Maths Olympiad Prep

Track / Stage 6 / 90 of 400 #1090 of 1964

Problem 1090

National olympiad, first round
Combinatorics Difficulty 6.1 Find the answer

3. (CZS1)IMO2(\mathbf{C Z S} 1)^{\mathrm{IMO} 2} On a circle, 2n1(n3)2 n-1(n \geq 3) different points are given. Find the minimal natural number NN with the property that whenever NN of the given points are colored black, there exist two black points such that the interior of one of the corresponding arcs contains exactly nn of the given 2n12 n-1 points.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

3. A segment connecting two points which divides the given circle into two arcs, one of which contains exactly nn points in its interior, we will call a good segment. Good segments determine one or more closed polygonal lines that we will call stars. Let us compute the number of stars. Note first that gcd(n+1,2n1)=gcd(n+1,3)\operatorname{gcd}(n+1,2 n-1)=\operatorname{gcd}(n+1,3). (i) Suppose that 3n+13 \nmid n+1. Then the good segments form a single star. Among any nn points, two will be adjacent vertices of the star. On the other hand, we can select n1n-1 alternate points going along the star, and in this case no two points lie on a good segment. Hence N=nN=n. (ii) If 3n+13 \mid n+1, we obtain three stars of [2n13]\left[\frac{2 n-1}{3}\right] vertices. If more than [2n16]=n23\left[\frac{2 n-1}{6}\right]=\frac{n-2}{3} points are chosen on any of the stars, then two of them will be connected with a good segment. On the other hand, we can select n23\frac{n-2}{3} alternate points on each star, which adds up to n2n-2 points in total, no two of which lie on a good segment. Hence N=n1N=n-1. To sum up, N=nN=n for 32n13 \nmid 2 n-1 and N=n1N=n-1 for 32n13 \mid 2 n-1.

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