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

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The answer is thus <math>15</math>.
 
The answer is thus <math>15</math>.
  
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== See also ==
 
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{{AIME box|year=1983|num-b=1|num-a=3}}
* [[1983 AIME Problems/Problem 1|Previous Problem]]
 
* [[1983 AIME Problems/Problem 3|Next Problem]]
 
* [[1983 AIME Problems|Back to Exam]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 17:35, 21 March 2007

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 $15$.

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions