The circumferences of two concentric disks are each divided into 200 sections.…

Concept

The averaging argument (a counting form of the pigeonhole principle): if a quantity is measured over a finite collection of n cases and its total over all cases is T, then at least one case attains a value of at least the average T/n. No collection of numbers can have every member strictly below its own mean.

The companion technique is double counting: the same total T is computed in two different orders — here, once by summing over the alignments and once by summing over the individual sections — and the two computations must agree.

Application

1. Set up the alignments. Each disk carries 200 equal sections, so rotating the inner disk by a whole number of section-widths gives exactly 200 distinct alignments (rotate by 0, 1, 2, …, 199 sections). In every alignment each inner section faces exactly one outer section.

2. Count what one inner section contributes. Fix any single inner section s and let c be its colour. As the inner disk runs through all 200 alignments, s faces each of the 200 outer sections exactly once. The outer disk is painted with exactly 100 red and 100 blue sections, so exactly 100 of those encounters are with a section of colour c — whichever colour c happens to be. So s produces exactly 100 matches, summed over all 200 alignments.

3. Double count the grand total. That per-section count of 100 does not depend on the inner colouring at all, so summing it over all 200 inner sections gives the total number of matches over all alignments: T = 200 × 100 = 20000.

4. Take the average. Spread over the 200 alignments, the mean number of matches per alignment is T/200 = 20000/200 = 100.

5. Apply the averaging argument. The 200 alignment-scores are non-negative integers with mean 100, so at least one alignment scores 100 or more. This holds for every arbitrary red/blue colouring of the inner disk, because step 2 never used that colouring.

Cross-check: can the guarantee be pushed higher?

A guarantee must survive the worst case, so test the bound against a deliberately awkward inner colouring. Paint every one of the 200 inner sections red. In any alignment whatsoever, the matches are exactly the outer sections that are red — and there are exactly 100 of those. So this colouring scores exactly 100 at every one of the 200 alignments, and never more.

• That single colouring shows no guarantee above 100 can be claimed: 125, 150 and 175 all fail against it, so none of them is achievable for every inner colouring.

• It also confirms the averaging bound is tight — the mean of 100 is actually attained, and cannot be improved in general.

Result

Every colouring admits an alignment with at least 100 matched sections, and the all-red inner disk shows that no larger number can be guaranteed. The correct completion of the blank is therefore “100 or more”.