Why can I not tile any rectangle without gaps with the given shape? enter image description here


You can mirror the shape (i.e. turn it around an axis in its own plane by $\pi$).


Now that's a nice question indeed. At first glance, it is not clear at all why the shape can't tile a rectangle. After all, it can obviously tile the plane. Moreover, it can even tile a rectangular shell: tiling 1

That's where parity kicks in.

Quite apparently, a rectangle has none of those diagonal ears dangling outside, which means they all must be coupled with similar ears of other shapes. In this manner, each shape is "connected" to two others, and the whole figure (rectangle or whatever else) becomes a cycle, or maybe a union of several disjoint cycles. Let's look closer at one such cycle.

Let's mark the midpoints of all $\sqrt2$-length sides and join them sequentially. That would give us a cyclic path, or a polygon. Being composed of diagonal sides, it is a union of $\sqrt2\times\sqrt2$ squares, and as such, has even area. Let's remember that for later use.

tiling 2

Now, each of the shapes forming the cycle sits on the path and is cut by it in two unequal parts with areas $1\over4$ and $7\over4$. Some have their greater part inside the path, and some outside; let's call them "inner" and "outer", respectively. How many of each kind do we have?

Let's sum up the angles of the path, as if it were a polygon. Say, it has $n$ edges. For obvious parity reasons, $n$ must be even: $n=2k$. (This, BTW, does indeed enforce the checkerboard pattern envisioned by Dan Rust, albeit only within one cycle; not that it helped us much in evading the corners of a rectangle!) Let's look from the inside. Then every inner shape touches two vertices(*) and contributes $3\pi\over2$ to the sum, while every outer one touches just one vertex and contributes $\pi\over2$. On the other hand, the sum of all angles is known to be $\pi(n-2)$.

(*) Other points where the shapes touch or cross the path do not count. On the contrary, the vertices where the path goes straight (which, strictly speaking, are not corners) still do count, as if they were the angles of $180^\circ$.

The rest is simple: to match the sum of angles, we must have $k-2$ inner shapes and $k+2$ outer ones. Thus the sum of their parts inside the path takes up ${7\over4}(k-2)+{1\over4}(k+2)=2k-3$, which is odd. So the unoccupied space inside the path has odd area, and therefore can't be empty.



In the picture below, you can see that we've placed the red tile in the only place that any tile can go in order to fill the bottom left corner of the rectangle. We have then placed the blue tile in the only place that any tile can go in order to fill the square which is directly below the green frame. Now notice that there is no way to place a tile so that the green frame is filled without overlapping tiles.

enter image description here

  • $\begingroup$ @Dan_Rust Sorry, I forgot to add it but you can mirror the shapes. $\endgroup$ – Tamiel Farken Sep 12 '15 at 8:45
  • $\begingroup$ In that case, I'd propse trying to prove that the 'square' part of tiles have to be placed in such a way so as to land on a checkerboard pattern. You can then use the fact that each tile has even area, so the area (hence at least one side length) of the rectangle must be even. This means that one of the corners can not be covered, due to the checkerboarding forcing a corner that has to be covered by triangles - but that can't be done as at least one of the tiles would have to intersect the outside of the rectangle. $\endgroup$ – Dan Rust Sep 12 '15 at 12:28
  • $\begingroup$ Keep in mind, this relies on being able to show the checkerboarding property is necessary, which I don't know how to check without doing an exhaustive (but finite) check of local configurations of the 'first corona' of a tile. $\endgroup$ – Dan Rust Sep 12 '15 at 12:29
  • $\begingroup$ Take your own configuration and attach to it from the right a copy of itself. Now the square parts are not in a checkerboard pattern. $\endgroup$ – Ivan Neretin Sep 12 '15 at 13:02
  • $\begingroup$ @IvanNeretin that's going beyond the 1-corona, that is, the set of tiles which intersect some central tile. If you can show that the only 1-coronas with no gaps and which can be extended beyond the 1-corona, have checkerboard patterns, then by induction any tiling with no gaps of a rectangle must be covered by a checkerboard of the square parts. $\endgroup$ – Dan Rust Sep 12 '15 at 13:46

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