2021 AMC 12A Problems/Problem 2

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, $0=2ab\implies ab=0$ . Also, it is clear that both sides of the equation must be nonnegative. The answer is $\boxed{\textbf{(D)}}$ .

~Jhawk0224

Solution 2 (Process of Elimination)

The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need $a+b\ge 0,$ which refutes $\textbf{(B)}$ and $\textbf{(E)}.$ Next, picking $(a,b)=(0,0)$ refutes $\textbf{(A)},$ and picking $(a,b)=(1,2)$ refutes $\textbf{(C)}.$ 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 union of:

positive $x$ -axis

positive $y$ -axis

origin

Therefore, the answer is $\boxed{\textbf{(D)}}.$

The graph of $\sqrt{x^2+y^2}=x+y$ is shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),E); label("$y$",(0,yMax),N); draw((xMax,0)--(0,0)--(0,yMax),red+linewidth(1.5)); [/asy]$ ~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

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

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