Difference between revisions of "1983 AIME Problems/Problem 2"
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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>, of which the minimum value is attained when <math>x=\boxed{ | + | 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=\boxed{015}</math>. |
Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are | Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are | ||
− | <math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math> | + | <math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>15</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. |
Revision as of 15:41, 4 October 2016
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 |