Difference between revisions of "2018 AMC 12A Problems/Problem 22"
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By the same process, we conclude that <math>\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.</math> | By the same process, we conclude that <math>\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.</math> | ||
</ol> | </ol> | ||
| − | Note that the problem is equivalent to finding the area of a | + | Note that the problem is equivalent to finding the area of a parallelogram with consecutive vertices <math>(x_1,y_1)=\left(\sqrt{10}, \sqrt{6}\right),(x_2,y_2)=\left(\sqrt{3},1\right),(x_3,y_3)=\left(-\sqrt{10},-\sqrt{6}\right),</math> and <math>(x_4,y_4)=\left(-\sqrt{3}, -1\right)</math> in the coordinate plane. By the Shoelace Theorem, the area we seek is <cmath>\frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = 6\sqrt2-2\sqrt{10},</cmath> so the answer is <math>6+2+2+10=\boxed{\textbf{(A) } 20}.</math> |
| − | ~ | + | ~Rejas (Fundamental Logic) |
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
Revision as of 03:08, 30 August 2021
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
We solve each equation separately:
Let
for some real numbers 
and 
Substituting and expanding, we have
Equating the real parts and the imaginary parts, respectively, we get
We rearrange the first equation and square the second equation:
Substituting 
into 
we obtain 
Since 
either inspection or factoring gives 
Substituting this into either or 
produces 
Since
we conclude that 
from which 
for some real numbers 
Equating the real parts and the imaginary parts, respectively, we get
We rearrange the first equation and square the second equation:
Substituting 
into 
we obtain 
Since 
either inspection or factoring gives 
Substituting this into either 
produces 
we conclude that 
from which 
By the same process, we conclude that 
Note that the problem is equivalent to finding the area of a parallelogram with consecutive vertices 
and 
in the coordinate plane. By the Shoelace Theorem, the area we seek is ![\[\frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = 6\sqrt2-2\sqrt{10},\]](http://latex.artofproblemsolving.com/8/1/d/81da780d80e9ef9dc8007ebfb2364aaaa1a54e43.png)
so the answer is 
~Rejas (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
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 not too hard 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 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
