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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.
 
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 $\sqrt{a^2+b^2}=a+b$ 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

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

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