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Algebra Difficulty 5.9 AIME, harder Prove it Croatia

Let x1,x2,…,x100x_1, x_2, \dots, x_{100} be real numbers such that
∣2xk−xk+1∣=xk+2for all k∈{1,2,…,98},∣2x99−x100∣=x1,∣2x100−x1∣=x2. \begin{align*} |2x_k - x_{k+1}| &= x_{k+2} \quad \text{for all } k \in \{1, 2, \dots, 98\}, \\ |2x_{99} - x_{100}| &= x_1, \\ |2x_{100} - x_1| &= x_2. \end{align*}
Prove that x1=x2=⋯=x100x_1 = x_2 = \dots = x_{100}.
(Tonći Kokan)

Solution

Squaring the given equations we get
4xk2−4xkxk+1+xk+12=xk+22for k∈{1,2,…,98},4x992−4x99x100+x1002=x12,4x1002−4x100x1+x12=x22. \begin{gathered} 4x_k^2 - 4x_kx_{k+1} + x_{k+1}^2 = x_{k+2}^2 \quad \text{for } k \in \{1, 2, \dots, 98\}, \\ 4x_{99}^2 - 4x_{99}x_{100} + x_{100}^2 = x_1^2, \quad 4x_{100}^2 - 4x_{100}x_1 + x_1^2 = x_2^2. \end{gathered}
Adding these equations we get
4(x12+x22+⋯+x1002)−4(x1x2+⋯+x99x100+x100x1)=0. 4(x_1^2 + x_2^2 + \dots + x_{100}^2) - 4(x_1x_2 + \dots + x_{99}x_{100} + x_{100}x_1) = 0.
After regrouping and dividing by 2 we get
(x1−x2)2+⋯+(x99−x100)2+(x100−x1)2=0. (x_1 - x_2)^2 + \dots + (x_{99} - x_{100})^2 + (x_{100} - x_1)^2 = 0.
Since all the terms are non-negative, they all have to be equal to zero, so we have x1=⋯=x100x_1 = \dots = x_{100}.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.