Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
m (→‎Solution 1)
 
(9 intermediate revisions by 5 users not shown)
Line 23: Line 23:
 
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>:
 
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=\boxed{9}</math>,  
+
<math>4+3+2=9</math>,  
  
−
<math>4+3-2=\boxed{5}</math>,  
+
<math>4+3-2=5</math>,  
  
−
<math>4-3+2=\boxed{3}</math>,  
+
<math>4-3+2=3</math>,  
  
−
<math>-4+3+2=\boxed{1}</math>,  
+
<math>-4+3+2=1</math>,  
  
−
<math>4-3-2=\boxed{-1}</math>,  
+
<math>4-3-2=-1</math>,  
  
−
<math>-4+3-2=\boxed{-3}</math>,  
+
<math>-4+3-2=-3</math>,  
  
−
<math>-4-3+2=\boxed{-5}</math>,  
+
<math>-4-3+2=-5</math>,  
  
−
<math>-4-3-2=\boxed{-9}</math>
+
<math>-4-3-2=-9</math>
  
−
Therefore, the only possible values of <math>|a-d|</math> are 9, 5, 3, and 1. Their sum is <math>\boxed{18}</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 ==
 
== Solution 2 ==
Line 47: Line 47:
 
From <math>|a-b|=2</math> we get that <math>|b|=2</math>, hence <math>b\in\{-2,2\}</math>.
 
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. (This is due to the fact that we are calculating |d| at the end ~Williamgolly) Hence we can [[WLOG]] 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>.
 
From <math>|b-c|=3</math> we get that <math>c\in\{-1,5\}</math>.
Line 53: Line 53:
 
From <math>|c-d|=4</math> we get that <math>d\in\{-5,1,3,9\}</math>.
 
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 ==  
 
== 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>
+
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>)
 
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{\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==
 
==Video Solution==

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

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_16&oldid=194718"