2021 AMC 12A Problems/Problem 2

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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 1

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.

Solution 2 (Quick Inspection)

The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need $a+b\ge 0,$ which eliminates $\textbf{(B)}$ and $\textbf{(E)}.$ Next, picking $(a,b)=(0,0)$ reveals that $\textbf{(A)}$ is incorrect, and picking $(a,b)=(1,2)$ reveals that $\textbf{(C)}$ is incorrect. By POE (Process of Elimination), the answer is $\boxed{\textbf{(D)}}.$

~MRENTHUSIASM

Solution 3 (Graphing)

If we graph $\sqrt{x^2+y^2}=x+y,$ then we get the positive $x$-axis and the positive $y$-axis, plus the origin. Therefore, the answer is $\boxed{\textbf{(D)}}.$

Graph in Desmos: https://www.desmos.com/calculator/0p3w7auwde

~MRENTHUSIASM (credit given to TheAMCHub)

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=40

Video Solution by Hawk Math

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

Video Solution by OmegaLearn (Using logic and analyzing answer choices)

https://youtu.be/Yp2NfDk_D-g

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/rEWS75W0Q54?t=71

~IceMatrix

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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