Mathematical Games Hosted by Ed Pegg Jr. [Episode 3: Algebraic Number Magic]

Sa. Oh, Okay. Hello, I'm Ed Pegg Jr. And welcome to episode three of Mathematical Games. Mathematical Games was originally a column hosted by Martin Gardner in Scientific American, and it was very influential. So I've been trying to make some talks in sort of the spirit of Martin Gardner. And this is the latest one. Welcome, everybody. Welcome to Jimmy. And I'm ready to get started here. So we'll start with what is an algebraic number? So here's an example. X squared equals x plus 1. And the solutions to that are the golden ratio, these numbers here, golden ratio here, and one minus the golden ratio or one over the golden ratio. So this is an example of a quadratic polynomial. In these polynomials, basically everything has to be either a rational or algebraic coefficient. And if that happens, then you can get an algebraic number out back. If the Highest exponent is 2, the number is geometrically constructible. And there's a few other higher order polynomials which we'll get to, such as this one for the poly for the 17GON, which are also constructible. But cubics and above are not constructible. And here's two examples. Here's x to the third equals x plus one. This is the plastic constant. And X to the third equals X squared plus one. This is the super golden ratio. These are not constructible and you Euclidean geometry. However, they are constructable in origami because there's an extra trick you can use in origami which allows the construction of cubics. So here's one of the valuable things within algebraic algebraic numbers is the discriminant. For example, the discriminant of the square root of 5 is 5. And for quadratics it's usually pretty easy. Discriminant of square root of 5 is 5, 13 is 13. Discriminant of square root of 21 is 21. And similarly, the square root of 17 discriminant is 17. It's either the root number or four times the root number. And you're probably familiar with the quadratic formula. The discriminant is this part on the inside. Now, funny story, I've been using discriminants in number theory for a number of years, but it wasn't until I started putting this talk together that I came back to the actual definition of discriminant and realized that this was the discriminant in the quadratic formula. So I didn't really understand discriminants until I started researching for this talk. You don't necessarily need to understand a tool in mathematics in order to use it. And I've used discriminants a lot without having any idea what they actually were. But now I know, since I looked it up, the discriminant of the cubic is this thing. And you can, you can kind of get that out from looking at the solution for the general cubic. And if you do this operation on a, on the, this part of the cubic, you get back this item here, which is what the discriminant is for a cubic. And we can make a function out of it like so. So for x to the third minus x minus one, you get that the discriminant of the plastic constant, which is from this Polynomial here is -23. There's also a whole bunch of other polynomials that also have the same discriminant as this one. There's an infinite number of polynomials that all have a root which is in the algebraic field of the plastic constant. And basically, whenever you see a polynomial, if you look at the discriminant, you can find out what number field you're in. Another definition is it's the product of the square of the differences of the roots. So here's another way. If you take a look at the roots of the plastic constant and do the differences in times and reduce everything, you get back minus 23. I mentioned the 17 gone a little bit ago. If you look at this, this radical statement, or this root which was on the previous page, or the cosine of PI over 2, the discriminant in all of these cases is equal to 17 to the 7th. All of these are the same thing. So even though these are in completely different forms, once you look at the discriminant, you can see that, oh, well, these are all the same thing. So how did I start looking into number theory? Well, it started with this little experiment. There's something called Parborth's conjecture where you have a planar graph and the conjecture is that you can set up the lengths of this planar graph so that they're all integers. And this is a famous unsolved problem. And I figured, well, one thing you could do as a piece of, that is to, is to set up all the possible wheels of the, of the vertices around a point. So, for example, this could be a section of a planar graph. And all these lengths here are integers, all the sides and all the distances from the central point. And when I started looking at this problem, I noticed that for every single wheel graph, and I looked at about a trillion of them, if you look at the areas of the triangles, there's always a certain set of radicals in the set of areas. And the maximum number of radicals you can have is two. And I'm after looking at this again, I like this is something I should probably prove. So I'm, I'm working on a proof of this. Shouldn't be too hard, but I think this is a good step in resolving Harbor's conjecture that once, once you've set up your planar graph and triangularized it, you can color all the triangles and you're restricted by a, a radical of of how the graph can be filled in. So I'm going to try to basically build the octahedral graph using every single possible radical and see if there's any restricted radicals for the octahedral graph. So after this experiment with a wheel, my next question was, well, what similar triangles can make a wheel graph? And that led to this experiment here. So for example, these sides here, these are all powers of the square root of 2. So if you look at square root of 2 to the 0, square root of 2 to the 1 to the 2. So this is just basically just means that this length here actually is two, this one is two, two times the square root of two, this one is actually four. I basically started building all these wheel graphs and for example, all these have discriminate five, which means these are in. These are built with the algebraic space of the golden ratio and these are minus 23. These are in the algebraic space of the plot constant. I just basically set up a computer just to start finding these solutions and the plastic constant and here's the super golden ratio here happen to have a lot of different solutions where all the lengths, vertices and everything else were in the space of a particular algebraic space. -44. This is the Tribonacci constant. And it goes on. Most of these solutions were in items with a small discriminant and I felt that that was something valuable. So I one thing I noticed was the values that turned up a lot were 2, 5 or the golden ratio minus 31 or the super golden ratio minus 283 or the chi constant and minus 23 the plastic constant. And I noticed that all of them were basically sums of themselves. If you took A look at 2 is equal to 1 plus 1 half plus 1 fourth plus 1, 8 and so on. The golden ratio is equal to 1 over the golden ratio plus 1 over the golden ratio squared plus 1 over the golden ratio to the third and so on. And that happens to be working for all of those. So I thought these five constants might be very interesting in geometry for tiling purposes. And that was just a suspicion I had. So I started looking at other polynomials. There's the Passat numbers. This is the, the plastic constant here. And these are basically positions of the complex roots. Here's the chi constant. There's one root outside and all the rest are inside. And the important thing to notice for basalt number is that the complex, the real complex root, there's only one of them and it's less than two. So here's the third best basalt number, and here's the super golden ratio here. They're all small basalt numbers. So I started looking for various basalt numbers which, which happen to have nice solutions. Salem polynomials. Those have one complex, one real complex root outside and one inside. And all the rest of the roots are on the unit circle. And also there's the cyclotomic polynomials where all the roots are on the unit circle. So I figured I should look at all of these types of polynomials for special geometries. And one thing I noticed right off is while working with these geometries was that, for example, here's something in the, in the root space of the square root of 33. But the vertices had these root objects. They, they didn't look nice. And I couldn't get things to work out for a while. But then I remember this old adage, when something doesn't work, try 0, 1 and the square root. So I tried squaring this number and suddenly it worked. If I remove the square here, this is not in the, in the number field of square root of 33 anymore, but here it is, it's 772 plus one over 72 times the, the square root of 33. And once you were in an algebraic space with the squaring in geometry, everything happened to be, happened to work, right? So eventually, here's the vertices up here. I just list them. I created this function square root space, which basically squared things, but retained sign. So the, for example, minus three and square root space, it would be minus the square root of three and vice versa, minus the square to square root of three squared would be minus three, because you're basically looking at the sign, then doing the squaring or square root operation on the positive and then putting the sign back. And with this little simple function, basically everything started work. I could, I could represent all the vertices in terms of the square root of 33 space. But everything is a nice rational number or nice pair of rational numbers since it's a quadratic. So once I had this this function in place, I started looking at each quadratic space and each number field space. Yes, the, the square root space. Just squaring things was very effective, and I'll get to that later, but I could basically say I looked at 150 different geometries and the squaring trick simplified things. In all 150 cases, it just really, really simplifies things. So the simplest algebraic field is the rational space. So here's an example here. All the vertices are simple integer pairs. And this is a solution for 12 trees and 19 rows of three. And this, this is a solution that wasn't known in the literature for some reason. The Heilbronn problem arranges points in a unit square so that the smallest triangle is as large as possible. And all the vertices in these solutions are nice rational numbers. And let's go to an extreme. There's something called the hexagram mysticum, which is looked at by, by Pascal, Kirkman, Cayley, Steiner, Plukers and Salmon. And basically, once you start with six numbers, making a conic section, for example, here you get 60 Pascal lines, 60 Kirtland points, 20 K lines, 20 Steiner points, 15 Pluker lines and 15 Salmon points. And all of them are in the space of rationals. If you start with rationals, if you start with say, points that are in the algebraic space of the golden ratio, everything will still be in the golden ratio. If you start with the plastic constant, everything will be involved in the, the plastic constant. So basically, once you're in an algebraic space, you're basically stuck inside of it for many of these construction techniques. And I found this to be very valuable. One curious thing I found out later, these six guys, Pascal, well, not so much Pascal because he died, but the later five, Kirkman, Kaylee, Steiner, Pluker and Salman, apparently they all hated each other. There's kind of weird that they all got stuck together with this particular geometric construction. So the next in the list of algebraic spaces is the silver ratio, or the square root of 2. Hipposis should probably be mentioned with this too, because he died for the square root of two since he was the one that proved it was irrational. But the continued fraction for 1 over square root of 2 is basically a series of twos. And once you set up this octagon and the square root space of square root of two, everything is nice and tidy. They're all rationals. If you look at the algebraic space, you can also look at A4 paper. It has the property when you fold it in half, you Get a rectangle which has the same proportions as the original rectangle, which is why they preferred in Britain for making things like fires and stuff, because once you fold it in half, you still have this nice rectangle which is the same size as the original. One weird property of A4 paper is if you started with a piece of paper with area of 200, you can split it up into other A4 rectangles that are all either squares or twice as square. And this is a perfect dissection. These are all either the squares or twice squares under 50. Making up this dissection of a larger A4 rectangle. We can also look at something we saw in the similar triangle things. We get this diagram here. If you look at these, these sides, you basically make this wheel and you get this object here. But if you extend it, you get the 4, 5, 6 triangle, and it dissects into five similar triangles, which wasn't known before. So this is a nice little puzzle. Give somebody a 4, 5, 6 triangle and ask them to divide it into five similar triangles, and this is the solution. And I found that I had those, those five numbers. Two golden ratio, super golden ratio, classic constant, and the chi constant. Well, it turns out that 2 has this substitution piling. You start with this, this triangle, and all these are similar triangles inside. You get this substitution tiling that wasn't known before. So two is one of the magic numbers. And I actually proved that it had this nice tiling property in geometry because of that similar triangle thing I showed earlier. It turns out you get this nice substitution tiling pattern out of it. And then I found this, this next one, these individual triangles here, this is basically the, the 1, 2, 2 triangle. And this is a 1, 1, 1 half triangle. Well, it turns out that if you add a line here, you get a, A2 triangle. So that leads to a fractal substitution system. And this is what it looks like after a number of substitutions. This was also not known in the literature. This, this simple little thing with the 1, 2, 2 triangle. Once I had the intuition that there was a tiling system that could be brought out of it, I just found it. The next up in algebraic numbers is the square root of three. Theodorus was the one who was responsible for that. And more recently, this pentagon tiling based on the square root of three was found. Here's a way of dissecting a square into 45, 60, 75 triangles. This seems to be the best solution. And all of the positions of vertices are based on the square root of 3. The dodecagon is in this space. And all the vertices and intersections and everything else within the root space of square root of three have nice rational values. Let's see what's the Next one here? 5 PHI space or the golden ratio space. So basically, almost all of the five related polyhedra have constructions entirely within the space of the golden ratio. And you can kind of see that here by. I'm using square root space here. These are all of the values needed for this demonstration. And they're all nice rational numbers. We basically go back and forth to square root space, which is this simple function here. And you get things like this. This is just basically a usage of properties of the golden ratio. The Almond chair. You have this chair object, you divide it into two chairs which are similar. They're proportioned like the golden ratio. My friend Carl Scherer independently discovered this object at about the same, same time as Almond did. The best four in a row solution for 16 lines with 15 points is in the space of the golden ratio. And all the vertices are in the root of the root space of the golden ratio. There's also other three in a row problems which have. And that one's not working correctly. All everything is in that particular algebraic space. A triangle with sides one, this, the square, the golden ratio. And the golden ratio squared is known as a Kepler triangle. So that's, that's here. If you, if you do more than one power of that, if you take three of these triangles in series, you can, you can get what's called Brugner's three piece Tangram. I decided to extend it to six triangles and you could get this nice rectangle out of it. You can also build the club surface entirely with the space of the golden ratio. And some friends of mine that work with ZOME just recently built this object using nothing but zone pieces. And I mentioned that the golden ratio is one of the magical values. So I had this intuition that there was a unknown substitution tiling pattern. And it turns out I was right. There. This, this section here is vertices around a vertex. They're all similar triangles. This is one of the things that was in my experiment earlier. I managed to expand that to get back the original triangle. And here's how the substitution pattern grows. So you can see that we've got a lot of similar triangles in a chaotic pattern to make up the original triangle. And that all the triangles here are similar to the outer triangle. Next, square root of six. I don't have too much for this particular root space, but I'm always Looking though, if, if you, if you know something that's in a particular algebraic space that isn't in my demonstrations or in this talk, please let me know about it. I'd love to know the bronze space. So if you look at the, the continued fraction for the square root of 13 with add 3 and divide by 2. The continued fraction is basically a whole bunch of threes. The golden ratio is a bunch of ones. The silver constant is a bunch of twos. The bronze space is a bunch of threes, and the continued fractions. The BBC had a contest a few years back. There was no good geometry known for showing the bronze ratio. There's a contest for somebody to make a geometric picture which showed why the bronze space was interesting. And I sent this in these two segments and the best Heilbronn solution for eight. These have the ratio of the bronze constant and everything is in the space of the is in the square to 13 space. So BBC actually wrote back to me that I was the only person that sent in anything and they didn't think it was interesting enough so they canceled the contest. But I still think it's interesting even if BBC didn't, didn't like it. I showed the Golem graph earlier. That's this object here, the Moser spindle is also in this, in this algebraic space. And I have a demonstration which shows a whole bunch of different, different ways of coming up with either the Mosul spinder or the Golem graph. But it turns out that Aubrey De Gray came up with a better solution. And so he, he advanced the chromatic number of the plane problem. I was very close though. I just didn't use enough roots. My, my faith in algebraic space was not yet sufficient to solve the problem. For the square root of 57. You get this, this other pentagon tiling and basically all the vertices are in the root space of the square root of 57. For65. There's another Heilbron solution. Everything here is in the root space of square root of 65. And we can also do the reverse. Look at the complex numbers. Turns out that the, this, the standard complex numbers. Gauss studied those and they got named after them the Gaussian integers. Eisenstein extended it to this square root of minus three. And you get what these objects here, which are known as the Eisenstein integers. And you can see these here. Klein looked at square root of -7 and those are the Kleining integers. Well, nobody looked at the square root of two all that much. Hipposis in, in Greece was killed about 3,000 years ago because he proved that the square root of minus 2 was irrational. So I, I figure for in his honor, that this particular space should be called the Hippasys integers. And let's see if we can show those here. These are the Hipposis integers that are primes, and then these are the Gaussian primes here. So let's, there's, there's many more that can be shown in quadratic spaces, but let's move on to cubics now. So the cube root of two is also known as the Delian constant. There was a long unsolved question as to whether the cube root of two could be constructed for the doubling of the cube problem. Well, one other doubling you can do is if you take a cuboid with one point at the origin and the other point at powers 0, 1 and 2 of the cube root of 2, you get this cuboid here, which can be divided into two cuboids, each similar to the original cuboid. And I call this the Delian brick after the Delian problem. This is an object that has been discovered many times. Nobody knows who this first discovered it. So it's probably good to just call it after Delian, since that's where the doubling of the cube problem came from. And now we get to the plastic constant. So here's, here's what the plastic constant looks like under radicals. And it has this nice property that you can divide a square into three similar rectangles. And the Padavan sequence and parent sequence both have numbers that grow like the plastic constant. The ratios between adjacent terms goes increasingly towards the plastic constant. So this triangle over here, Andrew Zach was the first one to find that. He basically showed that you could divide this particular triangle here into five similar triangles. I extended this solution to add this extra triangle, and you get back the original triangle. So now it's a substitution pattern. You can also arrange these same triangles to get this pattern down here, which Andrew Zach had not found. Here's a geometric scene solving this problem. And we can also build up a similar fractal. So here's another way of dividing up two objects. With the ratio of the plastic constant. You have this big weird fractal object on the outside. These two things on the inside are both similar to the original shape, but scaled in terms of the plastic constant. And with that, you can create an infinite series showing that if you look at the different shapes here, you basically get a whole bunch that sum to the original value. These are powers of the square root of the plastic constant as distances and you get the numbers 0, 1, 2, 3, 4, 5, 6, 7 and 8, and the 3 is repeated. So this is a very weird object here. All the distances are powers of the square root of the planet. And when I found this in 2D, I wondered, well, what happens if you look at 3D? Well, here's 19 points. And every single pair of points is a power of the square root of the plastic constant. I call this the plastic pyramid is apparently not, not known. But it's a way of constructing K19 so that every set of points is a power of the square root of the plastic constant. And there's no other algebraic space. This is possible. And with, with the golden ratio, I think you're limited to nine. And other, other algebraic spaces are even worse. So the fact that this one works so well, the plastic constant is something I still don't understand yet. I'm still not quite sure what you can do with this pyramid. Here's a polynomial x to the fourth plus X to the third plus the square of the plastic constant plus this root of the plastic constant plus this plastic constant minus one. Well, it turns out that the coefficients of this polynomial are also the solutions. They're also the roots. And this is something that's very rare with polynomials. But it turns out that the plastic constant is able to provide a solution. And here is a rectangle that is built with seven triangles in series, all based on the sides having the powers of the plastic constant. Let's see the underlying. Let's see, what is that? So here's, here's kind of a reminder of the so numbers. And the polynomial for the plastic constant is at the top. The next one is the chi space. I forget what this polynomial is called. And then the fourth one is the super golden ratio. Not sure why that's not working. Let's see what happens here. This wants. Oh, I need to open this. Okay, here's a tiling. You've got this triangle here. And this is a substitution pattern for building the, the triangle I showed earlier with a series of triangles, not, not skipping any values. And you can continue to make these substitutions and you get this, this growing pattern. You can also build this substitution pattern using the plastic constant. And this was apparently not known. Let's see here. And now we get to the super golden ratio. You may recall that Fibonacci did something with rabbits. Well, later Narayana did the same thing with cows. A cow gives birth to a calf every year. In turn a calf gives birth to another calf, only gets to three years old. What is the number of progeny that it produces. This became as the Narayana cow problem. And here's how these things progress. And the ratio is equal to the super golden ratio based on x squared plus 1 equals x to the third. So this quadrilateral here is something I discovered. You take the psi is is the constant used for the super golden ratio. And if you have psi to the 0 or 1 side to the 1. And for these values of psi, I'm basically using, I solve X squared plus one equals x to the third and I take one of the complex roots instead of the real root. So these are these values here on the complex plane are powers of the complex root. And the first power, second power, third power, fourth power and negative the second power. You get this. Well with 0, 1, 2 and 3 you get this quadrilateral. And when you add these two other powers, you get a way of dividing the original quadrilateral into two similar quadrilaterals almost. You're basically have one piece missing up here and another piece added over here. But this leads to a way of doing a substitution system. These numbers in blue are the lengths in terms of powers of the super golden ratio. So psi squared plus one equals psi to the third. Just using that, you can, you can use power properties of the complex plane to self construct itself. And it leads to one of the best substitution systems available. And I'll show that in a little bit. We can also build this using what's called the Chaos game where you randomly pick a point from two options. And if you repeat this 3,000 times, you basically get this same dissection up here where you've got a fractal quadrilateral divided into two fractal shapes which are similar. And we can use this, this set of powers to build this, this wriggly edge to any desired level. Here's a way of coming up with a triangle dissection where you have one triangle divided into similar copies of itself. We can use geometric scene and random instance to basically solve this. And all the vertices here are based on the super golden ratio or psi. And when you expand this out further, you can basically build a fractal nautilus as shown here. The code is surprisingly simple for something so complicated. So we can expand this out some. I was flabbergasted when this popped out of my searches. And with this fractal object we can basically make infinite number of copies. And when sum them up, this is one big one. And if you take this big one is equal to psi or the super golden ratio. This one here is equal to one over the super golden ratio squared. One over the super golden ratio to the third, one over super golden ratio to the fourth. And this infinite series sums up to itself. We can also make seven of these, these triangles where the sides are powers of the super golden ratio. And you can make this nice rectangle. So I showed the triangle section earlier. Here's one way of building up that triangle and you basically get this substitution system which was unknown. And we can also use the quadrilateral I had at the start and you get this nice substitution system here which was unknown. This, this one particularly should have been one since it's, it's just basically built directly off of powers of the complex root. And once you just look at that, you basically get the solution automatically. So that should have been found years ago. Let's see, the next is the Tribonacci constant. So with that we can build the snub cube and it's anti amorphous. And here's a nice A ROM substitution pattern. The heptagon is next, also known as the silver constant. Different from the silver ratio. And all the points are nicely in the space of the silver constant. The last known of the in a row problems. All the vertices are in this the space of the solver constant for the next discriminant. As far as cubics are concerned, I wasn't able to find all that much. I found these two similar triangles solutions, but I don't have anything else for it. However, it is a nice solution in the, in the ABC conjecture. This particular route is, is used for this particular solution. And for minus 76. There's two interesting things in geometry. There's the Melson disc covering and here's a way of covering a circle with 12 discs. This is the optimal way of doing it with the 12 Heilbronn points. This is the best way of doing it. And it's based on this particular strange root. And these numbers are oddly associated with E to PI times the square root of 19. So here's, here's, here's the algebraic space we were looking at. Raised to the 24th. Minus 24. You get 80, 800 and 885,479 plus 7.777. If you look at this value, it's the same thing for nine digits. And there's also this, this root object where you have the, the coefficients of the polynomial that are also giving the roots of the polynomial. This is one of the other strange solutions that's Possible using something with a small discriminant. The horseshoe and medallion is in the space of the nonagon or nine sided polygon. You can get all of these are unit lines as this is a nice object here with 81 edges that are all, all have unit length. And this is the Hirshhorn medallion, a way of arranging 18 identical pentagons in order to make an 18 gon. The next Interesting discriminant is 316 which can make the Calabi triangle. And after that you get real hard cubic. This little matrix here in the 60s was known to crash systems because the eigenvalue had strange properties. If you look at the continued fraction you get these huge values unexpectedly. Well it turns out that's because it's, it's equivalent to E to the square root of 163 times PI, which is a very weird number in number theory. And if you look at the, the root of the, of the Brillhart cubic to the 24th power -24 you get these digits here, these 30 digits here. And if you look at E to the PI squared of 163 you get these 30 digits and they're identical. And these are both almost equal to this number here. This weirdness was enough to break many computer systems because it made no sense that these, these numbers had agreement to 30 digits. So that's something that, that, that fell out of this simple little matrix. And now it's, it's called Brillhart cubic. The Snobedodecahedron. If you take the Fibonacci constant and combine it with the golden ratio, this little polynomial here is everything you need to build the snub dodecahedron which I show here. And now the discriminant minus 1 minus 283. I mentioned the chi constant earlier and here it is. Here's a way of dividing up a a equilateral or a isosceles triangle into similar triangles that all are based on the chi constant. And that's based on, on this object here. I found in my similar triangle study. And I also found this in a look at degenerate simplex Simplices. There's a demonstration that has these. And it turns out that the chi constant is one of the values that gives solutions to to this where everything is is powers of the chi constant. And with the triangle above I came up with this substitution pattern where this big triangle on the outside, all the triangles on the inside are similar to it. So this is a another similar triangle dissection. And this was the strangest one yet, you get this, this really weird series of, of of triangles all, all working together to make a substitution pattern. So I mentioned these self summing infinite series at the start. Two golden ratio, super golden ratio, the chi constant, the plastic constant. Well, it turns out here's 2. This triangle has an area of 2. This similar triangle here has an area of 1. This one has 1/2, 1/4, 1/8 and so on. So 2 is equal to the sum of its reciprocal powers starting with zero. Here's the same with the golden ratio, but we skipped the first one and now we have this. This whole thing is equal to the golden ratio here. This triangle is equal to one over the, the golden ratio, one over the golden ratio squared, and so on. This is also an infinite series. Here's the super golden ratio. We basically start with two. This time you get this, this series and with the plastic constant you start with four. I haven't found a, a infinite series proof for the chi constant yet. So if anybody can find that, I'd love to see it. And it turns out that I missed a really nice dissection. This was found by Andrew Hudson just a few months ago. Here's another way of dividing up a triangle into similar copies of itself. My search through similar triangles basically had to have complete edge matching. So this point here, since it's, he's basically stopping in the middle of the vertex here, I didn't go far enough. So this is a very nice solution that I didn't find. You, you get to this, this really nice substitution system which, which I still need. So I managed to find all of these things by asking the question what similar triangles surround the point? And that led to at least 10 nice discoveries. But there's, there's still a lot I haven't solved. And one of the questions I'm really wondering about is what similar tetrahedra surround a segment or point. And so far I have no idea what the solutions for that are. It shouldn't be too hard to solve, but I haven't had time to focus on it yet. If anybody can solve that, I'm sure it will lead to all sorts of nice and interesting discoveries. So that's what I had prepared. I'd love to see any sort of comments in the chat that, that people might want to ask and otherwise many thanks for listening and I'll stick around and ask questions to anything that gets posted. Yeah, Petra? Yeah, I was wondering. I've been wondering about the tetrahedron for a while now. I've looked at it several times, but usually I run out of time after a day or so of exploration. And so far I haven't found anything worth worth posting. But I'm getting starting to get closer to how to attack that problem. Another thing to look at is the harbor conjecture I had at the start. There's a. Let's go back here. This is a demonstration integer wheel graphs. It's on the, on, it's on demonstrations. And basically every single integer wheel graph is only built out of two different radicals. So this is, this is probably something worth looking at. I'm. I'm pretty sure there's a solution for this unsolved problem using an application of this, of this little thing. And many thanks for everyone who watched.