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 | + | Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>. |
== Solution == | == Solution == | ||
− | It is best to get rid of the [[absolute value]] first. | + | === Solution 1 === |
+ | It is best to get rid of the [[absolute value]]s 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>. | ||
− | Adding these together, we find that the sum is equal to <math>30-x</math>, | + | Adding these together, we find that the sum is equal to <math>30-x</math>, which attains its minimum value (on the given interval <math>p \leq x \leq 15</math>) when <math>x=15</math>, giving a minimum of <math>\boxed{015}</math>. |
− | + | === Solution 2 === | |
− | <math> | + | Let <math>p</math> be equal to <math>15 - \varepsilon</math>, where <math>\varepsilon</math> is an almost neglectable value. Because of the small value <math>\varepsilon</math>, the [[domain]] of <math>f(x)</math> is basically the [[set]] <math>{15}</math>. plugging in <math>15</math> gives <math>\varepsilon + 0 + 15 - \varepsilon</math>, or <math>15</math>, so the answer is <math>\boxed{15}</math> |
− | == See | + | == See Also == |
{{AIME box|year=1983|num-b=1|num-a=3}} | {{AIME box|year=1983|num-b=1|num-a=3}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 20:16, 14 January 2023
Problem
Let , where
. Determine the minimum value taken by
for
in the interval
.
Solution
Solution 1
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that ,
, and
.
Adding these together, we find that the sum is equal to , which attains its minimum value (on the given interval
) when
, giving a minimum of
.
Solution 2
Let be equal to
, where
is an almost neglectable value. Because of the small value
, the domain of
is basically the set
. plugging in
gives
, or
, so the answer is
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 |