Difference between revisions of "2021 AMC 12A Problems/Problem 2"

(Video Solution by Hawk Math)
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==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
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== Video Solution (Using logic and analyzing answer choices) ==
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https://youtu.be/Yp2NfDk_D-g
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~ pi_is_3.14
  
 
==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 17:29, 11 February 2021

Problem

Under what conditions does $\sqrt{a^2+b^2}=a+b$ hold, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true.

$\textbf{(B) }$ It is true if and only if $ab=0$.

$\textbf{(C) }$ It is true if and only if $a+b\ge 0$.

$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$.

$\textbf{(E) }$ It is always true.

Solution

Square both sides to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$. Then, $0=2ab\rightarrow ab=0$. Then, the answer is $\boxed{\textbf{(B)}}$. Consider a right triangle with legs $a$ and $b$ and hypotenuse $\sqrt{a^{2}+b^{2}}$. Then one of the legs must be equal to $0$, but they are also nonnegative as they are lengths. Therefore, both $\textbf{(B)}$ and $\textbf{(D)}$ are correct.

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution (Using logic and analyzing answer choices)

https://youtu.be/Yp2NfDk_D-g

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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