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Algebra Difficulty 5.1 AIME, harder Find the answer

Find the largest real number cc such that ∑i=1101xi2≥cM2\sum_{i=1}^{101} x_{i}^{2} \geq c M^{2} whenever x1,…,x101x_{1}, \ldots, x_{101} are real numbers such that x1+⋯+x101=0x_{1}+\cdots+x_{101}=0 and MM is the median of x1,…,x101x_{1}, \ldots, x_{101}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Suppose without loss of generality that x1≤⋯≤x101x_{1} \leq \cdots \leq x_{101} and M=x51≥0M=x_{51} \geq 0. Note that f(t)=t2f(t)=t^{2} is a convex function over the reals, so we may "smooth" to the case x1=⋯=x50=−51rx_{1}=\cdots=x_{50}=-51 r and x51=⋯=x101=50rx_{51}=\cdots=x_{101}=50 r for some r≥0r \geq 0, and by homogeneity, CC works if and only if C≤50(51)2+51(50)2502=51(101)50=515150C \leq \frac{50(51)^{2}+51(50)^{2}}{50^{2}}=\frac{51(101)}{50}=\frac{5151}{50}.

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