Difference between revisions of "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, $0=2ab\rightarrow ab=0$. 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)

~ pi_is_3.14

~IceMatrix