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2002 USA TST Problems/Problem 4

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Problem

(David Savitt) Let \displaystyle n be a positive integer and let \displaystyle S be a set of \displaystyle 2^n+1 elements. Let \displaystyle f be a function from the set of two-element subsets of \displaystyle S to \{0, \dots, 2^{n-1}-1\}. Assume that for any elements \displaystyle x, y, z of \displaystyle S, one of \displaystyle f(\{x,y\}), f(\{y,z\}), f(\{z, x\}) is equal to the sum of the other two. Show that there exist \displaystyle a, b, c in \displaystyle S such that \displaystyle f(\{a,b\}), f(\{b,c\}), f(\{c,a\}) are all equal to 0.

Solution

We will prove that for any integer 0 \le k \le n-1, there exists a set S_k \subseteq S of cardinality at least \displaystyle 2^{n-k} + 1 such that for any a, b \in S_k, 2^k \mid f(\{a,b\}).

Our base case is \displaystyle S_0 = S. Now, asume that there exists such an \displaystyle S_{k-1} (1 \le k \le n). We will consider two elements \displaystyle a, b of \displaystyle S_{k-1} to be related if \displaystyle a=b, or if \displaystyle f(\{a,b\}) is divisible by \displaystyle 2^k. Any two elements of \displaystyle S_{k-1} which are related to some a \in S_{k-1} must also be related to each other; similarly, any two elements of \displaystyle S_{k-1} which are not related to \displaystyle a must be related to each other. Thus the relation partitions \displaystyle S_{k-1} into two equivalence classes; we pick the larger one, which must have at least \displaystyle 2^{n-k} + 1 elements, to be \displaystyle S_k.

Thus there must exist an \displaystyle S_{n-1} with at least three elements \displaystyle a,b,c such that 2^n-1 \mid f(\{a,b\}), f(\{b,c\}), f(\{c,a\}). But the only possible value of \displaystyle {} f(\{x,y\}) which is divisible by \displaystyle 2^{k-1} is zero.


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Resources

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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