## 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 2018 Mathematics A To Z: Volume

Ray Kassinger, of the popular web comic Housepets!, had a silly suggestion when I went looking for topics. In one episode of Mystery Science Theater 3000, Crow T Robot gets the idea that you could describe the size of a space by the number of turkeys which fill it. (It’s based on like two minor mentions of “turkeys” in the show they were watching.)

I liked that episode. I’ve got happy memories of the time when I first saw it. I thought the sketch in which Crow T Robot got so volume-obsessed was goofy and dumb in the fun-nerd way.

I accept Mr Kassinger’s challenge only I’m going to take it seriously.

# Volume.

How big is a thing?

There is a legend about Thomas Edison. He was unimpressed with a new hire. So he hazed the college-trained engineer who deeply knew calculus. He demanded the engineer tell him the volume within a light bulb. The engineer went to work, making measurements of the shape of the bulb’s outside. And then started the calculations. This involves a calculus technique called “volumes of rotation”. This can tell the volume within a rotationally symmetric shape. It’s tedious, especially if the outer edge isn’t some special nice shape. Edison, fed up, took the bulb, filled it with water, poured that out into a graduated cylinder and said that was the answer.

I’m skeptical of legends. I’m skeptical of stories about the foolish intellectual upstaged by the practical man-of-action. And I’m skeptical of Edison because, jeez, I’ve read biographies of the man. Even the fawning ones make him out to be yeesh.

But the legend’s Edison had a point. If the volume of a shape is not how much stuff fits inside the shape, what is it? And maybe some object has too complicated a shape to find its volume. Can we think of a way to produce something with the same volume, but that is easier? Sometimes we can. When we do this with straightedge and compass, the way the Ancient Greeks found so classy, we call this “quadrature”. It’s called quadrature from its application in two dimensions. It finds, for a shape, a square with the same area. For a three-dimensional object, we find a cube with the same volume. Cubes are easy to understand.

Straightedge and compass can’t do everything. Indeed, there’s so much they can’t do. Some of it is stuff you’d think it should be able to, like, find a cube with the same volume as a sphere. Integration gives us a mathematical tool for describing how much stuff is inside a shape. It’s even got a beautiful shorthand expression. Suppose that D is the shape. Then its volume V is:

$V = \int\int\int_D dV$

Here “dV” is the “volume form”, a description of how the coordinates we describe a space in relate to the volume. The $\int\int\int$ is jargon, meaning, “integrate over the whole volume”. The subscript “D” modifies that phrase by adding “of D” to it. Writing “D” is shorthand for “these are all the points inside this shape, in whatever coordinate system you use”. If we didn’t do that we’d have to say, on each $\int$ sign, what points are inside the shape, coordinate by coordinate. At this level the equation doesn’t offer much help. It says the volume is the sum of infinitely many, infinitely tiny pieces of volume. True, but that doesn’t give much guidance about whether it’s more or less than two cups of water. We need to get more specific formulas, usually. We need to pick coordinates, for example, and say what coordinates are inside the shape. A lot of the resulting formulas can’t be integrated exactly. Like, an ellipsoid? Maybe you can integrate that. Don’t try without getting hazard pay.

We can approximate this integral. Pick a tiny shape whose volume is easy to know. Fill your shape with duplicates of it. Count the duplicates. Multiply that count by the volume of this tiny shape. Done. This is numerical integration, sometimes called “numerical quadrature”. If we’re being generous, we can say the legendary Edison did this, using water molecules as the tiny shape. And working so that he didn’t need to know the exact count or the volume of individual molecules. Good computational technique.

It’s hard not to feel we’re begging the question, though. We want the volume of something. So we need the volume of something else. Where does that volume come from?

Well, where does an inch come from? Or a centimeter? Whatever unit you use? You pick something to use as reference. Any old thing will do. Which is why you get fascinating stories about choosing what to use. And bitter arguments about which of several alternatives to use. And we express the length of something as some multiple of this reference length.

Volume works the same way. Pick a reference volume, something that can be one unit-of-volume. Other volumes are some multiple of that unit-of-volume. Possibly a fraction of that unit-of-volume.

Usually we use a reference volume that’s based on the reference length. Typically, we imagine a cube that’s one unit of length on each side. The volume of this cube with sides of length 1 unit-of-length is then 1 unit-of-volume. This seems all nice and orderly and it’s surely not because mathematicians have paid off by six-sided-dice manufacturers.

Does it have to be?

That we need some reference volume seems inevitable. We can’t very well say the area of something is ten times nothing-in-particular. Does that reference volume have to be a cube? Or even a rectangle or something else? It seems obvious that we need some reference shape that tiles, that can fill up space by itself … right?

What if we don’t?

I’m going to drop out of three dimensions a moment. Not because it changes the fundamentals, but because it makes something easier. Specifically, it makes it easier if you decide you want to get some construction paper, cut out shapes, and try this on your own. What this will tell us about area is just as true for volume. Area, for a two-dimensional sapce, and volume, for a three-dimensional, describe the same thing. If you’ll let me continue, then, I will.

So draw a figure on a clean sheet of paper. What’s its area? Now imagine you have a whole bunch of shapes with reference areas. A bunch that have an area of 1. That’s by definition. That’s our reference area. A bunch of smaller shapes with an area of one-half. By definition, too. A bunch of smaller shapes still with an area of one-third. Or one-fourth. Whatever. Shapes with areas you know because they’re marked on them.

Here’s one way to find the area. Drop your reference shapes, the ones with area 1, on your figure. How many do you need to completely cover the figure? It’s all right to cover more than the figure. It’s all right to have some of the reference shapes overlap. All you need is to cover the figure completely. … Well, you know how many pieces you needed for that. You can count them up. You can add up the areas of all these pieces needed to cover the figure. So the figure’s area can’t be any bigger than that sum.

Can’t be exact, though, right? Because you might get a different number if you covered the figure differently. If you used smaller pieces. If you arranged them better. This is true. But imagine all the possible reference shapes you had, and all the possible ways to arrange them. There’s some smallest area of those reference shapes that would cover your figure. Is there a more sensible idea for what the area of this figure would be?

And put this into three dimensions. If we start from some reference shapes of volume 1 and maybe 1/2 and 1/3 and whatever other useful fractions there are? Doesn’t this covering make sense as a way to describe the volume? Cubes or rectangles are easy to imagine. Tetrahedrons too. But why not any old thing? Why not, as the Mystery Science Theater 3000 episode had it, turkeys?

This is a nice, flexible, convenient way to define area. So now let’s see where it goes all bizarre. We know this thanks to Giuseppe Peano. He’s among the late-19th/early-20th century mathematicians who shaped modern mathematics. They did this by showing how much of our mathematics broke intuition. Peano was (here) exploring what we now call fractals. And noted a family of shapes that curl back on themselves, over and over. They’re beautiful.

And they fill area. Fill volume, if done in three dimensions. It seems impossible. If we use this covering scheme, and try to find the volume of a straight line, we get zero. Well, we find that any positive number is too big, and from that conclude that it has to be zero. Since a straight line has length, but not volume, this seems fine. But a Peano curve won’t go along with this. A Peano curve winds back on itself so much that there is some minimum volume to cover it.

This unsettles. But this idea of volume (or area) by covering works so well. To throw it away seems to hobble us. So it seems worth the trade. We allow ourselves to imagine a line so long and so curled up that it has a volume. Amazing.

And now I get to relax and unwind and enjoy a long weekend before coming to the letter ‘W’. That’ll be about some topic I figure I can whip out a nice tight 500 words about, and instead, produce some 1541-word monstrosity while I wonder why I’ve had no free time at all since August. Tuesday, give or take, it’ll be available at this link, as are the rest of these glossary posts. Thanks for reading.

## Calculus For Breakfast

Robert Austin, of the RobertLovesPi blog, got to thinking about one of those interesting mathematics problems. It starts with the equations that describe the volume and the surface area of a sphere.

If the sphere has radius r, then the surface area of the sphere is 4πr2. And the volume is (4/3)πr3. What’s interesting about this is that there’s a relationship between these two expressions. The first is the derivative of the second. The derivative is one of the earliest things one learns in calculus. It describes how much a quantity changes with a tiny change in something it depends on.

And this got him to thinking about the surface area of a cube. Call the length of a cube’s side s. Its surface is six squares, each of them with a side of length s. So the surface area of each of the six squares is s2, which is obvious when you remember we call raising something to the second power “squaring”. Its total surface area then is 6s2. But its volume is is s3. This is why we even call raising something to the third power “cubing”. And the derivative of s3 is 3s2. (If you don’t know calculus, but you suspect you see a pattern here, you’re learning calculus. If you’re not sure about the pattern, let me tell you that the derivative of s4 would be 4s3, and the derivative of (1/3)s2 would be (2/3)s.)

There’s an obvious flaw there, and Austin’s aware of it. But it got him pondering different ways to characterize how big a cube is. He can find one that makes the relationship between volume and surface area work out like he expects. But the question remains, why that? And what about other shapes?

I think that’s an interesting discussion to have, and mean to think about it some more myself. And I wanted to point people who’d be interested over there to join in.

## 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.

## Reblog: Why are Hot Dogs So Inexpensive?

Here’s a cute little observation about presentation and the power of those volume formulas that kind of get looked at when we’re in the chapter about the volumes of basic solids (circular cylinders, in this case) and not afterwards. It’s also for hot dog fans.

Memorial Day is the unofficial start of grilling season. According to the National Hot Dog and Sausage Council (yes, it exists), Americans will consume about 7 billion hot dogs between now and Labor Day–that’s about 818 per second! The estimated cost of all this is about \$1.7 billion, or less than 25 cents per serving.

A large part of this low price is probably due to the quality of the ingredients, but I want to focus on hot dogs purchased from vendors rather than at supermarkets. Street corner hot dog stands have been cropping up around Durham for the last several weeks, and while I haven’t purchased from any, I get the impression that they are quite inexpensive.

A nice stylized example for us to consider comes from a new book entitled X and the City: Modeling Aspects of Urban Life by John Adam. In chapter 4, “Eating in…

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