2010 AMC 12A Problems/Problem 22
Contents
Problem
What is the minimum value of ?
Solution 1
If we graph each term separately, we will notice that all of the zeros occur at , where is any integer from to , inclusive: .
The minimum value of occurs where the absolute value of the sum of the slopes is at a minimum , since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some .
The sum of the slopes at is
Now we want to minimize . The zeros occur at and , which means the slope is where .
We can now verify that both and yield .
You can also think of the slopes playing 'tug of war', where the slope of each absolute function upon passing its -intercept is negated, positively tugging on the remaining negative slopes.
The sum of the slopes is
So we need to find the least integer such that
This "exactly" means that the slope is ZERO between the whole interval . We can explicitly evaluate both to check that they are both equal to the desired minimum value of :
Thus the minimum value of is .
Solution 2
Rewrite the given expression as follows: Imagine the real line. For each imagine that there are boys standing at the coordinate . We now need to place a donut on the real line in such a way that the sum of its distances from all the boys is minimal, and we need to compute this sum.
Note that there are boys in total. Let's label them from 1 (the only boy placed at ) to (the last boy placed at .
Clearly, the minimum sum is achieved if the donut's coordinate is the median of the boys' coordinates. To prove this, place the donut at the median coordinate. If you now move it in any direction by any amount , there will be boys such that it moves away from this boy. For each of the remaining boys, it moves at most closer, hence the total sum of distances does not decrease.
Hence the optimal solution is to place the donut at the median coordinate. Or, more precisely, as is even, we can place it anywhere on the segment formed by boy and boy : by extending the previous argument, anywhere on this segment the sum of distances is the same.
By trial and error, or by solving the quadratic equation we get that boy number is the last boy placed at and the next boy is the one placed at . Hence the given expression is minimized for any .
Common part of both solutions
To find the minimum, we want to balance the expression so that it is neither top nor bottom heavy. .
Now that we know that the sum of the first 84 's is equivalent to the sum of 's 85 to 119, we can plug either or to find the minimum.
Note that the terms to are negative, and the terms to are positive. Hence we get: and Hence the total sum of distances is .
Solution 3
By the triangle inequality, However, we may change signs of some of these terms to cancel out the 's. Since the minimum exists, we want all the s to cancel out. Thus, we want to find some such that
Then, . The answer(expression's value) is then , which becomes .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
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