+(12-x)+1 \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(0,6)$ to $(12,6)$. If $P$ is on the segment from $(13,0)$ to $(13,5)$, then $P$ 's coordinates are $(13, y)$, with $0 \leq y \leq 5$, and thus

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-13|+|5-y|+|12-13|+|0-y| \\
& =13+(5-y)+1+y \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(-1,0)$ to $(-1,5)$.

Finally, if $P$ is on the segment from $(12,-1)$ to $(13,0)$, then $d(A, P)+d(B, P)$ is constant:

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-x|+|5-y|+|12-x|+|0-y| \\
& =x+(5-y)+(x-12)+(-y) \\
& =2 x-2 y-7
\end{aligned}
$$

and because the line segment has equation $x-y=13$, this expression reduces to

$$
\begin{aligned}
d(A, P)+d(B, P) & =2(x-y)-7 \\
& =2(13)-7 \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(B, P)=19$ on the segments joining $(13,5)$ and $(12,6),(0,6)$ and $(-1,5)$, and $(-1,0)$ to $(0,-1)$. The shape of the "ellipse" is given below.





The simplest way to compute the polygon's area is to subtract the areas of the four corner triangles from that of the enclosing rectangle. The enclosing rectangle's area is $14 \cdot 7=98$, while each triangle has area $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$. Thus the area is $98-4 \cdot \frac{1}{2}=\mathbf{9 6}$.

Final answer: 96