Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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</math> | </math> | ||
− | + | ==Solution 1== | |
− | |||
From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math> | From <math>|a-b|=2</math> we get that <math>a=b\pm 2</math> | ||
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<math>-4-3-2=\boxed{-9}</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>. | + | 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{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>. | 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>. | ||
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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. | + | 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>. | ||
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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>. | ||
+ | |||
+ | == 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+7 = \boxed{18}.</cmath> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/z_W-Z9CHPR4 | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:45, 8 June 2021
Problem
Let , , , and be real numbers with , , and . What is the sum of all possible values of ?
Solution 1
From we get that
Similarly, and .
Substitution gives . This gives . There are possibilities for the value of :
,
,
,
,
,
,
,
Therefore, the only possible values of are , , , and . Their sum is .
Solution 2
If we add the same constant to all of , , , and , we will not change any of the differences. Hence we can assume that .
From we get that , hence .
If we multiply all four numbers by , we will not change any of the differences. (This is due to the fact that we are calculating at the end ~Williamgolly) WLOG we can assume that .
From we get that .
From we get that .
Hence , and the sum of possible values is .
Solution 3
Let Note that we have from which it follows that
Note that must be positive however, the only arrangements of and signs on the RHS which make positive are (There are no cases with or more negative as )
Thus, the answer is
Video Solution
~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 |
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