2011 AMC 12B Problems/Problem 24: Difference between revisions

Revision as of 21:49, 29 January 2017

Problem

Let $P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)$ . What is the minimum perimeter among all the $8$ -sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)$ ?

$\textbf{(A)}\ 4\sqrt{3} + 4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2} + 3\sqrt{6}$

$\textbf{(D)}\ 4\sqrt{2} + 4\sqrt{3} \qquad \textbf{(E)}\ 4\sqrt{3} + 6$

Solution

Answer: (B)

First of all, we need to find all $z$ such that $P(z) = 0$

$P(z) = \left(z^4 - 1\right)\left(z^4 + \left(4\sqrt{3} + 7\right)\right)$

So $z^4 = 1$ or $z = e^{i\frac{n\pi}{2}}$

or $z^4 = - \left(4\sqrt{3} + 7\right)$

$z^2 = \pm i \sqrt{4\sqrt{3} + 7} = e^{i\frac{(2n+1)\pi}{2}} \left(\sqrt{3} + 2\right)$

$z = e^{i\frac{(2n+1)\pi}{4}} \sqrt{\sqrt{3} + 2} = e^{i\frac{(2n+1)\pi}{4}} \left(\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)$

Now we have a solution at $\frac{n\pi}{4}$ if we look at them in polar coordinate, further more, the 8-gon is symmetric (it is an regular octagon) . So we only need to find the side length of one and multiply by $8$ .

So answer $= 8 \times$ distance from $1$ to $\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 + i\right)$

Side length $= \sqrt{\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^2} = \sqrt{2\left(\frac{3}{4} + \frac{1}{4}\right)} = \sqrt{2}$

Hence, answer is $8\sqrt{2}$ .

Easier method: Use the law of cosines. We make $a$ the distance. Now, since the angle does not change the distance from the origin, we can just use the distance. $a^2 = (\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 + 1^2 -2 \times \Big( \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \Big)\times 1 \times \cos \frac{\pi}{4}$ , which simplifies to $a^2= 2 + \sqrt3 +1 - 1 - \sqrt3$ , or $a^2=2$ , or $a=\sqrt2$ . Multiply the answer by 8 to get $\boxed{ (B) 8\sqrt2}$

@@ Line 37: / Line 37: @@
 Easier method: Use the law of cosines.
 We make <math>a</math> the distance.
-Now, since the angle does not change the distance from the origin, we can just use the distance. <math>a^2 = (\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 + 1^2 -2 \times \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \times 1 \times \cos \frac{\pi}{4}</math>, which simplifies to <math>a^2= 2 + \sqrt3 +1 - 1 - \sqrt3</math>, or <math>a^2=2</math>, or <math>a=\sqrt2</math>.
+Now, since the angle does not change the distance from the origin, we can just use the distance. <math>a^2 = (\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 + 1^2 -2 \times \Big( \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \Big)\times 1 \times \cos \frac{\pi}{4}</math>, which simplifies to <math>a^2= 2 + \sqrt3 +1 - 1 - \sqrt3</math>, or <math>a^2=2</math>, or <math>a=\sqrt2</math>.
 Multiply the answer by 8 to get <math>\boxed{ (B) 8\sqrt2}</math>

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2011 AMC 12B Problems/Problem 24: Difference between revisions

Revision as of 21:49, 29 January 2017

Problem

Solution

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