Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

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, none of which I can now access:

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.

share|improve this answer
add comment

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" :).

share|improve this answer
Are you Dmytro Karabash, as I identified in my posting? –  Joseph O'Rourke Mar 16 at 0:32
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.