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Difference between revisions of "1983 AIME Problems/Problem 2"
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== Solution ==
 
== Solution ==
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It is best to get rid of the absolute value first.  
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It is best to get rid of the [[absolute value]] first.  
  
 
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>.
 
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>.

Revision as of 11:37, 31 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 $15$.

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