HMMT November, 2014

Algebra Difficulty 5.0 AIME Prove it United States

Problem:
Let $a, b, c, x$ be reals with $(a+b)(b+c)(c+a) \neq 0$ that satisfy
$\frac{a^{2}}{a+b}=\frac{a^{2}}{a+c}+20, \quad \frac{b^{2}}{b+c}=\frac{b^{2}}{b+a}+14, \quad \text{ and } \quad \frac{c^{2}}{c+a}=\frac{c^{2}}{c+b}+x$

Compute $x$ .

Solution

Solution:
Answer: $-34$ Note that
$\begin{aligned} \frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}-\frac{a^{2}}{c+a}-\frac{b^{2}}{a+b}-\frac{c^{2}}{b+c} & =\frac{a^{2}-b^{2}}{a+b}+\frac{b^{2}-c^{2}}{b+c}+\frac{c^{2}-a^{2}}{c+a} \\ & =(a-b)+(b-c)+(c-a) \\ & =0 \end{aligned}$
Thus, when we sum up all the given equations, we get that $20+14+x=0$ . Therefore, $x=-34$ .

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