2009 AMC 10A Problem 16

P16-20~3 minPrint

Absolute Value & Inequalities Split into exhaustive cases and handle each

Let $a$ , $b$ , $c$ , and $d$ be real numbers with $|a-b|=2$ , $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$ ?

Show solution without answering

Key insight

Put a at 0 and step 2, then 3, then 4, each to the left or right. The possible distances from the start are 1, 3, 5 and 9.

Solution

Think of $a, b, c, d$ as points on a number line. $|a - b| = 2$ means $b$ is $2$ away from $a$ , on either side.

Put $a$ at $0$ . The distance from $a$ to $d$ is then just how far $d$ is from $0$ .
From $a$ , take a step of $2$ to reach $b$ , then a step of $3$ to reach $c$ , then a step of $4$ to reach $d$ . Each step goes left or right.
Flipping every direction just gives the mirror image, with the same distance from $0$ . So we may take the first step to the right, to $b = 2$ , and try the $4$ choices for the other two steps:

Step of 3	Step of 4	Where $d$ lands	Distance $\lvert a - d \rvert$
right	right	$2 + 3 + 4 = 9$	$9$
right	left	$2 + 3 - 4 = 1$	$1$
left	right	$2 - 3 + 4 = 3$	$3$
left	left	$2 - 3 - 4 = -5$	$5$

The possible values are $1, 3, 5, 9$ . Their sum is $1 + 3 + 5 + 9 = 18$ .

The answer is $\boxed{\textbf{(D)}\ 18}$ .

Why this works

An absolute value hides a direction. The differences chain together, $a - d = (a - b) + (b - c) + (c - d)$ , so $a - d$ is $2$ , $3$ and $4$ added with some plus and some minus signs. Listing the sign choices is a very small amount of casework, and the mirror-image idea cuts it in half.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The three conditions say $a - b = \pm 2$ , $b - c = \pm 3$ , $c - d = \pm 4$ , with the signs chosen independently. Adding,
$a - d = (a - b) + (b - c) + (c - d) = \pm 2 \pm 3 \pm 4.$

List the eight sign patterns (the negatives of each pattern give the same absolute value, so four suffice):
- $2 + 3 + 4 = 9$
- $2 + 3 - 4 = 1$
- $2 - 3 + 4 = 3$
- $-2 + 3 + 4 = 5$

The other four patterns are the negatives $-9, -1, -3, -5$ . So $|a - d|$ can be $1, 3, 5$ or $9$ , and each really occurs (e.g. $a = 0$ and the appropriate $b, c, d$ ).

Sum: $1 + 3 + 5 + 9 = 18$ .

The answer is $\boxed{\textbf{(D)}\ 18}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Prealgebra, Chapter 14: Counting (§14.3)