Tetrahedron Center Functions

Greetings everyone. We're going to wait for the audience, for people to come in and for the audience to stabilize. Up on screen are 140,000 tetrahedron centers found in with various ways. So I'm going to talk about some of that. These are. There's a community post that talks about these. Look for tetrahedron centers and you'll see the community post where I talk about how these 140,000 tetrahedron centers were calculated. And I'll go over that a bit in my talk as well. And I think we can start here. So I'm going to talk about some tetrahedron center functions. I'm Ed Pegg Jr. I run the Wolfram Demonstrations project. I also work on the Wolfram Functions repository and we've added many triangle and tetrahedron related functions recently to the Wolfram Functions repository and also quite a few triangle demonstrations. So I'm going to talk about some of them that apply to both two dimensions and three dimensions. One example of that built in are the N sphere and circumsphere functions which can find the N circle of triangles and the N spheres of tetrahedra as same with circumcircles and circumspheres if you use the circumsphere function. And I really like those two functions because they worked in multiple dimensions. So I wanted to try to develop more of those. So here are some of the functions currently in the function repository now that work in multiple dimensional spaces. Kimberlain Center Clark Kimberlain runs the Encyclopedia of Triangle Centers and there's over 54,000 triangle centers defined at the moment. So that works up to about 80 Kimberly Center 80 and I plan to expand that to perhaps go up to 10,000 sometime soon. Tetrahedron center is a new function. I'll just drag this over here and it has many different tetrahedron centers and I'll talk about some of these in the talk. Subtriangle and subtetrahedron. This one does special triangles based on a triangle. Subtetrahedron does special tetrahedra based on a tetrahedron. And we'll discuss some of these isogonal conjugate flips. Does a mirror based on the angle bisectors or the dihedral angle bisectors. Isotomic conjugates Does a reflection based on medians. Triangle edge angles if you know the edges of a triangle, you'd find the angles. Well, this does that automatically. Tetrahedron edge angles does the same thing. X spheres is pretty much like N Spheres. But spheres on the outside reflect points, will reflect on a line or reflect on a plane. Cesians will find either based on a simplex and a given point. It will find the intersections with the opposite face or line. Find extraordinary lines. If you have a group of points, it will find all sets of three or more points which form a line. Arial and barycentric coordinates work in both two dimensions and three dimensions and so on. So I like functions that are multi dimensional. We've also done a lot of demonstrations. We've got almost 200 tetrahedron demonstrations and 1300 triangle demonstrations at the wolfram demonstrations.wolfram.com so let me start with the N center. That's the first of the Kimberling centers. And it's the place where angle bisectors intersect. This was known to the Greeks that this point here was a concurrence point between the three lines. If you extend out to the opposite sides, you get what's called the central triangle. And that makes this green triangle here. The incircle itself is tangent at the contact points. So the red triangle is the contact triangle. And we use sub triangle, Kimberline, center Cesians to make this diagram. And in sphere we use pretty much the same functions for doing the same with a tetrahedron. In each one of these gray lines here is the intersection of three edge bisecting dihedral planes. And where three intersect you get one of these gray lines. The green triangle. Our green tetrahedron is the in central tetrahedron. The red tetrahedron, where the sphere is tangent, is the contact tetrahedron. And I need to solve things like this regularly. So I finally gave up and wrote a function that would just solve all these sorts of problems for me. The centroid for a line, the centroid is the midpoint. And we can see various ways of connecting up the centroids, which is here in the gray triangle. The red is the medial triangle. And if you wanted a triangle that had the original triangle as the medial triangle, that's called the anti complementary and that's shown in green. The concurrence of the centroid also happens in three dimensions. And the medians, the medial points for the faces are the centroids. So if you connect up the centroids of a tetrahedron, you get the medial tetrahedron. You can also do another anticomplementary tetrahedron in green here. The circumcenter is where the perpendicular bisectors intersect. So in gray here we can find this point here, which is at the center by looking at the perpendicular bisectors. The, the tangential triangle is tangent to the circle at the given vertex points. And we can do the same thing in three dimensions. Here we have a tetrahedron at the center with a sphere around it going through all four of the vertices. And the red tetrahedron is tangent to the sphere at those four points. So the subtetrahedron function calculates the tangential tetrahedron. The orthocenter is where the altitudes intersect and you get the altitude triangle in red. However, there's not a corresponding center this place where the altitudes intersect, they can intersect at what's something called the Monge point. And if you take a look at a tetrahedron, it has opposing edges. And if you go through, if you, if you build a plane that's perpendicular to one edge and goes through the midpoint of the opposing edge, you get what's called a midplane. And there are six midplanes each going through the midpoint of one edge and perpendicular to the opposing edge. And all six of those planes intersect at what's called the Monge point, which is a green sphere. In this case, the altitudes will intersect if the opposing edges have one of three properties. One, if the distances between the midpoints of the opposing edges are all the same, then they will all intersect at a point and also the altitudes will intersect also. If the sum of the square of the lengths of opposing edges reach the same value, then you'll have the same property. It will be an orthocentric tetrahedron. Most of the time though, the altitudes do not intersect and you don't have those three properties. However, the altitudes still exist. So you can still get an altitude tetrahedron from any tetrahedra, The X spheres or sorry, this is the nine point circle. You can look at the altitude points, the feet of the altitudes in green. You can look at the midpoints in red. You can also go to the orthocenter here and pick points that are 1/2 of the way from the vertices to the orthocenter. That's something Euler noticed. And all nine of these points are what's on the nine point circle, which is tangent to the X spheres or X circles. And that's what's shown here. We have the. The. The three generated the. The altitude triangle, the medial triangle and the Euler triangle all have points on the nine point circle tangent to the X spheres. There's also something similar in three dimensions. You get the medial points again shown here. The Euler tetrahedron Instead of being half the distance to the orthocenter, which doesn't necessarily exist, you go one third of the distance to the Monge point and that gives four points. You can also look at the feet of the projection of each point to the opposite face and that will give four more points. So that's the Euler projected tetrahedron. And all 12 of these points are on the 12 point sphere. The tangency quality is lost though that that doesn't happen with the third dimension. However, we have these nice functions now which generate all these. The X spheres and all these other tetrahedra can all be generated automatically. Now the median point uses in green we have the medians. In cyan we have the angle bisectors. If you do a mirroring operation around the angle bisectors, they intersect at this point K, which is called the Simidian point. This is also how the isogonal conjugate is set up. So the simidian is the isogonal conjugate of the centroid, this point K here. The same thing can be done in three dimensions. Reflections of the median planes or the dihedral angle bisectors. It turns out for any given point for the 12, for the six dihedral bisecting planes, they will still intersect as shown here. And that's the isogonic conjugate. In three dimensions we can take a look at, say the mean here we can look at the isogonic conjugate of the of the centroid and we get the point and it's also the Simidian point. So this is two ways of getting the same point. The symmedial tetrahedron. We can use the subtetrahedron function and do that. Or we can look at the CE vians of the point of the of the median point and we get the same values again. So there's usually more than one way of doing something. And now there's multiple ways of doing all these calculations using the Wolfram function repository. X circles and X spheres. The X circles are tangent to the infinite lines of the triangle on the outside. The N circle is tangent to the lines of the triangle on the inside. The contact points of the X circles forms the X touch triangle. And the X central connects the centers. We can do the same thing in three dimensions. The red tetrahedron is the original tetrahedron. The X central tetrahedron in gray connects the centers. And the blue tetrahedron at the center connects the tangency points. There's also the X tangents tetrahedron shown here. And make this a bit Bigger here the xtangents tetrahedron is is tangent to the X spheres. Each face of the tetrahedron tangent to three of the X spheres. And it turns out that if you connect up the vertices of the original tetrahedron at the center with the vertices of the X tangents tetrahedron, they all meet at a central point. They all intersect. Now, I've generated a few thousand different tetrahedra based on given tetrahedron. Almost none of them have this property that they form a perspector with the original tetrahedron. So I found this last week. It seems to be a new property that wasn't known before. That's why it's somewhat handy to have a way of calculating all these things quickly. The X tangents tetrahedron is difficult to calculate and it turns out that these points always intersect. So that prompted me to add the X tangents point of a tetrahedron as one of the important centers. The parallelian's point. Say given a point here, if we, we pretty much just generated there. And my claim is that if we go through that point in red here with planes parallel to the faces of the tetrahedron, that all the four generated triangles have the same area. So this is the equal area parallelian's point. And here are the calculations here. They all do have the same area. So this is another, another point for the tetrahedron. We can look at the dihedral angle constant. If you look at opposing edges of a tetrahedron and use this operation, the sum of the squares times twice the cotangents of the dihedral angles, the values of all three of these for opposing, for the three pairs of opposing edges are the same. So that gives the dihedral angle constant. So I decided I would add a dihedral angle constant function to the function repository. And here's a few tetrahedra and their constants. We can also look at special tetrahedra such as the ortho schemes. Each of these tetrahedra here is divided into eight copies of itself that are smaller. There are two known ortho schemes. Here's one of them and here's the other. I'm not sure why they're called ortho schemes, but there's these two of them. There are only five known space filling tetrahedra and these were. It took a few hundred years to compile this list of five tetrahedra. And this is also a demonstration space filling tetrahedra. John Conway asks, well, what other? So those space filling ones, they have the property that you can cut them up and make a cube. What other tetrahedra have that property? Now, a regular tetrahedron, you cannot cut up into pieces a finite number of pieces and make a cube out of it. That's due to what's called the scissors congruence problem. If the DIN invariance of a polyhedron isn't zero, you can't make a cube out of it. So what tetrahedra have a dinner invariance of zero. To do that, you basically all the dihedral angles of the tetrahedron have to have a rational fraction of PI. And it turns out this question was answered just a few years ago with this paper, space vectors forming rational angles. And it turns out there are 59 sporadic tetrahedra that have this property. And there's also two infinite families that are quite limited. But the 59 are all related to either the icosidodecahedron or to the 3x3x3 cubic grid. And we recently or will soon add all 59 of these sporadic tetrahedra to the Demonstrations project. And eventually we'll also make a data repository for them. One thing to notice here, I give these fractions here. On the edges are the dihedral angles, and at the vertices are the solid angles in blue. And one thing to notice is that twice the sum of the dihedral angles minus the sum of the solid angles equals 4. If I added PI to these to represent that this edge here is say 1/5 PI, then these would all add up to 4pi. For a triangle you're looking for the sum of angles is equal to 2 PI. For a tetrahedron, it's twice the dihedrals minus the solid angles is equal to 4pi. We can look at some exact coordinates for some of these. For example here, they're somewhat nasty. These sporadic usually require either fourth degree polynomials, eighth degree polynomials are higher. And I still haven't found exact solutions for all of them. But you get this set of dihedral angles for this particular one. So we can do face reflections with the polyhedron face face reflect function. And we could just do those in series. So here's 14 of these in a circle around one of the edges. And this last wedge would fill it up. But since it's an odd number, the faces don't match up. Here's another one where you need 14, and they do match up precise. So with the Kimberling centers, we have almost 54,000 of them. Here's the website for them. Years ago I made a animation for how these look like as you, as you twisted a triangle around, how the centers would move around. And so when I built this years ago, I wondered how tetrahedra would behave. And that's where this comes from. This is just one tetrahedron. But you can see that these generated centers are all over the place. And if you look at the Community post, you can take a closer look at this particular animation. So the way I did it, as I mentioned in this, in this article, I used 10,000 triangle center functions. I got the center of each triangle, each of the four triangles on the tetrahedron's face, and that defined a tetrahedron. And then I could apply these tetrahedron centers to that set of four points. And I could also find the isogonal and isotomic conjugates. And then I could combine all of them together. Now, when I've done similar experiments on triangles, you'd get a tremendous amount of overlap. Pretty much everything with triangle centers is interrelated. So whenever you try one series of steps on something, it usually will hit another triangle center. With these, I expected to find lots of overlap between these 140,000 centers, but there was almost none. There are quite a few ways of getting back to the circumcenter or to the centroid, but pretty much nothing intersected. That was a new point. I decided to be more careful when doing these explorations. Otherwise I could get a list of a million items that was pretty much meaningless. The xtangent point I know is valuable because it's a perspector. And those are, those are almost vanishingly rare with tetrahedra. And I want to find more of these that are actually meaningful points. Let's see here. So one thing that is a sort of a constant with triangle centers is the Euler line. There are various important points on the Euler line, such as the circumcenter, centroid, nine point center, and orthocenter. And we can plot the isogonal conjugate of the Euler line. It's the Jarabec hyperbola, which I show here. So this is a diagram of the Euler line with a particular triangle where all of these have nice positions. The circumcenter at minus 2,0, the centroid at 00, and the orthocenter at 4,0. So here's a. In three dimensions, here's another special tetrahedron, and we get the same thing. These points are on the Euler line, but we substitute the Monge point for the orthocenter. Since the Orthocenter doesn't usually exist for tetrahedra. For all the centers in the tetrahedron center function, we can calculate what these centers are. And quite a few of them are on the Euler line. The circumcenter at -2,00, the monge point at 2,00, and a few of the new points I added. If you look at the medial tetrahedron of the Euler tetrahedron, you get a tetrahedron which has a perspector with the original tetrahedron with the lines intersecting at this particular point. The Euler and negative and medial tetrahedra all seem to work very nicely together to build new perspective points. And here's a map of them here at the center, which is somewhat hard to see. You've got a cyan line, which is the Euler line, and most many points are on that. I also get the isogonal conjugates and isotomic conjugates of the points. There are a few points off, but most of them are very near the Euler line. So one thing I hope to do is come up with new important tetrahedral points that are off the Euler line. So that is pretty much what I wanted to cover. There's adpwolfram.com if you would like to write to me after the talk. There's a lot that's unsolved with tetrahedra, which is pretty much virtually everything. Many of the tetrahedral formulas are unknown, such as with triangle centers that there's special formulas for many of them, but most of those are not known for tetrahedra. We still don't know which triangle centers have tetrahedral equivalence, but most do not have equivalence. One project I plan to do is to take some of the scaling tetrahedra I've mentioned, such as the Conway tetrahedra, and generate points based on particular values. For example, I could look at the solid angles and see whether those generate a interesting point based on say the cosine of the solid angles, whether the if that's used at the very center coordinates, whether that leads to a nice point or not. See how those work with all these various tetrahedra and then try to match up to see what's going on. That's my talk. I hope everyone has enjoyed it. If there's any questions anywhere, let me see if I've got any questions here. If not, feel free to write to me and I'll be glad to answer any questions.