Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

The real numbers x,y,zx, y, z satisfy 0≤x≤y≤z≤40 \leq x \leq y \leq z \leq 4. If their squares form an arithmetic progression with common difference 2, determine the minimum possible value of ∣x−y∣+∣y−z∣|x-y|+|y-z|.

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

Solution

Clearly ∣x−y∣+∣y−z∣=z−x=z2−x2z+x=4z+x|x-y|+|y-z|=z-x=\frac{z^{2}-x^{2}}{z+x}=\frac{4}{z+x}, which is minimized when z=4z=4 and x=12x=\sqrt{12}. Thus, our answer is 4−12=4−234-\sqrt{12}=4-2 \sqrt{3}.

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