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2021 AMC 12A Problems/Problem 2: Difference between revisions

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


Graph in Desmos: https://www.desmos.com/calculator/rlz6actjrw
Graph in Desmos: https://www.desmos.com/calculator/0p3w7auwde


~MRENTHUSIASM (credit given to TheAMCHub)
~MRENTHUSIASM (credit given to TheAMCHub)

Revision as of 13:44, 14 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 1

Square both sides to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$. Then, $0=2ab\rightarrow ab=0$. Also, it is clear that both sides of the equation must be nonnegative. The answer is $\boxed{\textbf{(D)}}$.

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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