2018 AMC 12A Problems/Problem 22
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,
![\[(a + bi)^2 = 4 + 4\sqrt{15}i\]](http://latex.artofproblemsolving.com/c/3/c/c3c6af2da0378d12759c8bf761f98548064179e0.png)
![\[a^2 + 2abi - b^2 = 4 + 4\sqrt{15}i\]](http://latex.artofproblemsolving.com/3/6/e/36ebf050dc1bb6b2b61afbbb90bbee4aeba28a4a.png)
Setting the real parts equal and imaginary parts equal, we have:
![\[a^2 - b^2 = 4\]](http://latex.artofproblemsolving.com/3/2/b/32b6ed211fcc4baeeb28bd0b5c09b81dc8ea8644.png)
![\[ab = 2\sqrt{15}\]](http://latex.artofproblemsolving.com/1/e/2/1e2853fc1825df726cc45780884c6ffb4e27a700.png)
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 
:
![\[(-\sqrt{10}, -\sqrt{6})\]](http://latex.artofproblemsolving.com/c/b/4/cb4c307ac3a0570020aaeca6195fd1286e83f846.png)
![\[(\sqrt{10}, \sqrt{6})\]](http://latex.artofproblemsolving.com/3/4/d/34da7a3e48704117fb02fcad245d85ec63643800.png)
Repeating the process for the second equation, we have two solutions:
![\[(-\sqrt{3}, -1)\]](http://latex.artofproblemsolving.com/0/8/6/0863dc31f6a604bc1218eadd188f1d1ca5b0e20f.png)
![\[(\sqrt{3}, 1)\]](http://latex.artofproblemsolving.com/2/2/5/2258ec2f31e502d0f13e5c10e4c998d94eeda15d.png)
In a clockwise direction, the points are 
. Now we can use the shoelace theorem. The area is , so the answer is .
See Also
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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 | |
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