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A regular tetrahedron has this property:

For any two of its vertices exists a third vertex, which forms a regular triangle with these 2 vertices.(It doesn't necessarily have to be a face of it).

Are there any other polyhedrons that have the same property?

I think there isn't such a polyhedron. But not sure how to prove it. I've tried proving that there isn't an irregular tetrahedron with this property, assuming that there is a pair of unequal sides.

Current answers are a little confused with the conditions of the problem.

Jackie Poehler
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  • Rollback to revision 1. After a question receive an answer, one should not change the question to the point that make existing answers meaningless. If you have another question, please ask a new one. – achille hui Oct 09 '14 at 15:47
  • Current answers are a little confused with the conditions of the problem. – Jackie Poehler Oct 15 '14 at 13:53

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The regular tetrahedron is not the only polyhedron with this property. A triangular bipyramid can be formed such that the line segment between any two of its vertices is part of an equilateral triangle of three vertices. For the base, take an equilateral triangle with side length $3/2$. Then add an apex with altitude $1/2$ above the center, and another apex at distance $1/2$ below the center. Then the apices are at distance $1$ from each vertex of the base triangle, and also distance $1$ from eachother.

Here's one set of coordinates for the 5 vertices of this bipyramid: $$ \left(\frac{\sqrt{3}}{2},0,0\right), \left(\frac{-\sqrt{3}}{4},\frac 3 4,0\right), \left(\frac{-\sqrt{3}}{4},\frac{-3}{4},0\right), \left(0,0,\frac 1 2\right), \left(0,0,\frac{-1}{2}\right). $$ Here's an animation of the line segments between these vertices.

enter image description here

Each of the $\binom{5}{2} = 10$ line segments between two vertices is either on one of the three blue equilateral triangles, or on the green equilateral triangle.

Nick Matteo
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Outline proof (assuming 'regular triangle' = equilateral triangle):

Consider a polyhedron with the stated property. It will contain three vertices A, B and C at the corners of an equilateral triangle.

A fourth vertex which forms an equilateral triangle with each pair of vertices taken from this set can occupy only two positions. Both of these positions - call them D and E - form a regular tetrahedron with A, B and C, so the regular tetrahedron is the only 4-vertex polyhedron with the property.

To preserve the property in relation to A, B and C, a fifth vertex must occupy position D or E (whichever of these positions was not chosen for the fourth vertex).

But the bipyramid with vertices at A, B, C, D, E does not satisfy the property, because ADE is not an equilateral triangle.

Hence there are no polyhedrons with the property having 5 or more vertices.

MartinG
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  • Good proof for 5 vertices, but how does it show no polyhedron with a greater number (after all, the vertices could be related to other vertices with are none of those of the tetrahedron you start from) ? Consider also the comments below about an "infini-hedron". – Tom Collinge Sep 28 '14 at 17:31
  • Sure - this needs more thought. – MartinG Sep 28 '14 at 21:23
  • One 'infinihedron' that meets the criterion is a 2-dimensional lattice of equilateral triangles. – MartinG Sep 28 '14 at 21:30
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This isn't a very good answer: it only applies if the regular triangles are all faces. I'll leave it here if it might help anyone. Please be good enough not to downvote

(1) Let F, E and V be the number of faces, edges, and vertices on a polyhedron.

(2) Refer to http://www.math.ku.edu/~jmartin/courses/math409-S13/polyhedra.pdf for proof that E <= 3V - 6.

(3) Note that an even more fundamental property which is implied in your statement is that every pair of vertices forms an edge (only true if the triangle must be a face). That being so, $E = ^VC_2$ = $V(V-1)/2$.

(4) Put (3) and (2) together so that 0 >= $V^2 - 7V +12$. Equality holds for V = 4 (tetrahedron) but cannot be satisfied for V > 4.

Tom Collinge
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    I don't see (3) as being obvious. How do you justify it? – Blue Sep 28 '14 at 12:27
  • @Blue. OP says "For any two of its vertices exists a third vertex, which forms a regular triangle with these 2 vertices" I take it that this regular triangle is a face (otherwise clearly true for any polyhedron), so every two vertices must be an edge. Therefore take the all the vertices two at a time to get the edges. – Tom Collinge Sep 28 '14 at 12:46
  • "I take it that this regular triangle is a face". Then you're right: every pair of vertices determine an edge, so that the skeleton is the complete graph on $n$ vertices ($K_n$), which is only a polyhedron for $n=4$ (since $K_n$ for $n\geq 5$ is non-planar). But that's too easy! Note that in the extreme (non-serious) case of a spherical "infini-hedron", any two "vertices" admit a third such that the (non-edge) chords joining them form an equilateral triangle; it seems to me that the real challenge here is to determine whether this can happen with "finite" polyhedra. – Blue Sep 28 '14 at 13:21
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    @Blue. Yes the bit I missed was the regular triangle - haven't seen the term before = equilateral. With that constraint then sure you can make a triangle from any three points of any polyhedron, but not necessarily a regular one. – Tom Collinge Sep 28 '14 at 13:26