Difference between revisions of "2018 AMC 12A Problems/Problem 22"

Revision as of 12:49, 20 August 2021

Problem

The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$

$\textbf{(A)} 20 \qquad \textbf{(B)} 21 \qquad \textbf{(C)} 22 \qquad \textbf{(D)} 23 \qquad \textbf{(E)} 24$

Solution 1

The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$ , so the answer is $\boxed{20}$ . (trumpeter)

Solution 2 (No DeMoivre's)

Write $z$ as $a + bi$ . For the first equation, $(a + bi)^2 = 4 + 4\sqrt{15}i$ $a^2 + 2abi - b^2 = 4 + 4\sqrt{15}i$ Setting the real parts equal and imaginary parts equal, we have: $a^2 - b^2 = 4$ $ab = 2\sqrt{15}$ Squaring the second equation gives $a^2b^2 = 60$ . We now need two numbers that have a difference of $4$ and a product of $60$ . By inspection, $10$ and $6$ work, so $a^2 = 10$ and $b^2 = 6$ . Since $ab$ is positive, $a$ and $b$ must have the same sign. Thus we have two solutions for $(a, b)$ : $(-\sqrt{10}, -\sqrt{6})$ $(\sqrt{10}, \sqrt{6})$ Repeating the process for the second equation, we have two solutions: $(-\sqrt{3}, -1)$ $(\sqrt{3}, 1)$ In a clockwise direction, the points are $(-\sqrt{10}, -\sqrt{6}), (-\sqrt{3}, -1), (\sqrt{10}, \sqrt{6}), (\sqrt{3}, 1)$ . Now we can use the shoelace theorem. The area is $6\sqrt{2}-2\sqrt{10}$ , so the answer is $\boxed{20}$ .

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 $ab\sin \theta$ where $a$ and $b$ are the side lengths.

The side lengths are easily found since we are given the squares of $z$ . Thus, the magnitude of $z$ in the first equation is just $\sqrt{16} = 4$ and in the second equation is just $\sqrt{4} = 2$ . Now, we need $\sin \theta$ .

To find $\theta$ , 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 $\cos$ of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are $\langle 4, 4\sqrt{15}\rangle$ and $\langle 2, 2\sqrt{3}\rangle$ , so their dot product is $8 + 24\sqrt{5}$ . Dividing by the magnitudes yields: $\dfrac{8+24\sqrt{5}}{4 \cdot 16} = \dfrac{1 + 3\sqrt{5}}{8}$ . This is $\cos 2\theta$ , and recall the identity $\cos 2\theta = 1 - 2\sin^2 \theta$ . This means that $\sin^2 \theta = \dfrac{7 - 3\sqrt{5}}{16}$ , so $\sin \theta = \dfrac{\sqrt{7- 3\sqrt{5}}}{4}$ . Now, notice that $\sqrt{7- 3\sqrt{5}} = \dfrac{3\sqrt{2}-\sqrt{10}}{2}$ (which is not too hard to discover) so $\sin \theta = \dfrac{3\sqrt{2}-\sqrt{10}}{8}$ . Finally, putting everything together yields: $2\times 4 \times \dfrac{3\sqrt{2}-\sqrt{10}}{8} = 3\sqrt{2} - \sqrt{10}$ 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 $6\sqrt{2} - 2\sqrt{10}$ . Then, the answer is $\boxed{20}$ .

~ Aathreyakadambi

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc12a/472

~ dolphin7

@@ Line 36: / Line 36: @@
 ~ Aathreyakadambi
-=== Video Solution by Richard Rusczyk ===
+== Video Solution by Richard Rusczyk ==
 https://artofproblemsolving.com/videos/amc/2018amc12a/472

2018 AMC 12A (Problems • Answer Key • Resources)
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