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Problem 1214

AMC 12 late, AIME early

Algebra Difficulty 5.0 Find the answer HMMT February

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 $a^{3}+b^{3}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official 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 $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=5^{3}-3 \cdot 5^{2}=50$

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.