Difference between revisions of "2011 AMC 12B Problems/Problem 24"
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Let <math>P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)</math>. What is the minimum perimeter among all the <math>8</math>-sided polygons in the complex plane whose vertices are precisely the zeros of <math>P(z)</math>? | Let <math>P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)</math>. What is the minimum perimeter among all the <math>8</math>-sided polygons in the complex plane whose vertices are precisely the zeros of <math>P(z)</math>? | ||
− | <math>\textbf{(A)}\ 4\sqrt{3} + 4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2} + 3\sqrt{6} | + | <math>\textbf{(A)}\ 4\sqrt{3} + 4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2} + 3\sqrt{6} \qquad \textbf{(D)}\ 4\sqrt{2} + 4\sqrt{3} \qquad \textbf{(E)}\ 4\sqrt{3} + 6</math> |
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== Solution == | == Solution == | ||
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<math>P(z) = \left(z^4 - 1\right)\left(z^4 + \left(4\sqrt{3} + 7\right)\right)</math> | <math>P(z) = \left(z^4 - 1\right)\left(z^4 + \left(4\sqrt{3} + 7\right)\right)</math> | ||
− | So <math>z^4 = 1 | + | So <math>z^4 = 1 \implies z = e^{i\frac{n\pi}{2}}</math> |
or <math>z^4 = - \left(4\sqrt{3} + 7\right)</math> | or <math>z^4 = - \left(4\sqrt{3} + 7\right)</math> | ||
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<br /> | <br /> | ||
− | Now we have a solution at <math>\frac{n\pi}{4}</math> if we look at them in polar coordinate, further more, the 8-gon is symmetric (it is an | + | Now we have a solution at <math>\frac{n\pi}{4}</math> 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 <math>8</math>. |
So answer <math>= 8 \times</math> distance from <math>1</math> to <math>\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 + i\right)</math> | So answer <math>= 8 \times</math> distance from <math>1</math> to <math>\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 + i\right)</math> | ||
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Hence, answer is <math>8\sqrt{2}</math>. | Hence, answer is <math>8\sqrt{2}</math>. | ||
+ | == Solution 2 == | ||
− | + | Use the law of cosines. | |
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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 \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> | Multiply the answer by 8 to get <math>\boxed{ (B) 8\sqrt2}</math> | ||
Latest revision as of 18:12, 22 August 2021
Contents
Problem
Let . What is the minimum perimeter among all the
-sided polygons in the complex plane whose vertices are precisely the zeros of
?
Solution
Answer: (B)
First of all, we need to find all such that
So
or
Now we have a solution at 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
.
So answer distance from
to
Side length
Hence, answer is .
Solution 2
Use the law of cosines.
We make the distance.
Now, since the angle does not change the distance from the origin, we can just use the distance.
, which simplifies to
, or
, or
.
Multiply the answer by 8 to get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.