CombinatoricsDifficulty 5.5AIME, harderProve itUnited States
Problem:
Let G be a weighted bipartite graph A∪B, with ∣A∣=∣B∣=n. In other words, each edge in the graph is assigned a positive integer value, called its weight. Also, define the weight of a perfect matching in G to be the sum of the weights of the edges in the matching. Let G′ be the graph with vertex set A∪B, and contains the edge e if and only if e is part of some minimum weight perfect matching in G. Show that all perfect matchings in G′ have the same weight.
Solution
Solution:
Let m denote the minimum weight of a matching in G. Let G′′ be the (multi)graph formed by taking the union of all minimum weight perfect matchings, but keeping edges multiple times. Note that G′′ is regular. Now, assume that some perfect matching in G′ (equivalently G′′) has weight M>m. Delete this matching from G′′, and call the resulting graph H. H is still regular, so it can be decomposed into a union of perfect matchings. Now using pigeonhole directly gives us a contradiction.
To show that all regular bipartite graphs can be decomposed into a union of perfect matchings, use Hall's marriage lemma to take out one perfect matching and use induction.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.