ISL 2003 C1

by Wolstenholme, Aug 4, 2014, 5:22 PM

Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.

Proof:

Let each element of $ S $ be the vertex of a graph where two vertices $ u, v $ are connected by an edge if and only if the sets $ A + u $ and $ A + v $ are disjoint. Consider an arbitrary vertex $ v $. Since the $ |A + v| = 101 $ the maximum number of vertices $ w $ such that sets $ A + v $ and $ A + w $ are not disjoint is $ 100 * 101 $. Therefore every vertex of the graph has degree at least $ 10^6 - 100*101 - 1. $ Therefore the graph has at least $ \frac{10^6(10^6 - 100*101 - 1)}{2} $ edges. It suffices to show that this graph contains $ K_{100} $ as a subgraph.

Now, by Turan's Theorem, the maximum number of edges a graph with $ 10^6 $ vertices that does not contain $ K_{100} $ may contain is obtained when the graph is a complete $ 99 $-partite graph with $ 98 $ independent sets of size $ 10101 $ and $ 1 $ independent set of size $ 10102 $. It is easy to compute that this graph has $ \binom{98}{2}10101^2 + 98 \cdot 10101 \cdot 10102 = 494949494949 $ edges. But since $ \frac{10^6(10^6 - 100*101 - 1)}{2} = 494949500000 > 494949494949 $ we have the desired result.

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