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>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>. | ||
== Solution == | == Solution == | ||
+ | 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>. |
− | * [[1983 AIME Problems]] | + | |
+ | 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>. | ||
+ | |||
+ | The answer is thus <math>015</math>. | ||
+ | |||
+ | ---- | ||
+ | |||
+ | * [[1983 AIME Problems/Problem |Previous Problem]] | ||
+ | * [[1983 AIME Problems/Problem |Next Problem]] | ||
+ | * [[1983 AIME Problems|Back to Exam]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 22:45, 23 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 .