# 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. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); draw((xMax,0)--(0,0)--(0,yMax),red+linewidth(1.5)); [/asy]$ ~MRENTHUSIASM (credit given to TheAMCHub)

~ pi_is_3.14

## Video Solution by TheBeautyofMath

~IceMatrix

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