[Origin unknown]

Given 2n points in the plane with no three collinear, show that it is possible to pair them up in such a way that the n line segments joining paired points do not intersect.

Since the n line segments joining the paired points are not intersecting, let us draw n horizontal lines. They could be n vertical lines too but we pick horizontal so that it is easier to visualize.

Next assume that the linear segments are of equal lengths, then we have all the n points on one side collinear and all their paired counterpart also collinear.  This forms a ladder.

Since the points are not supposed to be collinear, we increase the distance between each pair gradually as we go up from the bottom of the n pairs to the top leading to a two divergent curves that are narrower at the bottom and flared up at the top. We can adjust the shape of the curve depending by evenly distributing the range between the n-points. If the curves flare up too apart, they tend to form semicircles. Instead we take the curves as divergent quarter circles and then space the n points on them.

Clearly we can picture the solution so the proof exists.

[Russia 2005]

100 people from 25 countries, four from each country, sit in a circle. Prove that one may partition them onto 4 groups in such way that no two countrymen, nor two neighboring people in the circle, are in the same group.

If no two countrymen are in the same group, then they are each in each group. That means that each group has one member for each country. Therefore all the four groups are identical

Now we arrange the four groups one after the other as a quarter circle to form a circle, no two countrymen are in the same group. However, no two neighboring people in the circle should be in the same group.

Assume that each group consists of members 1 through 25  where each number belongs to a country.

Then we interleave twelve from one group with twelve from a reversed group and leave the number 13 from one group at one end.

1 14 2 15 …12 25 13

Now we can create another similar group with the remaining halves and ending with 1 and 13.

We can do the same for the other two groups as we did with this pair. Now the ends of the semi-circles will be 1 and 13 for the group pairs

Now we connect the two pairs into a circle with the 1 and the 13 together thus leading to a solution.