Difference between revisions of "1996 AIME Problems/Problem 11"
(→Solution 3) |
(→Solution 3) |
||
Line 21: | Line 21: | ||
=== Solution 3 === | === Solution 3 === | ||
− | Divide through by <math>z^3</math>. We get the equation <math>z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0</math>. Let <math>x = z + \frac {1}{z}</math>. Then <math>z^3 + \frac {1}{z^3} = x^3 - 3x</math>. Our equation is then <math>x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0</math>, with solutions <math>x = 1, \frac { - 1\pm\sqrt {5}}{2}</math>. For <math>x = 1</math>, we get <math>z = \text{cis}60,\text{cis}300</math>. For <math>x = \frac { - 1 + \sqrt {5}}{2}</math>, we get <math>z = \text{cis}{72},\text{cis}{292}</math> (using | + | Divide through by <math>z^3</math>. We get the equation <math>z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0</math>. Let <math>x = z + \frac {1}{z}</math>. Then <math>z^3 + \frac {1}{z^3} = x^3 - 3x</math>. Our equation is then <math>x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0</math>, with solutions <math>x = 1, \frac { - 1\pm\sqrt {5}}{2}</math>. For <math>x = 1</math>, we get <math>z = \text{cis}60,\text{cis}300</math>. For <math>x = \frac { - 1 + \sqrt {5}}{2}</math>, we get <math>z = \text{cis}{72},\text{cis}{292}</math> (using exponential form of <math>\cos</math>). For <math>x = \frac { - 1 - \sqrt {5}}{2}</math>, we get <math>z = \text{cis}144,\text{cis}216</math>. The ones with positive imaginary parts are ones where <math>0\le\theta\le180</math>, so we have <math>60 + 72 + 144 = \boxed{276}</math>. |
== See also == | == See also == |
Revision as of 11:02, 23 June 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.