Difference between revisions of "1996 AIME Problems/Problem 11"
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m (→Solution 1) |
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
− | Thus <math>z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72 | + | Thus <math>z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288</math>, |
− | or <math>z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60 | + | or <math>z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 30</math> |
(see [[cis]]). | (see [[cis]]). |
Revision as of 20:20, 16 August 2016
Problem
Let be the product of the roots of
that have a positive imaginary part, and suppose that
, where
and
. Find
.
Solution 1
Thus ,
or
(see cis).
Discarding the roots with negative imaginary parts (leaving us with ), we are left with
; their product is
.
Solution 2
Let the fifth roots of unity, except for
. Then
, and since both sides have the fifth roots of unity as roots, we have
. Long division quickly gives the other factor to be
. The solution follows as above.
Solution 3
Divide through by . We get the equation
. Let
. Then
. Our equation is then
, with solutions
. For
, we get
. For
, we get
(using exponential form of
). For
, we get
. The ones with positive imaginary parts are ones where
, so we have
.
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.