Can a row of five equilateral triangles tile a big equilateral triangle?

Question

Can rotations and translations of this shape

perfectly tile some equilateral triangle?

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I've now also asked this question on mathoverflow.

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Notes:

• Obviously I'm ignoring the triangle of side $0$.
• 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. has 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 $25$. Lengths of $30$ or more will take a very long time to check.
• This kind of problem can often be solved be a colouring argument but I've failed to find a suitable colouring. (See below.)
• 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 $\left<a,b,c\;\middle|\;a^3ba^{-2}c=a^{-3}b^{-1}a^2c^{-1}=b^3cb^{-2}a=b^{-3}c^{-1}b^2a^{-1}=c^3ac^{-2}b=c^{-3}a^{-1}c^2b^{-1}=e\right>$. 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.

Answer 1

I suppose I should post: I solved this on MathOverflow. The answer is YES: a size-45 triangle can be tiled.

I thank two insights from Josh B here: first that a rhombus with side length 15 can be tiled, and second the strategy to "select a different shape which does tile a triangle, then tile that shape with our $5$ triangle trapezoid."

This $15-15-15-30$ trapezoid can be tiled, and three such trapezoids can tile a triangle with side length $45$.

Answer 2

Some people were interested in heptiamond tiling. I found that the shortest trapezoid that can be tiled has 15 rows.

(click for larger image)

I think that this trapezoid plus parallelograms are sufficient to tile an equilateral triangle. [Edit: Nevermind, see "Tiling a triangle" below]

How I found this:

We start with an empty trapezoid, with 15 rows, and add heptiamonds left-to-right one at a time. We can only add a heptiamond if it doesn't leave a "gap" in any row. This way, we can represent our current state with only one number per row: the number of triangles in it.

Implicitly, our rows start with some triangles already. The bottom row has 1 triangle, the next row up has 3 triangles, then 5, 7, 9, ...

Also, we only need to keep track of the number of triangles in each row modulo 7, since we can always add a "flat" heptiamond, affecting only a single row.

So our state space has size 7^15 (~4.7 trillion).

From the starting configuration of {1,3,5,0,2,4,6,1,3,5,0,2,4,6,1} we do a breadth-first search and reach {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1} -- meaning that each row has the same number of triangles (modulo 7).

Here is the progression of states corresponding to adding one yellow heptiamond at a time:

{1,3,5,0,2,4,6,1,3,5,0,2,4,6,1}
{1,5,0,2,3,4,6,1,3,5,0,2,4,6,1}
{1,5,0,2,4,6,1,3,3,5,0,2,4,6,1}
{1,0,2,4,5,6,1,3,3,5,0,2,4,6,1}
{1,0,2,4,6,1,3,5,3,5,0,2,4,6,1}
{1,2,4,6,0,1,3,5,3,5,0,2,4,6,1}
{1,2,4,6,1,3,5,0,3,5,0,2,4,6,1}
...
{1,1,1,1,1,1,1,1,2,2,2,2,2,2,2}
{1,1,1,1,1,1,1,1,2,2,2,3,4,4,4}
{1,1,1,1,1,1,1,1,4,4,4,4,4,4,4}
{1,1,1,1,1,1,1,1,4,4,4,5,6,6,6}
{1,1,1,1,1,1,1,1,6,6,6,6,6,6,6}
{1,1,1,1,1,1,1,1,6,6,6,0,1,1,1}
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1}

Note that each state changes exactly 4 adjacent rows. For example, the first change is {3,5,0,2} -> {5,0,2,3}. There are only 28 such transitions possible, corresponding to the 4 non-flat heptiamond orientations, at 7 possible offsets each.

Pentiamond:

The pentiamond requires 8 rows

Tiling a triangle:

Initially, I thought a 15-row trapezoid would be useful for tiling a triangle, but my initial idea didn't work. Instead, I found a 21-row trapezoid:

This was found by doing a similar search, but switching the search space to consider only the first 15 rows. Except, when the first 6 rows are "0", only the last 15 rows are considered. (With this trick, my program could run within my machine's 16GB of memory).

These trapezoids can be stacked and padded to tile a 1-1-1-2 ratio trapezoid. 3 of these mega-trapezoids tile an equilateral triangle.

Note that it's important for the trapezoid to have 21 rows (a multiple of 7) instead of 15 rows. Then the trapezoid can be "extended" by any integer length, by appending 21x1 parallelograms.

Answer 3

SAT-solver (glucose) + requiring rotational symmetry very quickly finds solutions for side-30. For example:

Without requiring symmetry, glucose took ~1 hour to find a solution. (Note that rotational symmetry requires side-length to be a multiple of 3 -- otherwise there is a central triangle).

I didn't have much luck with heptiamonds. But, glucose reported side-49 or fewer is impossible (no symmetry requirement).