1983 AIME Problems/Problem 2

Problem

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

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 $x=\boxed{015}$ .

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.

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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