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

Revision as of 09:41, 12 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, $0=2ab\rightarrow ab=0$ . 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 $\textbf{(A)}$ and $\textbf{(B)}.$ Next, picking $(a,b)=(1,2)$ reveals that $\textbf{(C)}$ and $\textbf{(E)}$ are both 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 $\boxed{\textbf{(D)}}.$ ~MRENTHUSIASM (credit given to TheAMCHub)

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==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>. Also, it is clear that both sides of the equation must be nonnegative. The answer is <math>\boxed{\textbf{(D)}}</math>.
 ==Solution 2 (Quick Inspection)==

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