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Difference between revisions of "2018 AMC 12A Problems/Problem 22"

(Created page with "== Problem == The solutions to the equations <math>z^2=4+4\sqrt{15}i</math> and <math>z^2=2+2\sqrt 3i,</math> where <math>i=\sqrt{-1},</math> form the vertices of a parallelo...")
 
(→‎Solution)
Line 9: Line 9:
 
\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 $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$

$\textbf{(A)} 20 \qquad \textbf{(B)} 21 \qquad \textbf{(C)} 22 \qquad \textbf{(D)} 23 \qquad \textbf{(E)} 24$

Solution 1

The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$, so the answer is $\boxed{20}$. (trumpeter)

Solution 2 (No DeMoivre's)

Write $z$ as $a + bi$. For the first equation, \[(a + bi)^2 = 4 + 4\sqrt{15}i\] \[a^2 + 2abi - b^2 = 4 + 4\sqrt{15}i\] Setting the real parts equal and imaginary parts equal, we have: \[a^2 - b^2 = 4\] \[ab = 2\sqrt{15}\] Squaring the second equation gives $a^2b^2 = 60$. We now need two numbers that have a difference of $4$ and a product of $60$. By inspection, $10$ and $6$ work, so $a^2 = 10$ and $b^2 = 6$. Since $ab$ is positive, $a$ and $b$ must have the same sign. Thus we have two solutions for $(a, b)$: \[(-\sqrt{10}, -\sqrt{6})\] \[(\sqrt{10}, \sqrt{6})\] Repeating the process for the second equation, we have two solutions: \[(-\sqrt{3}, -1)\] \[(\sqrt{3}, 1)\] In a clockwise direction, the points are $(-\sqrt{10}, -\sqrt{6}), (-\sqrt{3}, -1), (\sqrt{10}, \sqrt{6}), (\sqrt{3}, 1)$. Now we can use the shoelace theorem. The area is $6\sqrt{2}-2\sqrt{10}$, so the answer is $\boxed{20}$.

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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