# Which convex deltahedra are inscribable in the sphere?

A convex deltahedron in $$\mathbb{R}^3$$ is a convex polyhedron whose faces are all equilateral triangles. There exist precisely 8 convex deltahedra. Some examples are the regular tetrahedron, the regular octahedron, and the regular icosahedron. As regular polyhedra, these three can be inscribed in the sphere $$\mathbb{S}^2$$, meaning all of their vertices lie in the sphere. Are there any other inscribable deltahedra?

## 2 Answers

Assuming that you ask about being inscribed (rather than being inscribable) the answer is: no, there aren't any others.

You can take a look at the non-regular deltahedra over here (all of them are Johnson solids). And over here you will find all the Johnson solids that are inscribed, and none of these is a deltahedron.

Asking about inscribable, i.e. having an inscribed realization, I would say all of them are inscribable (just my intuition, given the pictures), though I am not absolutely sure for the snub disphenoid.

• Thanks for your answer. Are you aware of a proof of those that can be inscribed with all vertices in the sphere? Oct 7 '20 at 15:00
• @user3816 You mean a proof that the regular polyhedra are inscribed? This is almost by definition: they are vertex-transitive, which implies being inscribed. On what lever do you need a proof? I would argue that it is folklore. Oct 7 '20 at 15:03
• @user3816 No, I do not have a source. It is quite clear for the two bipyramids though. Oct 7 '20 at 16:02
• @user3816 If my answer was helpful, please consider an upvote or an accept, though for the latter I suspect you want to have a proof for being inscribable. Oct 7 '20 at 22:40
• @RavenclawPrefect By inscribable (in contrast to inscribed) I mean that the polyhedron has a (combinatorially equivalent) realization with all vertices on a common sphere. After OPs comments I had the feeling that this is what he is actually after. Oct 31 '20 at 8:55

None of the non-regular convex deltahedra can be inscribed in a sphere.

Proofs:

• Triangular bipyramid: Consider inscribing an equilateral triangle in a sphere (this will be the "equatorial triangle" consisting of the vertices of degree 4 in the bpyramid). There are many ways to do this, but up to symmetry, they only depend on how far "up" or "down" an axis of the sphere the plane of the triangle moves. Because the "polar" vertices of the bipyramid have to be on this axis at equal distances from the plane of the triangle, we know that they have to be at antipodal points of the sphere, and so the center of the sphere aligns with the center of our equatorial triangle. But then (if we normalize to unit edge length) the triangle's equatorial vertices are at distance $$\sqrt{3}/3$$ from the center while its polar vertices are at distance $$\sqrt{6}/3$$ from the center.

• Pentagonal bipyramid: Exactly the same argument as above, but now the equatorial vertices are further away from the center than the polar vertices (which you can easily verify with some trigonometry or coordinate-bashing). Note that the square bipyramid, AKA the regular octahedron, hits the happy medium between these two scenarios, so you can inscribe it in the sphere.

• Snub disphenoid: Consider the two most distant edges of the shape. These four vertices form a tetrahedron which can only be inscribed in the sphere in one way, forcing the center of the sphere to coincide with the center of this tetrahedron by symmetry. Then the center of the sphere must line up with the origin in the Wikipedia list of coordinates for the shape. But you can easily check that the vertices are not all at the same distance from the origin, so they do not all lie on the surface of the sphere.

• Triaugmented triangular prism: This polyhedron contains square pyramids, each of which can only be inscribed in the sphere in one way up to rotation about the surface of the sphere. But we know from the octahedron (which is a union of two square pyramids) that this inscription aligns the center of the square base of the pyramid with the center of the sphere. But the triaugmented triangular prism has three such square pyramids, each of whose square bases have different centers. So there is no consistent positioning of the center of any circumscribing sphere.

• Gyroelongated square bipyramid: Same reasoning as above with its two opposing square pyramids.

• This states the same as my answer (though with explicit arguments instead of references). But still, this is about being inscribed rather than being inscribable. for example, the pentagonal bipyramid is inscribable. OP should have been clearer here. Oct 31 '20 at 8:57