It is relatively easy to show that if we have two equilateral triangles of side 1 in $R^3$, such that their union has diameter 1, then they must share a vertex. I wonder whether we have an analog of this in higher dimensions. To start with 4 dimensions, the question is whether the following statement is true:

If two regular tetrahedra of side length 1 are placed in $R^4$ so that the diameter of their union is 1, then the tetrahedra must share a vertex.

(here 'tetrahedron' is the convex hull of four points with equal pairwise distances and 'diameter' of a set is the maximum distance between two points of the set)

One can imagine how would a straightforward generalization to all dimensions sound.

It's easy to construct a tetrahedron and a triangle in $R^4$ (with similar conditions) that do not share a vertex.