Mathematical Games Hosted by Ed Pegg Jr. [Episode 8: Polygon Dissections]

Do I have to put the q in on rocket chat? Hello everyone. We're just waiting for people to come in. Everything should start within a minute or so. Feel free to post into the chat area. Okay, I'm Ed Pegg Jr. And I plan to give a talk today on polygon dissections. And we're still waiting for people to come in for another few seconds. This is a series of talks I'm given in the spirit of Martin Gardner's Mathematical Games, which ran in scientific American for 25 years. And I took a look at one of his old columns and decided I would try to base a talk on that given that a lot of work has been done in this area over the past hundred years. So today we're going to talk about polygon dissections. And at the demonstration site demonstrations.wolfram.com we have 422 different Mathematica demonstrations dealing with polygon dissections and related items. So I figured that would be a good basis for a talk. Most of the dissections there are by Isidore Hafner and we'll be seeing quite a few of those dissections today. So here's a snippet of Martin Gardner's column from 1961. In it he listed the records that were known at the time. And I thought I would compare to the latest records from Gavin Theobald site. If you look for Gavin Theobald you'll find his website.uk and it has all the best known polygon dissections and pictures of each one. But it doesn't have animations like I'll be able to show for the, say, heptagon here. All of them have been improved since 1961 and there's quite a few more sections that are known that have been discovered since 1961. Things get weird if you look at fractal dissections. For example, here's a six piece dissection of a square into a circle. You just basically divide the, the square up into these six pieces, these six fractal pieces and you get a circle. And I'm not exactly sure how this was found, but I don't have a demonstration for this particular item. However, I'd be very curious to see whether this sort of technique could be used on simpler polygons. So from, from the, the column for, by Martin Gardner, he mentions two main methods for finding these polygon dissections. One is to start with a tessellation, for example, a tessellation of squares, big squares and little squares. Here's a big square. Here's a little square overlaid with the dissection of octagons and squares. And with those two combined and overlaid, you can get a five piece dissection of a square to an octagon. Similarly, if you have two dissections of three crosses and overlay them, you can get a, a five piece dissection of, of two Greek crosses into one large cross. And here's a, a Latin cross to square, where basically they divide the Latin cross into two pieces to make a strip and then they overlay with a strip of squares. And so strips and tessellations are two of the main methods used for these polygon dissections. Here's a sample strip for a heptagon. You basically cut it into these five pieces. You move this piece over to here. This piece can then rotate around to here, this piece rotates to here, and this piece moves over to here and you get a rectangle. So with this sort of cut of a heptagon, you can make a tiling rectangular strip of these, which can be used for, for further strips. This is by Gavin Theobold and I'm, I'm planning on putting up code sometime soon. So when we're working with these, we basically want polygons to have the same area. So this is something I'll put up as a demonstration where you can compare polygons of the same area. Area. So one question is what are minimal numbers of cuts you need to make a particular item? For example, if you have a triangle and a square, what is the minimal number of pieces you need to cut the triangle into to make the square? And we'll see a solution for four pretty soon. But can you prove that four is minimal? Well, it turns out we can prove that you need at least three pieces and think for a while. Anybody have any clues how can we prove minimal 3? Does anybody have any ideas? You can go ahead and post if you want. So the answer is the area of an equilateral triangle is square root of three times the side squared divided by four. It turns out the side of this unit area equilateral triangle is 2 over the fourth root of 3, which is about 1.5. The longest we can get in a square is the square root of two. So if we plug that in, we get a triangle which isn't big enough. This edge here is basically longer than the diagonal of the square by the pigeonhole principle. If we only divide this triangle into two pieces, we can basically only cut two of the edges. We have to Leave one edge uncut. And because of that, because of that, that edge that isn't cut, it won't fit on the square. Therefore, it's impossible to cut a triangle into two pieces that will make up a square. This is one of the few minimality proofs that is known for these dissection problems. Are there any other minimality proofs that can be brought to bear on this? If you have any ideas, let me know, because there's various people that are interested in finding more minimality proofs for this sort of item. So the best known is four pieces. So it's also called the haberdasher problem. So here we have a square, and it goes to a triangle. And here's another version of it. And we can also do interesting rotations. Instead, We basically rotate on these four points, and we get a square one way and a triangle the other way. This is what's called a hinge dissection. We can also look at the same problem in spherical coordinates. And with that, we have a six. What is that? I think it's six. Yeah. Six piece dissection of a spherical triangle into a spherical square. Spherical dissections are not as well known as the regular Euclidean plane dissections. We can also do a. A pentagon to a triangle. And it's. It's possible that because I've got so many different related demonstrations here, it's possible that things are going to get wild in this talk due to overlaps of. Of programs. So if we move the pieces around this way, we get a parallelogram, and we can change it to a near equilateral triangle. And for this one, this is Goldstein's triangle to pentagon. All right. We can also do a. A Oops. Like I said, things are going to get wild because there's so many different versions of this overlapping code. So hopefully. Right. So we move the pieces like. So we get a triangle one way and a hexagon the other way. That is Lindgram's dissection. And here's another version. Going from one piece to another. Now the. You can sort of see how strips were used in this and this particular one. We can also go to a heptagon. I showed a heptagon earlier in a strip. You can sort of see the pieces of the strip here. We've got these two pieces and the third. This is the start of putting a heptagon to a strip. And then we move them like this to get a parallelogram. And then we finish off to get an equilateral Triangle. And that's a Theobald's dissection of a heptagon to triangle. This is one of the first examples that broke the records known to Martin Gardner. And we can do it with a octagon as well. All right, I think this is the same one. Oops. Let me see if I can recover this one. Oh, this doesn't look like that one wants to work. Like I said, the. This one can get wild. It might be that I just. Oh, yes. I just needed to move the pieces. So we move these pieces to there, that piece down to there, move those around, and then everything falls into place. So that is Theobald's dissection of an octagon into a triangle. We can also do nonagon to triangle. Those move down there and those move. Looks like this one needs to be reevaluated. Let's try that again. Those move there, then those move that goes up there. And then we rotate down. So that is a nonagon to triangle. Let's see if we got. Let's see if we can make this one work. Here we go. We rotate these up and then these fold up. That's a hinged section. All right, We can go to a decagon. And. And here are the pieces. We don't have an animation for this one yet. And there's the site. May as well show that site. Here's. Here it is at the site. And triangle to dodecagon. All of these. All of these solutions I'm showing can be laser cut or 3D printed. Here are some examples from the collection of Isadore Haithner. Instead of triangles, we can do squares. For example, here's a dissection of a pentagon to square. I think this one causes more problems than usual. So I'll just go right to this dissection here, showing a pentagon to square. And this one hopefully works another dissection of a pentagon to square. I think, oh, this is a pentagram. And a near dissection. This one is very close to being a square, but it's only got five pieces. This isn't quite a square. It's almost a square, though. It turns out that with just four pieces, you can turn a pentagon into a rectangle. And here's one example of how to do that. This would be a good envelope design, I think. Boost off hexagon to square. We start with the square. Everything translates to make a hexagon. We can also do a hexagram. Move that piece there, that over there, and then the two triangles. This is basically something I've wanted to see for myself for a while, how all these interplay with each other. But I never had a chance to just sit down and find each one in the demonstration site and. And put them all together. So this is sort of the first time I'm seeing all of these together as well. So we start with a square and we cut it into those seven pieces and we get a heptagon, Like so. You can see these two pieces are sort of repeated. And all of the heptagon dissections. Because it's a particularly good script strip found by Theobol. The octagon to square dissection I showed at the start. It's squares and octagons overlaid with a bigger square and a square. I wasn't able to get this one to work so that it didn't break everything else. So I'm just leaving that as a image. But this one should work. Let's see what I've got here. I think I can just move the pieces. Is that right? I have notes to myself on how I got each one of these to work with everything else. Here we go. Octagram, the square like so. And a nonagon to square. Let's see if we can get this one to work. Do I have it in my other. Other notebook? Nope. Doesn't seem like this one wants to work. But this is the idea behind it. These are two strip dissections from Theobald site showing how to cut the nonagon onto a strip and then overlay it with a strip of squares. This is two different ways of doing this dissection. Both get rather elaborate. Let me see if I can get this one to work. I'll just bring it over from this other copy of Mathematica. We basically move the pieces around like so. And we get a square. And going back, We get the decagon. Let's see if we get this one to work. Here we go. We move these pieces like so. We get the Doge Dodecagon, 12 cider. And here's another version which I wasn't able to get to work well with the others. Let's see if this one will work. Here's a way of doing two regular decagons into a square. This is by freeze. Isn't that gorgeous? All right. We can go from a pentagon to a hexagon. Theoretically. First we go to a trapezoid, like so. Then to a parallelogram, then to a hexagon. This is by Lindgren. Let's see if we can get this one to work. Now. Here we go. You can see these. These two pieces are reused. Since it's a heptagon dissection, we rotate it into the standard heptagon strip, change it to a parallelogram, and then a few more moves to get it to the pentagon. Here's how Gavin Theobald did the pentagon to octagon. Things get stranger with this one. We don't have a demonstration for this one yet. And the same thing for the nonagon or an aagon. And I think this is the last of the regular polygon sections I wanted to show. We move it to the standard heptagon strip and then move the pieces to get a hexagon. And obviously you could keep going. But a lot of these solutions for the stay 11 gone are not known. So would be handy if I didn't have to refresh the code for all of these. See if I have this one. Nope. Looks like this one wants to. Well, I think it's further enough into the talk that I can use some other tricks to get things to work. All right. Okay. We translate the pieces like so, and then we rotate the two triangles into place. We get a. A six piece dissection for the hexagon to dodecag, Like so. I. I think that these. A lot of these are discordant. Let's see. We can also do shapes going into themselves. For example, a heptagon into seven heptagons, Like so. When you're moving copies of a shape into themselves, usually the angles will work out much more nicely. A heptagon into eight pieces. Oops. That one doesn't want to work. But it's sort of similar. Let's see if we can do this one. The section of a dodecagon into three decagons. There's three dodecagons transform into one dodecagon. Three into one. We can also do three or six into one. I think this one is known as the fruit salad dissection. And this one doesn't want to work. Now. We can also go into 12. Isadore plans to do this one in glass. So here's 12 dodecagons moving pieces around. You get one dodecagon, and these are. This is done in 3D. There are many books on the subject that you can take a look at. I particularly recommend the sections Plain and Fancy, which kind of covers the whole field. But then he went on to write four more books on the topic, but usually with more restrictions such as hinges and twisting, swinging and other items. This one is Kind of the. Here's all the classic methods with hints at some of these stranger restrictions. So. So this one is particularly good to get if you want overview of the whole field. There's many strange properties of polygon that come into play. For example, we can take a look at a regular polygon. And an annulus is the inner inscribed circle and the X circle the outer circle. The annulus is the, is the between part for a polygon. So we can get, for any regular polygon, we can get the circle annulus. And we're going to restrict these polygons everything on this page to have sides of length two. So the question is, what is the area of the annulus for a regular polygon with edge length 2? Well, it turns out that the annulus always has area PI. The answer is PI in all cases, and it's not too hard to prove this. The circle in the center here is a unit circle with, with area PI. It's just a strange property of polygons in the annulus that these always have area PI if you restrict it to edge length two. Here's another with edge length two. If you take these various strips and square the lengths after this dividing line, what is the sum of the strips? You can sort of see it here. For an octagon, if you add up the squares, you get eight. With the hexagon, if you add up the squares, you get six. For the square, it's four. Well, it turns out for all the polygons, if you use a slide length of 2 and add up these squares, the total is equal to the number of sides of the polygon. So if you add up 1, 3, 4, 3 and 4, you get 12, which is the number of sides of the dodecagon. That's a unusual polygon property. If you use side lengths of two, And Dugan asked if does that work with all ngons? I think that's a different problem. I'll have to come back to that. I believe it does work, but there's different dissections that you need to use. So here's a, here's a dissection problem that I worked on a while ago. You, you basically want a, any polygon and you've got a central point for which of these polygons can you divide them up into similar triangles? So here's various ways of dividing up a polygon polygon into similar triangles. The numbers on the sides here are powers of a given values. For example, for this, if X is this, these powers X to the zero, x to the one x to the seven will get this particular figure. And some are stranger than others. For example, here the side lengths are powers of the square root of two and you get this strange figure. Here's a few more. Here's the plastic constant. And I don't remember what that one is, what that constant is called, but it's. I believe this is the. I'm not sure this is the chi constant here. And here's where things got interesting. This is the square root of the plastic constant. When I, when I did this little experiment of just playing with similar triangles, I had a suspicion that some of these might be usable to make substitution tilings. And it turns out that my suspicion was correct. And we got all these ways of dividing a triangle into copies of themselves. So these are three new substitution tilings that came out directly from this demonstration up here. However, I haven't figured out how to do a 3D equivalent. Can polyhedra be divided into similar tetrahedra so that all of the tetrahedra meet at a point? And I say similar there, not contru it. If you take, say an icosahedron, you can put a point at the center and you get 12 pyramids, which will make the icosahedron. Are there any other shapes that aren't regular polygons where you can dissect the polyhedron into similar tetrahedra? And that's an unsolved question. I'm not sure how to solve it. So the Wallace Bolyer Gerwin theorem. Two polygons of the same area can be sected into each other. This was proved three times by Wallace in 1807, Bollier in 1833, and Gerwin in 1835, all independently. So it's called this, and that's all of the dissections here fall under that theorem. In the 1900s, Hilbert asked a whole series of questions and his third problem was basically, does this work in 3D? And his student Max Den proved that it doesn't apply. He managed to prove that it's impossible to dissect a regular tetrahedron to make a cube. So out from that came what's called the Den invariant. And it turns out that polyhedra, the same volume, are scissors congruent if and only if they have the same Den invariant. So we can take a look at some of the Stranger examples. There are exactly five known tetrahedra that fill space, and none of these are regular tetrahedra. This is a, a skewed prism that can be divided into three tetrahedra this is found by Somerville, Baumgartner and Hill all discovered this one. Here's another tetrahedron that works. And this is the demonstration space filling tetrahedra on the demonstration site. This one was found by those three people. This is a way of dissecting up a cube that works. Baumgarten Gardner missed this in his listing. All of these people listed up here, they were all trying to enumerate all of the ways a tetrahedron could could fill space. And all three of them managed to miss one or more until it finally discovered that there are five. And the complete list of these five was was made. So these are a another way of doing it. And this is the strangest one. A a skew prism divides into six tetrahedra. So figuring out that these work is a little bit tricky. And I we haven't even started to make a complete list of all the space space filling polyhedra. But it would be very interesting to have those. Those would be the polyhedra that have a den environment of zero and which fill space. It's possible to have a polyhedron that has thin invariant zero, which means it could be cut up to make a cube, but which doesn't fill space. And one example of that is. I've forgotten the name, but it's one of the bigger Archimedean solids made with triangles, squares and hexagons. I've forgotten the name of it. Here's an extreme example of a space filling polyhedron. Not sure why that's not working. Now this is the 38 sided angle polyhedron. So here's a animation of it moving. If you look for angle 38 space filler, you'll see my community write up on this particular shape. The object in red at the center is being touched by 38 copies of itself. This is the most faces known for a space filling polyhedra that's con convex. It's believed that that's the maximal number of faces a space filling convex polyhedron can have. It would be good to have a list of all these in mathematics, obviously. Here are some of the den invariants of various simple polyhedra. The tetrahedron and octahedron are basically inverses. You can't fill space with them alone. But if you make combinations of say the tetrahedron and an octahedron, you can fill space. So when you have a list like this and you've got items that pair up, it's possible that they work together to fill space. See here's an example of a a a domino, basically a, a two by one by one cuboid. And we can move the pieces around to get a cube like so. This is on the demonstration site and it's. I don't know how to say that person's name. This is one of the first forays into 3D dissections. So there's a number of open questions in this field. For example, what is the minimal number of polygonal pieces for a regular N gon to a square? Here's the sequence addresses that is known. However, it has not been proven that many of these are minimal except for this one. You only need one piece for a square to a square. However, it's not been proven that four pieces is needed to make a triangle. I showed earlier a minimality proof that you need at least three pieces for a triangle to a square. But it's not yet been proven how to do it. How to prove that four pieces is necessary. Here's a new sequence that Neil Floan is working on. Minimal number of polygonal pieces in a section of a regular end on to a rectangle. And here's the sequence as it is currently known. Neil hasn't added the sequence to OEIS yet, but that's going to come out soon when he publishes a paper on it. Let me skip down to this bottom one here. Can it be proven that three pieces are necessary for transforming a hexagon to a rectangle? Here's a proof that three pieces is enough. You just basically cut it into these three pieces and move these two triangles over and you get a rectangle. But so far we haven't come up with a way of proving that two pieces isn't enough. It seems kind of obvious that you can't. But we don't have a proof of that. So if you can think of a nice proof, I'm sure Neil Sloan would like to see that proof. Let's see. I showed at the start that a circle can be cut into six fractal pieces to make a square. What are the fractal records for regular polygons? I'd be curious to see if there's any interesting methods for doing that. And I wouldn't mind having a program that did that automatically between two similar shapes. And as I mentioned before, most dissections lack minimality proofs. Can anybody come up with a general method for at least giving lower bounds of three or greater for for many of these? For example, can you prove that at least three pieces is needed to cut a 11 gone into a 12 gone so far, we don't know how to show that minimality. So that is what I had prepared. Many thanks. You can write to me at the address there and I'll be glad to answer any questions, if anybody has any. And gons I believe that it does. Yes, you obviously could. If they have the same areas, you could make the cut. Whether it would be an nice dissection, I'm not sure, but it would be interesting to know. Say, the for what K is is the number of pieces you're cutting reasonably interesting. So I think that's all for now. I'll be glad to stick around for any other questions. And many thanks for attending my talk.