Odd Number of Cats?

You can't have an odd number of cats, not because of geometry, but something more fundamental: the handshaking lemma. Basically, if two cats seeing each other is mutual, then the number of cats that see an odd number of other cats must be even.

If we assume that "can see" is symmetric (no one-way mirrors, no cats prevented from turning their heads accordingly, etc.), then the situation describes a graph with $n$ vertices (the cats) and a number $e$ of edges, where each vertex has degree $3$. By the hand-shaking lemma, i.e., by counting the vertex-edge-incidences in two different ways. we find $3n=2e$, hence $n$ must be even.

This is a special case of theorem from graph theory: Given an undirected graph $G$, if $d(g)$ is the degree of each node, then $\sum_{g\in G} d(g)=2E$, where $E$ is the number of edges in the graph.

In this case, you have $d(g)=3$ for each cat $g$, so you'd want $3|G|=2E$, which means that $|G|$, the number of cats, must be even.

Basically, if $C$ is the number of cats, then $3C$ counts the each pair of cats that can see each other twice, so $3C$ must be even, so $C$ must be even.

This also means that if every cat can see exactly an odd number of cats, then the total number of cats must be even.

How about if there was a Triangle of cats. Each cat is a vertex. There are mirrors between the cats, so each cat can see itself.

Graphic Demonstration

Summary: In conclusion, I state that it is possible to see three cats in a triangle of three cats if you use mirrors.