Difference between revisions of "2000 AIME II Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
− | Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{i\theta}.</math> | + | Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.</math> |
− | + | <math>r</math> must be <math>1</math> (or else if you take the magnitude would not be the same). Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1</math>. Therefore, the least integer greater is <math>\boxed{000}.</math> | |
~solution by williamgolly | ~solution by williamgolly | ||
+ | |||
+ | == Solution 3 Intuitive == | ||
+ | For this solution, we assume that <math>z^{2000} + 1/z^{2000}</math> and <math>z^{2048} + 1/z^{2048}</math> have the same least integer greater than their solution. we have <math>z + 1/z = 2\cos 3</math>. Since <math>\cos 3 < 1</math>, <math>2\cos 3 < 2</math>. If we square the equation <math>z + 1/z = 2\cos 3</math>, we get <math>z^2 + 2 + 1/(z^2) = 4\cos^2 3</math>, or <math>z^2 + 1/(z^2) = 4\cos^2 3 - 2</math>. <math>4\cos^2 3 - 2</math> is is less than <math>2</math>, since <math>4\cos^2 3</math> is less than <math>4</math>. if we square the equation again, we get <math>z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2</math>. since <math>4\cos^2 3 - 2</math> is less than 2, <math>(4\cos^2 3 - 2)^2</math> is less than 4, and <math>(4\cos^2 3 - 2)^2 -2</math> is less than 2. However <math>(4\cos^2 3 - 2)^2 -2</math> is also less than <math>4\cos^2 3 - 2</math>. we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is <math>\boxed{000}.</math> | ||
+ | |||
+ | ~ PaperMath | ||
+ | |||
+ | == Solution 4 == | ||
+ | First, let <math>z = a+bi</math> where <math>a</math> and <math>b</math> are real numbers. We now have that <cmath>a+bi + \frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}</cmath> given the coniditons of the problem. Equating imaginary coefficients, we have that <cmath>b \left( 1 - \frac{1}{a^2+b^2}\right) = 0</cmath> giving us that either <math>b=0</math> or <math>|z| = 1</math>. Let's consider the latter case for now. | ||
+ | |||
+ | We now know that <math>a^2+b^2=1</math>, so when we equate real coefficients we have that <math>2a = 2 \cos{3^{\circ}}</math>, therefore <math>a = \cos{3^{\circ}}</math>. So, <math>b = \cos{3^{\circ}}</math> and then we can write <math>z = \text{cis}(3)^{\circ}</math> | ||
+ | |||
+ | By De Moivre's Theorem, <cmath>z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}</cmath>. The imaginary parts cancel, leaving us with <math>2 \cos{6000^{\circ}}</math> which is <math>240 \pmod{360}</math>. Therefore, we have it being <math>-1</math> and our answer is <math>\boxed{000}</math>. | ||
+ | |||
+ | Now, if <math>b=0</math> then we have that <math>a+\frac{1}{a} = 2 \cos{3^{\circ}}</math>. Therefore, <math>a</math> is not real violating our conditions set above. | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=8|num-a=10}} | {{AIME box|year=2000|n=II|num-b=8|num-a=10}} |
Latest revision as of 13:42, 4 April 2024
Problem
Given that is a complex number such that
, find the least integer that is greater than
.
Solution
Using the quadratic equation on , we have
.
There are other ways we can come to this conclusion. Note that if is on the unit circle in the complex plane, then
and
. We have
and
. Alternatively, we could let
and solve to get
.
Using De Moivre's Theorem we have ,
, so
.
We want .
Finally, the least integer greater than is
.
Solution 2
Let . Notice that we have
must be
(or else if you take the magnitude would not be the same). Therefore,
and plugging into the desired expression, we get
. Therefore, the least integer greater is
~solution by williamgolly
Solution 3 Intuitive
For this solution, we assume that and
have the same least integer greater than their solution. we have
. Since
,
. If we square the equation
, we get
, or
.
is is less than
, since
is less than
. if we square the equation again, we get
. since
is less than 2,
is less than 4, and
is less than 2. However
is also less than
. we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is
~ PaperMath
Solution 4
First, let where
and
are real numbers. We now have that
given the coniditons of the problem. Equating imaginary coefficients, we have that
giving us that either
or
. Let's consider the latter case for now.
We now know that , so when we equate real coefficients we have that
, therefore
. So,
and then we can write
By De Moivre's Theorem, . The imaginary parts cancel, leaving us with
which is
. Therefore, we have it being
and our answer is
.
Now, if then we have that
. Therefore,
is not real violating our conditions set above.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.