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

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=== Solution 3 ===
 
=== Solution 3 ===
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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>.
+
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 09:49, 23 June 2016

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 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 $60 + 72 + 144 = \boxed{276}$.

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

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