Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Ireland

Three real numbers $a$ , $b$ , $c$ satisfy the equations
$\begin{align*} a + 2b + 3c &= 12, \\ 2ab + 3ac + 6bc &= 48. \end{align*}$
Solve for $a$ , $b$ , $c$ .

Solution

From $(a + 2b + 3c)^2 = a^2 + 4b^2 + 9c^2 + 2(2ab + 3ac + 6bc)$ we obtain
$a^2 + 4b^2 + 9c^2 = 48.$
Hence $(a-2b)^2 + (2b-3c)^2 + (a-3c)^2 = 2(a^2+4b^2+9c^2) - 2(2ab+3ac+6bc) = 96 - 96 = 0$ . Therefore $a = 2b = 3c$ , and so $a = 4$ , $b = 2$ and $c = 4/3$ .

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