Question

Suppose that $y = \frac{3}{4}x$ and $x^y = y^x.$ The quantity $x + y$ can be expressed as a rational number $\frac{r}{s},$ where $r$ and $s$ are relatively prime positive integers. Find $r + s.$

Answer

Solution:

Substituting $y = \frac{3}{4}x$ into $x^y = y^x$ gives $x^{\frac{3}{4}x} = \left(\tfrac{3}{4}x\right)^{x}.$ Taking $x$ th roots (the quantities here are positive), $x^{3/4} = \frac{3}{4}x,$ so dividing by $x$ yields $x^{-1/4} = \frac{3}{4},$ that is, $x = \left(\frac{4}{3}\right)^4 = \frac{256}{81}.$

Then $y = \frac{3}{4} \cdot \frac{256}{81} = \frac{64}{27},$ and $x + y = \frac{256}{81} + \frac{192}{81} = \frac{448}{81}.$ Since $\gcd(448, 81) = 1,$ the answer is $448 + 81 = 529.$

2010 AIME I Problem 3

Solution:

Problem 3 in Other Years