Noli turbare circulos meos! (Annuli VII)

In science, our goal should always be to present with clarity. Since the discovery of perspective drawings, a realistic representation of 3-dimensional objects has become almost mandatory. However, very often these objects have an appeal beyond their scientific truth which gets lost if its is shown in full clarity. This blog has two series of posts titled “Spheres” and “Annuli” that both showcase images of simple 3-dimensional mathematical objects which deliberately forsake clarity in order to convey that other appeal. While accurate perspective renderings are used, the  perspective and textures are chosen as to emphasize the abstract aspect. The example above shows a triply orthogonal system of surfaces. An easy way to create such a system is by taking a doubly orthogonal system of curves in the plane, revolve them about a common axis to obtain two families of surfaces of revolution that intersect orthogonally, and add all planes through the axis of revolution. For instance, we can choose two families of touching circles that pass through a common point, as above. A single circle, rotated about the black axis, will revolve into a torus. To spice things up, let’s apply an inversion at a sphere centered at the intersection of the circles. This turns the tori into special cyclids like the one above, which all have the appearance of a plane with a handle. Using both a red and a green circle will invert-revolve in two such cyclids that intersect in a straight line and a circle: These are still attempts of realistic drawings, but we already get the feeling that things aren’t completely evident anymore. For instance, the two cyclids above should be equals: but where did the corresponding red handle go? Above is the same pair of objects from a different perspective. Now we can see the two handles and the intersection in a line, but where is the intersection circle? Also, where do we need to place the third surface family, which consists of inverted planes, i.e. spheres? The answer to that question is indicated below. Other perspectives allow amusing variations: For the top image, I have used several cyclids from each family, and several spheres, clipping them between two planes. To appreciate the image, all this knowledge might be irrelevant. To create it, it is essential.

Imaginary Simplicity

Multiplication with the imaginary number √-1 accomplishes a counterclockwise 90° rotation, and that’s what today’s puzzle is based on. As casual readers will know, I enjoy simple things that rapidly get complicated. This puzzle is played on a rectangular checker board, with checker pieces in various colors placed on the squares. A move consists of rotating a checker about another checker by 90° either way. Below is the puzzle graph for a 3×3 board using two adjacent checkers, showing all possible positions and the possible moves indicated by edges that are colored by the checker you rotate about: It’s clear that in this case the two checkers have to stay together, and this poses no real challenge: Dealing with just two checkers is simple. Let’s add another one. Below you can see all possible six moves from the central position. You can either rotate about green or blue, but not (yet) about red. The puzzle graph in this case is connected and has diameter 7. You will notice that the checkers cannot change their parity. As simple count will tell you that there are therefore 80=5x4x4 possible positions or vertices. Below is a simple example how a puzzle could look like, with an optimal solution: Use as few moves as possible to get from the left position to the right one. Naturally, things get trickier with more checkers and larger boards. Below is an optimal solution for a 4 checker problem that realizes the diameter 7 of the game graph (that has 240 vertices). Finally, here is a problem on a 3×4 board with 4 checkers. The shortest solution takes 9 moves, and takes place on a game graph with 900 vertices. Having choices is hard, isn’t it? There are many variations possible. One can, for instance, designate left and right handed checkers that can only rotate one way, making the puzzle graph directed. One can also turn this into a two person game by letting two players take turns. More about this at a later point.

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. 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. 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: 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. This object can be squeezed together in two different planes. Ideal for people who can’t keep their hands to themselves. 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. Cut, score, and fold as shown above. 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. 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.

Napoleon’s Theorem

Coming fall semester  I’ll be teaching our geometry undergraduate course, and to prepare myself a little, I am going to try some of the ideas I’d like to explain. One emphasis is on simplicity. If we just concern ourselves with points in the (say Euclidean) plane, a first fundamental question is whether they can be in an interesting position, and if so, how we can tell? There is (almost) nothing to say here for one or two points, but for three points it gets interesting. Three points can be collinear, a condition from projective geometry, about which I know what to say. Another natural choice is to place the three points at the vertices of an equilateral triangle

That is already a fairly advanced concept. How to you explain what an equilateral triangle is to your friend’s children in elementary school? As a full orbit of a cyclic group of order 3? That is, in fact, the simplest definition I know, avoiding conceptually much more difficult measurements of lengths or angles. It also has the advantage that it generalizes to other geometries.

How do we tell? An elegant answer in the spirit of the orbit definition can be given be denoting the vertices A, B, C as complex numbers. They lie at the vertices of an equilateral triangle if and only if A+𝛇B+𝛇²C=0, where 𝛇 denotes a third root of unity: 𝛇 = (-1+i√3)/2. This is easy to see, because it holds for your favorite equilateral triangle, and all equilateral triangles are similar, i.e. differ by a complex linear transformation which changes the left hand side of the test equation by a nonzero factor. I know I will have to say a little more, but not much. What is this good for? One amusing application is Napoleon’s Theorem:

Let ABC be any triangle. Construct equilateral Napoleon triangles CBA’, ACB’ and BAC’ outside of ABC. Then the centroids A”, B”, C” of the Napoleon triangles form another equilateral triangle.

The proof is very simple now: We just express the centroid A’’ of CBA’ as (C+B+A’)/3, and likewise the other two centroids as B’’=(A+C+B’)/3 and C’’=(B+A+C’)/3, and apply the test.

Could there possibly a better proof? Yes and no. Here is one that is more a revelation than a proof. The first key observation is that there is a tiling of the plane, using the given triangle and the equilateral Napoleon triangles, as shown above. This tiling becomes periodic with a hexagonal period lattice. This allows to overlay the Napoleon tiling with a tiling by equilateral triangles that have their vertices at the centroids of the Napoleon triangles.

The reader will find it amusing to fit this all together. How general is all this? For instance, can we do this in a plane over a field without a third root of unity?

It is not clear whether Napoleon has actually proven this theorem. He was interested in Mathematics, and discussed science with Lagrange and Laplace. The first known written appearance is an article by W.
Rutherford in 1825, which doesn’t mention Napoleon.

Rationality 2

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

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: 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. 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?

Hidden Simplicity (Maybe-Ferns 5)

Mathematicians like to do things a little differently. An excellent example was the Mathematische Arbeitstagung, a yearly event held in Bonn, where the (mathematical) audience was asked to publicly suggest speakers. Friedrich Hirzebruch would write the suggested names on the board (he sometimes misheard…), and then create a list of speakers on the fly. Sometimes they ended up with unexpected results. One year, Michael Barnsley was suggested, who had been working on a new fractal image compression method. His talk was exciting for us graduate students, because we for once could understand something. The idea was to use special types of iterated function systems: Take a few linear maps that are all contractions, and use them to map a subset  of the plane to the union of the images of that set under all the linear maps. This becomes a contraction of the space of closed subsets of the plane to itself with respect to the Hausdorff distance, and hence has a fixed point, which is again a subset of the plane. It turns out that these subsets are highly complicated fractals, encoded just by a few numbers. For instance, all images on this page (except for the photo of Hirzebruch at the top) were made with just two linear maps, requiring 12 decimal numbers. Barnsley claimed that he could reverse engineer this: Start with an image, and find a small collection of linear maps that would produce the given image very accurately. If true, this would revolutionize image compression.  We went home and tried it out on our Atari ST computers and the likes. All we could produce were ferns, twigs, and leaves. Paul Bourke has a nice web site where he explains how one can design some simple fractals, and has also some very impressive images of ferns using four and more linear maps. Below are the two simple maps used to create the polypodiopsida psychedelica above. 