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Wonder whether you know where the following pearl of Topology first appeared:

Prove that at any instant of time you can find three isothermal points on the surface of the Earth that correspond to the vertices of an equilateral triangle.

According to somebody, all problems appeared once in the Monthly. Does that dictum applies to the above teaser, too?

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2 Answers

up vote 8 down vote accepted

Let me just elaborate a little on the references that Charlie Frohman listed (so this isn't really a separate answer, but it's too long for a comment).

The theorem for equilateral triangles is due to S. Kakutani (1942 Annals). He stated it just for triangles formed by orthonormal bases for $\mathbb R^3$, but the argument applies for all sizes of equilateral triangles. He deduced the interesting corollary that any compact convex set in $\mathbb R^3$ has a circumscribing cube, answering a question posed by Rademacher. Charlie Frohman's proof is similar in spirit to Kakutani's.

The n-dimensional generalization of Kakutani's theorem (and corollary) is due to H. Yamabe and Z. Yujobo (1950 Osaka Math J).

Returning to 3 dimensions, the corresponding result for 4 points at the vertices of a square inscribed in a great circle was shown by F. Dyson (1951 Annals), and G.R. Livesay generalized this to rectangles inscribed in a great circle (1954 Annals). The case of arbitrary triangles was done by E.E. Floyd (1955 PAMS).

It's interesting that this little topic produced three Annals papers, though each one was short -- just 3 pages. (When was the last 3-page Annals paper?)

The techniques used to prove the later theorems varied considerably. It might be interesting to see which ones can be proved by basic algebraic topology arguments as in Kakutani's theorem (which certainly deserves to be included in algebraic topology textbooks!).

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Oh the good ole days, when Annals published 3-page papers of general interest. –  Victor Protsak Jun 2 '10 at 1:43
Thanks a lot for taking the time to leave your reply, Professor Hatcher. –  J. H. S. Jun 3 '10 at 21:21
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I don't know if this ever appeared in the Monthly, but it is a version of the Ham Sandwich theorem. The Ham Sandwich theorem says that if you have three bounded regions of finite volume in space, for instance the ham, the swiss cheese, and the rye bread, then it is possible with one cut of a knife to divide it into two sandwiches having equal amounts of the three ingredients.

To prove the Ham Sandwich theorem you place the three sphere on top of $\mathbb{R}^3$ in four space. Every point on the three sphere defines a three dimensional hyperplane through the center of of the sphere, that cuts $\mathbb{R}^3$ in a plane, except for two that miss. The point in the sphere gives the preferred side of the plane. The differences in the measures of the three sets, amount on the preferred side minus the amount on the non preferred side extends to define a continuous function from the sphere to $\mathbb{R}^3$ that is symmetric about $\vec{0}$. By the Borsuk-Ulam theorem it takes on the value $\vec{0}$ and the corresponding plane cuts the sandwich, evenly in two. I read it in "Hilton and Wylie".


Allen Hatcher points out that the proof that I gave is incorrect as the map I constructed from $S^3$ lacks the appropriate symmetry.

Undaunted I present a new proof. To start with choose a congruence class of equilateral triangles with angle sum strictly less than $3\pi$. Such triangles have a natural orientation as they bound regions of different areas on the two sides.

The space of triangles with an ordering on the vertices in that congruence class is $\mathbb{R}P(3)$. To see this you characterize such a triangle by its first vertex and the tangent vector of the side going from the first to the second vertex. This is the unit tangent bundle of the sphere which is $\mathbb{R}P(3)$.

There is a fixed point free map of order three on the space of conjugacy classes that rotates a triangle taking the first vertex to the second, second to the third and third to the first. This is where we need equilateral.

Since the whole congruence class of triangles is the same as the space of oriented triangles, we have that the congruence class is $L(6,1)$. Thats the lens space with first homology $\mathbb{Z}_6$.

Let $T:S^2\rightarrow \mathbb{R}$ be temperature, and define

$$ f:\mathbb{R}P(3)\rightarrow \mathbb{R}^3 $$



Notice it is really a map to the plane perpendicular to $(1,1,1)$ as the sum of the entries is zero. If it did not take on the value $\vec{0}$ then we can homotope it to a map $f:\mathbb{R}P(3)\rightarrow S^1$ that takes the symmetry corresponding to rotation of the triangle to rotation of the circle by $2\pi/3$. This means that the generator of the first homology of $L(6,1)$ goes to a nontrivial homology class in the circle, which is ridiculous as the first space has homology $\mathbb{Z}_6$ and the second has homology $\mathbb{Z}$.

I get the last congruence class of triangles, those with angle sum $3\pi$ by taking the closure in the whole space of equilateral triangles.

OK. Its not the Ham Sandwich theorem its a scholium of the proof of Borsuk-Ulam. I stand corrected.

Is this the end of the story? Is there an analogous theorem for a class of simplices in a higher dimensional sphere?

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Beyer, W. A.; Zardecki, Andrew (2004), "The early history of the ham sandwich theorem" , American Mathematical Monthly 111 (1): 58–61 . –  Charlie Frohman May 29 '10 at 2:56
Could you please explain what it is that you refer to as a "half open interval of congruence classes of equilateral triangles"? Besides, are you sure you meant "conjugacy classes" there? What does the word equitemporal mean? I'm guesssing it means something akin to having the same temperature, but I am not sure about it. Finally, is this cute teaser mentioned in that note that you cited? Thanks in advance for your support. –  J. H. S. May 29 '10 at 9:53
The reason I said conjugacy class by accident is that each oriented equilateral triangle on the two sphere corresponds to an irreducible representation of the fundamental group of the complement of the trefoil, and vice verse. Two representations are conjugate if and only if the two oriented triangles are congruent. :) –  Charlie Frohman May 29 '10 at 10:38
The note in the Monthly doesn't mention your version of the Ham Sandwich theorem. It gives an elementary proof though. I never saw your version before. I put the Ham Sandwich theorem on a qualifying exam once, either that makes me a complete geek, or unusually cruel depending on your viewpoint. –  Charlie Frohman May 29 '10 at 13:04
Inspired by Alan Hatcher's excellent bibliographical comments, I dug through MathSciNet and found out that the general question for $n-1$-simplices on $S^{n-1}$ is the Knaster Problem (for $m=1$). Recently, counterexamples were constructed by Kashin and Szarek for very large $n$, and this was further improved to $n\geq 61, n\ne 62, 64, 66$ by Hinrichs and Richter. So the general answer is "no". Dimensions 4 through 60 may still be open. –  Victor Protsak Jun 7 '10 at 13:03
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