Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$, and the system of [equations](https://artofproblemsolving.com/wiki/index.php/Equation)

$2x + y = 2003 \quad$ and $\quad y = |x-a| + |x-b| + |x-c|$
has exactly one solution. What is the minimum value of $c$?
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$
Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$.
When $xb$, the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$. This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$, and thus the answer is $\boxed{\mathrm{(C)}\ 1002}$.
This indeed works for the two right segments of slope $1$ and $3$. We already know that the minimum is achieved between slopes $-3$ and $-1$ with $b+c=2003$:
\[2003-2x=-a-b+c+x\longrightarrow 3x\ne a+b-c+2003 \{ba\text{ (true)}\]
\[2003-2x=-a-b-c+3x\longrightarrow 5x\ne a+b+c+2003 \{x>c\}\rightarrow (5c,+5\infty)\ne a+2b+2c\rightarrow 3c>a+2b\text{ (true)}\]
Indeed, within the restricted domain of $x$ in each segment, these inequalities prove to be unequal everywhere. So $y=2003-2x$ is strictly below $y=|x-a|+|x-b|+|x-c|$ at these domains.