2010 AMC 10B Problem 13

An absolute value is never negative, so any solution has $x \ge 0$. The inside absolute value $|60 - 2x|$ is what makes the problem hard, so deal with it first. The expression $60 - 2x$ is zero at $x = 30$, positive before that and negative after, so split there.

Case 1: $x \le 30$. Here $60 - 2x \ge 0$, so $|60 - 2x| = 60 - 2x$.

1. The equation becomes
$$x = |2x - (60 - 2x)| = |4x - 60|.$$
2. A number equals $|4x - 60|$ when $4x - 60$ is that number or its negative.
- $4x - 60 = x$ gives $3x = 60$, so $x = 20$.
- $4x - 60 = -x$ gives $5x = 60$, so $x = 12$.
3. Both are at most $30$, so both belong to this case. Check them in the original equation:
- $x = 20$: $|40 - |60 - 40|| = |40 - 20| = 20$. Yes.
- $x = 12$: $|24 - |60 - 24|| = |24 - 36| = 12$. Yes.

Case 2: $x > 30$. Here $60 - 2x < 0$, so $|60 - 2x| = 2x - 60$.

4. The equation becomes
$$x = |2x - (2x - 60)| = |60| = 60.$$
5. So $x = 60$, which is indeed more than $30$. Check: $|120 - |60 - 120|| = |120 - 60| = 60$. Yes.

The solutions are $12$, $20$ and $60$, and their sum is $12 + 20 + 60 = 92$.

The answer is $\boxed{\textbf{(C)}\ 92}$.

With absolute values inside absolute values, remove them from the inside out. Find where the inside expression changes sign, split into cases there, and in each case the absolute value becomes ordinary parentheses. Always check that each answer really lies in the case that produced it.

Sketch the right side as a graph. For $x \le 30$ it is $|4x - 60|$, a V shape with its point at $x = 15$. For $x > 30$ it is the flat line at height $60$. The line $y = x$ crosses the V twice (at $12$ and $20$) and the flat part once (at $60$), giving the same three solutions.