Maths Olympiad Prep

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Combinatorics Difficulty 8.8 Shortlist Prove it IMO

In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals Q1,,Q24Q_{1}, \ldots, Q_{24} whose corners are vertices of the 100-gon, so that
- the quadrilaterals Q1,,Q24Q_{1}, \ldots, Q_{24} are pairwise disjoint, and
- every quadrilateral QiQ_{i} has three corners of one color and one corner of the other color.

Solution

Call a quadrilateral skew-colored, if it has three corners of one color and one corner of the other color. We will prove the following

Claim. If the vertices of a convex (4k+1)(4k+1)-gon PP are colored black and white such that each color is used at least kk times, then there exist kk pairwise disjoint skew-colored quadrilaterals whose vertices are vertices of PP. (One vertex of PP remains unused.)

The problem statement follows by removing 3 arbitrary vertices of the 100-gon and applying the Claim to the remaining 97 vertices with k=24k=24.

Proof of the Claim. We prove by induction. For k=1k=1 we have a pentagon with at least one black and at least one white vertex. If the number of black vertices is even then remove a black vertex; otherwise remove a white vertex. In the remaining quadrilateral, there are an odd number of black and an odd number of white vertices, so the quadrilateral is skew-colored.

For the induction step, assume k2k \geqslant 2. Let bb and ww be the numbers of black and white vertices, respectively; then b,wkb, w \geqslant k and b+w=4k+1b+w=4k+1. Without loss of generality we may assume wbw \geqslant b, so kb2kk \leqslant b \leqslant 2k and 2k+1w3k+12k+1 \leqslant w \leqslant 3k+1.

We want to find four consecutive vertices such that three of them are white, the fourth one is black. Denote the vertices by V1,V2,,V4k+1V_{1}, V_{2}, \ldots, V_{4k+1} in counterclockwise order, such that V4k+1V_{4k+1} is black, and consider the following kk groups of vertices:

(V1,V2,V3,V4),(V5,V6,V7,V8),,(V4k3,V4k2,V4k1,V4k) \left(V_{1}, V_{2}, V_{3}, V_{4}\right),\left(V_{5}, V_{6}, V_{7}, V_{8}\right), \ldots,\left(V_{4k-3}, V_{4k-2}, V_{4k-1}, V_{4k}\right)

In these groups there are ww white and b1b-1 black vertices. Since w>b1w > b-1, there is a group, (Vi,Vi+1,Vi+2,Vi+3)\left(V_{i}, V_{i+1}, V_{i+2}, V_{i+3}\right) that contains more white than black vertices. If three are white and one is black in that group, we are done. Otherwise, if Vi,Vi+1,Vi+2,Vi+3V_{i}, V_{i+1}, V_{i+2}, V_{i+3} are all white then let VjV_{j} be the first black vertex among Vi+4,,V4k+1V_{i+4}, \ldots, V_{4k+1} (recall that V4k+1V_{4k+1} is black); then Vj3,Vj2V_{j-3}, V_{j-2} and Vj1V_{j-1} are white and VjV_{j} is black.

Now we have four consecutive vertices Vi,Vi+1,Vi+2,Vi+3V_{i}, V_{i+1}, V_{i+2}, V_{i+3} that form a skew-colored quadrilateral. The remaining vertices form a convex (4k3)(4k-3)-gon; w3w-3 of them are white and b1b-1 are black. Since b1k1b-1 \geqslant k-1 and w3(2k+1)3>k1w-3 \geqslant (2k+1)-3 > k-1, we can apply the Claim with k1k-1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.