# 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 1

From $|a-b|=2$ we get that $a=b\pm 2$

Similarly, $b=c\pm3$ and $c=d\pm4$.

Substitution gives $a=d\pm 4\pm 3\pm 2$. This gives $|a-d|=|\pm 4\pm 3\pm 2|$. There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$: $4+3+2=9$, $4+3-2=5$, $4-3+2=3$, $-4+3+2=1$, $4-3-2=-1$, $-4+3-2=-3$, $-4-3+2=-5$, $-4-3-2=-9$

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

## 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. (This is due to the fact that we are calculating $|d|$ at the end ~Williamgolly) WLOG 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 $1+3+5+9 = \boxed{\textbf{(D) }18}$.

## Solution 3

Let $$\begin{cases} |a-b| = 2, \\ |b-c| = 3, \\ |c-d| = 4, \\ |d-a| = X. \\ \end{cases}$$ Note that we have $$\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies$$ $$\pm X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies X \pm 2 \pm 3 \pm 4 = 0,$$ from which it follows that $$X = \pm 4 \pm 3 \pm 2.$$

Note that $X = |a-d|$ must be positive however, the only arrangements of $+$ and $-$ signs on the RHS which make $X$ positive are $$(4, 3, 2) \ \ \implies \ \ X = 9$$ $$(4, 3, -2) \ \ \implies \ \ X = 5$$ $$(4, -3, 2) \ \ \implies \ \ X = 3$$ $$(-4, 3, 2) \ \ \implies \ \ X = 1.$$ (There are no cases with $2$ or more negative as $4-3-2<0.$)

Thus, the answer is $$1+3+5+9 = \boxed{\textbf{(D) }18}.$$

## Solution 4

Let $a=0$.

Then $\begin{cases} b=2 \\ b=-2 \\ \end{cases}$

Thus $\begin{cases} c=5 \\ c=-1 \\ c=1 \\ c=-5 \\ \end{cases}$

Therefore $\begin{cases} d=9 \\ d=1 \\ d=3 \\ d=-5 \\ d=5 \\ d=-3 \\ d=-1 \\ d=-9 \\ \end{cases}$

So $|a-d|=1, 9, 3, 5$ and the sum is $\boxed{\textbf{(D) } 18}$.

~JH. L

## Video Solution

~savannahsolver

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