1983 AIME Problems/Problem 2: Difference between revisions

Revision as of 23:14, 26 February 2011

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 $0 < x\leq15$ .

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 $x=\boxed{15}$ .

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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>.
-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>.
+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=\boxed{15}</math>.
-The answer is thus <math>15</math>.
 == See also ==

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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