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Romanian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Romania

Let xx, yy, zz be real numbers such that:
x2+4y2+9z2+20=4x+12y+24z. x^2 + 4y^2 + 9z^2 + 20 = 4x + 12y + 24z.

Prove that x∈[−1,5]x \in [-1, 5], y∈[0,3]y \in [0, 3] and z∈[13,73]z \in \left[\frac{1}{3}, \frac{7}{3}\right].

Solution

We rewrite the given equality as:
(x−2)2+(2y−3)2+(3z−4)2=9. (x-2)^2 + (2y-3)^2 + (3z-4)^2 = 9.
This implies (x−2)2≤9(x-2)^2 \le 9, (2y−3)2≤9(2y-3)^2 \le 9 and (3z−4)2≤9(3z-4)^2 \le 9, therefore ∣x−2∣≤3|x-2| \le 3, ∣2y−3∣≤3|2y-3| \le 3, ∣3z−4∣≤3|3z-4| \le 3 and the last inequalities are equivalent to the conclusion.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.