Mathematical Games Hosted by Ed Pegg Jr. [Episode 14: Hexagrammum Mysticum]

Sa. Hello everyone. I'm Ed Pegg Jr. And today we're going to be talking about something called the hexagram Mysticum, a obscure object in the history of mathematics. And welcome everyone. I'll wait for a few more people to arrive. And I'm talking today about the hexagram Mysticum. We'll be going into detail about this strange thing that got started off by Blaise Pascal. So there are many demonstrations that deal with conic sections, and here's a few of them. We've got 185 demonstrations dealing with conic sections. And at the bottom here you can see Pascal's mystic hexagon. So let me start with something else. A famous problem in recreational mathematics has been solved. Beggar My Neighbor is a card game similar to war, where there's two people have split up a deck of cards and based on whether a ace, king, queen or jack comes up, there's various penalties of four, three, two or one cards that are done. It's a WAR variant. But one question that wasn't known was whether an infinite game was possible. And Braden Casella just a few days ago solved the problem with this start. The game is infinite. So let me put up a link into the feed here. This is a, a link to the function I'll be talking about and I hope to come up with a demonstration that has bigger my neighbor games. Sometime before the next chat from last month I had this slide and I mentioned that five points define a con exception and there's various probabilities for whether an ellipse comes up. Well, it turns out I was wrong here. These odds come from something else. This is not the odds of an ellipse, but basically with five points. Five points defines a conic section. Here's an example of five points which define an ellipse. And here's the ellipse here. Here's an equation for it. Four points define a hyperbola. This is something, something else that was on that slide. In fact, the four points define two hyperbolas. And here's four points. Here's two different parabolic equations and here's an example of the parabolas. Well, it turns out these probabilities were wrong and I got them from Sylvester's four point problem. So you have some number of random points and depending on whether you've got a square region or a triangular region, you can get different, different holes, convex holes. This point is inside, so it's not a convex hole with all five points. Here it is a convex hole for four points. The odds are 25 out of 36 that you'll get a convex hole. It's possible for the one point to be inside. And with five points, it's 49 over 144 for rectangular and 11 over 36 for triangular. If, if everything is below the line for this triangle. So what I mentioned in the slide last month, those were not the odds of any lips, because it's possible for points to the. It's possible for something that's a convex hole to not make a ellipse. For example, here I, I highlight points in red which do not make an ellipse, and in green, those that do. But say these, these, these red points down here. If you use, say, this particular red point, that would be a convex hole, but it would not make an ellipse. It would make an hyperbola instead. In order for any lips to be possible, it has to be one of these green points. And one thing that you can pretty much show is if the point is within exactly one of the two parabolas, then the point will make an ellipse. And if it's in two or none, it is, it will make a hyperbola instead. So you can see these, these points that would be convex holes just barely over the line or over here, over here do not make ellipses. So this is an, actually an unsolved problem. What are the odds of an ellipse for a triangular region? Here it's about 0.2389 or about 1146. If you have a start with a triangular region and five random points with a rectangular region, it's about 0.28 or about 23 over 82. If you have five points that just randomly we can calculate those exactly. That's the Sylvester problem I just showed. And if you start with a five points that make a convex hole, random points that make a convex hole, then the odds that they will make an ellipse is about 0.78 from a triangle region, or about 0.82 if the random points are from a rectangular region. But these are just approximations based on about a million trials. So that's an unsolved problem that probably has a good chance of being solved if anybody wants to take a look at it. So I mentioned Pascal's Mystic Hexagon, and he discovered that if you take six points on a conic section, for example, these here, then you take the intersection points of the opposite sides. AB is opposite side, cd. In this case, then the three points of intersection are collinear. Collinear, and they make what's called a Pascal line, and we can move these around. And no matter how you warp the conic section or pick the six points, these three points of, of, of, of opposite sides intersecting will always make a straight line. So that, that's something that he published as a teenager and it's, it's basically started Pascal's mathematical fame. A similar question is, is. I'm not sure how to pronounce this person's name. Charles Julian Braetian. I'm guessing if instead you do points that have an inscribed conic section and inscribed ellipse, you can move around the points and let's start that one over again. The six points that make up the hexagon will still intersect. The, the opposite points with lines will intersect at a single point. So that's kind of the reverse of Pascal's hexagon is where you have a, a hexagon inscribed inside and you connect the edges. If you do the opposite, you start with a hexagon and inscribe a conic section inside and connect the vertices. The lines intersect rather than being collinear. And one person who looked for connections between the two things, the Pascal hexagon and the Brianchan hexagon was Dandelion, whose name I'm probably mispronouncing and part of his paper here, Hyperbole of Revolution. He noticed that if you take a cone and put the ellipse inside and then you add the two spheres above and below, then the spheres have the tendencies on the ellipse at the foci of the ellipse, and that's what's called the dandelion spheres for an ellipse that those that if you drop those two spheres in and make the ellipse, the, the points of the ellipse that are tangency points are the foci of the ellipse. Now, let's see here. So here's the main part. We've got this big object called the hexagram mysticum, 95 lines and 95 points. We've got these at the top, 95 lines and these at the bottom, 95 points. And here I show an example. The lines here are in Hessian line form. It's a little bit ridiculous, but, for example, here are the Steiner Steiner points. Here are the Pascal lines, here are the Cayley lines, the Kirkman lines, and so on. So let's get into it. First, we need to define what a hexagon is. So here is. We basically have a conic section and six points on it. So here's an example of a conic section, an ellipse with six points on it. And if we connect up all the points in all ways, there's 15 ways of choosing two points. Six, choose two, and that leads to 15 lines, and they intersect at 45 places. Unless there's parallelism. If there's parallelism, then things go wrong. But if, if things are random enough and no. 2, no 2 edges of the original hexagon are parallel, then we get all these intersections, the 45 generated. And I think I set these up to be random. Let's see how this works. So it was 17 before, 17 lines before, and now it is 11 lines. So it's not consistent what happens from these random points. Now we have 25 extra lines generated from the 45 points. And, and when I mentioned these lines, I'm talking about extraordinary lines, lines with more, three or more points. So it, it seems to be completely random. If you take six random points on a random conic section and you look at these 45 generated points, how many different extraordinary lines can you get? And that an extraordinary line is a, is a line with more than two points on it. And we've got a resource function called Find extraordinary Lines where you can put in any set of points and it'll find all the point sets that three or more points are on a line. And I found that's a very useful function. It's also a very fast function. It the, so there's a, in matrices, the linear reduce method is very fast, and that's a very good way of identifying all lines in a set of points. So how many hexagons are there with six given points? Well, if you just look at the 6 points, it's 6 factorial or 720. However, the dihedral group of order 6 has 12 orderings that are equivalent. For example, 1, 2, 3, 4, 5, 6, and 2, 3, 4, 5,. 6, 1. If you look at both of those, they're pretty much the same hexagon. We can look at list rotations and reversals to identify canonical hexagons. So from a, from a list, there's 12 different rotations of this list and one reversal combined, making 12 different variations on this list, which are pretty much the same. And when we eliminate all the equivalents, we're left with 60 different hexagons. We can use those 60 to generate all the Pascal lines for a given set of points. I've done that here. For example, 1, 2, 3, 4, 5, 6. The opposite edges are 4, 5 and 1, 2, in this case. And the Pascal line goes through these two intersections. And there's a green intersection way down here somewhere. With this one, you can actually see all three intersections. 1, 2, 3, 4, 5, 6. But this is the same. I believe that's the same. No, I guess it's different sets of points in each one. Oh, it's just that they're labeled differently, and the, the numbers show the different labelings for each set of points, and they're identical in each case. And I think I set this up to run on anything. So here's a different set of 6 points along with all 60 Pascal lines. So let's see here. For the 60 Pascal lines, if the opposing edges are extended, we get three intersections that are collinear, making what's called the Pascal line. So here's a demonstration that shows all 60 Pascal lines for these six given points. Let's go. For example, these two green lines intersect here, the two red lines intersect here, and the blue lines intersect over there. And we get a linear set and a Pascal line going through them. Here's easier to see two blue lines, two red lines, two green lines. And that is the Pascal one of the 60 Pascal lines for this set of six points. After Pascal made that discovery, more discoveries were made. And Kirkman, a famous clergyman, noticed that if you take one of these hexagons and then you find the other three hexagons that don't share any edges with the first one, those three will define three lines which all happen to intersect on the same point. So what does that mean? For example, here is a hexagon, and here are three other hexagons that don't share any edges with the original one. There's exactly three ways of doing it. If you start with a given hexagon and you look at the Pascal lines for these three, those three lines will intersect at a single point. So, for example, we can generate the hexagram of Mysticum for this set of points. We can look at this set of edges for making the hexagon 1, 2, 3, 4, 5, 6. And then look at, look for all the various hexagons that do not share. Basically, we're finding these three hexagons, and here's what they look like in list form. Then we use those hexagons to identify the three Pascal lines, and it's a line 135264. And these other two, we can pull those lines out of the hexagram of Mysticum. And here are the, Here are the Hessian forms for those three lines. And then we obtain a particular Kirkman point here. And here we show all that process here where these three red lines intersect at this single point. So that's. I did that a few dozen times and I got kind of tired of it. So I wondered, can we cheat? Is there a simpler way of doing this? So, yes, we've got these 95 lines and 95 points. So let's generate all 95 lines and 95 points. So we do that with this code here. Now, for every point, identify the index in 95, identify the index for every line containing that point. And here's code for doing that. And going the other way, for every line, identify the index of every point on that line. And here's code for doing that here. Well, it turns out those two things are identical. These, this list of indices is identical. And it looks basically like this. For point one, it's on lines 36, 41, 46 and 71. And also for line one, it has points 36, 41, 46 and 71 dot it's self dual. If index of a point is on an index of another line, the reverse is true that for any point on a line, the line is on the point when you look at the indices. So, so I was able to use that to speed up the process. So back for Kirkman points, we basically compute all these, and I use this HM graph to basically just look up the indices for the lines that are necessary. And then I can just scroll right through everything. Right off we pick out point 11, Kirkman point 11. And let's see, 11 would be. Let's see, these are five each. So it's this set of points here, 79 is an extra line that will come in later. But these are the index lines and they intersect. Those intersect. So this is a way of finding the Kirkman points and the associated Pascal lines that converge on at that given point. A bit more about Pascal. Here's one of the famous quotes he had. I would have written a shorter letter, but I did not have the time. That's something I. I feel all the time. He was also one of the first to develop a mechanical computer. And here's one of the ones that Pascal made here. One of the first calculators. He died at a young age. And you might be thinking, well, why? Well, here's one reason. Pascal's Law. He worked with mercury a lot. He did a lot of things with mercury and, well, that's not a healthy thing to do. But he did discover lots of things about siphons. And vacuums and pressure and pretty much revolutionized that field of physics. But when you work with that much mercury, that can have effects on your health. He also developed Pascal's theorem, where basically you have. You start with 1 1, 1 and then you start adding up what's above it. And this makes the binomial numbers. And since then it's been called Pascal's triangle, even though it was discovered by Indian mathematicians and Chinese mathematicians earlier. I forget who named it, but basically nothing in mathematics is named after the right person. And more on Thomas Kirkman. He's famous for something called the Kirkman School Girl problem, which is 15 young ladies in a school walk out three abreast for seven days in succession is required to arrange them daily so that no two shall walk twice abreast. So here's a solution for that. Red would be one day and none of the. The red groups have the same two students. Blue is the next day, cyan is another day, green, another black, another purple. We can go through the whole week with that. And this is based on 0 bit xor sums 2 plus 10 plus 8. The bit xor is 0. All the lines in this, this object, which is the smallest projective space is, is show. And more on that problem. It was in 1850 that Kirkman sent that question about the schoolgirls to the Ladies and Gentlemen's Diary, which is the leading publication for mathematics of the time. And it was solved by author Cayley and Jacob Signer, who are three of the people we're talking about today. There's all sorts of different relations and in general, this, what's called, produces what's called the social golfer problem. You've got a group of golfers who are meeting in a tournament for some number of days and they're meeting in threes or fours or fives. And you want to know how many different games they can play so that no two of them ever play a game with each other, play on the same team more than twice. And here's a solution for four golfers per group in seven groups playing for nine days. Here's another one with 32 golfers. They can play for up to 10 days without any repeats. And you can compare them and see that there's no overlaps in any of these. These use what are called Steiner systems, Kirkman systems. And there's a. There's actually about seven different types of things that can happen because some, sometimes these aren't perfect. And many of these solutions are not completely solved or proven anyways. The this is what's called the social golfer problem. And in general, the problem is still unsolved. And we can introduce Arthur Cayley at this point. He's the one that proved that 27 lines are in a cubic surface. And here we have the club surface here. We talked about these last time. Another person that proved that you can get 27 lines on a cubic surface is George Salman, who happens to be one of the other people in, involved with the hexagram of Mystica. Another thing Kaylee did was Kigraph. So here's an example of K graph of these two permutations. Oh, let's see. So I mentioned author Kaylee. Kaylee took a look at what Kirkman and Pascal did and discovered what are called Cayley lines. So consider the vertices of the hexagons as a cyclic permutation and square that. Well, that makes no sense to anybody now. So here's, here's what that means. You have these three permutations. And note that the, the first, third and fifth item are 1, 3 and 5. Here they are 1, 5 and 3. Here they are 1, 5and 3. So, so basically, if you, if you split up the, the odd elements, it's the same group in each case. Here they're 2, 4 and 6. 2, 6 and 4, 4, 2 and 6. So basically they are all based on these two items interleaved. So basically there's 20 different ways of doing that. And he found that if you take this set and find the items that are involved with that and, and find those lines, they will all intersect at a single point. Oh, wait a second, I have to get points and lines correct. Oh, okay. Basically, if you, if you find the corresponding Kirkman points, those Kirkman points are collinear and you get what's called the Cayley line. For example, this is one of the 20 Cayley lines. So we can do that with indexing. We find where this line is 1, 35246. It's index 71. And it turns out that these seven points are on line 71. But we're only interested in the first three. Those are the ones that have an index under 60 because all the Kirkman points are, have indices 1 to 60. So we're looking for three Kirkman points, and here they are, and they correspond with these three permutations. And we can check that up. We can check that These are the three points. Point 1, 2, 3, 4, 5, 6. Here it is point 1. Point 1, 2, 5, 6, 34 is here. And here's this one here. And these are some additional points that are on that line. So we can look at all the Cayley lines and their associated Kirkman points. In green is the Cayley line and in red are the Kirkman points. And since it's, it splits up basically the hexagon into two triangles. I've basically highlighted what those triangles look like. And here's all 20 of the Cayley lines. So that's what those look like. Steiner then got involved. Jacob Steiner. He put out a famous one page paper that basically found what's called Steiner points. However, his paper is riddled with errors and false assertions. So not sure how much you should pay attention to that. But this point about the Steiner points is one thing in the paper that was correct. So consider a hexagon as a cyclic permutation. Do the, basically the same thing as, as, as was done with the paley lines. There's 20 different possible sets of three. And in any of these sets the three corresponding Pascal lines concur at a point. And as a result they're called Steiner points. So basically just do the same thing that Kayleigh did. But instead of looking at Berkman points, look at Pascal lines and you'll find that they can, they intersect at a point. So they did that. If we look at that particular point, it has an index of 71. These are the lines that go through that point. Here are the permutations that have that same permutation square and line 1, 2, 3, 4, 5, 6. And these other two lines and these four extra lines all go through that point. And now we can look at, for a given set of three pascal lines, how they intersect at a, on a. Is it Steiner? Yeah, Steiner point. So here's, and again I, I indicate the hexagon as, as two triangles. So for example, for these two sets for 3, 5, 6 and 1, 2, 4, the three Pascal lines intersect at this Steiner point here. So Those are the 20 Steiner points. And Steiner did a whole bunch of other things. For example, here's a projective plane that produces what's called the Steiner surface. And we can also generate what's called Steiner chains. If we have two circles set up properly, you and you do tangent circles, they will always form a chain. No matter how you draw the initial first circle, you'll always get three tangent, three tangent circles. Once you've got this set up and it's possible to have say five circles. And we can change the radius. So these are the, this is what's called the Steiner chain. Let's see here. He also developed what's called Steiner trees. If you have a soap bubble, the edges of the soap bubble will intersect at a point which is approximately 120 degrees or exactly 120 degrees. I'm not sure if I can get there automatically by hand. But the, the two points between these four that minimizes the road needed is at the place where these two, these six angles are all 120 degrees. And you'll see the same thing in soap films. So after those four, a person by the name of Pluker noticed that if you take triplets of two cycles such as this set here, and then you found the four Steiner points that commute with that given cycle under permutation products, basically this 1, 2, 3, 4, 5, 6, like that. And 1 3, 5246 equals, you can the commutation of that if you reverse the order of the multiplication. These two are the same, but that's not always the case with permutations. And these also work. And in each case you, you, you're basically starting with the first two digits are 1 and 2. In each of these 1 and 2, 1 and 2. The next two digits are 3 and 4. We got 3, 4, 3 and 4, 3 and 4. And the last two digits are 6 and 5, 6, 5, 6, 5, 6 and 5. The, the Steiner points in these cases are collinear and you can get what's called the Pluker line. So if we want the four center points on this particular line, we can find the index of it. In this case, the index is 81. And the indices of the lines on point 81 are these as follows. And these four permutations all commute with this permutation. And we can look those up here. These are the four points and we can actually show these. We now connect up the hexagon in segments and we can look at these four points in red. Those are the, those are the four Steiner points. And in green, going through them is the Pluker line. And we can do show that for all 15 pluker lines for these four Steiner points, this green item is the Pluker line going through them. But we're not done yet. The other person who proved that the there are 27 lines on a cubic surface also proved that you can, you could extend this hexagram and mysticum to another set of points. So just like we showed with Pluker, we can identify triplets of two cycles and we get four corresponding Cayley lines. They all happen to intersect at the same point. And these are now called salmon points. So we do the same thing. We now look at the a given point, and we. We were just looking at a line. Now we're looking at the corresponding point. It has the same index and it has the same indices for the lines going through the point. We get the same set of permutation, but now they're on lines instead of points corresponding to lines of points. And we can take a look at the four Cayley lines intersecting at a given salmon point. So in red is the salmon point, and in green are the Cayley lines, as shown with these. And I think this is the first time that all 95 lines and all 95 points have actually been shown in one talk in a way that it's clear what's going on with. With each. Of course, I rushed through a whole bunch of these points. But I'll be making a demonstration soon that where you can look at all these connections back and forth with each other. And then John Conway got involved. He noticed that all six of the proofs that were generated back in the 1800s had lots of different flaws. So he basically came up with this mystic H, where he took out the 60 Pascal lines, the 20 Signer nodes, the 15 Pluker lines, the 60 Kirkman nodes, the 20 Halley lines, and the 15 Salman nodes, or points. Nodes and points are considered the same thing. And he basically proved all of these various properties using variations of Desarg's theorem, where if you have two triangles with the same perspective point, then if you extend their edges, they will intersect at a perspective line. And once you have a set of 10 points with that property, they will always make a perspective point and a perspector line. And he used that to prove all of these six parts of the Mysticum hexagram. It's a paper called the Hexagram Mysticum Demystified. And basically all you need to do is understand the desarg's theorem, and you can see the relation he did with each set to come up with these six different proofs. So we can pull all these together and, for example, show the 20 Cayley lines and the 15 Pluker lines together. Here's what they look like for a given set of six points. And you can kind of see some of the places where the red lines are intersecting at other sets of points. Here's where the. Here's a place for 4k force. It's easy to get these mixed up, but that's a Steiner point intersection there. Here's A configuration we obtain 20 Steiner points and 15 Salmon points. And we want to find the 20 Cayley lines and 15 Pluker lines. And we get those here. And this what's. What is produces what's called the Danzer35.4 configuration. For every one of these points there's four lines. And for every line there's four points. And here we show all the lines and all the points except for two points that are outside of the frame. But all the. All the other lines and points are shown here. This is the smallest dander configuration I was able to come up with. Usually the points are. Usually there's at least one point which is way far off. Whether there's a nice way of showing the Dancer configuration I'm not completely sure. Here's the original demo and I have. Different sets are possible for different conic sections and different sets of six points. And they usually look like a mess. There is no pictures of the hexagram of mysticum until 1921. Haley was one of the first to try to plot out all the Pascal lines, but he basically mentioned it looks completely unintelligible. And based on my work with computer plots of all the Pascal lines, usually they do look like a big mess. And that's when you have the luxury of trying a million different cases to see if there's any that look better than any others. And even then they. They tend to look like horrible messes. I'm not sure what's going on here, but it looks like a whole bunch of lines intersect at that point. So let's see what is unsolved. One thing that's unsolved is the chance of an ellipse. We still don't know what the probability is of five random points making an ellipse. If we have a triangle and rectangular region, we can get the exact odds of whether they make a convex hole. And that's the Sylvester problem. These are well known. In general, the odds are about 11 out of 46 if you have a triangular region, or 23 out of 62 if you have a rectangular. But those are just exact approximations. I'm not even sure whether these first two digits are correct. I believe they are because I've done a million trials, but I'm not totally sure. It might be 0.2799 something. But having. Doing a billion trials would probably be a good idea just to. To do a better approximation. And if you have. Oh, I think I mentioned these are the exact values down here. This is actually an approximation about. If you start with a convex hole. The odds of an ellipse are about that much for a rectangular and that much for a triangle. I'm pretty sure that someone who really understands how ellipses work, random ellipses work, basically start with four points in a rectangular area, generate the ellipses and do some sort of analysis on that. I'm pretty sure that that will solve the problem, but I don't know how to do that at this point. And I'll be. That's. I think that's all I had. Yes. Feel free to contact me at this address and I'll be glad to answer any questions that people might have. So feel free to post questions here if you have any questions. And that's a good idea of making a. A list of unsolved problems. I'm. We could probably update that on Math World, and I know there's a list of unsolved problems on Wikipedia, but that's hard to get sticky. But I could definitely try updating Wikipedia as well. Any questions from anybody? So I think that will wrap things up. Many thanks for attending my talk.