Difference between revisions of "2008 AMC 12B Problems/Problem 19"
Quaratinium (talk | contribs) (→Solution) |
Quaratinium (talk | contribs) (→Solution) |
||
Line 24: | Line 24: | ||
Solving, we get <math>\alpha=1-i</math>, <math>\gamma</math> can be anything, to minimize the value we set <math>\gamma=0</math>, so then the answer is <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. Thus, the answer is <math>\boxed{B}</math> | Solving, we get <math>\alpha=1-i</math>, <math>\gamma</math> can be anything, to minimize the value we set <math>\gamma=0</math>, so then the answer is <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. Thus, the answer is <math>\boxed{B}</math> | ||
+ | |||
By: Quaratinium | By: Quaratinium | ||
Revision as of 01:34, 30 December 2017
Problem 19
A function is defined by
for all complex numbers
, where
and
are complex numbers and
. Suppose that
and
are both real. What is the smallest possible value of
?
Solution
We need only concern ourselves with the imaginary portions of and
(both of which must be 0). These are:
Let and
then we know
and
Therefore
which reaches its minimum
when
by the Trivial Inequality. Thus, the answer is
Solution 2:
Since and
are both real we get,
Solving, we get ,
can be anything, to minimize the value we set
, so then the answer is
. Thus, the answer is
By: Quaratinium
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.