Difference between revisions of "2021 AMC 12A Problems/Problem 2"
Sugar rush (talk | contribs) |
|||
Line 14: | Line 14: | ||
==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. | 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. | ||
+ | |||
+ | ==Video Solution by Hawk Math== | ||
+ | https://www.youtube.com/watch?v=P5al76DxyHY | ||
==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 16:50, 11 February 2021
Problem
Under what conditions does hold, 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.
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
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.