Difference between revisions of "2008 AMC 12B Problems/Problem 7"
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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math> | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math> | ||
| − | ==Solution== | + | ==Solution 1 == |
<math>\left[ (x-y)^2 - (y-x)^2 \right]^2</math> | <math>\left[ (x-y)^2 - (y-x)^2 \right]^2</math> | ||
| Line 13: | Line 13: | ||
<math>0 \Rightarrow \textbf{(A)}</math> | <math>0 \Rightarrow \textbf{(A)}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | WLOG, let <math>x</math> and <math>y</math> both be 0. Thus,<math>0^2</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:23, 29 April 2020
Contents
Problem 7
For real numbers 
and 
, define 
. What is 
?
Solution 1
![$\left[ (x-y)^2 - (y-x)^2 \right]^2$](http://latex.artofproblemsolving.com/1/9/7/197c4c2e51a2ef944c7e964782f5a8df8a22b606.png)
![$\left[ (x-y)^2 - (x-y)^2 \right]^2$](http://latex.artofproblemsolving.com/f/d/7/fd7fd835e362057186d253e0724a1b0b5c07580f.png)
![$[0]^2$](http://latex.artofproblemsolving.com/b/7/e/b7e1d01b94c63fb026726b3c9b84d79c1aaed490.png)
Solution 2
WLOG, let 
and 
both be 0. Thus,
See Also
| 2008 AMC 12B (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 • 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. 
