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

## 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 $\boxed{20}$.

~ dolphin7