1983 AIME Problems/Problem 2

Revision as of 19:20, 2 October 2016 by Turtlepie (talk | contribs) (Aime 0 in front of the 15)

Problem

Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $p \leq x \leq 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $0 < x\leq15$.

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 $x=\boxed{15}$.

Edit: $|x-p-15|$ can equal $15+p-x$ or $x-p-15$ (for example, if $x=7$ and $p=-12$, $x-p-15=4$). Thus, our two "cases" are $30-x$ (if $x-p\leq15$) and $x-2p$ (if $x-p\geq15$). However, both of these cases give us $\boxed{015}$ as the minimum value for $f(x)$, which indeed is the answer posted above.


Also note the lowest value occurs when $x=p=15$ because this make the first two requirements $0$. It is easy then to check that 15 is the minimum value.

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