Difference between revisions of "2021 AMC 12A Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | + | Square both sides to get <math>a^{2}+b^{2}=a^{2}+2ab+b^{2}</math>. Then, <math>0=2ab\rightarrow ab=0</math>. Then, the answer is <math>\boxed{\textbf{(B)}}</math>. Consider a right triangle with legs <math>a</math> and <math>b</math> and hypotenuse <math>\sqrt{a^{2}+b^{2}}</math>. Then one of the legs must be equal to <math>0</math>, but they are also nonnegative as they are lengths. Therefore, both <math>\textbf{(B)}</math> and <math>\textbf{(D)}</math> are correct. | |
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==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2021|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:31, 11 February 2021
Problem
Under what conditions is true, where and are real numbers?
It is never true.
It is true if and only if .
It is true if and only if .
It is true if and only if and .
It is always true.
Solution
Square both sides to get . Then, . Then, the answer is . Consider a right triangle with legs and and hypotenuse . Then one of the legs must be equal to , but they are also nonnegative as they are lengths. Therefore, both and are correct.
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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.