2003 AMC 12B Problem 24

Question

Positive integers

a,

b,

and

c

are chosen so that

a \lt b \lt c,

and the system of equations

2x + y = 2003

and

\begin{aligned} &y = |x - a| + |x - b| \\ &\quad {}+ |x - c| \end{aligned}

has exactly one solution. What is the minimum value of

c?

668

669

1002

2003

2004

The graph of

y = |x-a| + |x-b| + |x-c|

is piecewise linear with slopes

-3, -1, 1, 3

The line has slope

-2;

a single intersection forces it to touch the corner at

x = a,

where

y = b + c - 2a

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

c \ge 1002.

This bound is attained: take

a = 1,

b = 1001,

and

c = 1002.

Therefore the minimum is

1002.

Thus, the correct answer is C.

Problem 24 in other years