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> for <math>x</math> in the [[interval]] <math> | + | 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> for <math>x</math> in the [[interval]] <math>p < x\leq15</math>. |
== Solution == | == Solution == |
Revision as of 19:20, 10 June 2018
Problem
Let , where
. Determine the minimum value taken by
for
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
.
Edit: can equal
or
(for example, if
and
,
). Thus, our two "cases" are
(if
) and
(if
). However, both of these cases give us
as the minimum value for
, which indeed is the answer posted above.
Also note the lowest value occurs when because this make the first two requirements
. 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 |