Library / /384 of 1394

, 2019

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $x$ and $y$ be positive real numbers. Define $a=1+\frac{x}{y}$ and $b=1+\frac{y}{x}$ . If $a^{2}+b^{2}=15$ , compute $a^{3}+b^{3}$ .

Solution

Solution:

Note that $a-1=\frac{x}{y}$ and $b-1=\frac{y}{x}$ are reciprocals. That is,
$(a-1)(b-1)=1 \Longrightarrow a b-a-b+1=1 \Longrightarrow a b=a+b$
Let $t=a b=a+b$ . Then we can write
$a^{2}+b^{2}=(a+b)^{2}-2 a b=t^{2}-2 t$
so $t^{2}-2 t=15$ , which factors as $(t-5)(t+3)=0$ . Since $a, b>0$ , we must have $t=5$ . Then, we compute
$a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=5^{3}-3 \cdot 5^{2}=50$

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.