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 < x | + | 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 < x\leq15</math>. |
== Solution == | == Solution == | ||
Revision as of 16:37, 14 February 2011
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 
.
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 | ||
