2015 IMO Problems/Problem 1
We say that a finite set in the plane is balanced if, for any two different points , in , there is a point in such that . We say that is centre-free if for any three points , , in , there is no point in such that .
- Show that for all integers , there exists a balanced set consisting of points.
- Determine all integers for which there exists a balanced centre-free set consisting of points.
Part (a): We explicitly construct the sets . For odd , can be taken to be the vertices of regular polygons with sides: given any two vertices and , one of the two open half-spaces into which divides contains an odd number of of vertices of . The vertex encountered while moving from to along the circumcircle of is therefore equidistant from and .
If is even, choose to be the largest integer such that Hence . Consider a circle with centre , and let be distinct points placed counterclockwise (say) on such that (for ). Hence for any line , there is a line such that (using the facts that , and odd). Thus , and form an equilateral triangle. In other words, for arbitrary , there exists equidistant to and . Also given any such that , is equidistant to and . Hence the points form a balanced set.
Part (b): Note that if is odd, the set of vertices of a regular polygon of sides forms a balanced set (as above) and a centre-free set (trivially, since the centre of the circumscribing circle of does not belong to ).
For even, we prove that a balanced, centre free set consisting of points does not exist. Assume that is centre-free. Pick an arbitrary , and let be the number of distinct non-ordered pairs of points () to which is equidistant. Any two such pairs are disjoint (for, if there were two such pairs and with distinct, then would be equidistant to , , and , violating the centre-free property). Hence (we use the fact that is even here), which means . Hence there are at least non-ordered pairs such that no point in is equidistant to and .
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