## 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.

## Reading the Comics, March 13, 2018: One Of My Assumptions Is Shaken Edition

I learn, from reading not-yet-dead Usenet group rec.arts.comics.strips, that Rick Stromoski is apparently ending the comic Soup To Nutz. This is sad enough. But worse, GoComics.com has removed all but the current day’s strip from its archives. I had trusted that GoComics.com links were reliable in a way that Comics Kingdom and Creators.com weren’t. Now I learn that maybe I need to include images of the comics I review and discuss here lest my essays become unintelligible in the future? That’s not a good sign. I can do it, mind you. I just haven’t got started. You’ll know when I swing into action.

Norm Feuti, of Retail, still draws Sunday strips for Gil. They’re to start appearing on GoComics.com soon, and I can talk about them from my regular sources after that. But for now I follow the strip on Twitter. And last Sunday he posted this one.

It’s sort of a protesting-the-problem question. It’s also a reaction a lot of people have to “explain how you found the answer” questions. In a sense, yeah, the division shows how the answer was found. But what’s wanted — and what’s actually worth learning — is to explain why you did this calculation. Why, in this case, 216 divided by 8? Why not 216 times 8? Why not 8 divided by 216? Why not 216 minus 8? “How you found your answer” is probably a hard question to make interesting on arithmetic, unfortunately. If you’re doing a long sheet of problems practicing division, it’s not hard to guess that dividing is the answer. And that it’s the big number divided by the small. It can be good training to do blocks of problems that use the same approach, for the same reason it can be good training to focus on any exercise a while. But this does cheat someone of the chance to think about why one does this rather than that.

Patrick Roberts’s Todd the Dinosaur for the 11th has mathematics as the thing Todd’s trying to get out of doing. (I suppose someone could try to argue the Y2K bug was an offshoot of mathematics, on the grounds that computer science has so much to do with mathematics. I wouldn’t want to try defending that, though.) I grant that most fraction-to-decimal conversion problems hit that sweet spot of being dull, tedious, and seemingly pointless. There’s some fun decimal expansions of fractions. The sevenths and the elevenths and 1/243 have charm to them. There’s some kid who’ll become a mathematician because at the right age she was told about $\frac{1}{8991}$. 3/16th? Eh.

Mark Anderson’s Andertoons for the 11th is the Mark Anderson’s Andertoons for the week. I don’t remember seeing a spinny wheel like this used to introduce probability. It’s a good prop, though. I would believe in a class having it.

Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 11th is built on the Travelling Salesman Problem. It’s one of the famous unsolved and hard problems of mathematics. Weinersmith’s joke is a nice gag about one way to “solve” the problem, that of making it irrelevant. But even if we didn’t need to get to a collection of places efficiently mathematicians would still like to know good ways to do it. It turns out that finding the shortest (quickest, cheapest, easiest, whatever) route connecting a bunch of places is great problem. You can phrase enormously many problems about doing something as well as possible as a Travelling Salesman Problem. It’s easy conceptually to find the answer: try out all the possibilities and pick the best one. But if there’s more than a handful of cities, there are so many possible routes there’s no checking them all, not before you die of old age. We can do very well finding approximate answers, including by my specialization of Monte Carlo methods. In those you take a guess at an answer. Then make, randomly, a change. You’ll either have made things better or worse. If you’ve made it better, keep the change. If you’ve made it worse, usually you reject the change but sometimes you keep it. And repeat. In surprisingly little time you’ll get a really good answer. Maybe not the best possible, but a great answer for how straightforward setting it up was.

Dan Thompson’s Brevity for the 12th is a Rubik’s Cube joke. There’s not a lot of mathematics to that. But I do admire how Thompson was careful enough to draw a Rubik’s Cube that actually looks like the real article; it’s not just an isometric cube with thick lines partitioning it. Look at the corners of each colored sub-cube. I may be the only reader to notice this but I’m glad Thompson did the work.

Mason Mastroianni’s The Wizard of Id for the 12th gets Sir Rodney in trouble with the King for doing arithmetic. I haven’t read the comments on GoComics.com. I’d like to enter “three” as my guess for how many comments one would have to read before finding the “weapons of math instruction” joke in there.

Jef Mallett’s Frazz for the 13th has mathematics homework given as the thing lost by the time change. It’s just a cameo mention.

Steve Moore’s In The Bleachers for the 13th features a story problem as a test of mental acuity. When the boxer can’t work out what the heck the trains-leaving-Penn-Station problem even means he’s ruled unfit to keep boxing. The question is baffling, though. As put, the second train won’t ever overtake the first. The question: did Moore just slip up? If the first train were going 30 miles per hour and the second 40 there would be a perfectly good, solvable question in this. Or was Moore slipping in an extra joke, making the referee’s question one that sounds like it was given wrong? Don’t know, so I’ll suppose the second.

## Reading the Comics, July 28, 2012

I intend to be back to regular mathematics-based posts soon. I had a fine idea for a couple posts based on Sunday’s closing of the Diaster Transport roller coaster ride at Cedar Point, actually, although I have to technically write them first. (My bride and I made a trip to the park to get a last ride in before its closing, and that lead to inspiration.) But reviews of math-touching comic strips are always good for my readership, if I’m readin the statistics page here right, so let’s see what’s come up since the last recap, going up to the 14th of July.