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1983 AIME Problems/Problem 2

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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 (ProblemsResources)
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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