Hot answers tagged

75

It seems that one can color a 15-15-15-30 trapezoid with the given tiles. Here is a picture (sorry about adjacent figures that are the same color, I used random colors so hopefully there are no ambiguities): In particular, OP pointed out that these scaled 1-1-1-2 trapezoids can tile any equilateral triangle whose side length is a multiple of three. So the ...


36

Q1: Yes. Any acute non-isosceles triangle can be tiled by three pairwise incongruent isosceles triangles, by connecting each vertex to the circumcenter. Start from some isosceles $T_0$ with repeated side $s$; inscribe $T_0$ into a larger triangle $T_1$ such that $T_1 - T_0$ is the union of three acute, non-isosceles triangles with circumradii distinct from ...


34

No. It is a famous problem. Suppose it were possible to cut the unit square into finitely many rectangles of sizes $a_i \times b_i$. This means that we have a decomposition $1 \otimes 1 = \sum_i a_i \otimes b_i$. If these rectangle could be reassembled into an $r \times 1/r$ rectangle, we would likewise have $r \otimes 1/r = \sum_i a_i \otimes b_i$. But $r \...


33

A positive answer to this question has just appeared in the arXiv: Tiling with arbitrary tiles; Vytautas Gruslys, Imre Leader, Ta Sheng Tan; http://arxiv.org/abs/1505.03697


32

Binary Tiling In fact, one can tile the hyperbolic plane with arbitrarily small tiles. There is a tiling of the hyperbolic plane (apparently due to Boroczky) by pentagons. The horizontal edges are horocycles in the upper half-space model of the hyperbolic plane, and the vertical lines geodesics. The edge at the top of each tile is half the length of the ...


31

Indeed, there is no solution for $n=5$, and also none for $n=7$. However, for $n=11$ there is a tiling of the requested form. I found it using a straightforward exact cover formulation and Knuth's original exact cover solver. I'm not sure how to best visualize a solution, here is an attempt using random colors for the tiles. It is a little hard to check that ...


30

Since nobody has posted it, here's the smallest triangle tilable by the 'straight pentiamond', ie a side-30 triangle. Simple backtracking program, takes 0.5 seconds to show no tilings of the side-20 triangle, 4:39 for the side 25. Took 12 minutes to find 120 tilings for the side-30 before I stopped it.


22

Not an answer. But permit me to draw attention to Robert J. MacG. Dawson's website on congruent sphere tilings, including this beautiful tiling by triangles:              


21

This is work by Guy Kindler, Ryan O’Donnell, Anup Rao, and Avi Wigderson, published in Spherical Cubes and Rounding in High Dimensions (2008), and in Spherical Cubes: Optimal Foams from Computational Hardness Amplification (2012). Foam problems are concerned with how one can partition space into "bubbles" which minimize surface area. We investigate the ...


19

The tilings mentioned by Ian Agol are related to an action of a Baumslag-Solitar group $\{ a,b \bigg| b^{-1}a^2b=a \}$ on the hyperbolic plane. They have arbitrarily small area, but diameter uniformly bounded away from $0$. It is possible to tesselate the hyperbolic plane with a single tile with arbitrarily small diameter, too. Let there be $n$ arcs on top ...


17

To follow up on the answer of dhy, there is a simple construction that works with simply connected regions: Take a $k\times k$ square with $k\geq 2$. Remove from the bottom left corner a staircase region with rows of $1,2,3,\ldots k-2$ cells, and from the top right corner a (suitably rotated) staircase region with rows of $1,2,3,\ldots k-3$ cells. The ...


17

Google soon finds that Q2 is problem C11 in Unsolved Problems in Geometry by Croft, Falconer, and Guy. But perhaps it's been solved during the intervening decades. URL


17

A tiling is not possible for $n=77$. Consider the polyomino of order 77 here. The two $4\times 19$ missing rectangles can be filled with only one tile, so this tile must be repeated. It should be possible to use this method to show that no tiling is possible for $n$ sufficiently large.


16

Simpler construction: a non-isosceles right triangle $T_0$ can be divided into two isosceles triangles not congruent to each other, or into two right triangles similar to $T_0$. Reversing the latter construction, let $T'_n$, $T_n$ ($n = 1, 2, 3, \ldots$) be right triangles similar to $T_0$ such that $T_n = T_{n-1} + T'_n$, and divide $T_0$ and each $T'_n$ ...


15

Presumably (though I don't have references handy) the behaviour of a quasiregular tiling in ${\mathbb Z}^d$ is essentially the same as that of ${\mathbb Z}^d$ itself. Essentially this should depend on the fact that the number of sites within $r$ steps of your starting point grows as $r^d$.


15

Using the main idea of one of the proofs of the famous Dehn's tiling problem (see for example here), assume that such a cutting of a square is possible. Let $S$ the unit square, and $S=\cup_{j=1}^n R_j$ and $$[0,r]\times[0,1/r]=T=\cup_{j=1}^n R_j',$$ were $R'_j$ is produced be translating and/or rotating by $\pi/2$ the $R_j$, and $r$ irrational - we assume ...


15

Infinitely many, even for a triangle. Let $T$ be the "Pythagorean" triangle with sides of length $3,4,5$. First, tile the $3\times 4$ rectangle $R$ by two copies of $T$. Tile the $60\times 60$ square $Q_1$ with $R$. Then, attach a $12$-times dilated copy of $T$ to each side of the square $Q_1$, forming a square $Q'_2$. Tile a square $Q_2$ with $25$ ...


15

The answer to the second question is yes, for any non-flat triangle $T$, the set of angles $A$ is dense in $(0, \pi)$. This follows from a stronger result of Barany et al. [Theorem 1, 1]: Theorem. Successive barycentric subdivisions of a non-flat triangle contain triangles which, to within a similarity, approximate arbitrarily closely any ...


14

You don't need an anchor tile. Jed Yang and I recently solved this on the way to stronger NP-completeness results (versions of this result were already known). We explain it all here. Similarly, the rectangular tileability problem is undecidable according to Yang's recent result here.


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Here is a better (possibly best) way of covering the plane with congruent regular pentagons:    The density of this covering is ${\sqrt5}/2 = 1.1180...$. This covering is generated by the maximum-area $p$-hexagon contained in $P$, and is the thinnest among all coverings with $P$ that are of the double-lattice type. I conjecture that this density ...


14

I do not know whether the triangular region with size a multiple of 5 is tileable in general, but I can address the question in the last paragraph: 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? Apparently, one can get extra mileage out ...


14

There was a conference in July 2007 at the University of Minnesota—Duluth in honor of Joseph Gallian's 65th birthday. At that conference, Michael Reid gave a talk about tilings, and among other things, he discussed this exact problem. Reid has shown that this "pentiamond" tiles all equilateral triangles with side length ≥30 (provided the side length ...


14

Non-convex solutions to Question 1 Consider the following polygon (the outward angle on the right is the same as the inward angle at the top) Since I didn't know any better way to show it does not tile the plane, I brute-forced my way through some case distinctions. The only non-convex corner of any tile must meet a corner of another tile. It can't be ...


13

Consider a $4\times 2k+1$ rectangle. Label its points by $(x,y)$, with $x$ from $0$ to $3$ and $y$ from $0$ to $2k.$ Remove from the rectangle the points $(1,y)$ and $(2,y)$ for $y$ odd. I claim that the resulting region has $k+1$ tilings. To see this, look at what happens at $(1,2k)$ and $(2,2k).$ If there is no domino containing both points, then it isn'...


13

Each four-celled animal tiles the plane! http://www.sciencedirect.com/science/article/pii/0097316585901050 The corresponding result for 1D three-celled "animals" holds as well. This was a recent problem in the German Math Olympiad (Problem 531046). Together with Wolfgang's comment this solves the original problem, since if all coordinates of the three ...


13

The Nick Baxter solution is actually Blanche's Dissection, published in 1971. I've outlined a general solution method at my Commuunity post Blanche Dissections. As a proof of concept, here are 16 dissections of a square into equal-area non-congruent rectangles: So far, none of these gives a rational solution. But many polyhedral graphs will give a ...


13

This is not a full answer to the question, but it is perhaps a start, and too long for a comment anyway. If $n$ distinct lines intersect at a single point, let's say the intersection is regular if the angle between any two lines is a multiple of $\pi/n$. (That is, all the angles are evenly spaced.) A fair cutting of the plane cannot contain a regular ...


12

Just to supplement Carlo's answer with a figure that illustrates this intermediate cube-sphere shape in $\mathbb{R}^3$, from the 2012 paper he cites—"we provide the most efficient known cubical foam in 3 dimensions":  


12

Q2: The answer is yes (unless we put some restrictions on tilings, see below), this has been demonstrated in this paper. In the above construction, there are three points such that any neighbourhood of these contains infinitely many triangles, which might make it not count as a "valid tiling". One way to exclude such situations is to enforce the triangles' ...


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