Difference between revisions of "2001 AMC 12 Problems/Problem 9"
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== Solution 2 == | == Solution 2 == | ||
The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \frac52</math>. | The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \frac52</math>. | ||
| − | + | ==Solution 3== | |
| + | Note that the equation given above is symmetric, so we have xf(x)=yf(y). Plugging in x=500 and y=600 gives f(y)=5/2 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=8|num-a=10}} | {{AMC12 box|year=2001|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:57, 26 January 2020
Problem
Let 
be a function satisfying 
for all positive real numbers 
and 
. If 
, what is the value of 
?
Solution 1
Letting 
and 
in the given equation, we get 
, or 
.
Solution 2
The only function that satisfies the given condition is 
, for some constant 
. Thus, the answer is 
.
Solution 3
Note that the equation given above is symmetric, so we have xf(x)=yf(y). Plugging in x=500 and y=600 gives f(y)=5/2
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
