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

Revision as of 11:19, 22 January 2007

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 $0 < p<15$ .

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 $15$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-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>.
+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 ==