Let be a positive integer and let be a strictly increasing sequence of positive real numbers with sum equal to 2. Let be a subset of such that the value of
is minimized. Prove that there exists a strictly increasing sequence of positive real numbers with sum equal to 2 such that
(New Zealand) Common remarks. In all solutions, we say an index set is -minimizing if it has the property in the problem for the given sequence . Write for the complement of , and for the interval of integers such that . Note that
so we may exchange and where convenient. Let
and note that is -minimizing if and only if it minimizes , and that if and only if .
In some solutions, a scaling process is used. If we have a strictly increasing sequence of positive real numbers (typically obtained by perturbing the in some way) such that
then we may put . 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 to form the (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 . Solution 3 then perturbs many of the to form the , together with scaling. Rather than using such perturbations, Solution 4 constructs a sequence directly from the set with the required properties. Solution 4 can be used to give a complete description of sets that are -minimizing for some .
Solution
Without loss of generality, assume , and we may assume strict inequality as otherwise works. Also, clearly cannot be empty. If , add to , producing a sequence of with , and then scale as described above to make the sum equal to 2 . Otherwise, there is some with and . Let . - If , add to and then scale. - If , then considering contradicts being -minimising. - If , choose any (possible since ), and any less than the least of and all the differences . If then add to and to , then scale; otherwise, add to and to , and subtract from , then scale.
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