1989 IMO Problems/Problem 3
Contents
Problem
Let and be positive integers and let be a set of points in the plane such that
i.) no three points of are collinear, and
ii.) for every point of there are at least points of equidistant from
Prove that:
Solution 1
Let be our set of points. We count the pairs such that . There are choices for , and for each there are choices for , so the number of such pairs is . On the other hand, there are choices for , and for each there are at most two choices for (because we can't have points on the perpendicular bisector of ). This means that the number of pairs is . We have thus found . If , then , contradicting the last inequality in , so we're done.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]
Solution 2
So obvious simplification and the fact that is an integer gives that what we want is Now for each point, draw a circle around it that contains at least points in . For any point in , let be the number of circles it's on. For any point in , let be the number of points on the defined circle with center , so for all . Since the function is increasing for an integer, for all . So . But notice that since any pair of two points shares at most 2 circles (otherwise we have 3 circles with centers on their perpendicular bisector, which is not allowed), , since the left side counts the number of point-pairs that share a circle (with possible circle multiplicity) in while the right side upper-bounds the number of point-pairs that can share a circle (with possible circle multiplicity). So now , which is what we want.
This solution was posted and copyrighted by antimonyarsenide. The original thread for this problem can be found here: [2]
See Also
1989 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |