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Based on my previous two questions (here and here) I found it safe to conjecture that:

There is no such regular polyhedra whose volume is equal to the difference between the volumes of its circumsphere and its insphere.

Now this should be pretty much straightforward to prove (or disprove) as there are only $5$ regular polyhedra. I have verified it myself for the cube but the others seem to be a bit too daunting, especially the icosahedron and the dodecahedron. I assume that the inspheres and circumspheres of the $5$ platonic solids are concentric spheres and it was true for the cube but that is as far as I have been able to go. My imaginative powers are incredibly weak and I would probably slap my head once I see how the parts of the proof involving the icosahedron and the dodecahedron are done.

By the way, is it necessary to try out all $5$ platonic solids or can it be proved by showing that the $n$ (denoting number of edges) required for this to be true is not a positive integer? Thanks in advance!

Bonus: I am almost sure that one or the other irregular polyhedra will not obey this conjecture (pardon my terminology if it is wrong). Can anyone find (at least) one such polyhedron?

Rules:
$(1)$ The circumsphere must pass through all the vertices of the polyhedron.
$(2)$ The insphere must be tangent to all the faces of the polyhedron.
$(3)$ The polyhedron can not be a spherical polyhedron.

I personally believe that a truncated pyramid will suffice easily. Can anyone find other counter-examples? Good luck!

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    $\begingroup$ The radii of the circumsphere and insphere of a regular polyhedron are given by $R=\left({a \over 2}\right)\tan {\frac {\pi }{q}}\tan {\frac {\theta }{2}}$ and $r=\left({a \over 2}\right)\cot {\frac {\pi }{p}}\tan {\frac {\theta }{2}}$, respectively, where $(p,q)$ is the Schäfli symbol, so it should be easy to check for the five polyhedrons. $\endgroup$ – anderstood Jan 19 '18 at 16:33
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    $\begingroup$ Or, even simpler, use the values from the table here. $\endgroup$ – anderstood Jan 19 '18 at 16:34
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The quotients of length and distance data describing all sorts of regular polyhedra, incl. the Archimedean ones, etc., are algebraic numbers. It follows that the volume of such a polyhedron is an algebraic multiple of $\rho^3$, where $\rho$ denotes the inradius of $P$. Similarly the circumradius $R$ of $P$ is an algebraic multiple of $\rho$. But the difference between the volumes of the circumsphere and the insphere of $P$ is given by $${4\pi\over3}(R^3-\rho^3)=\pi\>\lambda \rho^3$$ with $\lambda$ algebraic, and is not of this kind.

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  • $\begingroup$ Beware that this proof relies on the fact that sen(q\pi) with q rational is an algebraic number. This is because cos(pi/n) satisfy the chebycheff polynomial. In the irregular case, the surface of a polygon can be trascendent with respect to the side. $\endgroup$ – Andrea Marino Jan 19 '18 at 17:33
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Not as elegant as Christian Blatter's answer, but here is a Mathematica code to compare the values given in wikipedia's table.

comp[r_, V_, R_] := N@{4/3*Pi*R^3 - 4/3*Pi*r^3, V}
comp[1/Sqrt[6], Sqrt[8]/3, Sqrt[3/2]]
comp[1, 8, Sqrt[3]]
comp[Sqrt[2/3], Sqrt[128]/3, Sqrt[2]]
comp[phi^2/xi, 20 phi^3/xi^2, Sqrt[3]*phi]
comp[phi^2/Sqrt[3], 20 phi^2/3, xi*phi]

{7.41029, 0.942809}
{17.5768, 8.}
{9.5676, 3.77124}
{45.9338, 61.305}
{14.3614, 17.4536}

The three first ones have a volume smaller than the difference between the circum- and in-spheres volumes, and that is the reverse for the two last ones.

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  • $\begingroup$ When you have to wait for $7$ hours to upvote deserving answers :( $\endgroup$ – Mohammad Zuhair Khan Jan 19 '18 at 16:53
  • $\begingroup$ @MohammadZuhairKhan, it's been 8 hours since your comment, in case that was a subtle request for a reminder. $\endgroup$ – Wildcard Jan 20 '18 at 1:48
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    $\begingroup$ @MohammadZuhairKhan, good! Hey, it's not like I could tell if you had already voted or not. :) My comment was solely intended to be helpful, in case you were leaving the comment in hopes of returning to it later (like I sometimes do). $\endgroup$ – Wildcard Jan 20 '18 at 8:04

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