Maths Olympiad Prep

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Combinatorics Difficulty 6.9 National olympiad Prove it

All vertices of a regular 100-gon are colored in 10 colors. Prove that there exist 4 vertices of the given 100-gon which are the vertices of a rectangle and which are colored in at most 2 colors.

Solution

1. Applying the Pigeonhole Principle:
Since there are 100 vertices and 10 colors, by the Pigeonhole Principle, there must be at least one color that is used for at least 10010=10 \left\lceil \frac{100}{10} \right\rceil = 10 vertices. Let's denote this color as color C1 C_1 .

2. **Considering the 10 vertices of color C1 C_1 :**
We have 10 vertices all colored with C1 C_1 . We need to check if any of these vertices are diametrically opposite to each other.

3. Checking for diametrically opposite vertices:
If there exists a pair of diametrically opposite vertices among these 10 vertices, let's denote them as A A and C C . Since the 100-gon is regular, the vertex diametrically opposite to any vertex A A is exactly 50 positions away from A A . Thus, C C is 50 positions away from A A .

4. **Forming a rectangle with another vertex of color C1 C_1 :**
If we can find another vertex B B of color C1 C_1 , then the vertex diametrically opposite to B B (denote it as D D ) will also be 50 positions away from B B . The vertices A,B,C,D A, B, C, D will form a rectangle, and since A,B, A, B, and C C are of the same color C1 C_1 , the rectangle ABCD ABCD will have vertices of at most 2 colors.

5. Considering the case where no diametrically opposite vertices are of the same color:
If no pair of the 10 vertices of color C1 C_1 are diametrically opposite, then each of the 10 vertices of color C1 C_1 must have their diametrically opposite vertices colored with one of the remaining 9 colors. This gives us 10 vertices (each diametrically opposite to one of the 10 vertices of color C1 C_1 ) and 9 colors.

6. Applying the Pigeonhole Principle again:
By the Pigeonhole Principle, among these 10 vertices, at least two must share the same color. Let's denote these two vertices as E E and F F , and their diametrically opposite vertices as G G and H H , respectively. Since E E and F F are of the same color, and G G and H H are of color C1 C_1 , the vertices E,F,G,H E, F, G, H form a rectangle with vertices of at most 2 colors.

Thus, in both cases, we have shown that there exist 4 vertices of the 100-gon which form a rectangle and are colored in at most 2 colors.

\blacksquare

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.