HMMT November, 2018

Algebra Difficulty 4.5 AIME Prove it United States

Problem:
Let $a, b, c, n$ be positive real numbers such that $\frac{a+b}{a}=3$ , $\frac{b+c}{b}=4$ , and $\frac{c+a}{c}=n$ . Find $n$ .

Solution

Solution:
Answer: $\frac{7}{6}$
We have
$1=\frac{b}{a} \cdot \frac{c}{b} \cdot \frac{a}{c}=(3-1)(4-1)(n-1) .$
Solving for $n$ yields $n=\frac{7}{6}$ .

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