Difference between revisions of "1997 AIME Problems/Problem 14"

Revision as of 20:15, 7 March 2007

Problem

Let $\displaystyle v$ and $\displaystyle w$ be distinct, randomly chosen roots of the equation $\displaystyle z^{1997}-1=0$ . Let $\displaystyle \frac{m}{n}$ be the probability that $\displaystyle\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers. Find $\displaystyle m+n$ .

Solution

The solution requires the use of Euler's formula:

$\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)$

If $\displaystyle \theta=2\pi ik$ , where k is any constant, the equation reduces to:

$\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)$

$\displaystyle =1+0i$

$\displaystyle =1+0$

$\displaystyle =1$

Now, substitute this into the equation:

$\displaystyle z^{1997}-1=0$

$\displaystyle z^{1997}=1$

$\displaystyle z^{1997}=e^{2\pi ik}$

$\displaystyle z=e^{\frac{2\pi ik}{1997}}$

@@ Line 8: / Line 8: @@
 If <math>\displaystyle \theta=2\pi ik</math>, where k is any constant, the equation reduces to:
-<math>
-e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)\\
+<math>\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)</math>
-=1+0i\\
-=1+0\\
+<math>\displaystyle =1+0i</math>
-=1\\
-z^{1997}-1=0\\
+<math>\displaystyle =1+0</math>
-z^{1997}=1\\
-z^{1997}=e^{2\pi ik}\\
+<math>\displaystyle =1</math>
-z=e^{\frac{2\pi ik}{1997}}</math>
+Now, substitute this into the equation:
+<math>\displaystyle z^{1997}-1=0</math>
+<math>\displaystyle z^{1997}=1</math>
+<math>\displaystyle z^{1997}=e^{2\pi ik}</math>
+<math>\displaystyle z=e^{\frac{2\pi ik}{1997}}</math>
 == See also ==
 * [[1997 AIME Problems]]

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Difference between revisions of "1997 AIME Problems/Problem 14"

Revision as of 20:15, 7 March 2007

Problem

Solution

See also