Difference between revisions of "2008 AMC 10B Problems/Problem 5"
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==Solution== | ==Solution== | ||
Since <math>(-a)^2 = a^2</math>, it follows that <math>(x-y)^2 = (y-x)^2</math>, and | Since <math>(-a)^2 = a^2</math>, it follows that <math>(x-y)^2 = (y-x)^2</math>, and | ||
| − | < | + | <cmath>(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.</cmath> |
==See also== | ==See also== | ||
Revision as of 15:23, 10 October 2008
Problem
For real numbers 
and 
, define 
. What is 
?
Solution
Since 
, it follows that 
, and
![\[(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.\]](http://latex.artofproblemsolving.com/1/3/8/138e0ad8d731afe82997734cb68305608ba9622d.png)
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 10 Problems and Solutions | ||
