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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.

Michael Hardy
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filipm
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    @filipm, Have you done some work on this problem? Do you know whether the statement is provably true or provably false? The command tone in your problem statement "Prove or disprove" makes this look likes it's a homework assignment or problem-set with the "command" being given to the student to solve the problem. If so, then this is not the correct site for homework or unmotivated problems: see FAQs. If you could show what work you done, or some motivation behind this to show that it's a research problem & not homework you're trying to skip doing yourself, that would be better. – sleepless in beantown Nov 29 '10 at 12:59
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    To the best of my knowledge it is not known whether the statement is true or false. My guess is that it's true. I should have put maybe: "you're kindly asked to prove or disprove the following statement", so that it doesn't sound as a command. I read the faqs and it seems to me that the problem I asked fits well. – filipm Nov 29 '10 at 14:17
  • filipm: You have the option of changing the question so that it sounds less like homework. Just click the "edit" link below the question, and make the appropriate changes. – S. Carnahan Nov 29 '10 at 14:23
  • @filipm, what work have you done on the problem thus far? Have you played around with it restricted to $\mathbb{R}^3$ and seen what you might find? What kind of envelope would be required such that the restriction of a maximum diameter of $1$ exists? Think about that question in $\mathbb{R}^3$ and $\mathbb{R}^4$ and where that leads you. What have you gotten so far in your thoughts and efforts in this problem? The phrase "To the best of my knowledge it is not known" also encompasses the possibility that you have not thought about it or worked on it. Have you? – sleepless in beantown Nov 29 '10 at 14:26
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    Yikes; what on earth is wrong with the phrase "prove or disprove"? – Dr Shello Nov 30 '10 at 02:42
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    I'm not sure why this should generalize to higher dimensions, since it is not true for line segments in $\mathbb R^2$. In fact, using the bases of two bisecting line segments is the nicest way I can see to fit two equilateral triangles in $\mathbb R^3$ that share only one vertex. (E.g. five vertices $(\pm 1/2, \pm 1/2, 0)$ and $(0,0,\sqrt 3/2)$.) A similar construction for tetrahedra in $\mathbb R^4$ would make them share an edge.

    A possible attack: consider just the vertices of the two simplices, and study their convex hull. In my construction it's an irregular pryamid (half-octahedron).

     – Elizabeth S. Q. Goodman Dec 16 '10 at 06:13
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    The correct generalization appears to be the following: a unit regular $m$-simplex and a unit regular $n$-simplex embedded into $\mathbb R^d$ with union diameter 1 must have at least $m+n-d$ vertices in common. – Tracy Hall Jan 23 '11 at 22:06
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    @Tracy Hall: I indeed believe a stronger statement (than the one I proposed above) is true, namely that two $(d-1)$-simplices in $R^d$ must share at least $d-2$ vertices. On the other hand, I don't think your general statement is correct, I think we can embed a triangle and a tetrahedron in $R^4$ without sharing a vertex. Although, some similar generalization is certainly plausible. – filipm Jan 24 '11 at 19:34
  • Suppose we scale things to edge length $\sqrt{2}$, and place a tetrahedron with vertices at the four standard basis vectors of $\mathbb R^4$. A point whose four components sort to $a \le b \le c \le d$ lies within the allowed region if and only if $(a-1)^2 + b^2 + c^2 + d^2 \le 2$. Where are you proposing to put three such points, distinct from the basis vectors and at pairwise distance $\sqrt2$ from each other? – Tracy Hall Jan 31 '11 at 21:13
  • "It is relatively easy to show..." Filipm, can you give the argument? also can you explain your example of tetrahedron and triangle in R^4 responding to Tracy Hall's question? – Gil Kalai Feb 12 '11 at 18:41
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    Please check the edited version of the question, I tried to add more details. I'm not 100% sure about the construction, so you can check computations if you feel like that. – filipm Feb 15 '11 at 16:42
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    I did check your nice explicit example, and it works, so I was wrong: you have given an example with strictly fewer than $m+n-d$ vertices in common when $m=3$, $n=2$, and $d=4$. I wonder what the correct bound is. – Tracy Hall Feb 16 '11 at 02:00

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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 $d$.

polyanom
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