## Reading the Comics Follow-up: Where Else Is A Tetrahedron’s Centroid Edition

A Reading the Comics post a couple weeks back inspired me to find the centroid of a regular tetrahedron. A regular tetrahedron, also known as “a tetrahedron”, is the four-sided die shape. A pyramid with triangular base. Or a cone with a triangle base, if you prefer. If one asks a person to draw a tetrahedron, and they comply, they’ll likely draw this shape. The centroid, the center of mass of the tetrahedron, is at a point easy enough to find. It’s on the perpendicular between any of the four faces — the equilateral triangles — and the vertex not on that face. Particularly, it’s one-quarter the distance from the face towards the other vertex. We can reason that out purely geometrically, without calculating, and I did in that earlier post.

But most tetrahedrons are not regular. They have centroids too; where are they?

Thing is I know the correct answer going in. It’s at the “average” of the vertices of the tetrahedron. Start with the Cartesian coordinates of the four vertices. The x-coordinate of the centroid is the arithmetic mean of the x-coordinates of the four vertices. The y-coordinate of the centroid is the mean of the y-coordinates of the vertices. The z-coordinate of the centroid is the mean of the z-coordinates of the vertices. Easy to calculate; but, is there a way to see that this is right?

What’s got me is I can think of an argument that convinces me. So in this sense, I have an easy proof of it. But I also see where this argument leaves a lot unaddressed. So it may not prove things to anyone else. Let me lay it out, though.

So start with a tetrahedron of your own design. This will be less confusing if I have labels for the four vertices. I’m going to call them A, B, C, and D. I don’t like those labels, not just for being trite, but because I so want ‘C’ to be the name for the centroid. I can’t find a way to do that, though, and not have the four tetrahedron vertices be some weird set of letters. So let me use ‘P’ as the name for the centroid.

Where is P, relative to the points A, B, C, and D?

And here’s where I give a part of an answer. Start out by putting the tetrahedron somewhere convenient. That would be the floor. Set the tetrahedron so that the face with triangle ABC is in the xy plane. That is, points A, B, and C all have the z-coordinate of 0. The point D has a z-coordinate that is not zero. Let me call that coordinate h. I don’t care what the x- and y-coordinates for any of these points are. What I care about is what the z-coordinate for the centroid P is.

The property of the centroid that was useful last time around was that it split the regular tetrahedron into four smaller, irregular, tetrahedrons, each with the same volume. Each with one-quarter the volume of the original. The centroid P does that for the tetrahedron too. So, how far does the point P have to be from the triangle ABC to make a tetrahedron with one-quarter the volume of the original?

The answer comes from the same trick used last time. The volume of a cone is one-third the area of the base times its altitude. The volume of the tetrahedron ABCD, for example, is one-third times the area of triangle ABC times how far point D is from the triangle. That number I’d labelled h. The volume of the tetrahedron ABCP, meanwhile, is one-third times the area of triangle ABC times how far point P is from the triangle. So the point P has to be one-quarter as far from triangle ABC as the point D is. It’s got a z-coordinate of one-quarter h.

Notice, by the way, that while I don’t know anything about the x- and y- coordinates of any of these points, I do know the z-coordinates. A, B, and C all have z-coordinate of 0. D has a z-coordinate of h. And P has a z-coordinate of one-quarter h. One-quarter h sure looks like the arithmetic mean of 0, 0, 0, and h.

At this point, I’m convinced. The coordinates of the centroid have to be the mean of the coordinates of the vertices. But you also see how much is not addressed. You’d probably grant that I have the z-coordinate coordinate worked out when three vertices have the same z-coordinate. Or where three vertices have the same y-coordinate or the same x-coordinate. You might allow that if I can rotate a tetrahedron, I can get three points to the same z-coordinate (or y- or x- if you like). But this still only gets one coordinate of the centroid P.

I’m sure a bit of algebra would wrap this up. But I would like to avoid that, if I can. I suspect the way to argue this geometrically depends on knowing the line from vertex D to tetrahedron centroid P, if extended, passes through the centroid of triangle ABC. And something similar applies for vertexes A, B, and C. I also suspect there’s a link between the vector which points the direction from D to P and the sum of the three vectors that point the directions from D to A, B, and C. I haven’t quite got there, though.

I will let you know if I get closer.

## Reading the Comics, March 16, 2021: Where Is A Tetrahedron’s Centroid Edition

Comic Strip Master Command has not, to appearances, been distressed by my Reading the Comics hiatus. There are still mathematically-themed comic strips. Many of them are about story problems and kids not doing them. Some get into a mathematical concept. One that ran last week caught my imagination so I’ll give it some time here. This and other Reading the Comics essays I have at this link, and I figure to resume posting them, at least sometimes.

Ben Zaehringer’s In The Bleachers for the 16th of March, 2021 is an anthropomorphized-geometry joke. Here the centroid stands in for “the waist”, the height below which boxers may not punch.

The centroid is good geometry, something which turns up in plane and solid shapes. It’s a center of the shape: the arithmetic mean of all the points in the shape. (There are other things that can, with reason, be called a center too. Mathworld mentions the existence of 2,001 things that can be called the “center” of a triangle. It must be only a lack of interest that’s kept people from identifying even more centers for solid shapes.) It’s the center of mass, if the shape is a homogenous block. Balance the shape from below this centroid and it stays balanced.

For a complicated shape, finding the centroid is a challenge worthy of calculus. For these shapes, though? The sphere, the cube, the regular tetrahedron? We can work those out by reason. And, along the way, work out whether this rule gives an advantage to either boxer.

The sphere first. That’s the easiest. The centroid has to be the center of the sphere. Like, the point that the surface of the sphere is a fixed radius from. This is so obvious it takes a moment to think why it’s obvious. “Why” is a treacherous question for mathematics facts; why should 4 divide 8? But sometimes we can find answers that give us insight into other questions.

Here, the “why” I like is symmetry. Look at a sphere. Suppose it lacks markings. There’s none of the referee’s face or bow tie here. Imagine then rotating the sphere some amount. Can you see any difference? You shouldn’t be able to. So, in doing that rotation, the centroid can’t have moved. If it had moved, you’d be able to tell the difference. The rotated sphere would be off-balance. The only place inside the sphere that doesn’t move when the sphere is rotated is the center.

This symmetry consideration helps answer where the cube’s centroid is. That also has to be the center of the cube. That is, halfway between the top and bottom, halfway between the front and back, halfway between the left and right. Symmetry again. Take the cube and stand it upside-down; does it look any different? No, so, the centroid can’t be any closer to the top than it can the bottom. Similarly, rotate it 180 degrees without taking it off the mat. The rotation leaves the cube looking the same. So this rules out the centroid being closer to the front than to the back. It also rules out the centroid being closer to the left end than to the right. It has to be dead center in the cube.

Now to the regular tetrahedron. Obviously the centroid is … all right, now we have issues. Dead center is … where? We can tell when the regular tetrahedron’s turned upside-down. Also when it’s turned 90 or 180 degrees.

Symmetry will guide us. We can say some things about it. Each face of the regular tetrahedron is an equilateral triangle. The centroid has to be along the altitude. That is, the vertical line connecting the point on top of the pyramid with the equilateral triangle base, down on the mat. Imagine looking down on the shape from above, and rotating the shape 120 or 240 degrees if you’re still not convinced.

And! We can tip the regular tetrahedron over, and put another of its faces down on the mat. The shape looks the same once we’ve done that. So the centroid has to be along the altitude between the new highest point and the equilateral triangle that’s now the base, down on the mat. We can do that for each of the four sides. That tells us the centroid has to be at the intersection of these four altitudes. More, that the centroid has to be exactly the same distance to each of the four vertices of the regular tetrahedron. Or, if you feel a little fancier, that it’s exactly the same distance to the centers of each of the four faces.

It would be nice to know where along this altitude this intersection is, though. We can work it out by algebra. It’s no challenge to figure out the Cartesian coordinates for a good regular tetrahedron. Then finding the point that’s got the right distance is easy. (Set the base triangle in the xy plane. Center it, so the coordinates of the highest point are (0, 0, h) for some number h. Set one of the other vertices so it’s in the xz plane, that is, at coordinates (0, b, 0) for some b. Then find the c so that (0, 0, c) is exactly as far from (0, 0, h) as it is from (0, b, 0).) But algebra is such a mass of calculation. Can we do it by reason instead?

That I ask the question answers it. That I preceded the question with talk about symmetry answers how to reason it. The trick is that we can divide the regular tetrahedron into four smaller tetrahedrons. These smaller tetrahedrons aren’t regular; they’re not the Platonic solid. But they are still tetrahedrons. The little tetrahedron has as its base one of the equilateral triangles that’s the bigger shape’s face. The little tetrahedron has as its fourth vertex the centroid of the bigger shape. Draw in the edges, and the faces, like you’d imagine. Three edges, each connecting one of the base triangle’s vertices to the centroid. The faces have two of these new edges plus one of the base triangle’s edges.

The four little tetrahedrons have to all be congruent. Symmetry again; tip the big tetrahedron onto a different face and you can’t see a difference. So we’ll know, for example, all four little tetrahedrons have the same volume. The same altitude, too. The centroid is the same distance to each of the regular tetrahedron’s faces. And the four little tetrahedrons, together, have the same volume as the original regular tetrahedron.

What is the volume of a tetrahedron?

If we remember dimensional analysis we may expect the volume should be a constant times the area of the base of the shape times the altitude of the shape. We might also dimly remember there is some formula for the volume of any conical shape. A conical shape here is something that’s got a simple, closed shape in a plane as its base. And some point P, above the base, that connects by straight lines to every point on the base shape. This sounds like we’re talking about circular cones, but it can be any shape at the base, including polygons.

So we double-check that formula. The volume of a conical shape is one-third times the area of the base shape times the altitude. That’s the perpendicular distance between P and the plane that the base shape is in. And, hey, one-third times the area of the face times the altitude is exactly what we’d expect.

So. The original regular tetrahedron has a base — has all its faces — with area A. It has an altitude h. That h must relate in some way to the area; I don’t care how. The volume of the regular tetrahedron has to be $\frac{1}{3} A h$.

The volume of the little tetrahedrons is — well, they have the same base as the original regular tetrahedron. So a little tetrahedron’s base is A. The altitude of the little tetrahedron is the height of the original tetrahedron’s centroid above the base. Call that $h_c$. How can the volume of the little tetrahedron, $\frac{1}{3} A h_c$, be one-quarter the volume of the original tetrahedron, $\frac{1}{3} A h$? Only if $h_c$ is one-quarter $h$.

This pins down where the centroid of the regular tetrahedron has to be. It’s on the altitude underneath the top point of the tetrahedron. It’s one-quarter of the way up from the equilateral-triangle face.

(And I’m glad, checking this out, that I got to the right answer after all.)

So, if the cube and the tetrahedron have the same height, then the cube has an advantage. The cube’s centroid is higher up, so the tetrahedron has a narrower range to punch. Problem solved.

I do figure to talk about comic strips, and mathematics problems they bring up, more. I’m not sure how writing about one single strip turned into 1300 words. But that’s what happens every time I try to do something simpler. You know how it goes.

## My 2019 Mathematics A To Z: Platonic

Today’s A To Z term is another from goldenoj. It was just the proposal “Platonic”. Most people, prompted, would follow that adjective with one of three words. There’s relationship, ideal, and solid. Relationship is a little too far off of mathematics for me to go into here. Platonic ideals run very close to mathematics. Probably the default philosophy of western mathematics is Platonic. At least a folk Platonism, where the rest of us follow what the people who’ve taken the study of mathematical philosophy seriously seem to be doing. The idea that mathematical constructs are “real things” and have some “existence” that we can understand even if we will never see a true circle or an unadulterated four. Platonic solids, though, those are nice and familiar things. Many of them we can find around the house. That’s one direction to go.

# Platonic.

Before I get to the Platonic Solids, though, I’d like to think a little more about Platonic Ideals. What do they look like? I gather our friends in the philosophy department have debated this question a while. So I won’t pretend to speak as if I had actual knowledge. I just have an impression. That impression is … well, something simple. My reasoning is that the Platonic ideal of, say, a chair has to have all the traits that every chair ever has. And there’s not a lot that every chair has. Whatever’s in the Platonic Ideal chair has to be just the things that every chair has, and to omit things that non-chairs do not.

That’s comfortable to me, thinking like a mathematician, though. I think mathematicians train to look for stuff that’s very generally true. This will tend to be things that have few properties to satisfy. Things that look, in some way, simple.

So what is simple in a shape? There’s no avoiding aesthetic judgement here. We can maybe use two-dimensional shapes as a guide, though. Polygons seem nice. They’re made of line segments which join at vertices. Regular polygons even nicer. Each vertex in a regular polygon connects to two edges. Each edge connects to exactly two vertices. Each edge has the same length. The interior angles are all congruent. And if you get many many sides, the regular polygon looks like a circle.

So there’s some things we might look for in solids. Shapes where every edge is the same length. Shapes where every edge connects exactly two vertices. Shapes where every vertex connects to the same number of edges. Shapes where the interior angles are all constant. Shapes where each face is the same polygon as every other face. Look for that and, in three-dimensional space, we find nine shapes.

Yeah, you want that to be five also. The four extra ones are “star polyhedrons”. They look like spikey versions of normal shapes. What keeps these from being Platonic solids isn’t a lack of imagination on Plato’s part. It’s that they’re not convex shapes. There’s no pair of points in a convex shape for which the line segment connecting them goes outside the shape. For the star polyhedrons, well, look at the ends of any two spikes. If we decide that part of this beautiful simplicity is convexity, then we’re down to five shapes. They’re famous. Tetrahedron, cube, octahedron, icosahedron, and dodecahedron.

I’m not sure why they’re named the Platonic Solids, though. Before you explain to me that they were named by Plato in the dialogue Timaeus, let me say something. They were named by Plato in the dialogue Timaeus. That isn’t the same thing as why they have the name Platonic Solids. I trust Plato didn’t name them “the me solids”, since if I know anything about Plato he would have called them “the Socratic solids”. It’s not that Plato was the first to group them either. At least some of the solids were known long before Plato. I don’t know of anyone who thinks Plato particularly advanced human understanding of the solids.

But he did write about them, and in things that many people remembered. It’s natural for a name to attach to the most famous person writing them. Still, someone had the thought which we follow to group these solids together under Plato’s name. I’m curious who, and when. Naming is often a more arbitrary thing than you’d think. The Fibonacci sequence has been known at latest since Fibonacci wrote about it in 1204. But it could not have that name before 1838, when historian Guillaume Libri gave Leonardo of Pisa the name Fibonacci. I’m not saying that the name “Platonic Solid” was invented in, like, 2002. But traditions that seem age-old can be surprisingly recent.

What is an age-old tradition is looking for physical significance in the solids. Plato himself cleverly matched the solids to the ancient concept of four elements plus a quintessence. Johannes Kepler, whom we thank for noticing the star polyhedrons, tried to match them to the orbits of the planets around the sun. Wikipedia tells me of a 1980s attempt to understand the atomic nucleus using Platonic solids. The attempt even touches me. Along the way to my thesis I looked at uniform charges free to move on the surface of a sphere. It was obvious if there were four charges they’d move to the vertices of a tetrahedron on the sphere. Similarly, eight charges would go to the vertices of the cube. 20 charges to the vertices of the icosahedron. And so on. The Platonic Solids seem not just attractive but also of some deep physical significance.

There are not the four (or five) elements of ancient Greek atomism. Attractive as it is to think that fire is a bunch of four-sided dice. The orbits of the planets have nothing to do with the Platonic solids. I know too little about the physics of the atomic nucleus to say whether that panned out. However, that it doesn’t even get its own Wikipedia entry suggests something to me. And, in fact, eight charges on the sphere will not settle at the vertices of a cube. They’ll settle on a staggered pattern, two squares turned 45 degrees relative to each other. The shape is called a “square antiprism”. I was as surprised as you to learn that. It’s possible that the Platonic Solids are, ultimately, pleasant to us but not a key to the universe.

The example of the Platonic Solids does give us the cue to look for other families of solids. There are many such. The Archimedean Solids, for example, are again convex polyhedrons. They have faces of two or more regular polygons, rather than the lone one of Platonic Solids. There are 13 of these, with names of great beauty like the snub cube or the small rhombicuboctahedron. The Archimedean Solids have duals. The dual of a polyhedron represents a face of the original shape with a vertex. Faces that meet in the original polyhedron have an edge between their dual’s vertices. The duals to the Archimedean Solids get the name Catalan Solids. This for the Belgian mathematician Eugène Catalan, who described them in 1865. These attract names like “deltoidal icositetrahedron”. (The Platonic Solids have duals too, but those are all Platonic solids too. The tetrahedron is even its own dual.) The star polyhedrons hint us to look at stellations. These are shapes we get by stretching out the edges or faces of a polyhedron until we get a new polyhedron. It becomes a dizzying taxonomy of shapes, many of them with pointed edges.

There are things that look like Platonic Solids in more than three dimensions of space. In four dimensions of space there are six of these, five of which look like versions of the Platonic Solids we all know. The sixth is this novel shape called the 24-cell, or hyperdiamond, or icositetrachoron, or some other wild names. In five dimensions of space? … it turns out there are only three things that look like Platonic Solids. There’s versions of the tetrahedron, the cube, and the octahedron. In six dimensions? … Three shapes, again versions of the tetrahedron, cube, and octahedron. And it carries on like this for seven, eight, nine, any number of dimensions of space. Which is an interesting development. If I hadn’t looked up the answer I’d have expected more dimensions of space to allow for more Platonic Solid-like shapes. Well, our experience with two and three dimensions guides us to thinking about more dimensions of space. It doesn’t mean that they’re just regular space with a note in the corner that “N = 8”. Shapes hold surprises.

The essays for the Fall 2019 A To Z should be gathered here. And, in time, every past A to Z essay should be at this link. For now, it’s at least several years’ worth there. Thank you.

## Slowly Rotating Hyperdodecahedron

Here’s an engaging moving picture from RobertLovesPi. The Platonic solids — cubes, pyramids, octahedrons, icosahedrons, and dodecahedrons — are five solid shapes each with the same regular convex polygon as their face. This is a nice two-dimensional rendering of a three-dimensional projection of a “hyperdodecahedron”. It’s made of 120 dodecahedrons, in a four-dimensional space. And it’s got the same kind of structure that Platonic solids have, being made of the same regular convex polyhedron for each face.

Remarkably, I learn from Mathworld, the shape is three-colorable. That is, suppose you wanted to assign colors to each of the corners in this four-dimensional shape. They’re all green circles here, but they don’t have to be. There are a lot of these corners, and they’re connected in complicated ways to one another. But you could color in every one of them, so that none if them is connected directly to another of the same color, using only three different colors.

This is the hyperdodecahedron, or 120-cell, one of the six four-dimensional analogs of the Platonic solids. It’s been shown on this blog before, but this image has one major change: a much slower rotational speed. It is my hope that this will help people, including myself, with the difficult task of understanding four-dimensional objects.

This image was created using Stella 4d, a program you can try, as a free trial download, at this website.)

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