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worthawholebean
Navier-Stokes Equations
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#1
Unique Solution
AMC 12 2003B Problem 24
Positive integers a, b, and c are chosen so that a<b<c, and the system of equations
2x+y=2003\text{ and }y=|x-a|+|x-b|+|x-c|has exactly one solution. What is the minimum value of c?

\textbf{(A)}\ 668 \qquad \textbf{(B)}\ 669 \qquad \textbf{(C)}\ 1002 \qquad \textbf{(D)}\ 2003 \qquad \textbf{(E)}\ 2004
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PostPosted: Sun Jan 04, 2009 7:33 pm  Back to top 
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ocha
Yang-Mills Theory
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Is it suppose to be a\le b\le c or a<b<c?
If it is the first case then to minimuse c let a=b=c
So we have the equation

y=2003-2x=3|x-c|

Now, note that the graph of y=2003-2x has a gradient of -2 but the graph 3|x-c| has a gradient of \pm 3, and has a turning point at x=c. So if we sketch it we can easily see that in order for the equation to have only one solution, c must be greater than the x-axis intercept of y=2003-2x.

Now since c is an intiger c=\left\lceil \frac{2003}{2} \right\rceil = 1002

NOTE: if a<b<c then the answer is the same but we origionally set b=a+1, c=a+2 and the answer is 1004 but this is not an option Razz


PostPosted: Sun Jan 04, 2009 10:51 pm  Back to top 
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TZF
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Let's look at the slope of the absolute value equation as a function of x:
x < a \implies y' = -3
a < x < b \implies y' = -1
b < x < c \implies y' = 1
c < x \implies y' = 3

The slope of the other equation is -2, so the only way there can be exactly one solution is if the solution occurs at x=a.

The y value of the abs-value equation here is |a-b| + |a-c| = (b-a)+(c-a) = b+c-2a and the y value of the linear equation is 2003-2a.

Thus, 2003 - 2a = b+c-2a \implies b+c=2003.

Since b < c, the minimum c is produced by (b,c) = (1002,1003)
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PostPosted: Sun Jan 04, 2009 10:53 pm  Back to top 
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cdymdcool
Riemann Hypothesis
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#4
TZF wrote:
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Let's look at the slope of the absolute value equation as a function of x:
x < a \implies y' = - 3
a < x < b \implies y' = - 1
b < x < c \implies y' = 1
c < x \implies y' = 3

The slope of the other equation is - 2, so the only way there can be exactly one solution is if the solution occurs at x = a.

The y value of the abs-value equation here is |a - b| + |a - c| = (b - a) + (c - a) = b + c - 2a and the y value of the linear equation is 2003 - 2a.

Thus, 2003 - 2a = b + c - 2a \implies b + c = 2003.

Since b < c, the minimum c is produced by (b,c) = (1002,1003)

How can we conclude that there can be exactly on solution at x=a when we know that the slope of the line is -2?

PostPosted: Sat Nov 07, 2009 2:37 pm  Back to top 
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