Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Mongolia

Given nn different points in a plane, prove that among these points it is always possible to find 3 points forming an angle not exceeding πn\frac{\pi}{n}.

Solution

If 3 of points lie simultaneously on a line there is nothing to prove. Therefore suppose that, no two of them don't lie on a straight line and . We claim there is a convex polygon with knk \le n vertices such that all nn points lie inside of the polygon. It is obvious that there exists an angle of the polygon not greater than (n2)πn\frac{(n-2)\pi}{n}. Since all points lie inside the angle we can draw rays through all points from vertex of the angle. By the way the angle is divided in (n2)(n - 2) angles, the sum of which is not less than (n2)πn\frac{(n-2)\pi}{n}. It implies that there exists an angle not greater than πn\frac{\pi}{n}. If we take 3 points which form the angle we have done.

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