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

Latest revision as of 15:26, 23 June 2023

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

https://youtu.be/z_W-Z9CHPR4

~savannahsolver

@@ Line 15: / Line 15: @@
 </math>
-== Solution ==
+==Solution 1==
+From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math>
+Similarly, <math>b=c\pm3</math> and <math>c=d\pm4</math>.
+Substitution gives <math>a=d\pm 4\pm 3\pm 2</math>. This gives <math>|a-d|=|\pm 4\pm 3\pm 2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>\pm 4\pm 3\pm2</math>:
+<math>4+3+2=9</math>,
+<math>4+3-2=5</math>,
+<math>4-3+2=3</math>,
+<math>-4+3+2=1</math>,
+<math>4-3-2=-1</math>,
+<math>-4+3-2=-3</math>,
+<math>-4-3+2=-5</math>,
+<math>-4-3-2=-9</math>
+Therefore, the only possible values of <math>|a-d|</math> are <math>9</math>, <math>5</math>, <math>3</math>, and <math>1</math>. Their sum is <math>\boxed{\textbf{(D) } 18}</math>.
+== Solution 2 ==
 If we add the same constant to all of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, we will not change any of the differences. Hence we can assume that <math>a=0</math>.
@@ Line 21: / Line 47: @@
 From <math>|a-b|=2</math> we get that <math>|b|=2</math>, hence <math>b\in\{-2,2\}</math>.
-If we multiply all four numbers by <math>-1</math>, we will not change any of the differences. Hence we can assume that <math>b=2</math>.
+If we multiply all four numbers by <math>-1</math>, we will not change any of the differences. (This is due to the fact that we are calculating <math>|d|</math> at the end ~Williamgolly) [[WLOG]] we can assume that <math>b=2</math>.
 From <math>|b-c|=3</math> we get that <math>c\in\{-1,5\}</math>.
@@ Line 27: / Line 53: @@
 From <math>|c-d|=4</math> we get that <math>d\in\{-5,1,3,9\}</math>.
-Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>.
+Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{\textbf{(D) }18}</math>.
+== Solution 3 ==
+Let <cmath>\begin{cases} |a-b| = 2, \\ |b-c| = 3, \\ |c-d| = 4, \\ |d-a| = X. \\ \end{cases}</cmath> Note that we have <cmath>\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies</cmath> <cmath>\pm X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies X \pm 2 \pm 3 \pm 4 = 0,</cmath> from which it follows that <cmath>X = \pm 4 \pm 3 \pm 2.</cmath>
+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+9 = \boxed{\textbf{(D) }18}.</cmath>
+==Solution 4==
+Let <math>a=0</math>.
+Then <math>\begin{cases} b=2 \\ b=-2 \\ \end{cases}</math>
+Thus <math>\begin{cases} c=5 \\ c=-1 \\ c=1 \\ c=-5 \\ \end{cases}</math>
+Therefore <math>\begin{cases} d=9 \\ d=1 \\ d=3 \\ d=-5 \\ d=5 \\ d=-3 \\ d=-1 \\ d=-9 \\ \end{cases}</math>
+So <math>|a-d|=1, 9, 3, 5</math> and the sum is <math>\boxed{\textbf{(D) } 18}</math>.
+~JH. L
+==Video Solution==
+https://youtu.be/z_W-Z9CHPR4
+~savannahsolver
 == See Also ==
-{{AMC10 box|year=2009|ab=A|num-b=|num-a=}}
+{{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}}
+{{MAA Notice}}

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