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> | ||
− | ==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> | <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 | + | == See Also == |
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{{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}} | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:07, 22 March 2025
Problem
For real numbers and
, define
. What is
?
Solution
Since , it follows that
, and
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.