Difference between revisions of "2011 AMC 12B Problems/Problem 24"
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− | 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> |
Latest revision as of 17:12, 22 August 2021
Contents
[hide]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 |
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