TITLE: A Regular Tetrahedron is a cool Polyhedron.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.

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.