It is relatively easy to show (see below) that if we have two equilateral triangles of side 1 in $\mathbb R^3$,
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 $\mathbb R^4$ 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)


In general one is tempted to conjecture:


***If two regular $(d-1)$-dimensional simplices of edge length 1 live in $\mathbb R^d$ so that their union has diameter $1,$ then they must share a vertex.***


In fact, I believe they must share at least $d-2$ vertices, but let's start with just one vertex, probably it already contains the essence of the problem.


One may think of some even more general statements (what happens with $m$-simplex and $n$-simplex?), but I didn't come up with a reasonable conjecture.


It's easy to construct a tetrahedron and a triangle in $\mathbb R^4$ (with similar conditions) that do not share a vertex. Here I sketch it. I omit easy straightforward calculations, feel free to verify it on your own. Take a tetrahedron $abcd$ in $\mathbb R^4$ and take the midpoints $u$ and $v$ of the edges $ab$ and $cd$. We have $|uv|=1/\sqrt2$. Extend the segment $uv$ on both sides by an equal length to get a segment $xy$ of length $1.$  

Then the largest distance from $x$ and $y$ to a vertex of the tetrahedron is $\sqrt{\frac58+\frac{1}{2\sqrt2}}$ (e.g., the cosine law). Let the origin coincide with the center of the tetrahedron and let the tetrahedron lie in $\{x\_4=0\}$. Translate the points $x$ and $y$ by vector $(0,0,0,\frac{\sqrt{3}}{2}-\sqrt{5/8})$ to get points $p$ and $q$ and let $r=(0,0,0,-\sqrt{5/8})$. Now $pqr$ is an equilateral triangle of side 1 and all the distances among the points $p,q,r,a,b,c,d$ are at most 1 (some of them are trivially at most $1,$ while for the others we use the Pythagorean theorem).


Now let me sketch the proof of the statement for $d=3$. It's enough to prove that one cannot have a triangle and a unit segment which lies on one side of the triangle's plane (without increasing the diameter). Suppose this is possible. First translate the segment by a vector orthogonal to the plane until one endpoint gets into the plane (the diameter hasn't increased). Now rotate the segment around the endpoint that is in the plane, so that the other endpoint also falls in the plane (and so that the diameter doesn't increase). Now we are in a $2$-dimensional space and it's easy to show that the segment and triangle must share a vertex.


A less elementary approach is to just apply the so called Dolnikov's theorem: any two odd cycles in a graph of diameters in $\mathbb R^3$ must share a vertex, but that's an overkill.



two tetrahedra in $\mathbb R^4$
One can read the proof of the conjecture here: <http://arxiv.org/abs/1402.3694>


Also in this paper there is the proof of Schur's conjecture. This conjecture states that any diameter graph in $\mathbb R^d$ on $n$ vertices may have at most $n$ cliques of size $d$.


