AMC 10 · 2003 · #24

Grade 9 algebra
absolute-valuepiecewise-linear-functionoptimization convert-to-algebraextreme-principle ↑ Prerequisites: absolute-value
📏 Long solution 💡 3 insights
Problem
Choose positive integers a < b < c. One equation, 2x+y=2003, is a straight line; the other, y=|x-a|+|x-b|+|x-c|, bends at a, b and c. The two graphs must meet in exactly one point. Over every choice that makes this happen, find the smallest c can be.

Pick an answer.

(A)
668
(B)
669
(C)
1002
(D)
2003
(E)
2004
How to solve
Strategy Extreme Principle

The sum of three absolute values is a chain of segments whose slopes climb -3,-1,1,3. The line has slope -2, which sits strictly between the first two. That single fact is the whole problem: tool #16 (Change Focus) says to stop comparing two graphs and instead watch the gap between them, g(x)=|x-a|+|x-b|+|x-c|-(2003-2x). Adding 2x pushes every slope up by 2, so g has slopes -1,1,3,5: it falls, then rises, with its only turn at x=a. Tool #14 (Extreme Principle) then finishes it — a function that falls and then rises hits zero exactly once only when its lowest value is zero, which turns "exactly one solution" into a plain equation. Tool #1 (Draw a Diagram) keeps the four-piece shape in view, and tool #4 (Introduce a Variable) names that gap function so the slopes can be tracked.

1STEP 1

Collapse the system to one equation

Substituting the line leaves one equation in x, so meetings are its roots.

|x-a|+|x-b|+|x-c|=2003-2x
2STEP 2

Read the four slopes of the V-chain

The bent chain has slopes -3, -1, +1, +3 across its four stretches.

f'=-3 (x < a), -1 (a < x < b), +1 (b < x < c), +3 (x > c)
3STEP 3

Watch the gap instead of the two graphs

Adding the line lifts every slope by two, so the gap turns upward at the first corner.

g(x)=|x-a|+|x-b|+|x-c|+2x-2003; g'=-1, +1, +3, +5
4STEP 4

One root only when the bottom touches zero

Such a shape meets zero exactly once only when its minimum is zero.

#{x: g(x)=0}= 2,& g(a) < 0 ; 1,& g(a)=0 ; 0,& g(a) > 0
5STEP 5

Compute the bottom value

Evaluating there gives the clean condition b + c = 2003, with a dropping out.

g(a)=(b-a)+(c-a)+2a-2003=b+c-2003=0⇔ b+c=2003
6STEP 6

Push c as low as the condition allows

With b smaller than c, that forces c = 1002 at least, and it is attained, choice (C).

b+c=2003, b < c→ 2c > 2003→ c ≥ 1002; (a,b,c)=(1000,1001,1002)→(C) 1002
Answer
1002
Test (a,b,c)=(1000,1001,1002) by hand. At x=1000: the curve gives 0+1+2=3 and the line gives 2003-2000=3, so they meet at (1000,3). They meet nowhere else, since g falls with slope -1 before x=1000 and rises afterwards, so g=0 only at its bottom. Now check that c cannot be smaller: c ≤ 1001 forces b=2003-c ≥ 1002 > c, contradicting b < c. The wrong choices are recognisable traps — 668 and 669 come from splitting 2003 three ways as if a+b+c were the controlling sum, while 2003 and 2004 ignore that only b+c is pinned. A brute-force sweep over triples agrees: the system has exactly one solution precisely when b+c=2003, and the least such c is 1002.
💡Key takeaway

When a V-shaped graph meets a line, subtract the line first: the pair crosses exactly once only when the bottom of the V lands right on it.

  • Collapse the system to one equation
  • Read the four slopes of the V-chain
  • Watch the gap instead of the two graphs
  • One root only when the bottom touches zero
  • Compute the bottom value
  • Push c as low as the condition allows