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Difference between revisions of "2010 AMC 12A Problems/Problem 22"

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=== Solution 3 ===
 
=== Solution 3 ===
  
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Given that the minimum exists, we know that we want all <math>x</math>s to cancel out. Thus, we want to find some <math>n</math> such that <math>1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119</math>. It therefore follows that <math>\frac{n(n+1)\2}=\frac{119\cdot120\2}-\frac{n(n+1)\2}</math>. This gives <math>n^2+n-7140=0</math>, from which it follows that <math>n=84</math>.
+
Given that the minimum exists, we know that we want all <math>x</math>s to cancel out. Thus, we want to find some <math>n</math> such that <math>1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119</math>. It therefore follows that <math>\frac{n(n+1)}{2}=\frac{119\cdot120}{2}-\frac{n(n+1)}{2}</math>. This gives <math>n^2+n-7140=0</math>, from which it follows that <math>n=84</math>.
  
 
There are therefore <math>84</math> positive ones and <math>119-85+1</math>, or <math>35</math> negative ones, so <math>84-35=\boxed{49}</math>.
 
There are therefore <math>84</math> positive ones and <math>119-85+1</math>, or <math>35</math> negative ones, so <math>84-35=\boxed{49}</math>.

Revision as of 12:29, 1 January 2013

Problem

What is the minimum value of $\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$?

$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$

Solution

Solution 1

If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$, where $m$ is any integer from $1$ to $119$, inclusive.

The minimum value occurs where 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 $\frac{1}{m}$.


The sum of the slope at $x = \frac{1}{m}$ is

\begin{align*}&\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i\\ &=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i\\ &=-m^2-m+7140\end{align*}

Now we want to minimize $-m^2-m+7140$. The zeros occur at $-85$ and $84$, which means the slope is $0$ where $m = 84, 85$.

We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49\ \textbf{(A)}}$.

Solution 2

Rewrite the given expression as follows: \[1|x-1| + 2\left|x-\frac 12\right| + \cdots + 119\left|x-\frac 1{119}\right|\] Imagine the real line. For each $n\in\{1,\dots,119\}$ imagine that there are $n$ boys standing at the coordinate $\frac 1n$. We now need to place a girl on the real line in such a way that the sum of her distances from all the boys is minimal, and we need to compute this sum.

Note that there are $B=1+2+\cdots+119 = 119\cdot 60=7140$ boys in total. Let's label them from 1 (the only boy placed at $1$) to $B$ (the last boy placed at $\frac 1{119}$.

Clearly, the minimum sum is achieved if the girl's coordinate is the median of the boys' coordinates. To prove this, place the girl at the median coordinate. If you now move her in any direction by any amount $d$, there will be $B/2$ boys such that she moves $d$ away from this boy. For each of the remaining boys, she moves at most $d$ closer, hence the total sum of distances does not decrease.

Hence the optimal solution is to place the girl at the median coordinate. Or, more precisely, as $B$ is even, we can place her anywhere on the segment formed by boy $B/2$ and boy $(B/2)+1$: 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 $z(z+1)/2 = 7140/2$ we get that boy number $B/2$ is the last boy placed at $\frac 1{84}$ and the next boy is the one placed at $\frac 1{85}$. Hence the given expression is minimized for any $x\in\left[ \frac 1{85}, \frac 1{84} \right]$.

Common part of both solutions

To find the minimum, we want to balance the expression so that it is neither top nor bottom heavy. $\frac{119(120)}{2(2)}=\frac{7140}{2}=3570=\frac{84(85)}{2}=\frac{119(120)}{2}-\frac{84(85)}{2}$.

Now that we know that the sum of the first 84 $x$'s is equivalent to the sum of $x$'s 85 to 119, we can plug either $\frac{1}{84}$ or $\frac{1}{85}$ to find the minimum.

Note that the terms $x-1$ to $83x-1$ are negative, and the terms $85x-1$ to $119x-1$ are positive. Hence we get: \begin{align*} & |x-1| + |2x-1| + \cdots + |83x-1| \\ =~ & (1-x) + (1-2x) + \cdots + (1-83x) \\ =~ & 83 - x(1+2+\cdots+83) \\ =~ & 83 - \frac 1{84} \cdot \frac{83\cdot 84}2 \\ =~ & 83 - \frac{83}2 \\ =~ & \frac{83}2 \end{align*} and \begin{align*} & |85x-1| + |86x-1| + \cdots + |119x-1| \\ =~ & (85x-1) + (86x-1) + \cdots + (119x-1) \\ =~ & x(85+86+\cdots+119) - (119-84) \\ =~ & \frac 1{84} \cdot \frac{84\cdot 85}2 - 35 \\ =~ & \frac{85}2 - 35 \\ =~ & \frac{15}2 \end{align*} Hence the total sum of distances is $\frac{83}2 + \frac{15}2 = 49$.

Solution 3

Given that the minimum exists, we know that we want all $x$s to cancel out. Thus, we want to find some $n$ such that $1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119$. It therefore follows that $\frac{n(n+1)}{2}=\frac{119\cdot120}{2}-\frac{n(n+1)}{2}$. This gives $n^2+n-7140=0$, from which it follows that $n=84$.

There are therefore $84$ positive ones and $119-85+1$, or $35$ negative ones, so $84-35=\boxed{49}$.

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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