Fold Me!

Last year, Jiangmei Wu and I worked on some infinite polyhedra that can be folded into two different planes. Today, you get the chance to make your own (finite version of it). This is a simple craft that, time and energy permitting, will be featured at a fundraiser for the WonderLab here in Bloomington. You will need 3 (7 for the large version) sheets of card stock, scissors, a ruler and craft knife for scoring, and plenty of tape. A cup of intellectually satisfying tea will help, as always. 

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Begin by downloading the template, print the first three pages onto card stock, and cut the shapes out as above.  Lightly score the shapes along the dashed and dot-dashed line, and valley and mountain fold along them.  Note that there are lines that switch between mountain and valley folds, but all folds are easy to do.

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The letters come into play next. Tape the edges with the same letters together. Begin with the smaller yellow shape, and complete the two halves of the larger blue shapes, but keep them separate for a moment, like so:

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Stick the yellow piece into one of the blue halves, this time matching the digits. Complete the generation 2 fractal by taping the second blue half to the yellow generation 1 fractal and the other blue half.

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This object can be squeezed together in two different planes. Ideal for people who can’t keep their hands to themselves. 

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The next 2 pages of the template repeat the first three without the markings, if you’d like to build a cleaner model. You then need two printouts of page 5. The last page allows you to add on and build the generation 3 fractal. You need 4 printouts. 

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Cut, score, and fold as shown above.

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Again, tape edges together as before. There are no letters here, but the pattern is the same as before. Finally, wiggle the generation 2 fractal into the new orange frame, as you did before with the yellow piece into the blue piece.

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Here is how they now grow in our backyard. If anybody is willing to make a  generation 4 or higher versions of this, please send images.

All these polyhedra have as boundary  just a simple closed curve. Topologists will enjoy figuring out the genus.

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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

Formulas

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.

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…

Rationality

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:

Stereo

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:

Formulas

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:

HexaSwap2

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.

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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.

Rhomb2

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.

Rhomb3

 

 

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.

Rhomb4

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…

 

Roto-Tiler

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.

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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.

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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.

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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:

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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.

Swap

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. 

Graph0

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.

Graph2a

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.

Graph3

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.

Graph4

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. 

Swapme

 

Transdissections

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:

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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.