Page 25 - Science Focus (Issue 017)

Hungry Mathema ticians and The Ham Sandwich Theorem
Generalizing this logic, we can arrive at the
the Ham Sandwi ch statement of the Ham Sandwich Theorem itself: given
n objects in n-dimensional space, there exists an (n –
1)-dimensional hyperplane that simultaneously cuts
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肚餓的數學家與他 們的火腿三文治 all n objects in half. For our case, we are dealing with
day-to-day, 3D space (n = 3); the theorem says that
we can cut three objects in half by a 2D plane, in our
case the cutting plane created by our knife.

Some readers might have questions about the continuity We’ll test this statement out with a second
of the argument. Admittedly, the continuity of the scenario. Suppose Bob is craving cheese, and he
argument here is actually not guaranteed by IVT; it is adds in a piece of cheese to the sandwich. However,
more accurate, instead, to use the Borsuk-Ulam Theorem,
a famous result from algebraic topology that can be this seemingly insignificant act might make our task of
described as a sweeping generalization of IVT. I will not dividing the sandwich impossible. From the theorem’s
dive into the details here, as it involves quite a bit of statement, we know that only three objects, but not
difficult mathematics. For the intrepid reader, here is the four, can be divided by a two-dimensional plane.
alternate proof for our ham sandwich problem: As a reminder, there are only three variables we can
The Borsuk-Ulam Theorem guarantees some point where control about the knife.
f(ϕ) = f(–ϕ), the minus sign being our knife “switching One last note — so how exactly should we cut this
sides” (rotating by 180 degrees) — so the only case in
which this relation holds is when the solid is cut into sandwich up? We don’t know. IVT only tells us that such
half. In this case, setting up two volume functions for the a way to cut the sandwich exists; it doesn’t mention
ham and bread respectively, we can directly claim the the way you can cut it up at all. In other words, if your
existence of p(θ, ϕ) s.t. f2(p) = f3(p) = 0, so the two solids little brother has taken a bite from your sandwich, it’s
are bisected. probably a better idea to make a new one for the
friend you intended to share your sandwich with!

An Alternative Way of Thinking 1 For more mathematically inclined readers, the formal
mathematical statement says: If f(x) is continuous between a
If you don’t want to dive into the nitty-gritty of and b, then for any number γ satisfying f(a) < γ < f(b) or f(b) < γ <
the math here, it’s possible to view this problem as a f(a), there is a number c satisfying a < c < b and f(c) = γ. [2]
counting problem. The volume of each of the three 2 Some readers might note that not all sets of three equations
solids is made up of three dimensions — namely the in three unknowns can be solved (for example, the system x +
x, y and z axes, so the volume is a function of three y + z = 1, x + y + z = 2 and x + y + z = 3 is clearly unsolvable). This
is purely an analogy; the mathematical argument using IVT
variables — length (x), width (y) and height (z). There guarantees a solution.
are three different variables you can control. When 3 Hyperplane: A subspace which is one dimension less than its
solving simultaneous equations in three unknowns ambient space; in our case about bisecting the ham sandwich,
that depend on each other, one can at most impose it is just a regular 2D plane/surface.
three non-overlapping conditions in order to get a
solution . This is a similar logic to the puzzle above: we
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can only impose at most three conditions, in this case References 參考資料：
dividing each of the solids into half, for which there is a [1] Steinhaus, H. (1938). A note on the ham sandwich theorem.
Mathesis Polska, 9, 26-28.
solution. By comparing the number of dimensions (three [2] Ross, K. A. (2013). Elementary Analysis: The Theory of Calculus
unknowns, three constraints), we can solve the problem. (2nd ed.). New York, NY: Springer.

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