Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Belarus

Let x1,,x100x_1, \ldots, x_{100} be nonnegative real numbers such that xi+xi+1+xi+21x_i + x_{i+1} + x_{i+2} \le 1 for all i=1,,100i = 1, \ldots, 100 (we put x101=x1,x102=x2x_{101} = x_1, x_{102} = x_2).
Find the maximal possible value of the sum S=i=1100xixi+2S = \sum_{i=1}^{100} x_i x_{i+2}.

Solution

3. See IMO-2010 Shortlist, Problem A3.

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