# 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 $\pm 4\pm 3\pm2$: $4+3+2=\boxed{9}$, $4+3-2=\boxed{5}$, $4-3+2=\boxed{3}$, $-4+3+2=\boxed{1}$, $4-3-2=\boxed{-1}$, $-4+3-2=\boxed{-3}$, $-4-3+2=\boxed{-5}$, $-4-3-2=\boxed{-9}$

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

## Solution 3

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

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 $$1+3+5+7 = \boxed{18}.$$

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 