\magnification=1200 \hsize=4in \overfullrule=0pt \input amssym %\def\frac#1 #2 {{#1\over #2}} \def\emph#1{{\it #1}} \def\em{\it} \nopagenumbers \noindent % % {\bf Necklace Bisection with One Cut Less than Needed} % % \medskip \noindent % % {\bf G\'abor Simonyi} % % \vskip 5mm \noindent % % % % A well-known theorem of Goldberg and West states that two thieves can always split a necklace containing an even number of beads from each of $k$ types fairly by at most $k$ cuts. We prove that if we can use at most $k-1$ cuts and fair splitting is not possible then the thieves still have the following option. Whatever way they specify two disjoint sets $D_1, D_2$ of the types of beads with $D_1\cup D_2\neq\emptyset$, it will always be possible to cut the necklace (with $k-1$ cuts) so that the first thief gets more of those types of beads that are in $D_1$ and the second gets more of those in $D_2$, while the rest is divided equally. The proof combines the simple proof given by Alon and West to the original statement with a variant of the Borsuk-Ulam theorem due to Tucker and Bacon. \bye .