# A great Multivariable Calculus problem

Today I gave my multivariable calculus class a problem — a problem I give every year, that I found… somewhere. Maybe MIT, maybe an Exeter problem set, maybe a textbook. And if you ever want to see kids work together, and do some good problem solving, this is a prime problem for that.

Up to now, we’ve been working on vectors — and they learned vector basics (read: dot product and cross product). Here’s the question.

You have any tetrahedron. Sticking outwards from each face, orthogonal to each face, is a vector with magnitude equal to the area of face it is sticking out of. Prove that if you add these four vectors together, they sum to the zero vector.

It’s such a beautiful problem. I don’t have a totally geometric way to explain why this is true (we do lots of good vector algebra), but I do enjoy watching everything all come together. To me, it almost works like magic.

I then had a student ask an amazing extension question. (If they’re asking extension questions, you know it’s a rich problem.) He said: “Will this always work for any polyhedra? What about figures involving faces which aren’t triangles?” I, of course, decided I loved the problem. And I desperately want him to work on this for his final project.

I love the idea of this student taking this problem and seeing how far he can run with it. I mean: hello, gluing tetrahedrons together! (I expect him to make some stick models, if he does it!)

1. Alon Amit says:

I don’t know about totally geometric way either, but there’s a simple physical argument which explains why this must be true. Imagine a rigid polyhedron (convex or not, it doesn’t matter) filled with some gas under pressure. The gas exerts a force on each face, and this force acts perpendicular to the face and is proportional to its area (think “PSI”). If the vector sum of those forces was nonzero, the polyhedron would float away in the direction of the net force.

BTW this problem appears in Peter Winkler’s book “Mathematical Puzzles: A Connoisseur’s Collection”. The solution provided there is precisely this physical argument.

1. samjshah says:

Alon, thanks for sharing. This is one of the most stunning arguments for building my intuition. I love that it’s based on physics!

2. Robert Jones says:

Cool!

When I first read this, I thought you meant a regular tetrahedron. It seemed quite obvious. Then I realised what you area actually saying – ANY tetrahedron.

3. Jaime says:

My explanation is a little messy, but I think Alon’s argument can be turned into a proper proof by projecting the polyhedron facets onto a plane…

The area of the resulting polygon will be equal to the projection of the facet vector onto a line perpendicular to the plane. This of course means that you will have negative and positive areas, depending on wether the inside of outside of the polyhedron is facing the plane.

If you now consider the full polyhedron, it is sort of clear that the total projected area will sum up to zero. It is specially clear with convex polyhedra, but also so for non-convex ones. You have just proved (I think) that the component of the sum of the vectors along a given direction is zero. since this holds true for any direction, the sum is zero.

1. DavidC says:

Central (I think) to this really nice argument is that the contribution of a bit of face area to the area in the projection in a given direction is the same as the contribution of the vector pointing out of the face to the component of the vector sum in the given direction.

Let me know if I’m making this more complicated than it needs to be, but the relationship between cross products and area feels to me like the easiest way to show that: The region spanned by two vectors has area the magnitude of the cross product.

Let v be a vector in Jaime’s given direction, and let P be the plane perpendicular to V. Then I think the key fact about cross products will be: Projecting two vectors onto P and then taking their cross product is the same as taking the cross product before projecting, and then projecting that onto the line spanned by v.

I’m being a bit inarticulate, but maybe I’ll try to say this better later on.

Neat problem! Thanks!

4. Mimi says:

Jaime, the projection concept is a bit confusing to me. Are you arguing basically that any closed shape (even circular / elliptical / blobby) would have a net vector of zero? That makes sense if you think about just the vertical component alone. There’s as much surface facing upwards as there is facing downwards, in a closed 3-D shape. So, the net of those surfaces should sum to zero.

Now, you can turn the closed 3-D shape on its side to argue the same for the horizontal vector components, I think.

1. DavidC says:

Mimi: Your explanation sounds right on to me.

I believe the corresponding claim for curved (closed) surfaces is also true, but requires a bit of multivariable calculus to formulate: Instead of adding up vectors perpendicular to the faces, you integrate the ‘unit normal’ (take perpendicular vectors at each point on the curved surface, since there aren’t any ‘faces’).

5. Jaime says:

@Mimi, @DavidC,

Yes, you both have got the overall idea of what I had in mind.

Thinking of it more carefully, with continuous closed surfaces in mind, I believe this is just a particular case of the divergence theorem:

http://en.wikipedia.org/wiki/Divergence_theorem

1. samjshah says:

When reading through the comments, I started thinking: wow, this is exactly like what we do at the end of the year (divergence thm). Indeed it is.

I have never made that connection before!

Thank you all.

It all seemed like magic. But it’s less magic now. And I’m all for that.

6. Mimi says:

By the way, Sam, thinking about this problem made me leave 10 minutes later for work today than normal, so I was huffing and puffing by the time I got to work. ;)

7. Jake Wildstrom says:

I’d start out honestly in a brute-force way which nonetheless makes use of some multivariable concepts: if the vectors corresponding to the edges emanating from a vertex are u, v, and w, then the vectors in question will be u x v, v x w, and w x u for three of the faces. The other three edges (which bound the fourth face) are given by (v-w), (w-u), and (u-v), so this face’s vector is given by (w-u) x (v-u), and I think you can muck with the distributive law, anticommutative law, and the fact that u x u =0 to get a zero sum. But you do need to make sure all your orientations are correct.

The Divergence Theorem or physical approach is a lot cleverer.

1. samjshah says:

The brute force way is exactly the way we did it (my kids and I). It was great because it emphasized: orientation, basic laws of cross products, and properties of cross products.

8. Maya Incaand says:

In geometric algebra the wedge product gives an oriented area directly, the cross product is redundant (no need for vectors sticking up anywhere).

Then make use of the geometric product(dot plus wedge)and the solution falls out in a natural way.

In geometric algebra you can “see” the geometry in the algebra whereas vector algebra muddies the water.

1. peeterjoot says:

@maya.

Solving this with the wedge product or the cross product is pretty much identical, so I don’t think that GA adds anything to the problem (and could be viewed as making the problem slightly more complex). The main requirement is an enumeration of the vertices and using the fact that either a x b or a /\ b negate with reversal of the vectors.

9. The Virtuosi says:

Now, if I’m not mistaken, the generalized Stokes theorem
$\int_{V} d\omega = \int_{\partial V} \omega$

tells us that
$\int_{S} d\vec A = \int_V d^2 \vec A$
and since dA is closed, d2A is zero. QED.

I think this is what Maya was hinting at.

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