1983 AIME Problems/Problem 2

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 $\boxed{15}$ as the minimum value for $f(x)$ , which indeed is the answer posted above.

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1983 AIME Problems/Problem 2

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