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In discussions about symmetries, I found this problem-

The given pictures are of some tiles typically known as Persian tiles- enter image description here enter image description here The second picture is of one single tile while the first one gives the tiling using those tiles.

Each vertex in the Persian tile is surrounded by eight quadrilaterals. Show that given any two such quadrilaterals $A$, $B$ (similar to what marked in the second image), there is a unique symmetry which carries $A$ to $B$. Find a formula for the symmetry with $A$, $B$ as shown. You may assume that the lower left vertex in the given figure is the origin and the upper right one is $(2,2)$.

Fun problem: What proportion of area in $\mathbb R^2$ is covered by the inner slanted squares? You can assume that the octagon formed at the middle due to two adjacent tiles is a regular octagon.

I noticed that the first problem looks trivial if $A$ and $B$ are connected to the same vertex. In that case, you only need to perform a suitable rotation combined with a suitable reflection. But, it doesn't look very easy to prove that this symmetry is unique. Also, if $A$ and $B$ are connected to two different vertices, I think, we will have to do some more rotation and translations. But again, how do I prove that the symmetry is unique? Also, how do I find an explicit formula?

For the fun exercise, how can one even make an idea of how big the tiles are? Looking at the picture, I guessed that maybe each small quadrilateral is of equal area, and the central square is four times it. But, in that case, the answer should be $\frac 1 7 \times 100\% \approx 14.2\%$, while the given answer is $\approx 17\%$. I have no idea how to arrive at that. Please help me proceed.

Edit: Please note that the fun problem has an extra assumption that the octagon formed at the middle due to two adjacent tiles is a regular octagon.

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  • $\begingroup$ "Show that given any two such quadrilaterals A, B, there is a unique symmetry which carries A to B." Note that by symmetry they mean a symmetry of the tiling as a whole. "if A and B are connected to the same vertex [...] you only need to perform a suitable rotation.". A rotation by 45 degrees will bring one quadrilateral to an adjacent one, but that rotation is not a symmetry of the whole tiling. Instead you will have to use a reflection symmetry. $\endgroup$ Aug 16, 2021 at 14:28
  • $\begingroup$ @JaapScherphuis Oh yes! So we need a combination of rotation and reflection in general, right? $\endgroup$ Aug 16, 2021 at 14:46
  • $\begingroup$ Translation, reflection, rotation may all be used, but you won't need more than one of each. In fact, you can always do it using just two - some cases using a translation and a rotation, and other cases with a translation and a reflection. The easiest way is to put the origin of your coordinate system at B, and begin by translating A to the origin so it shares a vertex with B. $\endgroup$ Aug 16, 2021 at 15:05

2 Answers 2

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I think the best way to think about this type of problem is to find what is called a fundamental domain for the (group of) symmetries of the tiling. Intuitively, a fundamental domain is a smallest repeating unit from which the whole pattern is made. Of course the pattern can be built using the square tiles, but each square tile itself has symmetries, and can be built out of 8 smaller 45-45-90 triangles, each of which has no internal symmetries:

A fundamental domain

The key observation is that there exists a symmetry taking any copy of a fundamental domain to any other copy of the fundamental domain, and since this a fundamental domain has no symmetries itself, there this symmetry is unique.

Therefor there is a unique symmetry taking the green fundamental domain to the blue one, and it's not too hard to see that this symmetry is the reflection in the red line.

Fundamental domains containing A and B related by a reflection

More generally, here is how you might try to find a symmetry sending one copy of a fundamental domain to another. Notice that every copy to the fundamental domain has exactly one corner at a circle. So we can choose vertical and horizontal translations so that these circles match up. Now using your observation, we have apply rotations centred at that circle, and reflections in lines through that circle to make the fundamental domains line up. Applying this approach to the case of A and B in your question, first we would translate left by 2 and up by 2 to make the circles line up, and then reflect in a diagonal line:

Fundamental domains containing A and B related by a translation and a reflection

First we translate the green triangle to the purple one, and then reflect in the red line. I will leave it as an exercise to a) convince yourself that the two symmetries I have described are in fact the same symmetry in disguise, and b) use these descriptions to write down an explicit formula.

As for the final fun problem, it looks to me like there is insufficient information as you have stated the problem. Maybe it is possible if you assume that the pentagonal regions pair up to form regular octagons. Sorry I can't be of more help with this.

Edit: more details

First on the question of uniqueness. Above I gave an intuitive explanation of uniqueness in terms of the property that a fundamental domain has no internal symmetries. In fact we have the following general theorem, which I will not prove in general, although that's a fun, if challenging exercise.

Theorem: Let $\Delta ABC$ and $\Delta A'B'C'$ be two congruent triangles (in particular $A$, $B$, and $C$ are not colinear). Then there is a unique isometry which sends $A\mapsto A'$, $B\mapsto B'$, and $C\mapsto C'$.

Instead, suppose there are two symmetries $s_1$ and $s_2$ which the quadrilateral labelled $A$ to the quadrilateral labelled $B$. Then they must send the green fundamental domain to the blue fundamental domain. Now consider applying the first symmetry sending $s_1:A\mapsto B$, and then applying the reverse of the second symmetry which will send $s_2^{-1}:B\mapsto A$. The overall result is that $A$ is mapped to $A$, and the green fundamental domain is mapped to the green fundamental domain. But since the fundamental domain has no symmetries, this it has to be sent back to exactly how it started. But this means that $s_2^{-1}$ exactly undoes the symmetry $s_1$, so $s_1=s_2$.

Remark: this argument still isn't water-tight rigorous, but it can be made so using ideas from group theory and metric spaces. I can add that detail in another edit, but as the question was phrased in a completely elementary fashion, I wanted to reply in as elementary a fashion as possible.

Second, onto the question of how to write down a formula for the symmetry. I constructed the symmetry in the general case as a sequence of simpler symmetries. If you have formulae for each of these simpler symmetries, the overall symmetry is given by the composition of these formulae. Here is a dictionary of simple formulae for the symmetries in this problem.

  1. Translation sending the point $(0,0)$ to the point $(a,b)$: $T_{(a,b)}:(x,y)\mapsto(x+a,y+b)$.
  2. Rotation by $90^\circ$ anticlockwise centred at $(a,b)$: $R_{90^\circ,(a,b)}=T_{(a,b)}\circ R_{90^\circ,(0,0)}\circ T_{(-a,-b)}$, where $R_{90^\circ,(0,0)}:(x,y)\mapsto (-y,x)$. Other rotations can be achieved by applying this transformation several times.
  3. Reflection in the line through $(a,b)$ in the direction $(c,d)$: $M_{(a,b),(c,d)}=T_{(a,b)}\circ M_{(0,0),(c,d)}\circ T_{(-a,-b)}$, where $M_{(0,0),(c,d)}:(x,y)\mapsto (x,y)-2(yc-xd)(-d,c)$.

Again, simpler formulae can be given using some more complicated ideas from linear algebra, but I wanted to keep it as elementary as possible.

Edit 2: Some group theory and a more advanced viewpoint

A few brief remarks on a better way to think about all this if you have some background in linear algebra, group theory, and metric spaces. If we view the tiling as living in the Euclidean plane $\mathbb{E}^2$, then "symmetries" of the tiling are maps $\mathbb{E}^2\to \mathbb{E}^2$ which leave the Euclidean metric invariant, I'll call such maps isometries from now on. The set of all such isometries forms a group under composition $\textrm{Isom}(\mathbb{E}^2)$. This is quite easy to check, the hardest part is showing inverses, which follows from the observation that isometries must be bijections which takes a little work to prove.

It follows that the set of all isometries which fix the tiling pictured form a subgroup $G\le \textrm{Isom}(\mathbb{E}^2)$. Once this has been established, the proof of uniqueness I gave in the first edit can be made rigorous using the fact that inverses in a group are unique.

In my original answer I gave a way to find an isometry of the tiling taking an arbitrary quadrilateral labelled $A$ to another $B$ which involved first translating the tiling so that $A$ and $B$ share a vertex, and then applying an isometry fixing that vertex to make them line up exactly. This in fact follows from a general fact. Let $x_0\in\mathbb{E}^2$ be a fixed base point, then we can identify $\mathbb{E}^2$ with the vector space $\mathbb{R}^2$ by making $x_0$ the origin. Let $O(2)$ be the orthogonal group of $\mathbb{R}^2$, and let $T(2)\cong \mathbb{R}^2$ be the group of all translations of $\mathbb{E}^2$. Then the group of isometries of $\mathbb{E}^2$ can be written as a semi-direct product $$\textrm{Isom}(\mathbb{E}^2)=T(2)\rtimes O(2).$$ This is just a fancy way of saying that any isometry can be thought of as a translation followed by an isometry fixing a point. This decomposition descends to $G$, so we can write $$G=\mathbb{Z}^2\rtimes D_4,$$ where $\mathbb{Z}^2$ is the group of translations, and $D_4$ is the dihedral group of order 8 which is the group of symmetries fixing the origin $(0,0)$ as defined in the original question.

You can use the structure of $\mathbb{E}^2$ as an affine space two write any isometry as an affine transformation in terms of matrices (rather than the rather complicated formulae I give above). For details check out this page.

Finally, a small personal advertisement. I have actually taught an entire course (aimed at high school students) on isometry groups of the sphere, Euclidean plane, and hyperbolic plane, which explores these sorts of questions in more detail and generality. You can check it out for free online here: week 1, week 2, week 3, and week 4.

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  • $\begingroup$ This was quite an elaborate explanation (+1). It also makes the picture quite clear. But, how will we prove that this symmetry is unique? Also, how can we construct an explicit formula in terms of $A$ and $B$? $\endgroup$ Aug 16, 2021 at 19:38
  • $\begingroup$ Nice magazine by the way :) $\endgroup$ Aug 16, 2021 at 19:45
  • $\begingroup$ Thanks. I have added more details. Having a look at your profile it seems you might be familiar with some group theory/linear algebra/metric spaces, so let me know if you would prefer a higher level answer than the one I have provided. $\endgroup$ Aug 16, 2021 at 22:54
  • $\begingroup$ I loved your elementary approach. And yes, I'm a little familiar with groups and spaces (though I'm not that well versed in it). So, please add more details if you want to- I guess, I'll be able to manage. Thanks. $\endgroup$ Aug 17, 2021 at 3:42
  • $\begingroup$ I have added some more advanced stuff at the end, although I've only sought to give an overview, and not the details. Hope you find it interesting $\endgroup$ Aug 17, 2021 at 17:21
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As for the area of the quadrilaterals, I assume that the construction of the tile is based on the intersections of the angle bisectors represented by the dotted red lines in the diagram below, in which the length of the sides of the large square is taken to be $2$ units. If that's the case, then I get $\ 3-2\sqrt{2}\approx0.17\ $ square units for the area of the quadrilaterals A and B, and of the small squares in each quadrant of the diagram. I get $\ \frac{3}{\sqrt{2}}-2\approx0.12\ $ square units for the area of each of the other four quadrilaterals in each quadrant. If this is correct, then the area of one of the small central squares, and the quadrilaterals congruent to A and B, would be approximately $17\%$ of that of a quadrant. The slanted centrals squares in the tiling would therefore cover that proportion of $\ \mathbb{R}^2\ $

enter image description here

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  • $\begingroup$ +1. I agree that the construction seems to assume the use of angle bisectors (although this may not actually be so). With the minimal assumption that a vertex of a small square lies on angle bisectors from the big square's corners (so that the vertex is the incenter of a right triangular "quadrant" of the big square), you can get to the Fun Problem result simply by calculating the area-ratio of the small square to its quadrant: $\frac{(\sqrt{2}-1)^2}{2^2/4}=3-2\sqrt{2}$. There's no need to bother with the other quadrilaterals (whose shapes require additional assumptions). $\endgroup$
    – Blue
    Aug 17, 2021 at 5:12
  • $\begingroup$ This was a beautiful geometry. A great answer (+1)! $\endgroup$ Aug 17, 2021 at 18:55
  • $\begingroup$ On rereading the question, I see that the "fun problem" was in fact (effectively) to find the proportion of the tile covered by the central squares, and not, as I had assumed, to find the areas of the quadrilaterals A and B. As Blue noted, to get those, you need more than just the intersections of the angle bisectors, but not all that much more. $\endgroup$ Aug 17, 2021 at 23:31
  • $\begingroup$ I assumed that the pentagonal figure based on the centre of each side intersected the side at the same points where the extensions of the sides (i.e. the dotted blue lines in the diagram) of the small central square did, and its vertices were thus those two points and the intersections of the angle bisectors. The rest of the quadrant is then filled out by reflecting each half of this figure about its side which lies on an angle bisector. $\endgroup$ Aug 17, 2021 at 23:31
  • $\begingroup$ I knew there had to be something more to find the answer of the fun question. But, I just checked with the source of the question, and found out that the extra information is not what you assumed, but the fact that the octagon formed at the middle due to two adjacent tiles is a regular octagon. I'll add this to the question as well. Do you maybe want to place an edit in your question proving that the two are equivalent... (they're of course equivalent since the answer matched) $\endgroup$ Sep 18, 2021 at 14:37

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