Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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<math>b+1=0\rightarrow b=-1</math> so <math>|b|=1</math>. | <math>b+1=0\rightarrow b=-1</math> so <math>|b|=1</math>. | ||
− | <math>b^2+2a+1=0\rightarrow b^2=-1-2a</math>. This is a circle in the complex plane centered at <math>(-1,0)</math> with radius <math>2</math> since <math>|a|=1</math>. The maximum distance from the origin is <math>3</math> at <math>(-3,0)</math> and similarly the minimum distance is <math>1</math> at <math>(1,0)</math>. So <math>1\le b^2\le 3\rightarrow 1\le b\le \sqrt{3}</math>. | + | <math>b^2+2a+1=0\rightarrow b^2=-1-2a</math>. This is a circle in the complex plane centered at <math>(-1,0)</math> with radius <math>2</math> since <math>|a|=1</math>. The maximum distance from the origin is <math>3</math> at <math>(-3,0)</math> and similarly the minimum distance is <math>1</math> at <math>(1,0)</math>. So <math>1\le |b^2|\le 3\rightarrow 1\le |b|\le \sqrt{3}</math>. |
Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>. | Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/FU18x_LsTeQ | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==Note== | ||
+ | This problem is kinda similar to [[2002 AIME I Problems/Problem 12]] | ||
== See also == | == See also == |
Latest revision as of 15:49, 5 November 2023
Problem
Let and
, where
and
are complex numbers. Suppose that
and
for all
for which
is defined. What is the difference between the largest and smallest possible values of
?
Solution
By algebraic manipulations, we obtain
where
In order for
, we must have
,
, and
.
implies
or
.
implies
,
, or
.
implies
or
.
Since , in order to satisfy all 3 conditions we must have either
or
. In the first case
.
For the latter case note that
and hence,
.
On the other hand,
so,
. Thus
. Hence the maximum value for
is
while the minimum is
(which can be achieved for instance when
or
respectively). Therefore the answer is
.
Shortcut
We only need in
.
Set :
. Since
, either
or
.
so
.
. This is a circle in the complex plane centered at
with radius
since
. The maximum distance from the origin is
at
and similarly the minimum distance is
at
. So
.
Both solutions give the same lower bound, . So the range is
.
Video Solution
~MathProblemSolvingSkills.com
Note
This problem is kinda similar to 2002 AIME I Problems/Problem 12
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.