Difference between revisions of "2018 AMC 12A Problems/Problem 22"
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a^2b^2&=60. &&(2\star) | a^2b^2&=60. &&(2\star) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Substituting <math>(1\star)</math> into <math>(2\star)</math> | + | Substituting <math>(1\star)</math> into <math>(2\star),</math> we obtain <math>\left(b^2+4\right)b^2=60.</math> Since <math>b^2>0,</math> either inspection or factoring gives <math>b^2=6.</math> Substituting this into either <math>(1\star)</math> or <math>(2\star)</math> produces <math>a^2=10.</math> <p> |
− | </li><p> | + | Since <math>ab>0,</math> we conclude that <math>(a,b)=\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{10}, \sqrt{6}\right),</math> from which <math>\boldsymbol{z=\sqrt{10}+\sqrt{6}i,-\sqrt{10}-\sqrt{6}i}.</math></li><p> |
<li><math>z^2=2+2\sqrt 3i,</math></li><p> | <li><math>z^2=2+2\sqrt 3i,</math></li><p> | ||
+ | By the same process, we conclude that <math>\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.</math> | ||
</ol> | </ol> | ||
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<b>REFORMATTING IN PROGRESS</b> | <b>REFORMATTING IN PROGRESS</b> | ||
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In a clockwise direction, the points are <math>(-\sqrt{10}, -\sqrt{6}), (-\sqrt{3}, -1), (\sqrt{10}, \sqrt{6}), (\sqrt{3}, 1)</math>. Now we can use the shoelace theorem. The area is <math>6\sqrt{2}-2\sqrt{10}</math>, so the answer is <math>\boxed{20}</math>. | In a clockwise direction, the points are <math>(-\sqrt{10}, -\sqrt{6}), (-\sqrt{3}, -1), (\sqrt{10}, \sqrt{6}), (\sqrt{3}, 1)</math>. Now we can use the shoelace theorem. The area is <math>6\sqrt{2}-2\sqrt{10}</math>, so the answer is <math>\boxed{20}</math>. | ||
Revision as of 03:51, 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
By the same process, we conclude that
REFORMATTING IN PROGRESS
In a clockwise direction, the points are . Now we can use the shoelace theorem. The area is
, so the answer is
.
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.