# Ham Sandwich Theorem

The **ham sandwich theorem** states that given \(n\) objects floating in \(n\)-dimensional space, there exists a single \((n-1)\)-dimensional hyperplane that simultaneously cuts all \(n\) objects into two pieces of equal volume. In the case \(n=3\), this theorem states that if two pieces of bread and one piece of ham are floating in three-dimensional space, then there is a single plane that slices each of the three items into two equal-volume chunks (hence, the name "ham sandwich theorem").

## The Pancake Theorem: Proof When \(n=2\)

In the case \(n=2\), the ham sandwich theorem states that given two disjoint regions of the plane, there is a line that simultaneously divides both regions into two pieces of equal area. This special case is known as the **pancake theorem**, since regions of the plane can look a bit like pancakes.

Let \(K_1, K_2 \subset \mathbb{R}^2\) be pancakes, with \(K_1 \cap K_2 = \emptyset\). There exists a line that simultaneously slices both \(K_1\) and \(K_2\) into pieces of equal area.

The pancake theorem can be proved using the intermediate value theorem:

Each direction in the plane \(\mathbb{R}^2\) has a corresponding unit vector, which can be thought of as a point in the circle \(S^1 = \{(x,y) \in \mathbb{R}^2 \, : \, x^2 + y^2 = 1\}\). For each direction \(m = (\cos(\theta), \sin(\theta)) \in S^1\), there is a unique line \(\ell (m)\) with slope \(\tan(\theta)\) that divides the region \(K_1\) into two pieces of equal area.

Furthermore, the line \(\ell (m)\) divides \(\mathbb{R}^2\) into two regions, which we arbitrarily denote the

positive side\(P (m)\) and thenegative side\(N (m)\). Define a function \(f: S^1 \to \mathbb{R}\) by \[ f(m) := \text{Area} (K_2 \cap P (m))\] That is, \(f(m)\) equals the area of the part of \(K_2\) located on the positive side of \(\ell(m)\). Since \(f\) is a continuous function, we can apply the \(1\)-dimensional Borsuk-Ulam Theorem. This implies there is some \(n\in S^1\) such that \(f(n) = f(-n)\).Note that \(f(-n)\) is precisely \(\text{Area} (K_2 \cap N (n))\), since the positive side of \(\ell(-n)\) is precisely the negative side of \(\ell(n)\). Thus, \(\text{Area}(K_2 \cap P(n)) = \text{Area}(K_2 \cap N(n))\). This means \(\ell(n)\) bisects \(K_2\). But we constructed \(\ell\) so that \(\ell(m)\) bisects \(K_1\) for all \(m \in S^1\). Thus, the line \(\ell(n)\) simultaneously bisects \(K_1\) and \(K_2\)!

The proof of the ham sandwich theorem for \(n>2\) is essentially the same, but requires a higher-dimensional analog of the Borsuk-Ulam Theorem. Unfortunately, the intermediate value theorem does not suffice to prove these higher dimensional analogs; one needs to use the machinery of algebraic topology.

Let \(K_1\) denote the region of \(\mathbb{R}^2\) bounded by the ellipse with equation \[\frac{(x-9)^2}{9} + \frac{(y-9)^2}{16} = 1\] and let \(K_2\) denote the region bounded by the ellipse with equation \[\frac{(x+1)^2}{16} + \frac{(y+3)^2}{9} = 1,\] There is a unique line \(\ell\) which simultaneously bisects \(K_1\) and \(K_2\) into two pieces of equal area. What is the \(y\)-intercept of \(\ell\)?

**Cite as:**Ham Sandwich Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/ham-sandwich-theorem/