Maths Olympiad Prep

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Combinatorics Difficulty 5.4 AIME, harder Prove it United States

Problem:

On a distant planet, there are 20142014 cities, some pairs of which are connected by two-way roads. It turns out that the population of each city is the average of the populations of the cities to which it is connected by a single road, and moreover that it is possible to travel from every city to every other city by a sequence of roads.
Prove that all cities have the same population.

Solution

Solution:

Consider the city CmaxC_{\max} with the maximal population MM (breaking ties arbitrarily). Then MM is the average of the populations of the neighboring cities, say p1,p2,,pnp_{1}, p_{2}, \ldots, p_{n}, meaning that
p1+p2++pnn=M \frac{p_{1}+p_{2}+\cdots+p_{n}}{n}=M
But p1,p2,,pnMp_{1}, p_{2}, \ldots, p_{n} \leq M, and hence p1+p2++pnnMp_{1}+p_{2}+\cdots+p_{n} \leq n M. So this can only occur if p1=p2==pn=Mp_{1}=p_{2}=\cdots=p_{n}=M. Hence all neighbors of CmaxC_{\max} have population MM.

Proceeding in the same fashion, we find that all neighbors of neighbors of CmaxC_{\max} also must have population MM, and so on. Because the network of cities is connected, this implies that all cities must have population MM.

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