Maths Olympiad Prep

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Combinatorics Difficulty 6.2 National Olympiad Prove it South Africa

Sixty points, of which thirty are coloured red, twenty are coloured blue, and ten are coloured green, are marked on a circle. These points divide the circle into sixty arcs. Each of these arcs is assigned a number according to the colours of its endpoints: an arc between a red and a green point is assigned a number 1, an arc between a red and a blue point is assigned a number 2, and an arc between a blue and a green point is assigned a number 3. The arcs between two points of the same colour are assigned a number 0. What is the greatest possible sum of all the numbers assigned to the arcs?

Solution

Let the score of a red point be 00, the score of a green point be 11, and the score of a blue point be 22. Note that the number assigned to an arc is at most the sum of the scores of the endpoints. This means that the sum of all the numbers assigned to the arcs is at most twice the sum of all the sixty scores, which is
2(300+202+101)=100. 2(30 \cdot 0 + 20 \cdot 2 + 10 \cdot 1) = 100.
Equality holds if there are no arcs with two green or two blue endpoints. This can be achieved, for instance, by letting red and non-red points alternate. Hence the greatest possible sum is 100100.

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