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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. (But it doesn't mean any 3 vertices form a regular triangle)

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.

  • "it doesn't mean any 3 vertices form a regular triangle": all four faces are regular triangles ?! –  Oct 15 '14 at 10:39
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    Just to clarify: The two vertices do not need be adjacent, and the triangle does not need to be a facet of the polyhedron, right? – Sebastian Negraszus Oct 15 '14 at 10:47
  • @SebastianNegraszus No, for any two of them exists the third vertex, and it doesn't matter where in the polyhedron –  Oct 15 '14 at 11:11

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