Browse · MathNet
PrintTHE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS
Romania counting and probability
Problem
Consider a finite collection of 3-element sets no two of which share more than one element, whose union has cardinality . Show that the elements of this union can be coloured one of two colours, blue and red, so that at least elements are blue, and each contains at least one red element.
Solution
Let be the union of the . It is sufficient to show that a maximal (relative to set-theoretic inclusion) subset of containing no satisfies the required cardinality condition.
By maximality, for each element of , there exists a 2-element subset of such that is an .
If and are distinct elements of , and and are 2-element subsets of such that and are both amongst the , then and are also distinct, since two distinct 's share at most one element.
Therefore, choosing for each in a 2-element subset of such that is an defines an injection of into the collection of 2-element subsets of . Consequently, , so ; if , then .
By maximality, for each element of , there exists a 2-element subset of such that is an .
If and are distinct elements of , and and are 2-element subsets of such that and are both amongst the , then and are also distinct, since two distinct 's share at most one element.
Therefore, choosing for each in a 2-element subset of such that is an defines an injection of into the collection of 2-element subsets of . Consequently, , so ; if , then .
Techniques
Coloring schemes, extremal argumentsCounting two ways