Difference between revisions of "2009 AMC 10A Problems/Problem 16"

Revision as of 08:45, 1 August 2022

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) 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 $1+3+5+9 = \boxed{18}$ .

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 $1+3+5+9 = \boxed{18}.$

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

https://youtu.be/z_W-Z9CHPR4

~savannahsolver

@@ Line 61: / Line 61: @@
 Note that <math>X = |a-d|</math> must be positive however, the only arrangements of <math>+</math> and <math>-</math> signs on the RHS which make <math>X</math> positive are <cmath>(4, 3, 2) \ \ \implies \ \ X = 9</cmath> <cmath>(4, 3, -2) \ \ \implies \ \ X = 5</cmath> <cmath>(4, -3, 2) \ \ \implies \ \ X = 3</cmath> <cmath>(-4, 3, 2) \ \ \implies \ \ X = 1.</cmath> (There are no cases with <math>2</math> or more negative as <math>4-3-2<0.</math>)
-Thus, the answer is <cmath>1+3+5+7 = \boxed{18}.</cmath>
+Thus, the answer is <cmath>1+3+5+9 = \boxed{18}.</cmath>
 ==Solution 4==

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