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

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==Problem==
 
==Problem==
For [[real number]]s <math>a</math> and <math>b</math>, define <math>a * b=(a-b)^2</math>. What is <math>(x-y)^2 (y-x)^2</math>?
+
For [[real number]]s <math>a</math> and <math>b</math>, define <math>a * b=(a-b)^2</math>. What is <math>(x-y)^2*(y-x)^2</math>?
  
 
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy</math>
 
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy</math>

Revision as of 15:28, 19 May 2009

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
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