Difference between revisions of "1983 AIME Problems/Problem 2"

Revision as of 21:11, 20 October 2024

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 $\boxed{15}$

@@ Line 5: / Line 5: @@
 == Solution ==
+IT IS BETTER IF YOU PLAY SKIBIDI TOWER DEFENSE!!!! FREE UNITS IF YOU TRADE WITH damnyou_12!!!! exclusive deals!
 === Solution 1 ===
 It is best to get rid of the [[absolute value]]s first.

1983 AIME (Problems • Answer Key • Resources)
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