Difference between revisions of "2018 AMC 12B Problems/Problem 16"
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The graph of the original equation <math>(a+6+bi) = \sqrt{3}</math> is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation <math>(q+bi)^2 = 3</math>. Using Roots of Unity, we know that the roots of the equation lie at <math>0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} ... 2\pi</math> radians from the origin. | The graph of the original equation <math>(a+6+bi) = \sqrt{3}</math> is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation <math>(q+bi)^2 = 3</math>. Using Roots of Unity, we know that the roots of the equation lie at <math>0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} ... 2\pi</math> radians from the origin. | ||
− | We can quickly notice that the area of the roots will be smallest with points at <math>\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}</math>. Using trigonometry, we get the respective roots to be <math>(Re(z), Im(z)) \in \{(\sqrt{3},0), (\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}), (0,\sqrt{3})\}</math>. Using the shoelace method, the area | + | We can quickly notice that the area of the roots will be smallest with points at <math>\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}</math>. Using trigonometry, we get the respective roots to be <math>(Re(z), Im(z)) \in \{(\sqrt{3},0), (\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}), (0,\sqrt{3})\}</math>. Using the shoelace method, the area quickly comes out to be <math>\frac{3\sqrt{2}-3}{2} \implies \boxed{\textbf{B}}.</math> |
== See Also == | == See Also == |
Revision as of 03:31, 31 January 2021
Contents
Problem
The solutions to the equation are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled and . What is the least possible area of
Solution
The answer is the same if we consider Now we just need to find the area of the triangle bounded by and This is just
Solution 2 (Understanding the polygon)
The polygon formed will be a regular octagon since there are roots of . By normal math computation, we can figure out that two roots of are and . These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be and .
Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed).
The circumcircle of the octagon will have the equation . The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so . Solving gives (meaning that the root represented is ).
This way we can deduce the values of the roots of the equation to be .
To get the area, consecutive points such as and can be used. The area can be computed using different methods like using the shoelace formula, or subtracting areas to find the area. The answer you get is .
(This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using and multiplying your answer by ).
~OlutosinNGA
Solution 3 (Roots of Unity)
Now, we need to solve the equation where . We can substitute this as Now, let for some . Thus, the equation becomes . Taking it to the other side, we get the equation to be . Rearranging variables, we get . Plotting this in the complex place, this is a circle centered at the origin and of radius .
The graph of the original equation is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation . Using Roots of Unity, we know that the roots of the equation lie at radians from the origin.
We can quickly notice that the area of the roots will be smallest with points at . Using trigonometry, we get the respective roots to be . Using the shoelace method, the area quickly comes out to be
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.