Difference between revisions of "1996 AIME Problems/Problem 11"

Revision as of 16:17, 16 March 2009

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 $\theta$ .

Solution

Solution 1

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

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 $60 + 72 + 144 = 276$ .

@@ Line 10: / Line 10: @@
 \end{eqnarray*}</cmath>
-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, 300</math> (see [[cis]]). Discarding the roots with negative imaginary parts (leaving us with <math>\mathrm{cis} \theta,\ 0 < \theta < 180</math>), we are left with <math>\mathrm{cis}\ 60, 72, 144</math>; their product is <math>P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}</math>.
+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, 300</math>
+(see [[cis]]).
+Discarding the roots with negative imaginary parts (leaving us with <math>\mathrm{cis} \theta,\ 0 < \theta < 180</math>), we are left with <math>\mathrm{cis}\ 60, 72, 144</math>; their product is <math>P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}</math>.
 === Solution 2 ===

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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