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Brazilian Math Olympiad

Algebra Difficulty 4.7 AIME Prove it Brazil

Let $a$ , $b$ and $c$ be real numbers such that $a \neq b$ and $a^2(b + c) = b^2(c + a) = 2010$ . Compute $c^2(a + b)$ .

Solution

Since $a \neq b$ , $a^2(b+c) = b^2(c+a) \iff a^2b + a^2c - b^2c - ab^2 = 0 \iff ab(a-b) + c(a-b)(a+b) = 0 \iff ab+ca+bc = 0$ .

So $(a-c)(ab+bc+ca) = 0 \iff a^2b + abc + a^2c = abc + bc^2 + ac^2 \iff c^2(a+b) = a^2(b+c) = 2010$ .

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