2018 AMC 12A Problems/Problem 22
Contents
Problem
The solutions to the equations and where form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form where and are positive integers and neither nor is divisible by the square of any prime number. What is
Solution 1
The roots are (easily derivable by using DeMoivre and half-angle). From there, shoelace on and multiplying by gives the area of , so the answer is . (trumpeter)
Solution 2 (No DeMoivre's)
Write as . For the first equation, Setting the real parts equal and imaginary parts equal, we have: Squaring the second equation gives . We now need two numbers that have a difference of and a product of . By inspection, and work, so and . Since is positive, and must have the same sign. Thus we have two solutions for : Repeating the process for the second equation, we have two solutions: In a clockwise direction, the points are . Now we can use the shoelace theorem. The area is , so the answer is .
Solution 3 (Vectors)
Rather than thinking about this with complex numbers, notice that if we take two solutions and think of them as vectors, the area of the parallelogram they form is half the desired area. Also, notice that the area of a parallelogram is where and are the side lengths.
The side lengths are easily found since we are given the squares of . Thus, the magnitude of in the first equation is just and in the second equation is just . Now, we need .
To find , think about what squaring is in complex numbers. The angle between the squares of the two solutions is twice the angle between the two solutions themselves. In addition, we can find of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are and , so their dot product is . Dividing by the magnitudes yields: . This is , and recall the identity . This means that , so . Now, notice that (which is fairly easy to discover) so . Finally, putting everything together yields: as the area of the parallelogram found by treating two of the solutions as vectors. However, drawing a picture out shows that we actually want twice this (each fourth of the parallelogram from the problem is one half of the parallelogram whose area was found above) so the desired area is actually . Then, the answer is .
~ Aathreyakadambi
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc12a/472
~ dolphin7
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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