Rationality 2

Last time I was asking about cyclic polygons with rational vertices and rational edge lengths, like the one below.

10 gon

It is easy to find all points on the unit circle with rational coordinates as (cs(t), sn(t)) with rational t, using


Moreover, any pair of such points can be rotated into each other using a rational rotation R(t). Here t is again a rational parameter and not the angle of rotation. So we are interested in finding superrational rotations R(t) that are rational and displace points on the unit circle by a rational distance.

Surprisingly, the solution is quite simple: A rotation is superrational if and only if it is the square of a rational rotation.

Let’s first assume that u is rational so that R(u) is rational. The square R(u)² is then also rational. We can determine how it displaces points on the unit circle as follows: Suppose R(u) maps (1,0) into the point (c,s), which is rational. As R(Then  R(u)² maps (c,-s) into (c,s), and the distance between these two points is obviously 2s, R(u)² is indeed superrational.


Now assume that R(v) is superrational, and write it as R(u)². We need to show that u is rational. Again let (c,s) be the image of (1,0) under R(u). As R(v) is superrational, we see that at least 2s and hence s must be rational, as above. Now we compute, using the matrix form of R(u), the image of (1,0) under R(u)² as (c²-s², 2cs). As R(v) is rational, 2cs must be rational. As we already know that s is rational, it follows that also c is rational. Thus R(u) is a rational rotation, and we are done.

This shows that superrational rotations form a group, namely the group of squares of all rational rotations. And this implies that a superrational cyclic polygon has necessarily all its diagonals rational…

What’s next? Probably some dreary landscapes from wintry Indiana or heart warming pictures of tea leaves. But of course, there are more questions: Can we also have superrational spherical polyhedra? I don’t know yet…



Let’s do something simple today, something rational. We all know what a circle is, and most should know that there are points on the unit circle with rational coordinates, like (3/5, 4/5). This is because of the birational map called stereographic projection, known since antiquity:


Take a point t on the real axis, connect it to the north pole (0,1) on the unit circle with a straight line, and find the second intersection of that line with the circle. This is a rational expression in t. So when t is rational, you get a point on the circle with rational coordinates. This is, of course, also a quick way to get (all) Pythagorean triples. But today we are going elsewhere. Now that we have many points with rational coordinates on a circle, we can make rational polygons, like the one below.

9 gon

This 9-gon is not only rational, but super-rational in the sense that all its edge lengths are rational numbers. Try it out. Even better: All the diagonals are rational as well. Is this a miracle? Are there others? No and yes, of course. Let’s get started:


Using rational versions of the sine and cosine functions, we can write down rational rotation matrices. They will (for rational t) rotate any point with rational coordinates on a unit circle to another point with rational coordinates. What we are interested in are superrational rotations: Those that rotate a point to any other point. The example above suggests that there are many of those.

I will give the answer next time. For the moment, only a hint: The superrational rotations form a subgroup of the group of rotations. Which is it?

Unpacking the Hypercube

Last week we learned how Rototiler moves can unpack a cube.  As a warmup, below are the moves for a 2x2x2 cube projected parallel along a cube diagonal onto hexagons:


We start at the left, remove the frontmost cube, and keep going. The solution is far from being unique, but not too complicated. Today, we do the same with a  hypercube. The projection of a 1x1x1x1 into 3-space along a main diagonal is a rhombic dodecahedron, tiled by four rhomboids. These rhombic dodecahedra have 8 obtuse, 3-valent vertices at the corners of a cube, and 6 acute, 4-valent vertices at the corners of an octahedron.


There are two ways to tile the rhombic dodecahedron with these rhomboids, and changing one to the other corresponds to a rototiler move in space. Let’s do this with the 2x2x2x2 hypercube, whose projection is a rhombic dodecahedron tiled by 32 rhomboids.


At first it seems as if there is no swappable rhombic dodecahedron available, but if we remove three rhomboids and  look inside (which is the direction of the fourth dimension, after all), we can see it. After swapping it, we also remove the frontmost rhomboid of the  swapped dodecahedron.




We then see that the four removed rhomboids together make up another swappable dodecahedron. We replace it by its swap. The same can be done at three other places.


The next thing to do is to swap 6 more dodecahedra. One of them is the one which shows yellow and purple rhomboids in the right figure above, sitting between the red and blue “vertex”. All these dodecahedra correspond to the edges of the tetrahedron whose vertices are the already swapped four peripheral dodecahedra. Doing these six swaps leads to a tiling very much like the one above to the right, where now the other four obtuse vertices mark swappable dodecahedra. Swapping these and finally the hidden central dodecahedron  completely unpacks the hypercube. It took us 1+4+6+4+6+1 = 32 swaps, as expected.


Next week we’ll see what this is good for…



Today we look at a puzzle invented by Alan Schoen that he calls Roto-Tiler. He explained this to me a few years ago, and when I showed him notes I made for a class, he denied that this is the puzzle he described. I insist it is, and it is quite certainly not mine.


Things happen on a hexagonal board like the one above (it can but doesn’t need to be regular), tiled by hexagonal rhombi of equal size. The acute angles are marked by 1/3-circles, which occasionally happen to close up when three acute angles meet. In that case, a move consists of rotating the three involved rhombi by 120º either way.


Above you can see the possible four moves from the central position. At this point it is not clear at all that a move is always possible. The puzzle consists of transforming one given tiling by rhombi to another given tiling of the same hexagon. For instance, a simple example asks to find the smallest number of moves that takes the left tiling to the right tiling.


The clue to solve this puzzle is to view the hexagons as the parallel projection of a box subdivided into smaller cubes, and the rhombi as the projections of the faces of the smaller cubes. This becomes visually more intuitive if we color the rhombi by their orientation so that parallel cube faces have the same color:


Then the hexagon above becomes the projection of a box partially filled with cubes, and a move consists of adding or removing a frontmost cube. This step into the third dimension explains everything: We see that we can solve every Roto-Tiler puzzle by emptying and filling boxes with cubes. Last week’s first example was a 1-dimensional version of this, next week we will try to grasp a 3-dimensional version and practice our 4-dimensional intuition.


Enough of minimal surfaces, at least here, for a while. Time for a relaxing little game. Let’s put six tokens in a row, three showing a 1, the other three a 2. There are evidently 6 choose 3 = 20 possibilities for that. A move consists of swapping two neighboring tokens, which also evidently only makes sense if they are different. 


Above is the graph that shows all possible states, the connections indicating the possible moves. What is interesting here? Not much. The graph is connected, which is easy to explain, and I mention all this only because it will serve as the 1-dimensional example of what we’ll do in two dimensions next week and in even higher dimensions later. So, to make things more complicated we are allowing the six tokens to show three values 1, 2, 3. We begin with two tokens of each kind.


The graph is more complicated but offers no interesting insights. Let’s forbid to swap neighbors 12 or 21. The game graph will have six identical components. This usually makes for hideous puzzles, asking the innocent victim to find moves that change say 112233 to 332211. But here this is too obvious, as we clearly cannot change the order of the 1s and 2s while the 3s can move freely among them.

To fix this, we make one more change: We arrange the tokens in a circle, still use 112233, and disallow a 12-21 swap.


Two different components. Why can’t we move from 112233 to say 132213? Again, order plays a role: There is just no way to get from 1133 to 1313, even if we allow cyclic permutations.

Finally, an example with just four tokens, having values 1 through 4, arranged in a square. We allow the swaps 12-21, 23-32, 34-43, and 41-14, but forbid 13-31 and 24-42.


This time we get two components as well. Why can’t we move from 1234 to 4123, i.e. rotate all tokens clockwise by 90 degrees? This time, the order still plays a role, but involves the orders of both pairs 13 and 24 simultaneously, which is hard to see. So this makes, in all its simplicity, a pretty nice puzzle. 




A few weeks ago, I explained that two Euclidean polygons are scissor congruent if and only of they have the same area. A scissor congruence is a dissection of the two polygons into smaller polygons (“pieces”) so that the pieces of the first polygon can be translated and rotated into the pieces of the second polygon. Then I asked whether we really need to rotate the pieces, or whether translating them is enough. For instance, can one dissect a square into pieces that can be translated into a second square that is rotated by say 45º?


 Mahlo 01

That this is possible is a consequence of a famous dissection of a square into two squares, shown above, which uses translations only, but tilts the square by an angle which we can make 22.5º. Then we need to repeat this process, tilting the other way, using a mirrored dissection.  This increases the number of pieces needed, and the question arises with how few pieces one can do this. A few years back I ran a contest about this in our department. The best solution with six pieces was found by Seth, one of our (then) undergraduates:

SolpuzzleSince then, I learned that there are solutions with only five pieces. Check them out!

Another cool example is the trans-dissection of a single pentagon into four smaller ones, by Harry Lindgren:

Fourpentagons 01

So, do trans-dissections always exist? Not at all. Let’s try to trans-dissect an equilateral triangle into a copy of it that is rotated by 180º. Suppose we found such a dissection, like the one above. Look at its horizontal edges. There are two types, the ones at the bottom of the pieces, (pointing right), and the ones at the top of the pieces, pointing left. When we add these together (taking the direction into account) for the dissected pieces, the edges in the interior cancel, while the boundary edges add up to the length of the bottom edge of the initial triangle.



Transdiss 01


Thus the oriented length of all horizontal edges of the two tiles that we want to dissect into each other need to be equal. This eliminates the possibility to rotate a triangle at all. The same argument works of course for all directions.

Therefore, in order for two polygons to be trans-scissor-congruent, they not only need to have the same area, but also have the same oriented edge lengths for all edge directions. Remarkably, these conditions are also sufficient, as was proven in 1951 by Paul Glur and Hugo Hadwiger. The argument is a little tricker than the one for general dissections, but not too bad. Maybe I’ll come back to it later.

Dissections and Area

Whenever need to explain what Mathematics is about, one of my favorite examples is the concept of area. The existence of an elementary notion of area hinges on the fact that any two Euclidean polygons have the same area if and only if they are scissor congruent, meaning that they can be cut into congruent pieces using straight cuts. To see this, it suffices to show that any polygon can be dissected into a square. Rect2square

The example above shows how to dissect a well-proportioned rectangle into a square. Here, well-proportioned means that the rectangle is not more than twice as tall than wide. If a rectangle is not well-proportioned, a few cuts parallel to the edges will make it so. Thus any two rectangles of the same area can be dissected int each other. We will use this later.


Next we show that any polygon can be dissected into triangles. By induction, it suffices to find a secant inside the polygon. To find this secant, pretend that the polygon is actually the floor plan of a room, and we are standing at one vertex V . The two adjacent walls lead to two vertices A and B which we can see. If we can see yet another vertex W from our position, we have found our secant VW. If we can’t see another vertex, nothing obstructs our view in the triangular region formed by A, B and V , and thus A and B can be joined by a secant.

As a further simplification, we cut all triangles into two pieces along one of their heights so that all triangles become right triangles.

Now we have a collection of right triangles, which will need to be dissected into a single square.

Triangle2squareTo do so, we dissect each right triangle into a rectangle. This can be done as shown above by dissecting the triangle into two pieces along a segment parallel to one of the legs and dividing the other leg into equal parts.

This leaves us with a collection of rectangles that most likely have different dimensions. So we dissect them into new rectangles that all have all height 1, using the example at the beginning. 

Then, the new rectangles can be lined up edge to edge along their sides of length 1 to form one very wide rectangle that finally can be dissected into a square.


As this was nice and easy, here a challenge: In our dissections, we were allowed to translate and rotate the pieces arbitrarily. What about if we forbid rotations? Can you dissect an equilateral triangle into finitely many pieces and translate them so that the result is the same triangle upside down? Or, can you cut a square, translate the pieces, and thereby achieve a 45 degree rotation of the square?