For $\epsilon > 0$ sufficiently small, can a regular hexagon with sides of length $1 + \epsilon$ be covered by seven equilateral triangles with sides of length $1$?

Motivation: Conway and Soifer showed that an equilateral triangle with sides $n + \epsilon$ can be covered with $n^2 + 2$ triangles. They conjectured that this is best possible, i.e. that it can not be covered by $n^2 + 1$ such triangles. This is fairly clear for $n=1$ and $n=2$ but the problem seems to be open even for $n=3$. The hexagon I've asked about is a substructure of the $n=3$ case that might be more tractable, but might still capture some of the difficulties of the $n=3$ and larger cases.


3 Answers 3


The case of hexagon is exactly the problem that is Alexander and I posed in: Karabash, D., and Soifer, A., On Covering of Trigons, Geombinatorics XV(1) (2005), 13–17.

Hexagon is type of 6-trigon (n-trigon is n connected triangles from triangulation); n-trigons for n<6 are trivial counting of vertecies and hence 6-trigon is the simplest hard case.

I gave this question to several students as one of possible problems, but I consider it a hard problem even though I see a clear non-elegant solution to the hexagon problem; one just has to work hard: 7 covering triangles can be described by 21 variables (x_i,y_i,r_i) for i=1,...,7 where x_i,y_i are coordinates of the center and r_i is rotation. Then line intersections with sides define a clear regions that one has to analyze; one can write computer program that checks all the regions and that hexagon is never covered via checking conditions. This is of course not something I would expect a high-school student to do, that is why I gave this problem with 2 stars saying it is most likely not a good project unless you get some very new idea because I have given this quesiton to guys with IMO gold, putnam fellows, and top computer science guys without any progress.

If you get any progress I will be very interested--it is also a "50 dollar problem" :).


It seems that Dmytro Karabash might be a source on this topic. A 2010 NYU course of lectures included a talk by him whose description matches your problem statement exactly:
Several possibly relevant references:

Karabash, D., On The Soifer Fifty Dollar Problem, Part I: Construction, Geombinatorics XVII(2) (2007), 68–77.

Karabash, D., On The Soifer Fifty Dollar Problem, Part II: The Existence of the Counterexample to the Conjecture, Geombinatorics XVII(3) (2008), 124–128.

Karabash, D., and Soifer, A., On Covering of Trigons, Geombinatorics XV(1) (2005), 13–17.


This is not a complete solution, but some observations.

Consider the points on the corners of the hexagon, and the centre point, since these are all distance $1+\epsilon$ from each other. Clearly, a triangle cannot cover more than one of these points, so each triangle must cover exactly one of these special points.

The circumference of the hexagon is $6+6\epsilon$, but the maximum diameter of a triangle is one. Therefore, six triangles can not cover the entire boundary, and the seventh triangle that covers the middle point must cover the cap. (This gap is a single interval on one of the boundary edges.)

Therefore, the first six triangles must cover all six outer vertices, and leave a gap on the boundary filled by the last triangle.

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    $\begingroup$ A unit triangle can cover more than $1$ of the circumference! I think it can cover up to $2/\sqrt{3}$. $\endgroup$ Apr 29, 2017 at 16:39

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