2009 AMC 10A Problems/Problem 16

Problem

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 $|a-d|$ ?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution

Solution 1:

From $|a-b|=3$ we get that $a=b+-2$

Similarly, $b=c+-3$ and $c=d+-4$ .

Substitution gives $a=d+-4+-3+-2$ . This gives $|a-d|=|+-4+-3+-2|$ . There are $2^3=8$ possibilities for the value of $|+-4+-3+-2|$ :

$4+3+2=\boxed{9}$ , $4+3-2=\boxed{5}$ , $4-3+2=\boxed{3}$ , $-4+3+2=\boxed{1}$ , $4-3-2=\boxed{-1}$ , $-4+3-2=\boxed{-3}$ , $-4-3+2=\boxed{-5}$ , $-4-3-2=\boxed{-9}$

Therefore, the only possible values of $|a-d|$ are 9, 5, 3, and 1. Their sum is $\boxed{18}$ .

Solution 2:

If we add the same constant to all of $a$ , $b$ , $c$ , and $d$ , we will not change any of the differences. Hence we can assume that $a=0$ .

From $|a-b|=2$ we get that $|b|=2$ , hence $b\in\{-2,2\}$ .

If we multiply all four numbers by $-1$ , we will not change any of the differences. Hence we can assume that $b=2$ .

From $|b-c|=3$ we get that $c\in\{-1,5\}$ .

From $|c-d|=4$ we get that $d\in\{-5,1,3,9\}$ .

Hence $|a-d|=|d|\in\{1,3,5,9\}$ , and the sum of possible values is $1+3+5+9 = \boxed{18}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2009 AMC 10A Problems/Problem 16

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