# Covering a hexagon

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.

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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.

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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.)