Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

4. Let real numbers $a_{1}, a_{2}, \cdots, a_{2017}$ satisfy
$\begin{array}{l} a_{1}=a_{2017}, \\ \left|a_{i}+a_{i+2}-2 a_{i+1}\right| \leqslant 1(i=1,2, \cdots, 2015) . \end{array}$

Let $M=\max _{1 \leqslant i<j \leqslant 2017}\left|a_{i}-a_{j}\right|$ . Find the maximum value of $M$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

4. Let $\left|a_{i_{0}}-a_{j_{0}}\right|=\max _{1 \leqslant i1008$ when, then
$\begin{array}{l} \left|a_{j_{0}}-a_{i_{0}}\right|=\left|a_{i_{0}}-a_{1}\right|+\left|a_{2017}-a_{j_{0}}\right| \\ \leqslant \frac{\left(i_{0}-1\right)^{2}}{2}+\frac{\left(2017-j_{0}\right)^{2}}{2} \\ \leqslant \frac{\left(2016-j_{0}+i_{0}\right)^{2}}{2} \leqslant \frac{1008^{2}}{2} . \end{array}$

The above shows that $M \leqslant \frac{1008^{2}}{2}$ .
Taking $a_{n}=\frac{(1009-n)^{2}}{2}(n=1,2, \cdots, 2017)$ , then $a_{1009}-a_{1}=\frac{1008^{2}}{2}$ .
At this point, $M=\frac{1008^{2}}{2}$ .
Therefore, the maximum value of $M$ is $\frac{1008^{2}}{2}$ .

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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.