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

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== Problem ==
 
== Problem ==
 
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>.
 
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>.
  
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== Solution ==
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== Solution 1 ==
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=== Solution 1 ===
 
 
It is best to get rid of the [[absolute value]]s first.  
 
It is best to get rid of the [[absolute value]]s first.  
  
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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>.
 
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>.
  
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=== Solution 2 ===
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== Solution 2 ==
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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>
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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{015}</math>
  
 
== See Also ==
 
== See Also ==
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{{AIME box|year=1983|num-b=1|num-a=3}}
 
{{AIME box|year=1983|num-b=1|num-a=3}}
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{{MAA Notice}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 21:11, 4 February 2025

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 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 $\boxed{015}$

See Also

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
All AIME Problems and Solutions

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