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

(Another method)
m (fixed latex)
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We make <math>a</math> the distance.
 
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 \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>.
−
Multiply the answer by 8 to get $\boxed{ (B) 8\sqrt2}
+
Multiply the answer by 8 to get <math>\boxed{ (B) 8\sqrt2}</math>
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:34, 22 December 2015

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 equilateral 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. $a^2 = (\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 + 1^2 -2 \times \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \times 1 \times \cos \frac{\pi}{4}$, which simplifies to $a^2= 2 + \sqrt3 +1 - 1 - \sqrt3$, or $a^2=2$, or $a=\sqrt2$. Multiply the answer by 8 to get $\boxed{ (B) 8\sqrt2}$

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

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