2000 AIME I Problem 7

Question

Suppose that

x,

y,

and

z

are three positive numbers that satisfy the equations

xyz = 1,

x + \frac{1}{z} = 5,

and

y + \frac{1}{x} = 29.

Then

z + \frac{1}{y} = \frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Multiply the three expressions

x + \frac{1}{z},

y + \frac{1}{x},

z + \frac{1}{y}

together and expand

Because

xyz = 1,

the product equals

2

plus the sum of the same three expressions

Let

t = z + \frac{1}{y}.

Expanding the product of all three expressions,

\begin{aligned} &\left(x + \frac{1}{z}\right)\!\left(y + \frac{1}{x}\right)\!\left(z + \frac{1}{y}\right) \\ &= xyz + \frac{1}{xyz} + \left(x + \frac{1}{z}\right) \\ &\quad {}+ \left(y + \frac{1}{x}\right) + \left(z + \frac{1}{y}\right). \end{aligned}

Since

xyz = 1,

the left side is

5 \cdot 29 \cdot t = 145t

and the right side is

2 + 5 + 29 + t = 36 + t.

So

145t = 36 + t,

giving

t = \frac{36}{144} = \frac{1}{4}.

Thus

m + n = 1 + 4 = 5.

Problem 7 in other years