Problem (Geometry):
$\quad$ Compute the greatest real number $K$ for which the graphs of

$$
(|x|-5)^{2}+(|y|-5)^{2}=K \quad \text { and } \quad(x-1)^{2}+(y+1)^{2}=37
$$

have exactly two intersection points.

Solution:
The graph of the second equation is simply the circle of radius $\sqrt{37}$ centered at $(1,-1)$. The first graph is more interesting, and its behavior depends on $K$.

- For small values of $K$, the first equation determines a set of four circles of radius $\sqrt{K}$ with centers at $(5,5),(5,-5),(-5,5)$, and $(-5,-5)$. Shown below are versions with $K=1, K=4$, and $K=16$.

- However, when $K>25$, the graph no longer consists of four circles! As an example, for $K=36$, the value $x=5$ gives $(|y|-5)^{2}=36$; hence $|y|=-1$ or $|y|=6$. The first option is impossible; the graph ends up "losing" the portions of the upper-right circle that would cross the $x$ - or $y$-axes compared to the graph for $(x-5)^{2}+(y-5)^{2}=36$. The graph for $K=36$ is shown below.



- As $K$ continues to increase, the "interior" part of the curve continues to shrink, until at $K=50$, it simply comprises the origin, and for $K>50$, it does not exist. As examples, the graphs with $K=50$ and $K=64$ are shown below.




Overlay the graph of the circle of radius $\sqrt{37}$ centered at $(1,-1)$ with the given graphs. When $K=25$, this looks like the following graph.



Note that the two graphs intersect at $(0,5)$ and $(-5,0)$, as well as four more points (two points near the positive $x$-axis and two points near the negative $y$-axis). When $K$ is slightly greater than 25 , this drops to four intersection points. The graph for $K=27$ is shown below.



Thus for the greatest $K$ for which there are exactly two intersection points, those two intersection points should be along the positive $x$ - and negative $y$-axes. If the intersection point on the positive $x$-axis is at $(h, 0)$, then $(h-1)^{2}+(0+1)^{2}=37$ and $(h-5)^{2}+(0-5)^{2}=K$. Thus $h=7$ and $K=\mathbf{2 9}$

Final answer: 29