Difference between revisions of "1983 AIME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | It is best to get rid of the [[absolute value]] first. | + | 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>. | 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>, | + | 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=\boxed{015}</math>. |
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== See Also == | == See Also == | ||
Revision as of 18:04, 15 February 2019
Problem
Let 
, where 
. Determine the minimum value taken by 
for 
in the interval 
.
Solution
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that 
, 
, and 
.
Adding these together, we find that the sum is equal to 
, which attains its minimum value (on the given interval 
) when 
.
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 | ||
