AMC12 2003 B

Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$. When $x<a$, the slope is $-3$. When $a<x<b$, the slope is $-1$. When $b<x<c$, the slope is $1$. When $c<x$, the slope is $3$. Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$, so $(b,c-a)$ is a point on $f(x)$. In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$. Thus, graphing this will produce a figure that looks like a cup: From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$. Since the line where $a<x<b$ has slope $-1$, the positive difference in $y$-coordinates from $x=a$ to $x=b$ must be $b-a$. Together with the fact that $(b,c-a)$ is on $f(x)$, we see that $P=(a,c-a+b-a)$. Since this point is on $x=a$, the only intersection point with $2x+y=2003$, we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$. As $c>b$, 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 $\boxed{\mathrm{(C)}\ 1002}$.