Can rotations and translations of this shape

perfectly tile some equilateral triangle?

I originally asked this on math.stackexchange where it was well received and we made some good progress. Here's what we learnt:

- Because the area of the triangle has to be a multiple of the area of the tile, the triangle must have side length divisible by $5$ (where $1$ is the length of the short edges of the tile).
- The analogous tile made of
*three*equilateral triangles*can*tile any equilateral triangle with side length divisible by three. - There is a computer program, Burr Tools, which was designed to solve this kind of problem. Josh B. used it to prove by exhaustive search that there is no solution when the side length of the triangle is $5$, $10$, $15$, $20$ or $25$. The case of a triangle with side length $30$ would take roughly ten CPU-years to check using this method.
- Lee Mosher pointed me in the direction of Conway's theory of tiling groups. This theory can be used to show that if the tile can cover an equilateral triangle of side length $n$ then $a^nb^nc^n=e$ in the group $\left<a,b,c\;\middle|\;a^3ba^{-2}c,a^{-3}b^{-1}a^2c^{-1},b^3cb^{-2}a,b^{-3}c^{-1}b^2a^{-1},c^3ac^{-2}b,c^{-3}a^{-1}c^2b^{-1}\right>$. But sadly it turns out that we
*do*have that $a^nb^nc^n=e$ in this group whenever $n$ divides by $5$. - In fact one can use the methods in this paper of Michael Reid to prove that this tile's homotopy group is the cyclic group with $5$ elements. I think this means that the
*only*thing these group theoretic methods can tell us is a fact we already knew: that the side length must be divisible by $5$. - These group theoretic methods are also supposed to subsume all possible colouring arguments, which means that any proof based purely on colouring is probably futile.
- The smallest area that can be left uncovered when trying to cover a triangle of side length $(1,\dots,20)$ is $($$1$$,\,$$4$$,\,$$4$$,\,$$1$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$,\,$$6$$,\,$$4$$,\,$$4$$,\,$$6$$,\,$$5$$)$ small triangles. In particular it's surprising that when the area is $1\;\mathrm{mod}\;5$ one must sometimes leave six triangles uncovered rather than just one.
- We can look for "near misses" in which all but $5$ of the small triangles are covered and in which $4$ of the missing small triangles could be covered by the same tile. There's essentially only one near miss for the triangle of side $5$, none for the triangle of side $10$ and six (1,2,3,4,5,6) for the triangle of side $15$. (All other near misses can be generated from these by rotation, reflection, and by reorienting the three tiles that go around the lonesome missing triangle.) This set of six near misses are very interesting since the positions of the single triangle and the place where it "should" go are very constrained.

I'd also be interested in learning what kind of methods can be used to attack this sort of problem. Are there any high-level approaches other than the tiling groups? Or is a bare hands approach most likely to be successful?

signedtilings in which some copies of an "anti-tile" are allowed to appear. The tile homotopy groupalwaysdetects whether or not a signed tiling exists and can sometimes tell you more. I think all this is best explained in the Michael Reid paper which I linked to above. – Oscar Cunningham Apr 13 '17 at 11:51allsigned tilings that are near-tilings to see if this space exhibits some structure that might suggest a proof that a tiling is impossible. – Timothy Chow Apr 14 '17 at 20:10