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This MO answer discusses this table involving the maximal side lengths of the five Platonic solids $T,C,O,D,I$ inscribed in the other solids,

new table

This table is also found in Moritz Firsching's paper. I noticed that almost all are roots of equations with solvable Galois groups. For example, $d = 0.162631\dots$ is a root of,

$$4096d^{16} - 3701760d^{14} + 809622720d^{12} - 17054118000d^{10} + 79233311025d^8 - 94166084250d^6 + 31024053000d^4 - 3236760000d^2 + 65610000=0$$

The discriminant of $F(\sqrt{d})=0$ is a perfect square and Magma says the above has a solvable Galois group.


The exception seems to be $t=1.3474429\dots$ (the tetrahedron in icosahedron case) which is a root of,

$$5041 t^{32} - 1318386 t^{30} + 60348584 t^{28} - 924552262 t^{26} + 5246771058 t^{24}-15736320636 t^{22} + 29448527368 t^{20} - 37805732980 t^{18} + 35173457839 t^{16}-24298372458 t^{14} + 12495147544 t^{12} - 4717349124 t^{10} + 1256858478 t^8 - 217962112 t^6 + 21904868 t^4 - 1536272 t^2 + 160801 = 0$$

The discriminant of $F(\sqrt{t})=0$ is not a perfect square (it is divisible by the seemingly random prime $466369383062945371$), and Magma says $F(\sqrt{t})$ has group $16T1952$, order $2^{15}\cdot3^4\cdot5^2\cdot7^2$, hence is not solvable.

Q: Is there an a priori reason why the T in C case is not a radical, unique among the other cases? (Or is there something wrong with the polynomial?)

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    $\begingroup$ If I understand correctly, you are going from decimal approximations with around 7 digits, and fitting polynomials to them. If the polynomials are small, this is surely an appropriate thing to do. But your polynomials have coefficients with 11 digits in them, so how do you know you have not overfit your equations? If I have misunderstood your strategy, please forgive me, and elaborate? $\endgroup$ Jul 29, 2017 at 21:20
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    $\begingroup$ @TheoJohnson-Freyd: My apologies for that impression, but it's the other way round. In the paper I cited by Firsching, he first derives the polynomials, then the table gives the approximate roots. However, I believe he does use an integer relations algorithm to find the polynomials based on hundreds of decimal digits. $\endgroup$ Jul 30, 2017 at 0:51
  • $\begingroup$ I think, in the question you mean T in I, not T in C. I will try to answer... $\endgroup$ Aug 1, 2017 at 16:58
  • $\begingroup$ @TitoPiezasIII you are right, I use a good numerical approximation to determine the algebraic solutions. $\endgroup$ Aug 1, 2017 at 17:29
  • $\begingroup$ @TitoPiezasIII here is another answer with a more involved algebraic answer: mathoverflow.net/a/278340/39495 $\endgroup$ Aug 10, 2017 at 6:52

1 Answer 1

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I guess it would be difficult to prove that the answero your question is "no", since proving that "no a priori reason exists" might be hard.

More modestly, I can say that I don't really know a good reason. However, here are some pointers: One essential fact is that four of the five platonic solids are centrally symmetric; all but the tetrahedron. Croft observed: in an optimal inclusion two concentric polytopes will share a center. Therefore, an optimal inclusion for two platonic solids, both of which are not the tetrahedron will result in a concentric situations. This makes these configurations considerably easier. This (at least hopefully somewhat) explains that the degree of the algebraic numbers in the 12 cases not involving the tetrahedron. Here is a table with the exact values:

Table with exact values

Let's look at the cases involving the tetrahedron, that is the first row and the first columns of the table above. The pairs $(T\subset C)$ and $(O\subset T)$ are again concentric, and so are the pairs $(T\subset D)$ and $(I\subset T)$. The case $(T\subset O)$ is particulary simple to describe, since $T$ and $O$ share a face in the optimal solution. What remains are really just $(C\subset T)$, $(D\subset C)$ and $(T\subset I)$. And indeed those are the one with solutions that are more algebraically involved. (The case $(C\subset T)$ not so much, since in the optimal configuration edges align and the constraints to begin with are "only" coming from $T$ and $C$ and not, as in the other two cases from the more complicated $D$ and $I$.)

to summarize: There are really only two cases where one might expect more involved solutions, one of which is radical, the other not.


To address the second part of the question: "Or is there something wrong with the polynomial?": I hope not.

Let me give you a few more details, how one can calculate the polynomial in question. We assume we know the incidences, i.e. what vertex of the tetrahedron lies on what face of the icosahedron and deduce the optimal solution from here. The incidences are shown in the following picture: one vertex of T coincides with a vertex of I, one lies in an edges, the two remaining in two seperate faces of I.

Tetrahedron in Icosahedron

I fix an icosahedron with edge-length $1$ and vertices $$\begin{align} v_{0} & = \left(0,\,\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\ v_{1} & = \left(0,\,-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\ v_{2} & = \left(\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\ v_{3} & = \left(\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\ v_{4} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,\frac{1}{2}\right) \\ v_{5} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\ v_{6} & = \left(-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\ v_{7} & = \left(-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\ v_{8} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,\frac{1}{2}\right) \\ v_{9} & = \left(0,\,\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\ v_{10} & = \left(0,\,-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\ v_{11} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\ \end{align}$$ The four points of the tetrahedron are then given as follows: $$\begin{align}p_0 =& v_{11}\\p_1 =& f_0v_0 + f_1v_2 + (1-f_1-f_0)v_6 \\ p_2 =& g_0v_1 + g_1v_3 + (1-g_0-g_1)v_7 \\p_3 =& e_0v_5 + (1-e_0)v_{10}\end{align}$$ For some positive variables $f_0, f_1, g_0, g_1$ and $e_0$. For each $(i,j)\in\binom{[4]}{2}$, we have $(p_i-p_j)^2 = s^2$ for some real variable $s$. Putting it all together, we obtain the following system of 6 equations with 6 variables: $$\begin{align} 0 =& f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} f_{0} + \frac{1}{2} \, \sqrt{5} f_{1} - \frac{1}{2} \, f_{0} - \frac{1}{2} \, f_{1} + 1 \\ 0 =& g_{0}^{2} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} g_{0} + \frac{1}{2} \, \sqrt{5} g_{1} - \frac{1}{2} \, g_{0} - \frac{1}{2} \, g_{1} + 1 \\ 0 =& e_{0}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} - \frac{1}{2} \, e_{0} + 1 \\ 0 =& -\frac{1}{2} \, \sqrt{5} f_{0} g_{0} + f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - \frac{1}{2} \, f_{0} g_{0} - f_{1} g_{0} + g_{0}^{2} - f_{0} g_{1} - 2 \, f_{1} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} - f_{0} - g_{0} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\ 0 =& -\frac{1}{2} \, \sqrt{5} e_{0} f_{1} + e_{0}^{2} - e_{0} f_{0} + f_{0}^{2} - \frac{1}{2} \, e_{0} f_{1} + f_{0} f_{1} + f_{1}^{2} - s^{2} - e_{0} - f_{1} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\ 0 =& -\frac{1}{2} \, \sqrt{5} e_{0} g_{0} - \frac{1}{2} \, \sqrt{5} e_{0} g_{1} + e_{0}^{2} - \frac{1}{2} \, e_{0} g_{0} + g_{0}^{2} - \frac{1}{2} \, e_{0} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} + \frac{1}{2} \, \sqrt{5} g_{0} - \frac{1}{2} \, e_{0} - \frac{1}{2} \, g_{0} - g_{1} + 1 \\ \end{align}$$ This systems happens to have the solution $$\begin{align} f0=& 0.356785524577257 \text{..., zero of } 5751x^{16} + 54216x^{15} - 434286x^{14} - 466374x^{13} + 9452306x^{12} - 19323022x^{11} - 33022460x^{10} + 206565938x^{9} - 384738484x^{8} + 362774804x^{7} - 180708354x^{6} + 47907122x^{5} - 9497814x^{4} + 61518x^{3} + 895260x^{2} - 144318x + 401 \\ f1=& 0.352452740635196 \text{..., zero of } 5751x^{16} + 24732x^{15} - 16650x^{14} - 342012x^{13} - 901126x^{12} - 959712x^{11} + 743040x^{10} + 4747162x^{9} + 2877417x^{8} - 2704036x^{7} + 2767104x^{6} + 2220422x^{5} - 977942x^{4} - 2032764x^{3} - 504834x^{2} + 243738x + 64251 \\ g0=& 0.595049283356260 \text{..., zero of } 5751x^{16} + 46926x^{15} - 623124x^{14} + 3139674x^{13} - 7786156x^{12} + 2057864x^{11} + 49657348x^{10} - 189659290x^{9} + 400533046x^{8} - 535383644x^{7} + 467013828x^{6} - 298170008x^{5} + 167384576x^{4} - 74703516x^{3} + 17034188x^{2} - 698296x + 164 \\ g1=& 0.0729071548475811 \text{..., zero of } 5751x^{16} - 46224x^{15} - 194904x^{14} + 1789782x^{13} - 3522208x^{12} - 3388962x^{11} + 50328778x^{10} - 43454770x^{9} - 123966325x^{8} + 205675176x^{7} - 80109508x^{6} - 35043358x^{5} + 51387632x^{4} - 18740400x^{3} + 1053588x^{2} + 9072x - 369 \\ e0=& 0.645495309223693 \text{..., zero of } 71x^{16} + 972x^{15} - 2730x^{14} - 19898x^{13} + 64290x^{12} - 61466x^{11} + 46202x^{10} - 95276x^{9} + 136499x^{8} - 92310x^{7} + 51560x^{6} - 48088x^{5} + 35854x^{4} - 11920x^{3} - 2804x^{2} + 2406x - 41 \\ s=& 1.34744285033120 \text{..., zero of } 5041x^{32} - 1318386x^{30} + 60348584x^{28} - 924552262x^{26} + 5246771058x^{24} - 15736320636x^{22} + 29448527368x^{20} - 37805732980x^{18} + 35173457839x^{16} - 24298372458x^{14} + 12495147544x^{12} - 4717349124x^{10} + 1256858478x^{8} - 217962112x^{6} + 21904868x^{4} - 1536272x^{2} + 160801 \\ \end{align}$$ While the solution was found by using Newton's method combined with integer relations algorithms, the solution can be checked using exact calculations in $\mathbb{A}$ or in a smaller numberfield, which contains all the relevant numbers. Indeed one can take the number field $F$ with defining polynomial any of the ones defining the solutions (substituting s by sqrt(s)). They are all isomorphic and have discriminant $10637699079912558734361600000000 = 2^{16} \cdot 3^{4} \cdot 5^{8} \cdot 11 \cdot 466369383062945371$, divisible by the seemingly random prime.

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  • $\begingroup$ What is the meaning of the five-pointed stars next to some of the entries in the table? $\endgroup$ Aug 1, 2017 at 23:20
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    $\begingroup$ @GerryMyerson : I copied that from the paper. Those mark the new results, the other were known to Croft previously. $\endgroup$ Aug 2, 2017 at 6:42
  • $\begingroup$ @MoritzFirsching Ooops, I totally forgot to accept this answer. $\endgroup$ Jul 7, 2023 at 16:49

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