Difference between revisions of "2018 AMC 12A Problems/Problem 22"
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\textbf{(E)} 24 </math> | \textbf{(E)} 24 </math> | ||
− | == Solution == | + | ==Solution 1== |
The roots are <math>\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)</math> (easily derivable by using DeMoivre and half-angle). From there, shoelace on <math>\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)</math> and multiplying by <math>4</math> gives the area of <math>6\sqrt{2}-2\sqrt{10}</math>, so the answer is <math>\boxed{20}</math>. (trumpeter) | The roots are <math>\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)</math> (easily derivable by using DeMoivre and half-angle). From there, shoelace on <math>\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)</math> and multiplying by <math>4</math> gives the area of <math>6\sqrt{2}-2\sqrt{10}</math>, so the answer is <math>\boxed{20}</math>. (trumpeter) | ||
+ | |||
+ | ==Solution 2 (No DeMoivre's)== | ||
+ | Write <math>z</math> as <math>a + bi</math>. For the first equation, | ||
+ | <cmath>(a + bi)^2 = 4 + 4\sqrt{15}i</cmath> | ||
+ | <cmath>a^2 + 2abi - b^2 = 4 + 4\sqrt{15}i</cmath> | ||
+ | Setting the real parts equal and imaginary parts equal, we have: | ||
+ | <cmath>a^2 - b^2 = 4</cmath> | ||
+ | <cmath>ab = 2\sqrt{15}</cmath> | ||
+ | Squaring the second equation gives <math>a^2b^2 = 60</math>. We now need two numbers that have a difference of <math>4</math> and a product of <math>60</math>. By inspection, <math>10</math> and <math>6</math> work, so <math>a^2 = 10</math> and <math>b^2 = 6</math>. Since <math>ab</math> is positive, <math>a</math> and <math>b</math> must have the same sign. Thus we have two solutions for <math>(a, b)</math>: | ||
+ | <cmath>(-\sqrt{10}, -\sqrt{6})</cmath> | ||
+ | <cmath>(\sqrt{10}, \sqrt{6})</cmath> | ||
+ | Repeating the process for the second equation, we have two solutions: | ||
+ | <cmath>(-\sqrt{3}, -1)</cmath> | ||
+ | <cmath>(\sqrt{3}, 1)</cmath> | ||
+ | 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>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2018|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:55, 27 March 2018
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, Setting the real parts equal and imaginary parts equal, we have: 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 : Repeating the process for the second equation, we have two solutions: 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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