In a convex 100-gon all the vertices, as well as some other points inside the polygon, are selected. No three of them are collinear. The selected points are joined (by straight line segments) in such a way that the 100-gon is partitioned into 2011 convex polygons. Prove that at least one of these polygons has an even number of sides.
Solution
Compute a number , which is the sum of numbers of sides of all the polygons of the partition. It is even because it is the sum of 100 (boundary segments) and the doubled number of all segments that are inside the 100-gon. But there is an odd number of summands (exactly 2011), and so it cannot happen that all of them are odd as they have an even sum. So, at least one of is even, as required.
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