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Combinatorics Difficulty 6.5 National olympiad Prove it

Let n3n \geqslant 3 be a positive integer and let (a1,a2,,an)\left(a_{1}, a_{2}, \ldots, a_{n}\right) be a strictly increasing sequence of nn positive real numbers with sum equal to 2. Let XX be a subset of {1,2,,n}\{1,2, \ldots, n\} such that the value of
1iXai \left|1-\sum_{i \in X} a_{i}\right|
is minimized. Prove that there exists a strictly increasing sequence of nn positive real numbers (b1,b2,,bn)\left(b_{1}, b_{2}, \ldots, b_{n}\right) with sum equal to 2 such that
iXbi=1 \sum_{i \in X} b_{i}=1
(New Zealand) Common remarks. In all solutions, we say an index set XX is (ai)\left(a_{i}\right)-minimizing if it has the property in the problem for the given sequence (ai)\left(a_{i}\right). Write XcX^{c} for the complement of XX, and [a,b][a, b] for the interval of integers kk such that akba \leqslant k \leqslant b. Note that
1iXai=1iXcai, \left|1-\sum_{i \in X} a_{i}\right|=\left|1-\sum_{i \in X^{c}} a_{i}\right|,
so we may exchange XX and XcX^{c} where convenient. Let
Δ=iXcaiiXai \Delta=\sum_{i \in X^{c}} a_{i}-\sum_{i \in X} a_{i}
and note that XX is (ai)\left(a_{i}\right)-minimizing if and only if it minimizes Δ|\Delta|, and that iXai=1\sum_{i \in X} a_{i}=1 if and only if Δ=0\Delta=0.
In some solutions, a scaling process is used. If we have a strictly increasing sequence of positive real numbers cic_{i} (typically obtained by perturbing the aia_{i} in some way) such that
iXci=iXcci \sum_{i \in X} c_{i}=\sum_{i \in X^{c}} c_{i}
then we may put bi=2ci/j=1ncjb_{i}=2 c_{i} / \sum_{j=1}^{n} c_{j}. So it suffices to construct such a sequence without needing its sum to be 2.
The solutions below show various possible approaches to the problem. Solutions 1 and 2 perturb a few of the aia_{i} to form the bib_{i} (with scaling in the case of Solution 1, without scaling in the case of Solution 2). Solutions 3 and 4 look at properties of the index set XX. Solution 3 then perturbs many of the aia_{i} to form the bib_{i}, together with scaling. Rather than using such perturbations, Solution 4 constructs a sequence (bi)\left(b_{i}\right) directly from the set XX with the required properties. Solution 4 can be used to give a complete description of sets XX that are (ai)\left(a_{i}\right)-minimizing for some (ai)\left(a_{i}\right).

Solution

Without loss of generality, assume iXai1\sum_{i \in X} a_{i} \leqslant 1, and we may assume strict inequality as otherwise bi=aib_{i}=a_{i} works. Also, XX clearly cannot be empty. If nXn \in X, add Δ\Delta to ana_{n}, producing a sequence of cic_{i} with iXci=iXcci\sum_{i \in X} c_{i}=\sum_{i \in X^{c}} c_{i}, and then scale as described above to make the sum equal to 2 . Otherwise, there is some kk with kXk \in X and k+1Xck+1 \in X^{c}. Let δ=ak+1ak\delta=a_{k+1}-a_{k}. - If δ>Δ\delta>\Delta, add Δ\Delta to aka_{k} and then scale. - If δ<Δ\delta<\Delta, then considering X{k+1}\{k}X \cup\{k+1\} \backslash\{k\} contradicts XX being (ai)\left(a_{i}\right)-minimising. - If δ=Δ\delta=\Delta, choose any jk,k+1j \neq k, k+1 (possible since n3n \geqslant 3 ), and any ϵ\epsilon less than the least of a1a_{1} and all the differences ai+1aia_{i+1}-a_{i}. If jXj \in X then add Δϵ\Delta-\epsilon to aka_{k} and ϵ\epsilon to aja_{j}, then scale; otherwise, add Δ\Delta to aka_{k} and ϵ/2\epsilon / 2 to ak+1a_{k+1}, and subtract ϵ/2\epsilon / 2 from aja_{j}, then scale.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.