Mathematical Games Hosted by Ed Pegg Jr. [Episode 10: Special Triangles and Tetrahedra]

Sa. This is going to be a talk about special tetrahedra and special triangles. The what's showing right now is about 140,000 tetrahedron centers, which I discussed a few months ago. And. But we're, we're still waiting for this to start, so I'm going to give it another minute. But here's what a whole bunch of triangle centers look like. And if you go to. I think I forget which. Which mathematical games talk it is. But a few talks back, I talked about tetrahedron centers, and this is an animation from, from generated from those 140,000 points. So I think we can start here. We have episode 10, special triangles and Tetrahedra. Mathematical Games was a column written by Martin Gardner for Scientific American. He Martin Gardner wrote it for 25 years and developed a series of 15 books which were highly influential for myself and many other mathematicians. So I'm trying to give these talks basically in a style similar to something Martin Gardner might do. We have more than a thousand demonstrations involving triangles, and many of these deal with triangle centers and a few deal with tetrahedron centers. But today I'm going to talk about some special triangles and special tetrahedra. And we'll start with the incenter, which is a intersection of the angle bisectors. This was something known. It's been known for 4,000 years that if you bisect each angle of a triangle, they intersect at a point which is the center of a circle tangent to all three sides. So there's two triangles that you can get out of the special triangles. One is the. The of the angle bisectors. That's the triangle in green. So when you draw a line through the incenter to the opposite side, a a line through a particular point to an opposite side in a triangle is called a Cebian. And the three cens in green make what's called the N central triangle. The contact triangle is the points where the. Where the circle is tangent to the triangle where it makes contact. That's the contact triangle in red. So this is something you can do with a triangle. You can also do it with a tetrahedron. So here we have a tetrahedron here. And if we do bisections of the dihedral angles, basically each of the edges here, each of the six edges makes a dihedral angle. If you bisect each angle of the tetrahedron, all six planes will intersect at the end center point for the N sphere, which is tangent to all four sides of the tetrahedron. And again, you can get two, two special tetrahedron based on that. If they go through the end center, you take the CVNs where that where the vertex through the incenter intersects the opposite plane defined by the triangle, you get the N central triangle. And if you look at the places where the sphere is tangent to the four sides, you get the contact tetrahedron for the tech for the original tetrahedron. So those are two special triangles or two special tetrahedrons which have equivalence to each other in 2D and 3D geometry. And that's one thing I've been trying to do recently is what things in 2D geometry also apply in 3D geometry. And not everything does, which we will see. The next thing we can talk about is the centroid where the medians meet. And with that you get the medial triangle is the CE vians of the centroid in red. Here we've got the original triangle in gray. In red we've got the medial triangle where the medians intersect. The anti complementary triangle is the reverse. We have the original triangle in gray. And red is a triangle whose medial triangle is the original triangle. Median is basically a center of mass. With a normal triangle it's a midpoint, but for a triangle it's the centroid. So if you connect up the centroids of each of each face in a tetrahedron, you also get a medial triangle. The each each of these points is the centroid of the face. You can also do the anti complementary tetrahedron, which is the reverse. Black is the original tetrahedron and in red is a larger tetrahedron. And if you look at the centroids of each face, you get the original tetrahedron back. Another point is the circumcenter. And here it gets more complicated. In black we construct the circumcenter. The perpendicular bisectors intersect at a point which is of the circumcenter. And it is the center of a circle which goes through all three vertices of the original triangle. In green we have the medial lines and the circum medial point is the is the green lines going through the centroid to the opposite opposite arc of the circumcircle. And that makes the circle this the circum medial points or the circumdial triangle. The tangential triangle is tangent to the vertex points of the circumcenter. So we have the original triangle. Here we do tangency points. And the the red lines are tangent lines at the vertices. And you get the in red, what's called the tangential triangle, although you can do that with the tetrahedron as well. But instead of perpendicular bisectors, you get midplanes. There are six midplanes which intersect at the center of the circumsphere. And the circum medial points are the intersections of the sphere with the vertices going through the centroid point. So in black we have the original tetrahedron. In red we've got the circum medial tetrahedron. The tangential tetrahedron is tangent to the vertex points on the circumsphere. So in black we've got the original tetrahedron. In red we've got the circum or we've got the tangential tetrahedron. So all of these are inside of a new function on the WFR site, Wolfram Function Repository site, and it's called stubbed tetrahedron. There's also another function there called sub triangle, which we've had for a while. And those two programs will generate many of these special triangles and special tetrahedron. Special tetrahedron. Special tetrahedron. So here's where 2D and 3D break down the the fourth famous triangle center is the ortho center where the altitudes intersect. If we basically drop altitudes through the opposite sides. For from each vertex we get the feet and they intersect at the ortho center there's a no corresponding center in 3D. Sometimes it works in what's called an orthocentric tetrahedron, but it doesn't always work. When they do intersect, it's the same as the Monge point, which we'll cover on the next slide. Most of the time though, they do not intersect. However, we can still connect up the feet where these altitudes meet the opposite side and the connection of the altitude points gives the altitude tetrahedron. We can also do the circumange tetrahedron where the mon point intersects going through the vertices through the Monge point to the opposite edge. For the circumsphere, that's one I I tossed in. So the Monge point, if you connect up a midpoint of one edge so that it's perpendicular but to the opposite edge, you get what's called the midplane. And Gaspar Mange noticed that all six of these planes intersected at what's called the Monge point in any tetrahedron. And this is very similar to the orthocenter of a triangle. So this is considered the the fourth important tetrahedron center. The the Mange point. If you reflect a tetrahedron by the Monge point, you get this or by reflect by the centroid, you get this attractive figure. I'm not sure what this is supposed to be called, but I'm calling it the negative tetrahedron in the function. So if we have this property with opposing edges of a tetrahedron, so that the sum of squares for each opposing edge is the same, then we have the property that all four of the altitudes intersect and that's called the an orthocentric tetrahedron. So it's a special type of tetrahedron. Also if you join up the midpoints of each edge, what's called the bimedians, if the bimedians have the same length, then you also get the property that the tetrahedron is orthocentric. However, if either of those two properties is false, where the bi medians are not the same, or if the sum of the squares of opposite edges edge lengths is not the same, then it's not orthocentric and the the altitudes will not intersect. Let's see. The next important point in triangles is called the the center of the nine point circle. So if you have the, the feet, the feet where the altitudes are and the midpoints and the midpoints of the vertices to where the altitudes intersect. So in yellow we've got the midpoints to the ortho center. In green we have the altitudes and in red we have the medial triangle. So it's the Orthic triangle, the medial triangle and the Euler triangle. Euler was the one who made this yellow triangle famous. Those nine points are what's on the nine point circle and it's tangent to the three x circles. There's something similar with tetrahedra. The medial points of each triangle, the 1/3 distance points between the Monge point and vertices. So we'll call those Euler points. So this is the Euler tetrahedron, the medial tetrahedron and the altitude tetrahedron. We get these 12 points which are also on a sphere. And there's the 12 point sphere of a tetrahedron. And here's it showing in a different way as three tetrahedra all having the same circumsphere. This, this sphere is not tangent to the 4x spheres. And we can check that the Euler tetrahedron, the Euler projected tetrahedron and the medial tetrahedron all have the the, the same. These are the, the four tetrahedron, but these are the three tetrahedron. But they all lead to the same sphere. Oh, what is my comment here? The sphere tangent to the x sphere is close to the 12 point sphere. I call it the, the. I'm going to call it the Fuhr back sphere for now. Feuerbach was the one who proved that the nine point circle and the in the. The in circle and the. The nine point circle have a single tangentry point which could call the the fewer point back point. Since there is no name for this sphere, I'm just calling this the fewer back sphere for now. Just tangent to the 4x spheres. So there's, there's extra points that we could make out of this, but I haven't done that yet. I, I actually. But the tangency points in the sub tetrahedron function, I call that the fewer back tetrahedra tetrahedron for now until I come up with a better name for it. And these are the these in in the center there, that's the sphere tangent to all the x spheres. The median point we have the centroid here we can do the angle bisectors and if we reflect the lines we get what's called the Simidian point, the reflections of the medians in the angle bisectors. That point minimizes the sum of distances squared. And we can, we can check that. The reflections of the median median planes and the dihedral angle bisectors also works. So here's, here's the same thing. In the, in the tetrahedron, G is the centroid, K is the simidian point and we can get the the of the median point to get the stamedial tetrahedron. X circles and X spheres lead to a lot. So here we have a triangle in the middle here. That's the original triangle. We can make the X circles and if we connect up the points we get the X central triangle. If we look at the tangency points we get the X touch X touch triangle in green. We can do the same thing with 3D, but it's a little bit more complicated. Here we have the x spheres function generating the 3x spheres. The tetrahedron in red and in gray is the X central tetrahedron and in blue is the X touch tetrahedron. And these are all part of the subroutine function on the wolfram function repository 1 I wasn't able to get quite working yet because the code is hacky. Here's somewhat hacky code for picking out the the planes tangent to the to each of these planes is tangent to the three of the spheres. So however, the original tetrahedron also has planes that are tangent to three spheres. It's actually tangent to four spheres. But when we're generating solutions. For example, there are two planes. Tangent to three of these spheres, and there's probably more planes. It would be nice to have a full set of solutions. But for these particular planes on the outside, the four planes intersect to get the X tangent tetrahedron. And while I was making this, this talk, I happened to notice that if you connect up the points of the X tangent tetrahedron and the original tetrahedron, they all meet at a point right over here. You can barely see it. But at the center there we get a new tetrahedron center. Now I'm surprised to see this because with, with triangles, if you connect up three lines that are related to each other, there's a pretty good chance that those three lines will intersect at a point. And there's thousands of triangle centers where people have noticed. Well, there's these three lines correspond to each other. And when we connect up the three lines, they all meet at a point that happens all the time. But with tetrahedra that almost never happens. There's, there's like eight occurrences where that's known to happen. Well, this, this seems to be a new one. The, the concurrence point for the X tangents, tetrahedra and the root. And the original tetrahedra seems to produce a new point. And we can check that with a different tetrahedron. And here that is here, I've tried it with some others and it always seems to work with, with tetrahedra of any of any type. When you, when you make the X tangents tetrahedra and connect up the corresponding points with the original tetrahedron, they meet at a new point. But I haven't figured out what that point is yet or why this works. And I haven't proved it yet either. So that's a, a new tetrahedron center that was found last night. So now I'm going to switch over to special triangles and special tetrahedra of various sorts. So here we have a hexagon. Just take vertices 1, 2, 4, or any two neighboring points and skip over one to get a third point. Well, it turns out if you reflect each vertex on the opposite edge, for example, in green here, you get this point here. From this point, when you reflect on this edge, you get this point here. And when you reflect this edge on that edge, you get this point here. Well, it turns out you get another triangle which is similar to the original triangle, but twice its original size. Exactly twice the area. And it turns out that there are intersection points from the diagonals which lead to all but, all but one of the points of a new heptagon, which is twice the area of the original heptagon. And it turns out that this is the one of only two triangles that have this problem where the reflection triangle, this, this process of flipping all the vertices on the edge that gives what's called the reflection triangle. And there's only one other triangle that has a similar reflection triangle. That's this right here. And these have a size ratio of two over the golden ratio. And an unsolved question is, is there a tetrahedron that has a similar reflection tetrahedron? I only spent about an hour trying to solve this and this is the best I got. This is 0.009 away from being similar to each other. We've got an original tetrahedron over here and we've got the reflected tetrahedron over here where we reflect each vertex by its opposite triangle, the plane of the opposite triangle. And when you do those reflections, you get a tetrahedron which is very similar to the original. So this is an unsolved question. Is there a tetrahedron with a similar reflection tetrahedron? Another set of special tetrahedra is ones which will fill space. And for two cases we get something weird called an ortho scheme. So here is a tetrahedron divided into eight parts which are all similar to the original tetrahedron, all 18 size of the original. But here's one way of doing it, and that's ortho one. And here's a second way of doing it. This is Ortho 2. So these are the two Ortho schemes that I know about. There's only five tetrahedra that fill space. Here's the first one. We can divide this skew prism up into what is that, three parts to get what's called Somerville 1, Baumgartner T or Hill I 160. Here's another Somerville 2. This is a way of dividing up a cuboid into what is that? Six similar parts. Here's a third solution. A way of dividing up a. A cube into what is that? 12 parts, I think. And another prism dissection. Here is 12 tetrahedra making up a prism which will file space. And here's the last one generating another prism which will tile space. This, this is the, the weirdest of, of the space filling tetrahedra. So these are all special tetrahedra. There's also the, what came to be known as the Conway tetrahedra. So here's here's something else. First, we have three integer vertices. And for Zach, you can make a, a space filling fractal out of the ortho schemes. So here we have four points that are integer coordinates and the euclidean distances are 6, 7, 8, 9, 10, 11. So that's kind of a special tetrahedron. I'd like to find more of those because when you have, and I use this several times in this, in this talk, when the edge lengths are nice, sometimes the generated special tetrahedra and special tetrahedron centers also have nice coordinates. But not always. There's always ways of making things more complicated, but usually if the edge lengths and things are on integer coordinates, usually things work out more neatly. And another thing in addition to rational edge lengths is whether the dihedral angles are rational in terms of PI. And John Conway wondered what was the full set of tetrahedron where all of the dihedral angles were fractions, rational fractions of PI. And it turns out that a recently published paper, this paper here, Space Vectors forming Rational Angles 2020, did a complete compilation of all of the tetrahedra. And it turns out that if you use an icosi dodecahedron or a three by three lattice, you can get all but three of the tetrahedra that have these properties where all the dihedral angles are nice. And there's three more that can be derived from this tetrahedron. These are the points that generate it. And the, the dihedral angles are right here. If we go. This community page and I'll be posting this on the link, you'll be able to see all the links. This gives a the solving method for finding this particular tetrahedron. Let's see. Solid angles. Well, I might as well read this. During an eclipse, the moon and sun seem to have roughly the same size viewed from Earth. That's because the field of view is the same, even though they're very different distances apart. And that field of view is called the solid angle. The unit sphere around the observer has a solid angle of 4 PI steridians. And that's the same as the surface area of a sphere. And for a tetrahedron, the solid angle is defined as what an observer at a vertex sees when looking at the opposite triangle. And one thing I thought was that the smallest solid angle corresponded to the smallest of the four triangles. But that doesn't seem to be the case. There are weird tetrahedra that that violate that when two planes intersect, the angle is called the dihedral angle. So here we've got a tetrahedron for this special tetrahedron, the solid angles are 1 12, 1 6, 1 12th and 1/2, all times PI. And the dihedral angles are 1/2, 1/3, 1/2, 1-3, 1/2, 1/4. So this is a Conway tetrahedron. And there's a property of tetrahedron. First, for a circle, if you add up the three angles of a triangle, you get 2 PI. If you multiply by two. So twice the sum of the angles of a triangle is 2 PI the full circle. Twice the sum of the dihedral angles minus the solid angles is 4 PI the full sphere. That's the corresponding angle sum formula for a tetrahedron. Twice the dihedral angles minus the solid angles is 4 PI. And we can look at various special tetrahedra. Here's another one though. This is also a Conway tetrahedron. And also this. So I'm going to try to compile a complete list of all the 3059 Conway tetrahedron for a database that's on my to do list. Let's see here. Here's another special tetrahedron. We have these values here are powers of a given root. So for this one I've picked the root which is the square root of the Tetranacci constant 1.39 or 1.38837 and so on. So 1 point the detector Nazi constant to the power of zero gives this edge length. Power of one gives this edge length. The power of two gives this length and the power of three gives this edge length. So this is what's called a similar hedron because all four of these triangles are similar. The two smaller triangles with the zero edge, if you multiply each one of them by the root you get, you basically pump up 1, 1 for each. So instead of 0, 1, 2, you get 1, 2, 3. So this is a special tetrahedron built up of similar triangles. And you can change the root and you get different tetrahedra. But this particular one, if you put it on the ground and you roll it, there's only 12 orientations that you get and it makes this hexagon. So a Tetranachi tetrahedron, if you roll it will only have six orientations if rolled on a plane. And this was not known apparently that there are tetrahedra with this sort of property. Let's see the. There are currently 53,000 triangle centers and here's the the first six of them. And we covered them earlier. Some years ago I decided to see what these triangle centers look like. And this is an Animation of, of a triangle warping. And you basically see how the, the triangles, how the triangle centers move around. So this is just a few hundred triangle centers. And what I started with at the, at the start, I think. Somewhere here, Here's a whole bunch of tetrahedron centers for a given tetrahedron. It's harder to see what's going on when you're in 3D. So basically I had to stick to one tetrahedron to generate these 140,000 different tetrahedron centers. And I described how I got all those in the talk that got mentioned earlier. The Euler line goes through the circumcenter centroid, nine point center and orthocenter. And the isogonal conjugate of the Euler line is the Jerebec hyperbola. These have these positions in relationship to each other. And you can basically skew this line. This point is to here is half the distance from this point to here. It's always this -2 1, -2 014 relationship. We can do the same thing in tetrahedral. We get the Euler line. We also get a Jirabec hyperbola. But the, the incenter and simidian are off the Euler line, but they're on the hyperbola. However, the positions are slightly different instead of what they were before. Now it's minus 20232 and we use the Monge point instead of the orthocenter. I mentioned Stevens earlier. You have a given point and it goes to the opposite edges. I plan to make a Stevens function or that'll work for either triangle or tetrahedra. Here's a another new center I found a little while ago. Opposing edges. If you define them like this, you get a constant. It's the same for each pair of edges in a tetrahedron. And it's called the dihedral constant. And you can take another point so that if you connect up the triangles with that point, each of the four triangles with that point has a dihedral constant which is the same for all four faces. And that's called the dihedral constant center. And it turns out this point does exist. That's another point I plan to add to the database I'm working on. Here's the. Here's the theorem in action as you move, as you adjust the the tetrahedron, the dihedral constant of the tetrahedron is always the same. Another point which is shown to exist in the tetrahedron is Parallelian's point. And basically we're going through a single point. So that the areas of the four triangles is the same and the each triangle is is parallel to the given face and going through the equal parallelians point. There's not many tetrahedron centers known. So we've been trying to build up a little database for when they work. And I've got a demonstration that has many of these tetrahedron centers and I need to update that one to include many of these new points and to update the code to use WFRS that we've been putting together. We also have isogonal conjugate resource function. I'm going to be adding a isotomic conjugate resource function soon. And they work with both a tetrahedron and a triangle. It turns out there there are four tetrahedral points and eight self conjugate points. And this is a diagram of how they connect together. One thing I need to check is whether that special point from the X tangents tetrahedron, whether that given point has anything to do with any of these self conjugate points. And I need to describe everything in Ariel or very centered coordinates for all of these centers that important centers that have been found. This is kind of a discussion for how barycentric coordinates work. And the barista coordinates can also be done in a tetrahedron. For example, you've got this given tetrahedron here and a given point the areas of the forward generated tetrahedra compared to each other is basically the areal or barycentric coordinate. The volumes have the 44 volumes. Assuming that the original tetrahedron have very all volume 1. These would be the volumes of the 4 tetrahedral generated by that point. And once you have areal coordinates you can do fairly amazing things like here are generators for areal coordinates and when you connect them up you get 149 points and 241 lines of five points. And trilinear coordinates are the same as Ariel, but they're divided by the opposite edge I believe. And in a tetrahedron they're divided by the opposite faces. And there's some odd things with trilinear. So. So that's a project I'm working on is getting all of these defined in Arial and and trilinear coordinates for each point. Here's what they look like as triangle functions. The centroid is if you have the triangle with sides A, B and C, the centroid has trilinear coordinates of 1 over A, 1 over B, 1 over C. The centroid is well defined for a tetrahedron. But the formulas are much more complicated for a tetrahedron. So I need to basically get all these into Mathematica. That's a project that I'm working on and I hope to have within a few weeks. I'll have to have this animation that I've shown a few times as a much more dynamic object and that's what I had prepared. We can get into the unsolved questions which is basically everything. For example, what is the tetrahedral formula for the fermo point? We've generated many special tetrahedra, but there's various that are not listed yet. It'd be nice to have more special tetrahedra added. If anybody has code for any. I'd be glad to add that to the subtetrahedron function. And if there are special lines or parabolas and things like that, those would also be good to have as a function. I mention I'll show the sub triangles function and also special triangles. If there's other special triangles that could be added to this, let me know and we can try to add them to this. So I plan to make a database of the Conway tetrahedra. Plan to expand sub tetrahedron. I'm going to be putting together functions, isotopic conjugates, CVNs, plane line intersection and some other support programs. If anyone has anything that works in both 2 dimensions and 3 dimensions for triangles and tetrahedra that we don't have yet, let me know about it and I'll try to come up with a function for it. Here's my address and I will be glad to take any questions that people have. Go ahead and post in and YouTube or wherever you're listening from and I'll be glad to take. I'll be glad to try to answer them. Does anyone have any questions. For space filling? I'm trying to come up with things similar to. Let's see here. And I talked about this in a previous talk where. So here's. Here's a space filling triangle which was unknown. Shattering the plane is what you'd want to look for. And here are five space filling triangles that were unknown. Each is. Each is generated by. By similar copies of itself. One is based on two. One is based on the golden ratio, one is based on the super golden ratio. One is based on the chi constant and one is based on the plastic constant. And all four of these values have the property that they are. There are sums of infinite series of themselves. In reciprocal 2 is the sum of 1/1/1 + 1/4 + 1, 8 and so on. The golden ratio is the sum of the same sort of thing, but skipping the first power. The super golden ratio skips the first two powers. The chi constant skips the first three, and the plastic constant skips the first four. And you get these polynomials and values. And it turns out that all five of these lead to special fractals. I found all of these by asking the question is there a way of surrounding a point with similar triangles? And I wrote a program wheel powers. It's on the demonstration problem triangle wheels. And it wasn't too hard to solve. And when I found this, I figured I could extend that to come up with a similar triangle solution. And then it worked. And that's there. It is there. And that's what I did with all of these. There was a weird way the triangles interfaced and I could extend them to work out. So the question is, is there a way of surrounding a point with similar tetrahedra? And that's an unsolved question. I'm not sure how to solve it, but it shouldn't be too hard to solve. And if there are solutions to that where similar tetrahedra will make a can surround an angle, then there's probably ways of generating some really weird new fractals that have never been seen before. So that hopefully answers Zach's question. You'll probably want to look at Rousey fractals to see what's known about these self sum of things. Rouzi fractals R A, U, Z, Y have lots of weird properties and they're basically generated by simple matrices. So it's not too difficult to come up with new Rousey fractals. However, most of the interesting ones tend to go to those five constants. Any other questions? All right, let me see if we can find. All right. Many thanks for attending my talk.