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Can rotations and translations of this shape

enter image description here

perfectly tile some equilateral triangle?


I've now also asked this question on mathoverflow.


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.
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    $\begingroup$ Conway's Tiling Groups might be relevant; see Thurston's article cimat.mx/ciencia_para_jovenes/pensamiento_matematico/…. It might be fun to work out. $\endgroup$
    – Lee Mosher
    Apr 3, 2017 at 15:42
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    $\begingroup$ @LeeMosher I'll kick out [recreational-mathematics]. This is now serious business! :-) $\endgroup$ Apr 3, 2017 at 21:05
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    $\begingroup$ I used BurrTools to determine there is no tiling for a triangle of width 25. This took 21 hours on my computer. I'm not going to try width 30. $\endgroup$
    – Josh B.
    Apr 4, 2017 at 20:55
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    $\begingroup$ A couple close calls: one two $\endgroup$
    – Josh B.
    Apr 7, 2017 at 0:23
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    $\begingroup$ @OscarCunningham You can fill a rhombus with these pieces in all orientations. example Still, it seems that the orientations you point out do dominate the tiling. However, I'm not convinced this will remain the case as we examine larger rhombuses or triangles. But that's what I find interesting about this problem: my intuition isn't strongly pointing for or against. $\endgroup$
    – Josh B.
    Apr 7, 2017 at 18:13

4 Answers 4

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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 $45$.

enter image description here

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    $\begingroup$ Thanks to you! How did you find that tiling? In the MathOverflow post you mention an SAT solver. Any more details? I had attempted this very search with BurrTools but it was going to take far too long. I think BurrTools gets in trouble when there are many ways to tile a subset of the full shape. For example, a rhombus of side length 5 can be tiled in two different ways. A parallelogram of lengths 5 and 6 can be tiled 4 ways. It only gets worse from there. $\endgroup$
    – Josh B.
    Apr 17, 2017 at 1:57
  • $\begingroup$ Thanks again. I added a picture of the solution to your answer because math.stackexchange answers are supposed to be self-contained. $\endgroup$ Apr 17, 2017 at 7:06
  • $\begingroup$ Wonderful solution. Very entertaining how you reduced the problem to the case of three equilateral triangles. $\endgroup$ Apr 17, 2017 at 8:58
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    $\begingroup$ @Josh B. Nothing special: I had one True/False variable for each possible tile placement, with the constraints "every triangle must be filled" and "any two overlapping tiles can't both exist." I wonder what techniques Burr Tools uses under the hood. An off-the-shelf SAT solver (pycosat) solved this in ~20 seconds, but would take a long time to solve a side-length-25 equilateral triangle. I'm guessing the limited width of the shape helps. $\endgroup$
    – Lopsy
    Apr 17, 2017 at 15:57
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    $\begingroup$ It's a rep-tile. The above tiling can be extended with a side-15 rhombus to get a tiling of the original shape using 1125 tiles. $\endgroup$
    – nickgard
    Apr 19, 2017 at 7:56
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Here's the minimal solution, a side-30 triangle. Also posted to MathOverflow here https://mathoverflow.net/questions/267095/can-a-row-of-five-equilateral-triangles-tile-a-big-equilateral-triangle.enter image description here

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

enter image description here (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.

enter image description here

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

enter image description here

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: enter image description here

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).

enter image description here

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.

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  • $\begingroup$ Nice method! It's been a while since I've thought about this problem. I remember that being able to tile a trapezoid was important, but I don't quite remember how (of if) its existence lets us tile a triangle. $\endgroup$ Sep 23, 2021 at 13:16
  • $\begingroup$ Thanks! I just added a method to tile a triangle. $\endgroup$ Sep 23, 2021 at 22:30
  • $\begingroup$ Right. Tiling a trapezoid with height dividing by 7 lets us tile a triangle. That fits with what I remember. Amazing work! $\endgroup$ Sep 24, 2021 at 8:28
  • $\begingroup$ You might want to post this on mathoverflow as well. $\endgroup$ Sep 24, 2021 at 8:28
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SAT-solver (glucose) + requiring rotational symmetry very quickly finds solutions for side-30. For example:

enter image description here

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).

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