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

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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{18}</math>.
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== Solution 3 ==
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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 \ \ \pmX \pm 2 \pm 3 \pm 4 = 0,</cmath> from which it follows that <cmath>X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies \ \ X = \pm 2 \pm 3 \pm 4. </cmath>
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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>)
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Thus, the answer is <cmath>1+3+5+7 = \boxed{18}.</cmath>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 01:38, 19 January 2021

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

\[\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies \ \ \pmX \pm 2 \pm 3 \pm 4 = 0,\] (Error compiling LaTeX. ! Undefined control sequence.)

from which it follows that \[X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies \ \ X = \pm 2 \pm 3 \pm 4.\]

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

https://youtu.be/z_W-Z9CHPR4

~savannahsolver

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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