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

Revision as of 23:45, 23 July 2006

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 $015$ .

@@ 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>.
 == Solution ==
+It is best to get rid of the absolute value first.
-== See also ==
+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]]
+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>.
+The answer is thus <math>015</math>.
+----
+* [[1983 AIME Problems/Problem |Previous Problem]]
+* [[1983 AIME Problems/Problem |Next Problem]]
+* [[1983 AIME Problems|Back to Exam]]
+[[Category:Intermediate Algebra Problems]]