AMC12 2018 A

Answer (A): Let $z=a+bi$ be a solution of the first equation, where $a$ and $b$ are real numbers. Then $(a+bi)^2=4+4\sqrt{15}i$. Expanding the left-hand side and equating real and imaginary parts yields $$a^2-b^2=4 \quad \text{and} \quad 2ab=4\sqrt{15}.$$ AMC 12A Solutions 13 From the second equation, $b=\dfrac{2\sqrt{15}}{a}$, and substituting this into the first equation and simplifying gives $(a^2)^2-4a^2-60=0$, which factors as $(a^2-10)(a^2+6)=0$. Because $a$ is real, it follows that $a=\pm\sqrt{10}$, from which it then follows that $b=\pm\sqrt{6}$. Thus two vertices of the parallelogram are $\sqrt{10}+\sqrt{6}i$ and $-\sqrt{10}-\sqrt{6}i$. A similar calculation with the other given equation shows that the other two vertices of the parallelogram are $\sqrt{3}+i$ and $-\sqrt{3}-i$. The area of this parallelogram can be computed using the shoelace formula, which gives the area of a polygon in terms of the coordinates of its vertices $(x_1,y_1)$, $(x_2,y_2)$, $\ldots$, $(x_n,y_n)$ in clockwise or counter-clockwise order: $$\frac12\cdot\left|\,(x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1)-(y_1x_2+y_2x_3+\cdots+y_{n-1}x_n+y_nx_1)\,\right|.$$ In this case $x_1=\sqrt{10}$, $y_1=\sqrt{6}$, $x_2=\sqrt{3}$, $y_2=1$, $x_3=-\sqrt{10}$, $y_3=-\sqrt{6}$, $x_4=-\sqrt{3}$, and $y_4=-1$. The area is $6\sqrt{2}-2\sqrt{10}$, and the requested sum of the four positive integers in this expression is $20$.