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Difference between revisions of "2011 AMC 12B Problems/Problem 24"

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Hence, answer is <math>8\sqrt{2}</math>.
 
Hence, answer is <math>8\sqrt{2}</math>.
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=B}}
 
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=B}}

Revision as of 21:30, 26 September 2011

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 amont 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 a 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 $8\sqrt{2}$. $3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3<cmath>3</cmath>3$$3$

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_24&oldid=42418"