# 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 $S^1 = \big\{(x,y) \in \mathbb{R}^2 \, : \, x^2 + y^2 = 1\big\}$. For each direction $m = \big(\cos(\theta), \sin(\theta)\big) \in S^1$, 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 $f(m) := \text{Area} \big(K_2 \cap P (m)\big).$ 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 $\text{Area} \big(K_2 \cap N (n)\big)$, since the positive side of $\ell(-n)$ is precisely the negative side of $\ell(n)$. Thus, $\text{Area}\big(K_2 \cap P(n)\big) = \text{Area}\big(K_2 \cap N(n)\big)$. 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 $K_2$! $_\square$

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/