Proof that $n$ planes cut a solid torus into a maximum of $\frac16(n^3+3n^2+8n)$ pieces Question: How many pieces can a solid torus be cut into with three (affine) planar cuts?
A google search will quickly reveal that the answer is thirteen, as can be read about here. The picture below displays this solution. However, despite being able to look at the diagram and confirm that thirteen pieces is indeed possible, I do not see how one would prove this is the maximum number of cuts. Furthermore, I have found the following general formula for $n$ cuts: $f(n)=\frac{1}{6}(n^3+3n^2+8n)$, but have not been able to find a proof for this either.
Better Question: How does one prove that thirteen is the maximum number of pieces with three cuts, and that $f$ provides the number of pieces with $n$ cuts?

 A: Here's an intuitive, heuristic argument that I think makes it clear why the formula has to work [plausibly this could be made rigorous although the details are more elusive than I thought at first]: The maximum number of pieces is not going to change if you change the relative size of the donut and the hole, or if you stretch the donut in one direction. So you can think of a fat, sort of cigar-shaped, donut with a very skinny long hole in it. In fact, taking this equivalence to its logical extreme, you'll get the same number of pieces if your donut is just a cake with a slit cut through the middle of it (the slit does not touch any side of the cake, so from above it looks like $\Theta$ -- notice how the crossbar does not actually touch the outer circle).
Now, instead of starting with a slitted cake, we can think of cutting a cake with $n$ planar cuts, and then cutting the slit. But the slit is just like an $(n+1)$st planar cut, except that because it doesn't reach the edges of the cake, there must be (at least) two pieces of cake that it doesn't manage to separate. Therefore, the maximum slitted cake/torus number of pieces with $n$ cuts is just the maximum number of pieces you can cut a cake into with $n+1$ cuts, minus two. But the cake numbers $C(n)$ are already well known; see https://oeis.org/A000125 for a proof that $C(n) = (n^3+5n+6)/6$.
Finally, substituting gives us the torus formula $f(n) = C(n+1) - 2 = ((n+1)^3+5(n+1)+6-12)/6 = (n^3+3n^2+8n)/6$, as expected.
