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, 2011

Combinatorics Difficulty 4.8 AIME Prove it South Africa

Let a1,a2,...,an,b1,b2,...,bna_1, a_2, ..., a_n, b_1, b_2, ..., b_n be the numbers 1,2,...,2n1, 2, ..., 2n in some order. Suppose that a1<a2<...<ana_1 < a_2 < ... < a_n and that b1>b2>...>bnb_1 > b_2 > ... > b_n. Prove that
i=1naibi=n2. \sum_{i=1}^{n} |a_i - b_i| = n^2.

Solution

Note that for a given ii, aia_i and bib_i are never both in one of {1,2,...,n}\{1, 2, ..., n\} or {n+1,...,2n}\{n+1, ..., 2n\}. Suppose, without loss of generality, that ai,bina_i, b_i \le n. Then a1,a2,...,aina_1, a_2, ..., a_i \le n, and bi,bi+1,...,bnnb_i, b_{i+1}, ..., b_n \le n, giving n+1n+1 elements less than or equal to nn, which is impossible. Then every aibi|a_i - b_i| is a difference between an element in {n+1,...,2n}\{n+1, ..., 2n\} and an element in {1,2,...,n}\{1, 2, ..., n\}. Therefore:
i=1naibi=i=1n(n+i)i=1ni=n2. \sum_{i=1}^{n} |a_i - b_i| = \sum_{i=1}^{n} (n + i) - \sum_{i=1}^{n} i = n^2.

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