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

Revision as of 02:44, 8 February 2010

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 $|a-d|$ ?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

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 $\boxed{18}$ .

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

@@ Line 16: / Line 16: @@
 == Solution ==
+=== Solution 1 ===
-Solution 1:
+From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math>
-From <math>|a-b|=3</math> we get that <math>a=b+-2</math>
+Similarly, <math>b=c\pm3</math> and <math>c=d\pm4</math>.
-Similarly, <math>b=c+-3</math> and <math>c=d+-4</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>:
-Substitution gives <math>a=d+-4+-3+-2</math>. This gives <math>|a-d|=|+-4+-3+-2|</math>. There are <math>2^3=8</math> possibilities for the value of <math>|+-4+-3+-2|</math>:
+<math>4+3+2=\boxed{9}</math>,
-<math>4+3+2=\boxed{9}</math>,
 <math>4+3-2=\boxed{5}</math>,
 <math>4-3+2=\boxed{3}</math>,
 <math>-4+3+2=\boxed{1}</math>,
 <math>4-3-2=\boxed{-1}</math>,
 <math>-4+3-2=\boxed{-3}</math>,
 <math>-4-3+2=\boxed{-5}</math>,
 <math>-4-3-2=\boxed{-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>.
-Solution 2:
+=== 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>.

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