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

m (Aime 0 in front of the 15)
m (Fixed only answer being boxed without duplicates)
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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>.
 
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=\boxed{15}</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=\boxed{015}</math>.
  
 
Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are
 
Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are
−
<math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>\boxed{015}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above.
+
<math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>15</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above.
  
  

Revision as of 14:41, 4 October 2016

Problem

Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $p \leq x \leq 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $0 < x\leq15$.

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 $x=\boxed{015}$.

Edit: $|x-p-15|$ can equal $15+p-x$ or $x-p-15$ (for example, if $x=7$ and $p=-12$, $x-p-15=4$). Thus, our two "cases" are $30-x$ (if $x-p\leq15$) and $x-2p$ (if $x-p\geq15$). However, both of these cases give us $15$ as the minimum value for $f(x)$, which indeed is the answer posted above.


Also note the lowest value occurs when $x=p=15$ because this make the first two requirements $0$. It is easy then to check that 15 is the minimum value.

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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