AoPSWiki

Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!

Personal tools

Search

Toolbox

1996 AIME Problems/Problem 11

From AoPSWiki

Revision as of 21:17, 16 March 2009 by Boo (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

Problem

Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ .

Contents

1 Problem
2 Solution
3 See also

Solution

Solution 1

$\begin{eqnarray*}0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\0 &=& \frac{(z^5 -...$

Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$ ,

or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 300$

(see cis).

Discarding the roots with negative imaginary parts (leaving us with $\mathrm{cis} \theta,\ 0 < \theta < 180$ ), we are left with $\mathrm{cis}\ 60, 72, 144$ ; their product is $P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}$ .

Solution 2

Let $w =$ the fifth roots of unity, except for $1$ . Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$ , and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$ . Long division quickly gives the other factor to be $z^2 - z + 1$ . The solution follows as above.

Solution 3

Divide through by $z^3$ . We get the equation $z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0$ . Let $x = z + \frac {1}{z}$ . Then $z^3 + \frac {1}{z^3} = x^3 - 3x$ . Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$ , with solutions $x = 1, \frac { - 1\pm\sqrt {5}}{2}$ . For $x = 1$ , we get $z = \text{cis}60,\text{cis}300$ . For $x = \frac { - 1 + \sqrt {5}}{2}$ , we get $z = \text{cis}{72},\text{cis}{292}$ (using exponential form of $\cos$ ). For $x = \frac { - 1 - \sqrt {5}}{2}$ , we get $z = \text{cis}144,\text{cis}216$ . The ones with positive imaginary parts are ones where $0\le\theta\le180$ , so we have $60 + 72 + 144 = 276$ .

See also

1996 AIME (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1996_AIME_Problems/Problem_11"

Category: Intermediate Algebra Problems

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us