Skip to content
ProvedGeometry·1942

Ham Sandwich Theorem

In n dimensions, one hyperplane can simultaneously bisect n measurable bodies.

Formula

Theorem
μ1,,μn in Rnhyperplane H:  μi(H+)=μi(H)  i\mu_1,\dots,\mu_n\text{ in }\mathbb{R}^n\quad\Longrightarrow\quad\exists\,\text{hyperplane }H:\;\mu_i(H^+)=\mu_i(H^-)\;\forall\,i

One hyperplane can simultaneously bisect n finite Borel measures in Rⁿ — proved via the Borsuk–Ulam theorem.

Borsuk–Ulam proof sketch
F:Sn1Rn1,Fi(u)=μi(Hu+)12μiF(u0)=0F:S^{n-1}\to\mathbb{R}^{n-1},\quad F_i(u)=\mu_i(H_u^+)-\tfrac{1}{2}\mu_i\quad\Longrightarrow\quad F(u_0)=\mathbf{0}

Map each direction to the imbalance vector; Borsuk–Ulam forces a zero, which is the bisecting hyperplane.

Polynomial ham sandwich (Guth–Katz)
PR[x1,,xn],  degPd:  P=0 bisects (d+nn)1 measures\exists\,P\in\mathbb{R}[x_1,\dots,x_n],\;\deg P\le d:\;P=0\text{ bisects }\binom{d+n}{n}-1\text{ measures}

The polynomial generalization uses algebraic hypersurfaces instead of hyperplanes, key to the Erdős distinct-distances solution.

Summary

Slice a ham sandwich — two slices of bread and a slab of ham — with one straight cut so each ingredient is exactly halved. The theorem guarantees this is always possible: given n measurable bodies in Rⁿ, a single hyperplane bisects all n simultaneously. The proof is a direct application of Borsuk–Ulam — each direction on Sⁿ⁻¹ determines a family of hyperplanes, and the topological obstruction forces a bisecting cut to exist.

Sources

Videos