[Geometry]
Problem: 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.

<img_3562>



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