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](http://link.springer.com/chapter/10.1007%252F978-0-387-74652-4_15) 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.



Covering a hexagon
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" :).


