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

Latest revision as of 20:16, 14 January 2023

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

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 1: / Line 1: @@
+__TOC__
 == Problem ==
-Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>p \leq x \leq 15</math>. Determine the minimum value taken by <math>f(x)</math> by <math>x</math> in the interval <math>0 < p<15</math>.
+Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>.
 == Solution ==
-It is best to get rid of the absolute value first.
+=== Solution 1 ===
+It is best to get rid of the [[absolute value]]s first.
 Under the given circumstances, we notice that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.
-Adding these together, we find that the sum is equal to <math>30-x</math>, of which the minimum value is attained when <math>x=15</math>.
+Adding these together, we find that the sum is equal to <math>30-x</math>, which attains its minimum value (on the given interval <math>p \leq x \leq 15</math>) when <math>x=15</math>, giving a minimum of <math>\boxed{015}</math>.
-The answer is thus <math>015</math>.
-----
+=== Solution 2 ===
+Let <math>p</math> be equal to <math>15 - \varepsilon</math>, where <math>\varepsilon</math> is an almost neglectable value. Because of the small value <math>\varepsilon</math>, the [[domain]] of <math>f(x)</math> is basically the [[set]] <math>{15}</math>. plugging in <math>15</math> gives <math>\varepsilon + 0 + 15 - \varepsilon</math>, or <math>15</math>, so the answer is <math>\boxed{15}</math>
-* [[1983 AIME Problems/Problem |Previous Problem]]
+== See Also ==
-* [[1983 AIME Problems/Problem |Next Problem]]
+{{AIME box|year=1983|num-b=1|num-a=3}}
-* [[1983 AIME Problems|Back to Exam]]
 [[Category:Intermediate Algebra Problems]]

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