Difference between revisions of "2008 AMC 10B Problems/Problem 5"

m (Changed $ sign with * sign; doesn't affect the problem)
m (Changed $ sign with * sign; doesn't affect the problem; Also removed <center> tags)
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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  
<center><math>(x-y)^2\</math><math> (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.</math></center>
+
<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 16:23, 10 October 2008

Problem

For real numbers $a$ and $b$, define $a * b=(a-b)^2$. What is $(x-y)^2 * (y-x)^2$?

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy$

Solution

Since $(-a)^2 = a^2$, it follows that $(x-y)^2 = (y-x)^2$, and \[(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.\]

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 10 Problems and Solutions