Mathematical Games Hosted by Ed Pegg Jr. [Episode 5: Tetrahedral Centers]

Okay, I'm Ed Peg Jr. And I'm going to wait a few moments for other people to come in. And this is a talk about tetrahedral centers. And I found various new things with tetrahedron centers in the course of making this top because there's basically nothing known about them. But I'm going to tap dance for another few seconds while people stream in and then we'll get started and feel free to ask questions in the chat at any time and I'll try to address them as they come in. So let's see, we can go ahead and start this with the demonstrations project as many things on triangles and tetrahedra and I'll be using some of those in the course of the talk here. But for many of the items here, if it's based on a triangle, you could probably go to either demonstrations or to Math World or the Function Repository and there will be a, a demonstration or code relating to that particular topic. All right, and for initialization, I've got a whole bunch of different programs prepared that I'll be using in this talk. This notebook will be available later on the community site. All of this code will be listed for availability there. And you could play around with all these various things that I've shown here. So let's talk about a triangle center and then I'll mention the related tetrahedron center. For example, for the incenter, that is where the angle bisectors of a triangle intersect. And this was known to the ancient Greeks. The first four triangle centers were all known 4,000 years ago. So for example, if you take the angle bisectors, you get the point where a circle tangent, all three sides can be drawn. For the N sphere, it's a little bit more complicated. You need to take the, the dihedral angle bisectors as shown here, and those give the end sphere of a point. You can, you can think of these pairs of two lines as making the dihedral plane bisector. And it turns out they all intersect at a single point. If you reflect the incenter in the dihedral angle bisectors, you get the same point back and we'll get back to that later. The centroid is where the median center sectors. Median is where you take the midpoint of a side and connect it to the opposite vertex. So here's a, a triangle with the, the three intersecting lines meeting at the centroid, which is the center of gravity, the center of mass. And something similar happens with the tetrahedron. But instead of going to the midpoints, what you do is you go to the centroid of the opposite faces. And the centroid of a tetrahedron is the concurrence of the face centroid vertices pairings. And this is one of the few tetrahedron centers which is based on a. A point and opposite faces. These are the thieves of the centroid on each face. And it's. It's very rare that it actually works for a tetrahedron. As we shall see. The circumcenter of a triangle is where the perpendicular bisectors intersect. Here we have a triangle and we've drawn the perpendicular bisectors for each side. For almost all of these there's a really nice demonstration that, that you can play with. So here we have the circumsphere of this triangle. For the tetrahedron. What you need is the mid planes. You basically take a. A a segment of the tetrahedron and you make the plane perpendicular to the midpoint. And it turns out that all six of these planes will intersect at the. At the circumcenter of the tetrahedron. And here we have a. A picture showing that. The. The mid planes will all intersect at a. At a central point. And then we get to the first one where things start to break down with the ortho center. The, the altitudes will intersect in a triangle. And here we show that here. And the altitude is the line from a vertex perpendicular to the opposite side. There's, there are tetrahedra where it works. For example, here the, the. The. The four altitudes of this tetrahedron all meet at a single point. And this turns out to be the same as the mons point. When it works, however, there's. There's a rule that needs to be met for this to work. The opposite edges of the tetrahedron have to be perpendicular to each other. And that's an actual, actually a rare property. Most of the time the altitudes do not intersect. Here's another tetrahedron where I've drawn the four altitudes. They don't come anywhere near to intersecting the green point at the center there. That's the Monge point. So what is the Monge point? That is if you take the midplanes perpendicular to one edge and concurrent with the midpoint of the opposite edge, you get six midpoints. And these will intersect at what's called the Mongen point. So for example, here we have a tetrahedron. And here's one of the planes here. It's. It's perpendicular to this edge over here. And it goes through the midpoint of this opposite edge. So there's one of the midplanes, here's another midplane, the third, the fourth, the fifth, the sixth. So all of these six midplanes will intersect at the Monge point. The fifth well known triangle center is what's called the nine point center. And that is if you take the altitudes, for example, this line here is an altitude. You can get these three altitude points. And if you take the, let's see, the midpoints, these, these three points here are the midpoints of the edges. And also draw the 1/2 point from the intersection of the, the lines, the orthocenter. If you take these midpoints from there you get these nine points and they make the nine point center. And it turns out that it is tangent to the 3x circles of the triangle, which I will get into. However, it's more complicated. For a tetrahedron, you have the medial points of each triangle and those are the centroids of each face. One third distance point between the Mon point and the vertices, which are known as the Euler points. And then the three feet of the Euler points. And I illustrate all that here. So here we have, in red is the, Let's see, in red are the centroids of each face. In blue you can see there's, there's blue lines meeting at a, at a central point. The 1/3 points are given in blue. And then in yellow and green are the, are the feet of those of those points. When you combine those, you get these 12 points that are all on the sphere. So that in the center There is the 12 point sphere. It turns out it is not tangent to the 3x to the 4x spheres. So this is a rather complicated point, but it does still have a lot of properties that share with the nine point circle. And this is the 12 point sphere. And that's pretty much where research into tetrahedra centers stopped. You've got the four classical Greek points, the monge point and the 12 point center. And pretty much that's it for what's known about tetrahedral centers. But let's go on. There are a few that will include this. The six point, the median point for a triangle. It's the reflections of the medians in the angle bisector. So if we take the centroid here and we reflect lines through the angle bisector, we get this new point here, K, where all the lines meet. And this is what's known as the Simidian point. And that minimizes the sum of distances squared if you go to the faces. So it has a construction method and it has a nice property. It's a, it's a minimal solution to something. And we can, we can check that reflections of the median planes in the dihedral angle bisectors will also produce a six, a minimal least squares point. And that's also known as the median. So here I've drawn a picture here where I've taken G, which is the centroid, and K gives the median. And I've drawn all the various lines and whatnot that are, that are used for that. So this is another complicated point that's, that's known for tetrahedra. So I've mentioned six triangle centers and here they are. They're the incenter, centroid, circumcenter, orthocenter, nine point center and the median point. And the encyclopedia of triangle centers, which I'll show here. We've got thousands of centers. So this is. Oops. It takes a while for this site to work sometimes. Let me see if I can scroll down here. I'll just put it over here until it wants to cooperate. It doesn't want to cooperate. But at that site there are thousands of Friano centers listed. About 50,000 if I remember. 53,769 centers are currently listed there. And let me show you a picture of some of them here, if that works. Here's a triangle here with an animation of a few hundred centers. As the triangle moves, you can see these various points and these, these points correspond to various triangle centers. You can see outlines of a, of a circle here if you look closely. Let's see here. And my computer is freezing. Oops. Hopefully I'm still live. And these are some of the points given in, Given in the Encyclopedia of Triangle Centers. The Euler line goes through four of the points. It goes to the circumcenter, centroid, nine point center and orthocenter. The, the conjugate of that which we'll get into is what's known as the Jarabec hyperbole. And you can see it up here. Here's the formula for it. So the Euler line goes through the circumcenter, which is this, the center of the circle that goes through the vertices. Vertices this, the centroid, the orthocenter, which is the intersection of the altitudes. Another thing that Euler found. The, the incenter in yellow here and the median point are off the Euler line. The positions are interesting. This is one particular triangle that has everything at nice vertices. You have this relation. If you know the, if you know the, the centroid and the circumcenter then going in the opposite direction twice that gives the orthocenter the the distances relation between these four points is is fixed one twice that the opposite direction the next one four times that of the other direction is the next one. There's also an Euler line and a tetrahedron and it gets more involved to draw here are the four points again. The N center and the median point are off I and K but OGTNM are still in the same order and they still make a line. There is still a jurisdbach hyperbola given in brown and the points have a different distance relationship in the tetrahedron here the centroid is the midpoint of the Monge point and the circumcenter and the the 12 point center is at a 2/3 relation instead. We can also build CBNs for the points. If you. If we start with a triangle and put down a point and draw lines through that to the opposite point we get a. We get intersections with the opposite sides and those intersections are the CVN points based on whatever point. So if I had this set up as a manipulate I can move this point around that the black dots would move around. Let's see here. In a tetrahedron you can basically pick a point within the tetrahedron and generate the cebians which intersect the opposite faces. So this is generating the Stevian tetrahedron for a given point in a tetrahedron. So that's a bit of. That's the function I plan to add to the Wolfram function repository soon. The X circles. So here's a triangle. We can generate the tangent circles that are tangent to three sides. But instead of the end circle which would be on the inside, we generate them outside. And each triangle has 3x circles. For the tetrahedron we get 4x spheres. So here's a sample. A tetrahedron at the center and each one of these spheres is tangent to four planes. And there's a. We have a. Somewhere here we have a resource function called X spheres and that gives both X circles and X spheres. It'll actually work in any dimension. I think I know it works for dimensions two and three. We can also do what's called isogonal conjugates which I mentioned before. If you take the angle bisectors and reflect a point, you get a new point. So for example, if we take the blue point here and reflect the lines in the angles bisectors we get this new point here in green. And if you do the isogon to concrete again you get back to the blue point. If you take the ISO gonal conjugate twice, you basically return to the same point you started with. And you can do that for any point. For example, here we're doing it for the point zero one or actually this is the triangle and here's the point. But one of these is 3, 3. The other is the is this point here. The centers of the N circle and X circles are the only four self only four self conjugate points. So if we take the X spheres of a triangle and also add the N sphere. So here's the coordinate of the N sphere, these four points. If we look at the isogonal conjugates of these four points, we get the same points back. They are self conjugate points. And you can kind of see that because the encircle is based on the angle bisectors. If you basically take this point here and reflect it, you get the same point. And the X spheres have a similar property. In a tetrahedron, instead of reflecting by the angle bisectors, we're doing the angle bisectors of the dihedral edges. So there's actually six reflectors. But it happens that all six reflections will still intersect. And if we look at the X spheres and the N sphere, these five points are also still self conjugate. And here we have a test here. These are the for this tetrahedron here, These are the 5x circles and in and X centers and in center of the tetrahedron. And these are. These five are all self conjugate. But it turns out there's three other points. And here's a picture. So the green points are the original tetrahedron. The center red point is the in center. The outer at red points are the X centers. And in blue are the three other self conjugate points. So basically for any tetrahedron you can get a triangle of three self conjugate points which are not part of X spheres of the N sphere. And it turns out in I've added black lines. They line up curiously. This green point, this red point and this blue point are all on a line. And I've added lines for all the places where three of these points form a line. They make this curious structure for any tetrahedron. All right, edge angles. If you take twice the sun twice, this should be. This would be triangle angles. But also you can actually edges was correct. Based on the given size of a triangle, you can calculate what the angles are based on the edges. And we've got a resource function that could do that. But also twice the sum of the angles in a triangle is 2pi, which I'm sure everyone knows here. For example, if you take the 30, 60, 90 triangle, you get the angles 30, 60, 90, and these are the edge lengths of the 3060, 90 triangle. Or you can multiply by this by two and you get the same thing for a triangle in the plane. You could also say you got angles A, b, C. You get the full circle. In a tetrahedron, there's something similar happening twice the sum of the dihedral angles minus the solid angles is 4pi, or the entire sphere. And we can do the same thing where you have a given set of edges and that will return the the angles of that tetrahedron. And there's a nice demonstration where you can play around with different edge lengths and see what the dihedral angles are. For example, in this particular triangle and this particular tetrahedron, the values here on the edges, those are what the dihedral angles are in terms of PI. And at the corners, these are the solid angles. So if you add up these six values here and subtract the solid angle values, you basically get four, because we're dividing by PI here. So that brings us to how we track where all these points are. And usually aerial or barycentric coordinates are used. So for example, for a 3, 4, 5 triangle, we can, we can consider where is the point 1, 2 in relationship to that triangle. And the aerial or very centered coordinates are the following. So, so if we take this triangle and we add this point, we generate three triangles and the relative areas are 1, 6, 1, 3 and 1 half. If we and those sum up to one. If we take those barycentric coordinates and use the original triangle, we get the original point back very centric. The very. So this is our resource function. The very very centric coordinates function is the same, except it doesn't have this reversible property. However, a dot product here we basically set up the same triangle, same point, and you get the same very center coordinates. If we multiply this point by the original vertices, we get the point back. So the, the reversibility to reverse these, usually you just need to take a dot product. The minimal sign in aerial or barycentric coordinates determines if a point is inside, outside, or on the boundary of a given simplex or triangle. So here we have a triangle in yellow. The blue points are on the edge, the green points are inside, and the red points are outside. And basically all I've done here is looked at the sign of all the various barycentric coordinates or areal coordinates. And this is basically Pick's theorem. If we add up the number of interior points and half the boundary points and subtract one, you get the area of the triangle. In this case, this triangle has Area 14. Another thing we can do with very centric coordinates is make nice pictures like this. Here's 25 trees and 18 lines of five trees as found by. Can't remember who found this. And no melkeys. No melchized is the one that found this particular configuration. In a tetrahedron we can do the same thing. We have a given tetrahedron, we have a given point and that point will generate four tetrahedra based on the four faces and the relative area. Relative volumes of the four tetrahedra are these as follows. So if we take out this point and add in this one, these four points will give a tetrahedron which has relative volume minus 13 over 30, which means the point is on the outside of the tetrahedron. These are signed volumes. The the triangles use signed areas to determine whether a point is inside a triangle or out. Let's see the the volume and here. Here I show it the volume of this simplex divided by the volume of that is 13 over 30. And the very sacred coordinate functions works the same way. And you can also dot products. Pick's theorem does not apply in 3D. There's something called the Reeve tetrahedron or it's actually a whole infinite family of tetrahedron. You've got these three points and this one which is variable with an R. There's always a fixed number of points on the inside and edges. But you can get an arbitrarily high volume which is basically R over 6. And due to that, there's no equivalent of Pick's theorem in 3D. And we can do the the same trick for using very centric coordinates to make a whole bunch of lines of. Trilinear coordinates are similar to Arial and barycentric coordinates is that they're the relative distances to the side. For example, this green point here has relative distances to the sides of 4 over 49, 9 over 49 and 36 over 49. And you can basically get this since since the very center coordinates are the areas of the three triangles, you're just looking for the height you basically divide by the edge lengths to get these trilinear coordinates. In a tetrahedron, the relative distances to the faces provides the trilinear coordinate. So for this tetrahedron here we can look at the distances for this point and let's see, does this actually work? That's not the right response. Yes, these are the trilinear coordinates of this point in reference to this triangle. We can also get coordinates for the N sphere and X spheres. And in trilinear coordinates they have interesting values. The center point of the N sphere, the relative distances are all the same, 1/2 that, they're all 1/4 because there's four distances to the sides or to the faces. But the in sphere is right at the center and tangent, so they all have the same distance. So the relative distances are all the same. For the X spheres, there's always one face that's one sphere that's on the outside and three on the inside. So you have negative values for one of them. So we can look at areal or barycentric coordinates in triangle functions. And it turns out that the incenter is just based on the actual sides A, B and C. The centroid, everything is the same. The circumcenter, you can use this. And if you use this formula and go backwards, you get the circumcenter of the point orthocenter is based on the secant. The nine point center is based on the differences of angles times the edge. This median is the edge squared. The spiker center is the sum of the other two sides. So if you use these formulas, you can quickly build a whole bunch of triangle centers. And the Encyclopedia of Triangle centers with its 50,000 plus points has a lot of these. Triangle formula is given in a tetrahedron. Much less is known. The incenter works the same, the centroid works the same, the median point works the same. But these other three, they're much nastier. Let's see here. The trilinear coordinates is basically the same as the very centric coordinates, but we just divide by the edge lengths of the sides. So basically here we have ABC. If we divide ABC by ABC, we get 1, 1, 1. If we just divide 1, 1, 1 by ABC, we get 1 over A, 1 over B, 1 over C. All of these triangle centers for the trilinear coordinates are just basically dividing by A, B and C. But it turns out the trilinear coordinates tend to have better, slightly easier formulas than the barycentric coordinates because they usually have an extra A term, B term and C term in them. So usually the trilinear coordinates are preferred for triangle formulas. Again, for tetraplanar coordinates, instead of trilinear coordinates, there's four planes or so they're called Tetraplanar. Let's see here. Not much is known, however, we do have formulas for the circumcenter, but it's pretty nasty here. The. You basically need a diagram to see all the things that are going on. If you look up tetrahedron centers, you can see a post post I made on Stack Exchange that kind of describes that. But if I plan to write this up in a community post after this talk. So here's the somewhat nasty formula for the circumcenter. It's. It's complicated. The Monge point also has a complicated formula based on the edges. And the nine point center also has a complicated formula. Whether these are. Whether these formulas can be simplified is unknown because virtually nothing is known about tetrahedra centers or formulas. So Kimberly centers from the Encyclopedia of Triangle Center. The site was developed by Clark Kimberline. So the triangle centers with their numbering scheme is often named after Clark Kimberling. So the Ortho center x4 for a triangle is. We have a Kimberly center function or we should send this. I'm going to. I plan to add this to the Wolfram function repository soon. We can show the first 29 of these centers here. So somewhere here, here's the orthocenter here. These numbers correspond to which number they are in the numbering system by Clark Kimberly. For tetrahedron centers, we can basically get to 6. There's nothing equivalent. There may be something equivalent to X7, which is called the jargon point, but it's not well defined. There's a whole bunch of tetrahedron centers which don't correspond to. There's a whole bunch of triangle centers that don't correspond to anything in a tetrahedron. So there's nothing numbered yet. And I plan to get to that later. However, we can generate a whole bunch of centers. So for this I started with 11,000 triangle centers and basically calculated for each face what the point would be on. On the tetra. On. On the face of the tetrahedron. And then use that to generate. 11,000 different tetrahedra. And then I use my programs of tetrahedron centers to multiply that by six. So here's 60,000 or so different tetrahedron centers. And you can see there's all sorts of weird arcs and circles and lines popping up in strange places. None of these are understood. This is a brand new picture. No one has ever shown this before. So this is. This is a completely new thing here. But we can. We can go on from that. So if we have a point from tetraplanar coordinates, we can go from the trilinear to Arial multiplied by the side length and then normalize. For cartesian coordinates, we basically take that areal or barycentric coordinate and use the dot product with the original vertices for tetraplanar coordinates to arial or barycentric coordinates, we multiply by the face areas and normalize. And then for the cartesian coordinates, we multiply those areal or barycentric coordinates dot product with the original vertices. So here's an example. We have these points. We can get the areas like so we can get the solid angles. And one thing I tried to do once I generated all these points was here's a triangle function I made up. We take the. The. Oh, I forget what I did here. But it turns out that this particular point which is generated is part of the centroid. It's the end. It's the in. It's the end center of the centroid points. And that turns out to be the end center. And here's another formula I came up with that worked. I took the areas squared and the subsets and it turns out to be the, the median of the x2 point the. The medial tetrahedron. But I also tried about 50 other things and none of them worked. It turns out it's. It's hard to come up with a formula that corresponds to some of these hundred thousand or so points. So we can take a look at the isogonal conjugates of all these points and we get this new picture here. Let's see if I can make this move. And the isogonal conjugate points, all 50,000 or so of them, all also have strange properties. There is curiously very little intersection between the ISO gonal conjugates and the original center points that I generated. If I'd done the same experiment with the triangle, there would have been a lot of overlap. And that's one reason why there's only 50,000 known Triangle Centers. Because every time you try to generate something, it happens to be something that's already been found. With the tetrahedron, it seems like it's a much more chaotic. Every time you try something, it's something that's, that's new. Even if it's just a database of, of points. It seems like there's going to be millions of tetrahedron centers. If, if we do this little project, we can also look at isotopic conjugates which are basically reflections of the median. And here's a code for it. Here's a whole bunch of isotomic conjugate points based on the original 50,000 that I showed. And again we're getting these strange curves. And again, none of them happen to align with the original points or the isogonal conjugates. So I'm going to wait for this to finish rotating. So we can do all three of these. Hopefully this will work. Here we go. Here we have 140,618 triangle centers and it's a mess. I'm generating a a movie that will have a nice clean rotation of this and that'll be posted in the community center sometime soon at the header of this talk. And that will basically show this tetrahedron spinning smoothly with all the generated points. However, there's. Since there's 147,000 points, just rotating it by hand is slow. So we're going to make a nice movie of that to basically show how all these work. And this has never been seen before. There's a whole bunch of curves and whatnot here that aren't well understood. Let's see now. I need to wait for the finish rotating so I can click the forward button in the top. Looks like I clicked on the image a few too many times. And I'm still waiting for this to finish rotating. Okay. One weird property of a tetrahedron is we can take the opposite edges and use this formula here to define the inner dihedral angle. And what we get is the dihedral angle constant. So if we do that operation on opposite is, let's say B and F or C and D. The dihedral constant is the same for any, any tetrahedron. This operation on the edge angles, it's possible to add a new point so that it generates four tetrahedra that all have the same dihedral constant. And this was something I found a few years ago. And I called it the dihedral constant center. And one thing I asked was whether there's a formula for this. And a person by the name of Blue on Stack Exchange answered, yes, there is a formula. And here is. And he gave the gory details of it. It involves. Even for something simple, it involves. For the simple case that I gave, it involved order 27 polynomials. So it was really nasty. It would fill many, many pages. For the simple example, if I give a complicated example, it would be. It would. The formula would basically fill a book. So that's not something we want to actually show. But there is a way of calculating these based on the dihedral angle constant. So that's A new center. That's nice. We can also look at this equal area parallelians point. So here's a quick description. We basically want a point where you draw faces parallel to where you draw through the point parallel to the original faces. And what you want is four triangles that all have the same area. So let's see if we can do this. We have, we have a tetrahedron. And it turns out if we fall follow this formula here. These, these are the face areas. W is the face area of one face. These are the four face areas. If we use this formula to generate very centered coordinates, we get for, for this poly, for this tetrahedron we get this point. So here it is, that sphere of the center. That's the equal area parallelian point of the tetrahedron. And we've got the, the four triangles going through that point parallel to the original faces. And the area of these four triangles is all identical. So this is a nice tetrahedron center. It corresponds to something similar, the equal parallelian point of a triangle where you draw lines parallel to the original edges and all three segments through the point have the same length. It turns out you can do the same thing with triangles and a tetrahedron. So this is one of the cases where there's something known in triangle centers and there's something equivalent in tetrahedra. And I have a tetrahedron center demonstration that shows some of the well known triangle center points. So we can add in various points and they show up. And we can also look at the, the points corresponding on the, on the triangle or the points for the faces. And we can add in say, generating spheres and line. See if that works. So this is a function I plan to expand eventually with more centers and more points, more spheres, more lines. So what are the unsolved questions about tetrahedron centers? Well, it turns out it's virtually everything. There's pretty much nothing known about tetrahedron centers except for the simple cases corresponding to things that date back 4,000 years. One simple one. What is the formula for the Fermat point that's unsolved? I spent a while trying to solve that. I never got anywhere for this talk. I generated hundreds of thousands of points. Which ones have easy formulas? That's completely unknown. We can use simple formulas to generate points, for example, for the solid angles. We could take a look at this as either a very centric coordinate or as a tetraplanar coordinate. Does that correspond to anything or do they have useful points? Is it the minimal solution for something none of that is known. And you can substitute any trig formula. There's millions of different formulas you could come up with. Do any of them do anything interesting? So here's the big unsolved question. What tetrahedral centers deserve letters? So there's already a few letters that have been assigned. And I, I show, I showed a few of these back here. These are the tetrahedral centers with letters. There's only nine. What are, what are the other tetrahedral centers that can fill up the rest of the Alphabet with the triangle centers? The Alphabet is already pretty much filled. But there's so little known about tetrahedron centers that, that basically if you come up with something interesting enough, it might get a letter assigned to it permanently. What triangle centers have tetrahedral equivalents and which do not? That's a big unsolved question. What are the equivalent to the sub tetrahedra? Let me show some of that here. Let's see if I can get that. Let me see if I can show this here. All right. Sub triangles. We have, we have a triangle and we want a subtriangle. And these are some of the subtriangles. So for example, the anti complementary triangle is in green. Here we basically have this triangle in black. This is the anti complementary triangle, the Brocard triangle, tangential triangle, half altitude or thick triangle. And here's a whole bunch of different subtriangles for these. Which ones have equivalent, have equivalent sub tetrahedra? That's something I would like to know. So as a possible project that I'm considering is we would pick a few scaling few that have nice properties. We don't want to use say the regular tetrahedron because basically all the, all the centers would be at the center point of the regular tetrahedron. That's the same problem with the equilateral triangle. For that all centers are either the dead center of the equilateral triangle or it's a point at infinity. You basically need asymmetry for a center to be meaningful. But for the project we would have say four reference tetrahedra. Then for each one we generate millions and millions of points. And for each one we'd have a generation method listed. For example, the incenter of the X99 X777 point would be a generation method. Then we'd have thousands or millions of formula where we're just basically throwing things against the wall and see which formulas match up with with points for all four of the reference tetrahedra. It's quite possible that a formula would match up for a single tetrahedra, but not all tetrahedra. So basically you want four different reference tetrahedra to make sure that something that that seems to work will work in all of them, and then boil down a list of good tetrahedral centers and start building up a database similar to the encyclopedia of triangle centers. So that's what I had, and I'll be glad to take any questions if anybody had it has anything. And I do hope everyone enjoyed the talk.