Difference between revisions of "1983 AIME Problems/Problem 2"

 
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== Problem ==
 
== Problem ==
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Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>p \leq x \leq 15</math>. Determine the minimum value taken by <math>f(x)</math> by <math>x</math> in the interval <math>0 < p<15</math>.
  
 
== Solution ==
 
== Solution ==
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It is best to get rid of the absolute value first.
  
== See also ==
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Under the given circumstances, we notice that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.
* [[1983 AIME Problems]]
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Adding these together, we find that the sum is equal to <math>30-x</math>, of which the minimum value is attained when <math>x=15</math>.
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The answer is thus <math>015</math>.
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[[Category:Intermediate Algebra Problems]]

Revision as of 22:45, 23 July 2006

Problem

Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $p \leq x \leq 15$. Determine the minimum value taken by $f(x)$ by $x$ in the interval $0 < p<15$.

Solution

It is best to get rid of the absolute value first.

Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$.

Adding these together, we find that the sum is equal to $30-x$, of which the minimum value is attained when $x=15$.

The answer is thus $015$.