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".

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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

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

by

$$f(Q)=(T(v_2)-T(v_1),T(v_3)-T(v_2),T(v_1)-T(v_3)).$$

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?