Ham sandwich theorem

Remarks

The usual proof applies the Borsuk–Ulam theorem. A unit vector $u = (u_0, u_1, \dots, u_d)$ of $\R^{d+1}$ gives the half-space $H_u^+ = \set{x \in \R^d : u_0 + u_1 x_1 + \cdots + u_d x_d > 0}$. The map $f \colon S^d \to \R^d$ with $f_i(u) = \mu_i(H_u^+) - \mu_i(H_{-u}^+)$ is continuous and odd, so it has a zero $u$. At such a zero the hyperplane $u_0 + u_1 x_1 + \cdots + u_d x_d = 0$ bisects every $\mu_i$; the points $u = (\pm 1, 0, \dots, 0)$ are not zeros as long as the measures are nonzero. For $d = 2$: any two bounded regions of the plane, such as two pancakes, can be cut in half at once by one straight line. Absolute continuity is needed: for $d = 1$ and $\mu_1$ a single point mass, no point splits $\mu_1$ into two open halves of measure $\frac12$ each.

Going deeper

For a student

The theorem is a standard first application of the Borsuk–Ulam theorem to combinatorial geometry. It has a discrete version for finite point sets, where the hyperplane may contain some points and each open side has at most half of each set. It is used for partitioning point sets in computational geometry, and it is related to the necklace splitting problem.