AMC10 2010 10B, Problem 13

What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$?

Solution: We evaluate this in cases: Case 1 $x<30$ When $x<30$ we are going to have $60-2x>0$.  When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$.  Therefore we have $x=|2x-(60-2x)|$. $x=|2x-60+2x|\implies x=|4x-60|$ Subcase 1 $30>x>15$ When $30>x>15$ we are going to have $4x-60>0$. When this happens, we can express $|4x-60|$ as $4x-60$. Therefore we get $x=4x-60\implies -3x=-60\implies x=20$.  We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$. Therefore $20$ is one possible solution. Subcase 2  $x<15$ When $x<15$ we are going to have $4x-60<0$, therefore $|4x-60|$ can be expressed in the form $60-4x$. We have the equation $x=60-4x\implies 5x=60\implies x=12$.  Since $12$ is less than $15$, $12$ is another possible solution. $x=|2x-|60-2x||$ Case 2 :  $x>30$ When $x>30$, $60-2x<0$. When $x<0$ we can express this in the form $-x$.  Therefore we have $-(60-2x)=2x-60$.  This makes sure that this is positive, since we just took the negative of a negative to get a positive.  Therefore we have $x=|2x-(2x-60)|$ $x=|2x-2x+60|$ $x=|60|$ $x=60$ We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\fbox{92}$

Answer: 92