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]] 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>. |
Revision as of 10:37, 31 July 2006
Problem
Let , where . Determine the minimum value taken by by in the interval .
Solution
It is best to get rid of the absolute value first.
Under the given circumstances, we notice that , , and .
Adding these together, we find that the sum is equal to , of which the minimum value is attained when .
The answer is thus .