Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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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>) | 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+ | + | Thus, the answer is <cmath>1+3+5+9 = \boxed{18}.</cmath> |
==Solution 4== | ==Solution 4== |
Revision as of 09:45, 1 August 2022
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
Solution 4
Let .
Then
Thus
Therefore
So and the sum is
.
~JH. L
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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