2000 AIME I Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
{{ | Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | ||
<math> | |||
\begin{align*} | |||
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ | |||
&=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ | |||
&=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ | |||
&=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ | |||
&=2 + 5 + 29 + r\\ | |||
&=36 + r | |||
\end{align*} | |||
</math> | |||
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math> | |||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=6|num-a=8}} | {{AIME box|year=2000|n=I|num-b=6|num-a=8}} | ||
Revision as of 01:17, 3 December 2007
Problem
Suppose that 
and 
are three positive numbers that satisfy the equations 
and 
Then 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Let 
.
$\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Thus 
. So 
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
