Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem: Added spaces between answer choices.)
Line 12: Line 12:
 
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)==
+
==Solution 2==
−
Write <math>z</math> as <math>a + bi</math>. For the first equation,
+
We solve each equation separately:
−
<cmath>(a + bi)^2 = 4 + 4\sqrt{15}i</cmath>
+
<ol style="margin-left: 1.5em;">
−
<cmath>a^2 + 2abi - b^2 = 4 + 4\sqrt{15}i</cmath>
+
  <li><math>z^2=4+4\sqrt{15}i</math><p>
−
Setting the real parts equal and imaginary parts equal, we have:
+
Let <math>z=a+bi</math> for some real numbers <math>a</math> and <math>b.</math> <p>
−
<cmath>a^2 - b^2 = 4</cmath>
+
Substituting and expanding, we have
−
<cmath>ab = 2\sqrt{15}</cmath>
+
<cmath>\begin{align*}
 +
(a+bi)^2&=4+4\sqrt{15}i \\
 +
\left(a^2-b^2\right)+2abi&=4+4\sqrt{15}i.
 +
\end{align*}</cmath>
 +
Equating the real parts and the imaginary parts, respectively, we get
 +
<cmath>\begin{align*}
 +
a^2-b^2&=4, \\
 +
ab&=2\sqrt{15}.
 +
\end{align*}</cmath>
 +
We rearrange the first equation and square the second equation:
 +
<cmath>\begin{align*}
 +
a^2&=b^2+4, &&(1\star) \\
 +
a^2b^2&=60. &&(2\star)
 +
\end{align*}</cmath>
 +
Substituting <math>(1\star)</math> into <math>(2\star)</math> gives <math>\left(b^2+4\right)b^2=60.</math> Since <math>b^2\geq0,</math> either inspection or factoring produces <math>b^2=6.</math> Substituting this into either <math>(1\star)</math> or <math>(2\star),</math> we obtain <math>a^2=10.</math>
 +
</li><p>
 +
  <li><math>z^2=2+2\sqrt 3i,</math></li><p>
 +
</ol>
 +
 
 +
<b>REFORMATTING IN PROGRESS</b>
 +
 
 
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>:
 
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>

Revision as of 23:40, 29 August 2021

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

We solve each equation separately:

  1. $z^2=4+4\sqrt{15}i$

    Let $z=a+bi$ for some real numbers $a$ and $b.$

    Substituting and expanding, we have \begin{align*} (a+bi)^2&=4+4\sqrt{15}i \\ \left(a^2-b^2\right)+2abi&=4+4\sqrt{15}i. \end{align*} Equating the real parts and the imaginary parts, respectively, we get \begin{align*} a^2-b^2&=4, \\ ab&=2\sqrt{15}. \end{align*} We rearrange the first equation and square the second equation: \begin{align*} a^2&=b^2+4, &&(1\star) \\ a^2b^2&=60. &&(2\star) \end{align*} Substituting $(1\star)$ into $(2\star)$ gives $\left(b^2+4\right)b^2=60.$ Since $b^2\geq0,$ either inspection or factoring produces $b^2=6.$ Substituting this into either $(1\star)$ or $(2\star),$ we obtain $a^2=10.$

  2. $z^2=2+2\sqrt 3i,$

REFORMATTING IN PROGRESS

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 $\boxed{20}$.

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 $ab\sin \theta$ where $a$ and $b$ are the side lengths.

The side lengths are easily found since we are given the squares of $z$. Thus, the magnitude of $z$ in the first equation is just $\sqrt{16} = 4$ and in the second equation is just $\sqrt{4} = 2$. Now, we need $\sin \theta$.

To find $\theta$, 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 $\cos$ of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are $\langle 4, 4\sqrt{15}\rangle$ and $\langle 2, 2\sqrt{3}\rangle$, so their dot product is $8 + 24\sqrt{5}$. Dividing by the magnitudes yields: $\dfrac{8+24\sqrt{5}}{4 \cdot 16} = \dfrac{1 + 3\sqrt{5}}{8}$. This is $\cos 2\theta$, and recall the identity $\cos 2\theta = 1 - 2\sin^2 \theta$. This means that $\sin^2 \theta = \dfrac{7 - 3\sqrt{5}}{16}$, so $\sin \theta = \dfrac{\sqrt{7- 3\sqrt{5}}}{4}$. Now, notice that $\sqrt{7- 3\sqrt{5}} = \dfrac{3\sqrt{2}-\sqrt{10}}{2}$ (which is not too hard to discover) so $\sin \theta = \dfrac{3\sqrt{2}-\sqrt{10}}{8}$. Finally, putting everything together yields: $2\times 4 \times \dfrac{3\sqrt{2}-\sqrt{10}}{8} = 3\sqrt{2} - \sqrt{10}$ 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 $6\sqrt{2} - 2\sqrt{10}$. Then, the answer is $\boxed{20}$.

~ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_22&oldid=161215"