On maximal regular polyhedra inscribed in a regular polyhedron

Let T, C, O, D, or I be regular tetrahedron, cube, octahedron, dodecahedron, and icosahedron, respectively. Suppose that the outer polyhedron have edge-length 1.

For example, it's easy to prove that the length of each edge of the maximal T inscribed in C is $\sqrt2$. Let's call this problem T in C.

As far as I know, H. T. Croft proved 14 cases out of 20 cases, so the following 6 cases still remain unsolved. Two reciprocal pairs, C in I, D in O, and T in I, D in T, and (self-reciprocal) cases D in I, I in D.

Croft, HT : On maximal regular polyhedra inscribed in a regular polyhedron, Proc. London Math Soc(3), 41, 279-296, 1980.

I've tried to prove these six cases, but I'm facing difficulty. I need your help.

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Do you know how to deal with the 2-dimensional analog of your problem (n-gon inside m-gon)? I wouldn't attack the 3d problem before having completely understood its 2d baby-version. –  André Henriques Jul 31 '13 at 12:04
@AndréHenriques:No, but in my opinion we don't need to understand 2d version completely. Croft used some theorem, which is related with 'immobile polyhedra'. I think this is a key. –  mathlove Aug 3 '13 at 14:46
I feel your question is not precise enough. It sounds like you are saying "please do that on my behalf". Could you explain what you tried, and which difficulties you are facing? –  Benoît Kloeckner Jan 9 at 10:19
@BenoîtKloeckner: Well, to be honest, I don't have any good idea to start with to have a rigorous proof. I'm posting the question here just to get some hints to solve it. I don't have any intention to say "please do this on my behalf." I admit that it may sound like so, but I don't think the question is not precise. –  mathlove Jan 9 at 10:29

Edit: a preprint concerning this problem can now be found on the arXiv: http://arxiv.org/abs/1407.0683

Let me give an exhaustive answer. Croft closes his paper with a list of the unsolved cases:

Here $\kappa$ denotes the maximal edge length of the inner polyhedron. The ratio of the volumes can be easily computed when $\kappa$ is known.

Let's fill in the blank spaces that "denote ignorance"!

Let's say we want to find the largest polyhedron similar to a regular polyhedron $P$ that fits into a regular polyhedron $Q$ with edge length equal to $1$. We set up the following optimization problem.

real variables: $e$, and for every vertex $v$ of $P$: $(x_v,y_v,z_v)$

objective: maximize $e$

linear constraints: For each vertex $v$ in $P$ and each hyperplane $h^+$ defining $Q$: $$(x_v,y_v,z_v)\in h^+.$$

quadratic constraints: For each edge $(v,w)$ in $P$: $$e=(x_v-x_w)^2+(y_v-y_w)^2+(z_v-z_w)^2.$$ If $Q=C$ or $Q=D$ we have to include more constraints to make sure that the coordinates $(x_v,y_v,z_v)$ really give a polyhedron similar to $P$.

In other words: find coordinates of a polyhedron similar to $P$, which are contained in $Q$ with the square of the edge length being equal to $e$. Maximize $e$ to find the solution.

This non-linear programming problem with quadratic constraints can be effectively solved with SCIP, which uses branch-and-bound methods. (Many thanks to Ambros Gleixner for help with SCIP!). In order to run fast enough one has to set up the problem a bit more clever and remove redundant constraints. I ran the program to solve the unsolved cases and also checked to known ones. Here is a table with the results:

I emphasized new results (where Croft had blanks). Notice also that my values differ from Croft's in three places, which I marked with an exclamation mark.

• In the case $T$ in $D$ the number is off by a power of ten.
• In the case $O$ in $D$ the rounded decimal expansion should read $1.851$, not $1.1850$, since the number really is $1.85122958682191611960\ldots$.
• In the dual case $D$ in $C$ the correct decimal expansion reads $0.3942834797251374518168\ldots,$ so the rounded number should be $0.394$ not $0.395$.

However Croft gives correct exact results ($\tau$ denotes the golden ratio) and only makes these minor mistakes in the decimal approximation.

Using similar arguments about immobility as Croft, we can calculate exact results for the open pairs. Let me give you exact values for two open pairs.

For $D$ in $I$ the maximum is $$\frac{15-\sqrt{5}}{22}\approx 0.58017872829546410470867392\ldots.$$ For $C$ in $I$ the maximum is $$\frac{5+7\sqrt{5}}{22}\approx 0.93874890193175126703928253\ldots.$$

In some cases it is harder to find the algebraic values, but in all cases it can be done, see the arXiv preprint above for details.

I used sage to generate a few images. Click here to view a 3d animation.

The same optimization ansatz will also work in other dimensions and even for non-regular polyhedra.

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Congrats on filling in the unknown table entries! Can you describe the incidences in the new results, e.g., for C in I, how many cube vertices lie on {vertices, edge interiors, face interiors}? –  Joseph O'Rourke Jan 9 at 1:49
@Joseph O'Rourke Maybe these incidences are best described by this 3d animation. For C in I, two vertices of one edge in C lie in the interior of two adjacent edges in I. And the vertices of the antipodal edge of this edge in C lie in the interior of the corresponding antipodal edges in I. The other 4 edges of C lie in the interior of faces of I. Notice that there are cases where vertices of the inner polyhedron lie in the interior of the outer polyhedron. An example would be square in pentagon; see picture (a) in your answer. –  Moritz Firsching Jan 9 at 8:55
I couldn't get the 3D animation to work at first, but now I have it working---Beautiful! –  Joseph O'Rourke Jan 9 at 17:14
@Joseph O'Rourke: Thanks! It seems like java is getting less common, but sage uses it. –  Moritz Firsching Jan 9 at 18:02

To respond to André (rather than the OP's question), a 2010 paper settled the 2D problem whenever the number of sides have a common prime factor, among other results.

S. J. Dilworth, S. R. Mane. "On a problem of Croft on optimally nested regular polygons." Journal of Geometry, December 2010, Volume 99, Issue 1-2, pp 43-66. (Springer link)

For example, the largest square in a regular pentagon shares a vertex and two other corners lie on edges; the fourth corner does not touch the pentagon:

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Don't do this problem. Very boring and you will learn nothing from it.

If the first advise does not work:

• First guess what is the answer,
• Second prove that it is a strict local maximum in the space of relative orientations of these two bodies; it is essentially SO(3). (This part should be nearly a calculus problem.)
• Finally, show that it is global maximum by estimating the value at a $\varepsilon$-net of SO(3) for small enough $\varepsilon$. (While doing this you may also learn that your guess was wrong and it will suggest a better guess.)

Minkowski subtraction might help to do the last two steps.

The computers now are better than in 80's so all this might work.

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Thank you for your advice. –  mathlove Aug 3 '13 at 14:47