Library / /68 of 740

, 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 $n=\frac{7}{6}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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