Question

Positive integers $a,$ $b,$ and $c$ are chosen so that $a \lt b \lt c,$ and the system of equations $2x + y = 2003 \quad\text{and}\quad y = |x - a| + |x - b| + |x - c|$ has exactly one solution. What is the minimum value of $c?$

$668$

$669$

$1002$

$2003$

$2004$

Solution:

The function $y = |x-a| + |x-b| + |x-c|$ is piecewise linear with slopes $-3, -1, 1, 3$ and corners at $x = a, b, c.$ The line $2x + y = 2003$ has slope $-2.$

A line of slope $-2$ meets this graph exactly once only if it passes through the leftmost corner $(a,\, b + c - 2a),$ where the graph's slope jumps from $-3$ to $-1.$ Substituting, $2a + (b + c - 2a) = 2003,$ so $b + c = 2003.$

Since $b \lt c,$ we need $c \gt \dfrac{2003}{2},$ so the minimum is $c = 1002$ (with $b = 1001$ ).

Thus, the correct answer is C.

2003 AMC 12B Problem 24

Solution:

Problem 24 in Other Years