Are there any solids in $R^{3}$ for which, for any 3 points chosen on the surface, at least two of the lengths of the shortest curves which can be drawn on the surface to connect pairs of them are equal?
Is there a solid where all triangles on the surface are isosceles?
2 Answers
There can be no smooth surface with this property, because a smooth surface looks locally like a plane, and the plane allows non-isosceles triangles.
As for non-smooth surfaces embedded in $\mathbb R^3$ -- which would need to be everywhere non-smooth for this purpose -- it is not clear to me that there is even a good general definition of curve length that would allow us to speak of "shortest curves".
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2@Michael, I _don't know_ whether non-smoothness would make this possible or not. Only that my particular argument against it assumes smoothness. – 2011-09-08
If one has a plane 3-connected graph in the plane all of whose faces are triangles it is known that one can not always realize this graph by a convex 3-dimensional polyhedron all of whose faces are congruent strictly isosceles triangles. There are exactly 8 types of such graphs which can be realized with equilateral triangles - the so called convex deltahedra. It is however an open problem whether one can always realize such a graph so that all of the faces are isosceles but have faces with edges of different lengths. Your question goes beyond these considerations in its requirements.