1983 AIME Problems/Problem 2

Revision as of 21:12, 20 October 2024 by Aops student learner (talk | contribs) (Solution)

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

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Solution 1

It is best to get rid of the absolute values 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 $30-x$, which attains its minimum value (on the given interval $p \leq x \leq 15$) when $x=15$, giving a minimum of $\boxed{015}$.

Solution 2

Let $p$ be equal to $15 - \varepsilon$, where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$, the domain of $f(x)$ is basically the set ${15}$. plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$, or $15$, so the answer is $\boxed{15}$

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