Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are complex | + | <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are [[complex number]]s such that |
| − | <math>\begin{align*}\zeta_1+\zeta_2+\zeta_3&=1\\ | + | <center><math>\begin{align*}\zeta_1+\zeta_2+\zeta_3&=1\\ |
\zeta_1^2+\zeta_2^2+\zeta_3^2&=3\\ | \zeta_1^2+\zeta_2^2+\zeta_3^2&=3\\ | ||
| − | \zeta_1^3+\zeta_2^3+\zeta_3^3&=7\end{align*}</math> | + | \zeta_1^3+\zeta_2^3+\zeta_3^3&=7\end{align*}</math></center> |
Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | ||
==Solution== | ==Solution== | ||
| − | {{ | + | We let <math>e_1 = \zeta_1 + \zeta_2 + \zeta_3,\ e_2 = \zeta_1\zeta_2 + \zeta_2\zeta_3 + \zeta_3\zeta_1,\ e_3 = \zeta_1\zeta_2\zeta_3</math> (the [[elementary symmetric sums]]). Then, we can rewrite the above equations as |
| + | <center><math>\begin{align*}\zeta_1+\zeta_2+\zeta_3&=e_1 = 1\\ | ||
| + | \zeta_1^2+\zeta_2^2+\zeta_3^2&= e_1^2 - 2e_2 = 3</math></center> | ||
| + | from where it follows that <math>e_2 = -1</math>. The third equation can be factored as | ||
| + | <center><math>\begin{align*}7 =\zeta_1^3+\zeta_2^3+\zeta_3^3 &= (\zeta_1+\zeta_2+\zeta_3)(\zeta_1^2+\zeta_2^2+\zeta_3^2-\zeta_1\zeta_2-\zeta_2\zeta_3 -\zeta_3\zeta_1)+3\zeta_1\zeta_2\zeta_3\\ &= e_1^3 - 3e_1e_2 + 3e_3,</math></center> | ||
| + | from where it follows that <math>e_3 = 1</math>. Thus, applying [[Vieta's formulas]] backwards, <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are the roots of the polynomial | ||
| + | <center><math>\begin{align}x^3 - x^2 - x - 1 = 0 \Longleftrightarrow x^3 = x^2 + x + 1\end{align}</math></center> | ||
| + | Let <math>s_n = \zeta_1^n + \zeta_2^n + \zeta_3^n</math> (the [[power sums]]). Then from <math>(1)</math>, we have the [[recursion]] <math>s_{n+3} = s_{n+2} + s_{n+1} + s_n</math>. It follows that <math>s_4 = 7 + 3 + 1 = 11, s_5 = 21, s_6 = 39, s_7 = \boxed{071}</math>. | ||
| + | |||
| + | ==See also== | ||
| + | {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 22:36, 23 April 2008
Problem
and 
are complex numbers such that
\zeta_1^2+\zeta_2^2+\zeta_3^2&=3\\
\zeta_1^3+\zeta_2^3+\zeta_3^3&=7\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Compute 
.
Solution
We let 
(the elementary symmetric sums). Then, we can rewrite the above equations as
from where it follows that 
. The third equation can be factored as
from where it follows that 
. Thus, applying Vieta's formulas backwards, and are the roots of the polynomial
Let 
(the power sums). Then from 
, we have the recursion 
. It follows that 
.
See also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
