10
$\begingroup$

enter image description here

The figure shows a hexagon and triangle tiled by six tiles, which are pairwise congruent.

My question is:

What is the smallest number of polygonal tiles that will tile both a regular hexagon and an equilateral triangle in such a way that all the edges of the tiles are parallel to an edge of the tiled figure?

Small print:

  • I define a polygonal tile as a figure comprised of:

    A finite set $P$ of at least three points.

    The (straight) lines $X\pi(X)$ for each point $X\in P$ and some cyclic permutation $\pi$ of $P$ such that distinct lines do not intersect except at mutual end points.

    The region contained by the lines.

  • The hexagon, triangle and tiles are assumed to include the vertices, edges and interior.

  • The figures tiled must be the union of exactly one congruent copy of each tile (though the set of tiles itself is allowed to contain congruent figures). No copy of a tile can share an interior point with the copy of any other tile, but may share boundary points with one or more.

  • A congruent copy of a tile may be a "flipped" version (e.g. the cyan tiles in the above diagram are actually mirror images of each other - this is allowed).

$\endgroup$
  • $\begingroup$ I feel like the answer is four. $\endgroup$ – Ian MacDonald Feb 8 '17 at 2:20
  • $\begingroup$ @Ian MacDonald - Show me an example. $\endgroup$ – Martin Rattigan Feb 8 '17 at 5:20
  • 1
    $\begingroup$ Please note that I have corrected the definition of polygonal tile in the small print. The original would have allowed polygonal shapes with disconnected interiors (e.g. a "bow tie of two triangles connected at a single vertex) to count as a single tile, which was not the intention. My apologies to anyone who may have been relying on that. $\endgroup$ – Martin Rattigan Feb 8 '17 at 12:16
  • 1
    $\begingroup$ If I could show you an example, I would've answered the question instead of making a comment. ;) $\endgroup$ – Ian MacDonald Feb 8 '17 at 14:23
  • $\begingroup$ @Ian MacDonald - Ah. But note the question asks for the smallest number of tiles, so any example of an acceptable tiling would not be a complete answer without a valid argument that it is impossible with fewer tiles. $\endgroup$ – Martin Rattigan Feb 8 '17 at 14:38
1
$\begingroup$

Partial solution: I'm only trying to give upper and lower bounds of the solution.

First of all, let us notice, that

the area of the triangle and the area of the hexagon is obviously the same, so if the sidelength of the hexagon is $1$, then the sidelength of the triangle is $\sqrt{6}$.

With that it is easy to prove, that if such dissection exists it has to consist at least of

three tiles: The triangle contains three points - the three corners - which pairwise have a distance larger than $2$, which is the maximal distance of two points in the hexagon. So those three have to go in different tiles.

Let's go for an upper bound:

I might be totally wrong, but actually think this is a trick question, at least the tileset given in the question seems to be one of those which don't actually fit each other or the given larger shapes - like how 8x8 squares can be rearranged into 5x13 in the well known trick puzzle.
I've built a system of linear equations, where the equations came from:
1) The length of edges are known for both the hexagon and the triangle. This gives $9$ equations in total.
2) The length of 'internal' edges can be expressed two different ways. That is, one equation for each of these edges.
3) Each tile has edges parallel with one of the sides of the great polygon, which means that projections onto lines which are perpendicular to the sides again can be expressed two different ways with the help of the edges, and the coefficients are all the same in absolute value, because $\cos(\frac\pi3)=-\cos(-\frac\pi3)$. This gives $2$ linearly independent equations per tile.

I used the following notation:
enter image description here,
which gave the equations:
$A1=1$,
$F5=1$,
$C9=1$,
$C1=1$,
$C2+B7=1$,
$E2+A7=1$,
$A2=F4$,
$F2=D6$,
$A3+F3=D5$,
$A4=D4$,
$A5=D3+E4$,
$E3=A6$,
$D2=E5$,
$F1+E1+D1=2$,
$B1+C8=2$,
$C7=B2$,
$B3=C6$,
$B4=C5$,
$B5=C4$,
$B6=C3$,
$q=\sqrt{6}$,
$A2+E5+D2+F4=q$,
$F5+D6+C9+B2=q$,
$B1+A1=q$,
$B7=A7$,
$B3=C8$,
$B4=C7$,
$B5=C6$,
$B6+A6=C5$,
$A5=C4$,
$A4=C3+E3$,
$E4=A3$,
$E2=C2$,
$D1=C1+E1$,
$D3=F3$,
$D4=F2$,
$D5=F1$,
$A2=A4+A6+A7$,
$A1=A3+A4+A5+A6$,
$B2+B4+B6=B7$,
$B1=B3+B5+B7$,
$C9=C2+C3+C5+C7$,
$C1+C2=C4+C6+C8$,
$D2=D4+D6$,
$D1=D3+D4+D5+D6$,
$E2=E3+E5$,
$E1=E3+E4$,
$F5=F2+F4$,
$F1+F2+F3=F5$,
which does not have any solutions.

This means, that

we don't even have an upper bound of six for the answer. Furthermore, although I cannot yet prove it, I suspect there is no such tiling at all. But it can also happen, that I made some errors when transcribing the equations, and there is no trick at all.

$\endgroup$
  • $\begingroup$ The diagram is indeed a blind based on $\sqrt(\frac{3}{2})\approx\frac{6}{5}$ - well spotted. But my description is correct; the tiles are congruent and they do tile the hexagon and triangle shown. There is actually a tiling if you allow $\aleph_0$ tiles. $\endgroup$ – Martin Rattigan Feb 12 '17 at 18:12
  • $\begingroup$ The last sentence in the previous comment should have said "a tiling of regular hexagon and triangle satisfying the conditions in the question". $\endgroup$ – Martin Rattigan Feb 12 '17 at 18:32
  • $\begingroup$ Cool, @MartinRattigan. Indeed there was a point during organizing my equations, where an expression had to be equal to $\sqrt{6}$ looking from one point of view, but $\frac{12}{5}$ from an other. At that time I thought I had some wrong maths, and started it all over. When I failed the fourth time, I was still not sure that it was my bad equation-organizing skills, or there is in fact no solution. Are you aware of a simple proof, or at least some heuristics, that allows us to get to the conclusion that there is no solution using a finite number of tiles? $\endgroup$ – elias Feb 12 '17 at 21:39
  • $\begingroup$ No - I'm still looking at that (in between moving house). I don't say $\aleph_0$ is the answer (though I think it is), but it shows at least that there is an answer. I'll post a proof as soon as I have one unless someone beats me to it. If it's not the answer it could get complicated. Examples with $\aleph_0$ tiles are easy enough to produce. $\endgroup$ – Martin Rattigan Feb 12 '17 at 23:18
  • $\begingroup$ I haven't get round to looking for a proof yet but it struck me that my ℵ₀ solutions require a single point polygon, which doesn't conform with my definition of polygon in the question. So elias may be correct in suspecting there is no such tiling at all. $\endgroup$ – Martin Rattigan Mar 23 at 10:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.