Difference between revisions of "2000 AIME II Problems/Problem 9"

Revision as of 18:32, 11 November 2007

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ .

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 == Problem ==
-The system of equations
+Given that <math>z</math> is a complex number such that <math>z+\frac 1z=2\cos 3^\circ</math>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>.
-<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
-\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
-\log_{10}(2000zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
-\end{eqnarray*}</math></center>
-has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>.
 == Solution ==

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