## 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.

## Reading the Comics, September 16, 2015: Celebrity Appearance Edition

I couldn’t go on calling this Back To School Editions. A couple of the comic strips the past week have given me reason to mention people famous in mathematics or physics circles, and one who’s even famous in the real world too. That’ll do for a title.

Jeff Corriveau’s Deflocked for the 15th of September tells what I want to call an old joke about geese formations. The thing is that I’m not sure it is an old joke. At least I can’t think of it being done much. It seems like it should have been.

The formations that geese, or other birds, form has been a neat corner of mathematics. The question they inspire is “how do birds know what to do?” How can they form complicated groupings and, more, change their flight patterns at a moment’s notice? (Geese flying in V shapes don’t need to do that, but other flocking birds will.) One surprising answer is that if each bird is just trying to follow a couple of simple rules, then if you have enough birds, the group will do amazingly complex things. This is good for people who want to say how complex things come about. It suggests you don’t need very much to have robust and flexible systems. It’s also bad for people who want to say how complex things come about. It suggests that many things that would be interesting can’t be studied in simpler models. Use a smaller number of birds or fewer rules or such and the interesting behavior doesn’t appear.

Scott Adams’s Dilbert Classics from the 15th and 16th of September (originally run the 22nd and 23rd of July, 1992) are about mathematical forecasts of the future. This is a hard field. It’s one people have been dreaming of doing for a long while. J Willard Gibbs, the renowned 19th century physicist who put the mathematics of thermodynamics in essentially its modern form, pondered whether a thermodynamics of history could be made. But attempts at making such predictions top out at demographic or rough economic forecasts, and for obvious reason.

The next day Dilbert’s garbageman, the smartest person in the world, asserts the problem is chaos theory, that “any complex iterative model is no better than a wild guess”. I wouldn’t put it that way, although I’m not sure what would convey the idea within the space available. One problem with predicting complicated systems, even if they are deterministic, is that there is a difference between what we can measure a system to be and what the system actually is. And for some systems that slight error will be magnified quickly to the point that a prediction based on our measurement is useless. (Fortunately this seems to affect only interesting systems, so we can still do things like study physics in high school usefully.)

Maria Scrivan’s Half Full for the 16th of September makes the Common Core joke. A generation ago this was a New Math joke. It’s got me curious about the history of attempts to reform mathematics teaching, and how poorly they get received. Surely someone’s written a popular or at least semipopular book about the process? I need some friends in the anthropology or sociology departments to tell, I suppose.

In Mark Tatulli’s Heart of the City for the 16th of September, Heart is already feeling lost in mathematics. She’s in enough trouble she doesn’t recognize mathematics terms. That is an old joke, too, although I think the best version of it was done in a Bloom County with no mathematical content. (Milo Bloom met his idol Betty Crocker and learned that she was a marketing icon who knew nothing of cooking. She didn’t even recognize “shish kebob” as a cooking term.)

Mell Lazarus’s Momma for the 16th of September sneers at the idea of predicting where specks of dust will land. But the motion of dust particles is interesting. What can be said about the way dust moves when the dust is being battered by air molecules that are moving as good as randomly? This becomes a problem in statistical mechanics, and one that depends on many things, including just how fast air particles move and how big molecules are. Now for the celebrity part of this story.

Albert Einstein published four papers in his “Annus mirabilis” year of 1905. One of them was the Special Theory of Relativity, and another the mass-energy equivalence. Those, and the General Theory of Relativity, are surely why he became and still is a familiar name to people. One of his others was on the photoelectric effect. It’s a cornerstone of quantum mechanics. If Einstein had done nothing in relativity he’d still be renowned among physicists for that. The last paper, though, that was on Brownian motion, the movement of particles buffeted by random forces like this. And if he’d done nothing in relativity or quantum mechanics, he’d still probably be known in statistical mechanics circles for this work. Among other things this work gave the first good estimates for the size of atoms and molecules, and gave easily observable, macroscopic-scale evidence that molecules must exist. That took some work, though.

Dave Whamond’s Reality Check for the 16th of September shows off the Metropolitan Museum of Symmetry. This is probably meant to be an art museum. Symmetries are studied in mathematics too, though. Many symmetries, the ways you can swap shapes around, form interesting groups or rings. And in mathematical physics, symmetries give us useful information about the behavior of systems. That’s enough for me to claim this comic is mathematically linked.