AoPS - Problem

Source: IMO LongList 1959-1966 Problem 52

Tags: geometry, combinatorics, area, combinatorial geometry, dissection, IMO Shortlist, IMO Longlist

orl

02.09.2004 05:13

A figure with area $1$ is cut out of paper. We divide this figure into $10$ parts and color them in $10$ different colors. Now, we turn around the piece of paper, divide the same figure on the other side of the paper in $10$ parts again (in some different way). Show that we can color these new parts in the same $10$ colors again (hereby, different parts should have different colors) such that the sum of the areas of all parts of the figure colored with the same color on both sides is $\geq \frac{1}{10}.$

Here's a solution. call the areas on the second side $A_1,A_2,...,A_{10}$, let the colors be $c_1,c_2,...,c_{10}$. now, consider the 10 colorings coloring $A_i$ with $c_{i+k}$, $k=0,...,9$, where indices are mod 10. summing all the areas up which are in common for the two sides of the paper, we immediately see that each point is colored exactly once with any color, thus we get exactly 1(every point on the back has a certain color and this color occurs exactly once and the area of the figure is 1). it immediately follows that for some coloring, the areas with same color are at least $\frac {1}{10}$, it even follows that there is some coloring where it is at most $\frac {1}{10}$. Peter