Difference between revisions of "2018 AMC 12A Problems/Problem 22"
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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 parallelogram in the complex plane. The area of this parallelogram can be written in the form <math>p\sqrt q-r\sqrt s,</math> where <math>p,</math> <math>q,</math> <math>r,</math> and <math>s</math> are positive integers and neither <math>q</math> nor <math>s</math> is divisible by the square of any prime number. What is <math>p+q+r+s?</math> | 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 parallelogram in the complex plane. The area of this parallelogram can be written in the form <math>p\sqrt q-r\sqrt s,</math> where <math>p,</math> <math>q,</math> <math>r,</math> and <math>s</math> are positive integers and neither <math>q</math> nor <math>s</math> is divisible by the square of any prime number. What is <math>p+q+r+s?</math> | ||
− | <math>\textbf{(A)} 20 \qquad | + | <math>\textbf{(A) } 20 \qquad |
− | \textbf{(B)} 21 \qquad | + | \textbf{(B) } 21 \qquad |
− | \textbf{(C)} 22 \qquad | + | \textbf{(C) } 22 \qquad |
− | \textbf{(D)} 23 \qquad | + | \textbf{(D) } 23 \qquad |
− | \textbf{(E)} 24 </math> | + | \textbf{(E) } 24 </math> |
− | ==Solution 1== | + | ==Solution 1 (Complex Numbers in Rectangular Form)== |
− | + | We solve each equation separately: | |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>z^2=4+4\sqrt{15}i</math><p> | ||
+ | Let <math>z=a+bi</math> for some real numbers <math>a</math> and <math>b.</math> <p> | ||
+ | Substituting and expanding, we get | ||
+ | <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, &&(1) \\ | ||
+ | ab&=2\sqrt{15}. &&(2) | ||
+ | \end{align*}</cmath> | ||
+ | We rearrange <math>(1)</math> and square <math>(2):</math> | ||
+ | <cmath>\begin{align*} | ||
+ | a^2&=b^2+4, \hspace{4mm} &&(1\star) \\ | ||
+ | a^2b^2&=60. &&(2\star) | ||
+ | \end{align*}</cmath> | ||
+ | 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\geq0</math> for all real numbers <math>b,</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> Since <math>ab>0</math> from <math>(2),</math> we have <math>(a,b)=\left(\sqrt{10},\sqrt{6}\right),\left(-\sqrt{10},-\sqrt{6}\right).</math> <p> | ||
+ | The solutions to <math>z^2=4+4\sqrt{15}i</math> are <math>\boldsymbol{z=\sqrt{10}+\sqrt{6}i,-\sqrt{10}-\sqrt{6}i}.</math></li><p> | ||
+ | <li><math>z^2=2+2\sqrt{3}i</math></li><p> | ||
+ | By the same process, we have <math>(a,b)=\left(\sqrt3,1\right),\left(-\sqrt3,-1\right).</math> <p> | ||
+ | The solutions to <math>z^2=2+2\sqrt{3}i</math> are <math>\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.</math> | ||
+ | </ol> | ||
+ | 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> | ||
− | ==Solution 2 ( | + | ~Rejas (Fundamental Logic) |
− | + | ||
− | + | ~MRENTHUSIASM (Reconstruction) | |
− | < | + | |
− | + | ==Solution 2 (Complex Numbers in Polar Form)== | |
− | < | + | We solve each equation separately: |
− | <cmath> | + | <ol style="margin-left: 1.5em;"> |
− | + | <li><math>z^2=4+4\sqrt{15}i</math><p> | |
− | < | + | Let <math>z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,</math> where <math>r</math> is the magnitude of <math>z</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> <p> |
− | < | + | By De Moivre's Theorem, we have <cmath>z^2=r^2\operatorname{cis}(2\theta)=16\left(\frac14+\frac{\sqrt{15}}{4}i\right),</cmath> from which |
− | + | <ul style="list-style-type:square;"> | |
− | + | <li><math>r^2=16,</math> so <math>r=4.</math></li><p> | |
− | < | + | <li><math>\begin{cases} |
− | + | \begin{aligned} | |
+ | \cos(2\theta) &= \frac14 \\ | ||
+ | \sin(2\theta) &= \frac{\sqrt{15}}{4} | ||
+ | \end{aligned}, | ||
+ | \end{cases}</math> so <math>\cos\theta=\pm\sqrt{\frac{1+\cos(2\theta)}{2}}=\pm\frac{\sqrt{10}}{4}</math> and <math>\sin\theta=\pm\sqrt{\frac{1-\cos(2\theta)}{2}}=\pm\frac{\sqrt{6}}{4}</math> by Half-Angle Formulas. <br> | ||
+ | Since <math>\cos(2\theta)>0</math> and <math>\sin(2\theta)>0,</math> it follows that <math>2\theta\in\biggl(0,\frac{\pi}{2}\biggr)\cup\biggl(2\pi,\frac{5\pi}{2}\biggr),</math> or <math>\theta\in\biggl(0,\frac{\pi}{4}\biggr)\cup\biggl(\pi,\frac{5\pi}{4}\biggr).</math> We conclude that <math>(r,\operatorname{cis}\theta)=\left(4,\frac{\sqrt{10}}{4}+\frac{\sqrt{6}}{4}i\right),\left(4,-\frac{\sqrt{10}}{4}-\frac{\sqrt{6}}{4}i\right).</math> </li><p> | ||
+ | </ul> | ||
+ | The solutions to <math>z^2=4+4\sqrt{15}i</math> are <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> | ||
+ | By a similar process, we conclude that <math>(r,\theta)=\biggl(2,\frac{\pi}{6}\biggr),\biggl(2,\frac{7\pi}{6}\biggr).</math> <p> | ||
+ | The solutions to <math>z^2=2+2\sqrt{3}i</math> are <math>\boldsymbol{z=\sqrt3+i,-\sqrt3-i}.</math> | ||
+ | </ol> | ||
+ | We continue with the last paragraph of Solution 1 to get the answer <math>\boxed{\textbf{(A) } 20}.</math> | ||
+ | |||
+ | ~trumpeter (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
==Solution 3 (Vectors)== | ==Solution 3 (Vectors)== | ||
+ | Let <math>z_1</math> and <math>z_2</math> be the solutions to the equation <math>z^2=4+4\sqrt{15}i,</math> and <math>z_3</math> and <math>z_4</math> be the solutions to the equation <math>z^2=2+2\sqrt 3i.</math> Clearly, <math>z_1</math> and <math>z_2</math> are opposite complex numbers, so are <math>z_3</math> and <math>z_4.</math> <i><b>This solution refers to the results of De Moivre's Theorem in Solution 2.</b></i> | ||
+ | |||
+ | From Solution 2, let <math>z_1=4\operatorname{cis}\phi</math> for some <math>0<\phi<\frac{\pi}{4}.</math> It follows that <math>z_2=4\operatorname{cis}(\phi+\pi).</math> On the other hand, we have <math>z_3=2\operatorname{cis}\frac{\pi}{6}</math> and <math>z_4=2\operatorname{cis}\frac{7\pi}{6}</math> without the loss of generality. Since <math>\tan(2\phi)>\tan\frac{\pi}{3},</math> we deduce that <math>2\phi>\frac{\pi}{3},</math> from which <math>\phi>\frac{\pi}{6}.</math> | ||
+ | |||
+ | In the complex plane, the positions of <math>z_1,z_2,z_3,</math> and <math>z_4</math> are shown below: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(200); | ||
+ | |||
+ | int xMin = -5; | ||
+ | int xMax = 5; | ||
+ | int yMin = -5; | ||
+ | int yMax = 5; | ||
+ | int numRays = 24; | ||
+ | |||
+ | //Draws a polar grid that goes out to a number of circles | ||
+ | //equal to big, with numRays specifying the number of rays: | ||
+ | void polarGrid(int big, int numRays) | ||
+ | { | ||
+ | for (int i = 1; i < big+1; ++i) | ||
+ | { | ||
+ | draw(Circle((0,0),i), gray+linewidth(0.4)); | ||
+ | } | ||
+ | for(int i=0;i<numRays;++i) | ||
+ | draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); | ||
+ | } | ||
+ | |||
+ | //Draws the horizontal gridlines | ||
+ | void horizontalLines() | ||
+ | { | ||
+ | for (int i = yMin+1; i < yMax; ++i) | ||
+ | { | ||
+ | draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | //Draws the vertical gridlines | ||
+ | void verticalLines() | ||
+ | { | ||
+ | for (int i = xMin+1; i < xMax; ++i) | ||
+ | { | ||
+ | draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | horizontalLines(); | ||
+ | verticalLines(); | ||
+ | polarGrid(xMax,numRays); | ||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("Re",(xMax,0),(2,0)); | ||
+ | label("Im",(0,yMax),(0,2)); | ||
+ | |||
+ | pair Z1, Z2, Z3, Z4; | ||
+ | |||
+ | Z1 = (sqrt(10),sqrt(6)); | ||
+ | Z2 = (-sqrt(10),-sqrt(6)); | ||
+ | Z3 = (sqrt(3),1); | ||
+ | Z4 = (-sqrt(3),-1); | ||
+ | |||
+ | label("$z_1$", Z1, dir(Z1), UnFill); | ||
+ | label("$z_2$", Z2, dir(Z2), UnFill); | ||
+ | label("$z_3$", Z3, (0.75,-0.75), UnFill); | ||
+ | label("$z_4$", Z4, (-0.75,0.75), UnFill); | ||
+ | |||
+ | draw(Z1--Z3--Z2--Z4--cycle,red); | ||
+ | |||
+ | dot(Z1, linewidth(3.5)); | ||
+ | dot(Z2, linewidth(3.5)); | ||
+ | dot(Z3, linewidth(3.5)); | ||
+ | dot(Z4, linewidth(3.5)); | ||
+ | </asy> | ||
+ | Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices <math>z_1,z_2,z_3,</math> and <math>z_4,</math> we find the area of the triangle with vertices <math>0,z_1,</math> and <math>z_3,</math> then multiply by <math>4.</math> | ||
+ | |||
+ | Recall that <math>|z_1|=4, |z_2|=2, \sin\phi=\frac{\sqrt6}{4},</math> and <math>\cos\phi=\frac{\sqrt{10}}{4}</math> from Solution 2. The area of the parallelogram is | ||
+ | <cmath>\begin{align*} | ||
+ | 4\cdot\left[\frac12\cdot|z_1|\cdot|z_3|\cdot\sin\left(\phi-\frac{\pi}{6}\right)\right] &= 16\sin\left(\phi-\frac{\pi}{6}\right) \\ | ||
+ | &= 16\left[\sin\phi\cos\frac{\pi}{6}-\cos\phi\sin\frac{\pi}{6}\right] \\ | ||
+ | &= 16\left[\frac{\sqrt3}{2}\sin\phi-\frac12\cos\phi\right] \\ | ||
+ | &= 6\sqrt2-2\sqrt{10}, | ||
+ | \end{align*}</cmath> | ||
+ | so the answer is <math>6+2+2+10=\boxed{\textbf{(A) } 20}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 4 (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 <math>ab\sin \theta</math> where <math>a</math> and <math>b</math> are the side lengths. | 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 <math>ab\sin \theta</math> where <math>a</math> and <math>b</math> are the side lengths. | ||
The side lengths are easily found since we are given the squares of <math>z</math>. Thus, the magnitude of <math>z</math> in the first equation is just <math>\sqrt{16} = 4</math> and in the second equation is just <math>\sqrt{4} = 2</math>. Now, we need <math>\sin \theta</math>. | The side lengths are easily found since we are given the squares of <math>z</math>. Thus, the magnitude of <math>z</math> in the first equation is just <math>\sqrt{16} = 4</math> and in the second equation is just <math>\sqrt{4} = 2</math>. Now, we need <math>\sin \theta</math>. | ||
− | To find <math>\theta</math>, 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 <math>\cos</math> of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are <math>\langle 4, 4\sqrt{15}\rangle</math> and <math>\langle 2, 2\sqrt{3}\rangle</math>, so their dot product is <math>8 + 24\sqrt{5}</math>. Dividing by the magnitudes yields: <math>\dfrac{8+24\sqrt{5}}{4 \cdot 16} = \dfrac{1 + 3\sqrt{5}}{8}</math>. This is <math>\cos 2\theta</math>, and recall the identity <math>\cos 2\theta = 1 - 2\sin^2 \theta</math>. This means that <math>\sin^2 \theta = \dfrac{7 - 3\sqrt{5}}{16}</math>, so <math>\sin \theta = \dfrac{\sqrt{7- 3\sqrt{5}}}{4}</math>. Now, notice that <math>\sqrt{7- 3\sqrt{5}} = \dfrac{3\sqrt{2}-\sqrt{10}}{2}</math> (which is not too hard to discover) so <math>\sin \theta = \dfrac{3\sqrt{2}-\sqrt{10}}{8}</math>. Finally, putting everything together yields: <math>2\ | + | To find <math>\theta</math>, 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 <math>\cos</math> of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are <math>\Bigl\langle 4, 4\sqrt{15}\Bigr\rangle</math> and <math>\Bigl\langle 2, 2\sqrt{3}\Bigr\rangle</math>, so their dot product is <math>8 + 24\sqrt{5}</math>. Dividing by the magnitudes yields: <math>\dfrac{8+24\sqrt{5}}{4 \cdot 16} = \dfrac{1 + 3\sqrt{5}}{8}</math>. This is <math>\cos 2\theta</math>, and recall the identity <math>\cos 2\theta = 1 - 2\sin^2 \theta</math>. This means that <math>\sin^2 \theta = \dfrac{7 - 3\sqrt{5}}{16}</math>, so <math>\sin \theta = \dfrac{\sqrt{7- 3\sqrt{5}}}{4}</math>. Now, notice that <math>\sqrt{7- 3\sqrt{5}} = \dfrac{3\sqrt{2}-\sqrt{10}}{2}</math> (which is not too hard to discover) so <math>\sin \theta = \dfrac{3\sqrt{2}-\sqrt{10}}{8}</math>. Finally, putting everything together yields: <math>2\cdot 4 \cdot \dfrac{3\sqrt{2}-\sqrt{10}}{8} = 3\sqrt{2} - \sqrt{10}</math> 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 <math>6\sqrt{2} - 2\sqrt{10}</math>. Then, the answer is <math>\boxed{\textbf{(A) } 20}</math>. |
− | ~ Aathreyakadambi | + | ~Aathreyakadambi |
− | + | == Video Solution by Richard Rusczyk == | |
https://artofproblemsolving.com/videos/amc/2018amc12a/472 | https://artofproblemsolving.com/videos/amc/2018amc12a/472 |
Latest revision as of 11:45, 13 September 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 (Complex Numbers in Rectangular Form)
We solve each equation separately:
Let for some real numbers and
Substituting and expanding, we get Equating the real parts and the imaginary parts, respectively, we get We rearrange and square Substituting into we obtain Since for all real numbers either inspection or factoring gives Substituting this into either or produces Since from we have
The solutions to are
By the same process, we have
The solutions to are
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 so the answer is
~Rejas (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Complex Numbers in Polar Form)
We solve each equation separately:
Let where is the magnitude of such that and is the argument of such that
By De Moivre's Theorem, we have from which
- so
- so and by Half-Angle Formulas.
Since and it follows that or We conclude that
By a similar process, we conclude that
The solutions to are
We continue with the last paragraph of Solution 1 to get the answer
~trumpeter (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Vectors)
Let and be the solutions to the equation and and be the solutions to the equation Clearly, and are opposite complex numbers, so are and This solution refers to the results of De Moivre's Theorem in Solution 2.
From Solution 2, let for some It follows that On the other hand, we have and without the loss of generality. Since we deduce that from which
In the complex plane, the positions of and are shown below: Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices and we find the area of the triangle with vertices and then multiply by
Recall that and from Solution 2. The area of the parallelogram is so the answer is
~MRENTHUSIASM
Solution 4 (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 |
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