Given convex polyhedra $P, Q$, say that one can inscribe $P$ in $Q$ if we can find points on the surface of $Q$ whose convex hull is similar to $P$.

If we restrict $P, Q$ to be Platonic solids, we can achieve every case except inscribing the dodecahedron into the tetrahedron, cube, or octahedron; the other $17$ distinct pairs work.

For the $17$ working cases, there are "nice" constructions, where the solids are positioned in symmetric ways with respect to each other. Consequently, such constructions are pretty easy to verify by a combination of symmetry arguments, direct computation of one or two distances, and arguments from degrees of freedom / continuity. To see examples of these pairs, Moritz Firsching's 2018 paper "Computing maximal copies of polytopes contained in a polytope" (PDF link via arXiv.org) shows instances of maximal containment, most of which have "nice" symmetry (or are obviously given by e.g. taking duals). Perhaps the hardest case to see a "nice" construction for is the cube inside the tetrahedron, pictured below (bold edges tangent to the faces):

                                             Tetrahedron in a cube

For two of the three impossible cases, I have simple proofs of impossibility:

  • If a dodecahedron were inscribed in a tetrahedron, every face would need to have a pentagon on it, but the dodecahedron doesn't have four mutually disjoint faces.

  • If a dodecahedron were inscribed in a cube, then any face with at least three of the vertices of the dodecahedron would have to have an entire pentagon on it (and no more of the vertices), since three points determine a plane and the dodecahedron is convex. Then (by simple counting) at least three of the cube's faces must have such a pentagon on them, so two such are orthogonal to each other. But the dodecahedron has no orthogonal faces.

However, I know of no such proof that the dodecahedron cannot be inscribed in the octahedron; it seems fairly plausible, at first glance, that one could mount the dodecahedron with two opposite faces tangent to opposite faces of the octahedron and arrange the other $10$ points across the other $6$ faces of the octahedron somehow. The simplest proof I know involves doing some gross trigonometry involving the necessary side length of such a dodecahedron and the inradius of the regular $10$-gon formed by taking an equatorial cross-section of the dodecahedron in between these two opposite faces.

Is there a conceptually simple proof that this construction is impossible? I realize this is somewhat subjective, but I hope the intended style of argument is clear: reasoning from symmetry and combinatorial incidences, rather than bashing the coordinates of different vertices.

  • 1
    $\begingroup$ In the style of your dodecahedron-tetrahedron proof, "If a cube were inscribed in a tetrahedron, every face would need to have a square on it, but the cube doesn't have four mutually disjoint faces." This proof is wrong because it does not account for all the kinds of "inscribing" you allow. $\endgroup$
    – David K
    Commented Feb 14, 2021 at 1:01
  • $\begingroup$ @DavidK: Why would every face need to have a square on it? The reason every face must have a pentagon in the dodecahedron case is because the dodecahedron has $20$ vertices, and at most $5$ can be on a single face of the tetrahedron at once, so all four faces need to have a full allotment of $5$ vertices in order to get them all. This is not the case for the cube, because we can distribute $2$ vertices to every face (as shown). $\endgroup$ Commented Feb 14, 2021 at 4:21
  • $\begingroup$ So the argument is that each of the four faces would have to have at least $5$ vertices (since you can't put more than $5$ vertices of a dodecahedron on one face of a circumscribed polyhedron). And to have even three vertices you need an entire face of the dodecahedron. $\endgroup$
    – David K
    Commented Feb 14, 2021 at 4:39
  • $\begingroup$ @DavidK: Yep, exactly! $\endgroup$ Commented Feb 14, 2021 at 4:47

1 Answer 1


The regular dodecahedron has $20$ vertices; in order to place every vertex on a face of a regular octahedron, since the octahedron has only $8$ faces, by the pigeonhole principle we must put at least three vertices on some face of the octahedron. This implies one face $A$ of the dodecahedron lies on a face $B$ of the octahedron.

Consider the sphere $S$ inscribed in the dodecahedron. Sphere $S$ is tangent to face $B$ of the octahedron. Moreover, whenever three vertices of the dodecahedron lie on one face of the octahedron, a face of the dodecahedron lies on that face of the octahedron as well, and sphere $S$ also is tangent to that face of the octahedron. A face of the octahedron to which $S$ is not tangent can have at most two vertices of the dodecahedron. Moreover, if $S$ is tangent to face $F$ of the octahedron but not to the opposite face, then no vertices of the dodecahedron can be on the opposite face of the octahedron.

Now let's see how many faces of the octahedron $S$ is tangent to. By symmetries of the octahedron, $S$ can be tangent to exactly one face; tangent to two adjacent faces and no others; tangent to four faces around a vertex and no others with the center of $S$ on the diagonal between two vertices of the octahedron, or tangent to all eight faces and coincident with the inscribed sphere of the octahedron.

Exactly one tangent face is impossible, since $7$ faces with at most two vertices each is only $14$ vertices, whereas the dodecahedron has $15$ vertices that are not on face $B$ of the octahedron.

Exactly two tangent faces implies the two opposite faces have no vertices of the dodecahedron; since the two tangent faces can contain at most $10$ vertices of the dodecahedron, the remaining four faces must contain the other $10$ vertices, which cannot occur with only two vertices per face.

In the case of four tangent faces, each of those faces contains one face of the dodecahedron, and there is also a diametrically opposite parallel face of the dodecahedron. The planes of those diametrically opposite faces, together with the four tangent faces of the original octahedron, form a smaller octahedron tangent to $S$ on all eight faces, therefore with a face of the dodecahedron on each face of the octahedron. By inspection, it is not possible to place all five vertices of a regular pentagon on the sides of an equilateral triangle. We can place at most four vertices, none of them at a vertex of the triangle, leaving one inside the triangle. Since each vertex on an edge of a triangle lies on two faces of the octahedron, we have an average of at least three vertices per face, which comes to at least $24$ vertices. Since the dodecahedron has only $20$ vertices, this is impossible.

The case of eight tangent faces is the same as the case of four tangent faces, except that we start with an octahedron tangent to $S$ on all faces rather than having to construct a new one.

Hence it is not possible to inscribe a regular dodecahedron in a regular octahedron.

  • $\begingroup$ Nicely done! If no one finds a simpler proof, the bounty's yours. $\endgroup$ Commented Feb 15, 2021 at 16:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .