Difference between revisions of "2018 AIME II Problems/Problem 5"

m (Solution 2)
 
(2 intermediate revisions by the same user not shown)
Line 18: Line 18:
 
draw((0,0)--(4,-4),red);
 
draw((0,0)--(4,-4),red);
 
</asy>
 
</asy>
This allows us to see that the argument of <math>y</math> is <math>\frac{\pi}{4}</math>, and the argument of <math>z</math> is <math>-\frac{\pi}{4}</math>. We need to convert the polar form to a standard form. Simple trig identities show <math>y=10+10i</math> and <math>z=3-3i</math>. More division is needed to find what <math>x</math> is. <cmath>x=-20-12i</cmath> <cmath>x+y+z=-7-5i</cmath> <cmath>(-7)^2+(-5)^2=\boxed{074}</cmath>
+
This allows us to see that the argument of <math>y</math> is <math>\frac{\pi}{4}</math>, and the argument of <math>z</math> is <math>-\frac{\pi}{4}</math>. We need to convert the polar form to a standard form. Simple trig identities show <math>y=10+10i</math> and <math>z=3-3i</math>. More division is needed to find what <math>x</math> is. <cmath>x=-20-12i</cmath> <cmath>x+y+z=-7-5i</cmath> <cmath>(-7)^2+(-5)^2=\boxed{74}</cmath>
 
<cmath>QED\blacksquare</cmath>
 
<cmath>QED\blacksquare</cmath>
 
Written by [[User:A1b2|a1b2]]
 
Written by [[User:A1b2|a1b2]]
 
==Solution 2==
 
==Solution 2==
 
Dividing the first equation by the second equation given, we find that <math>\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)</math>. Substituting this into the third equation, we get <math>z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i</math>. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of <math>y</math> is the negative of that of <math>z</math>, and their magnitudes multiply to <math>60</math>. Thus, we have <math>z=\sqrt{-18i}=3-3i</math> and <math>3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i</math>. To find <math>x</math>, we can use the previous substitution we made to find that <math>x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i</math>.
 
Dividing the first equation by the second equation given, we find that <math>\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)</math>. Substituting this into the third equation, we get <math>z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i</math>. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of <math>y</math> is the negative of that of <math>z</math>, and their magnitudes multiply to <math>60</math>. Thus, we have <math>z=\sqrt{-18i}=3-3i</math> and <math>3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i</math>. To find <math>x</math>, we can use the previous substitution we made to find that <math>x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i</math>.
Therefore, <math>x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{074}</math>
+
Therefore, <math>x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}</math>
  
 
Solution by ktong
 
Solution by ktong
Line 29: Line 29:
 
==Solution 3 ==
 
==Solution 3 ==
  
We are given that <math>xy=-80-320i</math>. Thus <math>y=\frac{-80-320i}{x}</math>. We are also given that <math>xz= -96+24i</math>. Thus <math>z=\frac{-96+24i}{x}</math>. We are also given that <math>yz</math> = <math>60</math>. Substitute <math>y=\frac{-80-320i}{x}</math> and <math>z=\frac{-96+24i}{x}</math> into <math>yz</math> = <math>60</math>. We have <math> \frac{(-80-320i)(-96+24i)}{x^2}=60</math>. Multiplying out <math>(-80-320i)(-96+24i)</math> we get <math>(1920)(8+15i)</math>. Thus  <math>\frac{1920(8+15i)}{x^2} =60</math>. Simplifying this fraction we get <math>\frac{32(8+15i)}{x^2}=1</math>. Cross-multiplying the fractions we get <math>x^2=32(8+15i)</math> or <math>x^2= 256+480i</math>. Now we can rewrite this as <math>x^2-256=480i</math>. Let <math>x= (a+bi)</math>.Thus <math>x^2=(a+bi)^2</math> or <math>a^2+2abi-b^2</math>. We can see that <math>a^2+2abi-b^2-256=480i</math> and thus <math>2abi=480i</math> or <math>ab=240</math>.We also can see that <math>a^2-b^2-256=0</math> because there is no real term in <math>480i</math>. Thus <math>a^2-b^2=256</math> or <math>(a+b)(a-b)=256</math>. Using the two equations <math>ab=240</math> and <math>(a+b)(a-b)=256</math> we solve by doing system of equations that <math>a=-20</math> and <math>b=-12</math>. And <math>x=a+bi</math> so <math>x=-20-12i</math>. Because <math>y=\frac{-80-320i}{x}</math>, then <math>y=\frac{-80-320i}{-20-12i}</math>. Simplifying this fraction we get <math>y=\frac{-80(1+4i)}{-4(5+3i)}</math> or <math>y=\frac{20(1+4i)}{(5+3i)}</math>. Multiplying by the conjugate of the denominator (<math>5-3i</math>) in the numerator and the denominator and  we get <math>y=\frac{20(17-17i)}{34}</math>. Simplifying this fraction we get <math>y=10-10i</math>. Given that <math>yz</math> = <math>60</math> we can substitute <math>(10-10i)(z)=60</math> We can solve for z and get <math>z=3+3i</math>. Now we know what <math>x</math>, <math>y</math>, and <math>z</math> are, so all we have to do is plug and chug. <math>x+y+z= (-20-12i)+(10+10i)+(3-3i)</math> or <math>x+y+z= -7-5i</math> Now <math>a^2 +b^2=(-7)^2+(-5)^2</math> or <math>a^2 +b^2 = 74</math>. Thus <math>074</math> is our final answer.(David Camacho)
+
We are given that <math>xy=-80-320i</math>. Thus <math>y=\frac{-80-320i}{x}</math>. We are also given that <math>xz= -96+24i</math>. Thus <math>z=\frac{-96+24i}{x}</math>. We are also given that <math>yz</math> = <math>60</math>. Substitute <math>y=\frac{-80-320i}{x}</math> and <math>z=\frac{-96+24i}{x}</math> into <math>yz</math> = <math>60</math>. We have <math> \frac{(-80-320i)(-96+24i)}{x^2}=60</math>. Multiplying out <math>(-80-320i)(-96+24i)</math> we get <math>(1920)(8+15i)</math>. Thus  <math>\frac{1920(8+15i)}{x^2} =60</math>. Simplifying this fraction we get <math>\frac{32(8+15i)}{x^2}=1</math>. Cross-multiplying the fractions we get <math>x^2=32(8+15i)</math> or <math>x^2= 256+480i</math>. Now we can rewrite this as <math>x^2-256=480i</math>. Let <math>x= (a+bi)</math>.Thus <math>x^2=(a+bi)^2</math> or <math>a^2+2abi-b^2</math>. We can see that <math>a^2+2abi-b^2-256=480i</math> and thus <math>2abi=480i</math> or <math>ab=240</math>.We also can see that <math>a^2-b^2-256=0</math> because there is no real term in <math>480i</math>. Thus <math>a^2-b^2=256</math> or <math>(a+b)(a-b)=256</math>. Using the two equations <math>ab=240</math> and <math>(a+b)(a-b)=256</math> we solve by doing system of equations that <math>a=-20</math> and <math>b=-12</math>. And <math>x=a+bi</math> so <math>x=-20-12i</math>. Because <math>y=\frac{-80-320i}{x}</math>, then <math>y=\frac{-80-320i}{-20-12i}</math>. Simplifying this fraction we get <math>y=\frac{-80(1+4i)}{-4(5+3i)}</math> or <math>y=\frac{20(1+4i)}{(5+3i)}</math>. Multiplying by the conjugate of the denominator (<math>5-3i</math>) in the numerator and the denominator and  we get <math>y=\frac{20(17-17i)}{34}</math>. Simplifying this fraction we get <math>y=10-10i</math>. Given that <math>yz</math> = <math>60</math> we can substitute <math>(10-10i)(z)=60</math> We can solve for z and get <math>z=3+3i</math>. Now we know what <math>x</math>, <math>y</math>, and <math>z</math> are, so all we have to do is plug and chug. <math>x+y+z= (-20-12i)+(10+10i)+(3-3i)</math> or <math>x+y+z= -7-5i</math> Now <math>a^2 +b^2=(-7)^2+(-5)^2</math> or <math>a^2 +b^2 = 74</math>. Thus <math>74</math> is our final answer.(David Camacho)
  
 
==Solution 4 ==
 
==Solution 4 ==
Line 49: Line 49:
 
Case 2: <math>a=-20, b=-12,</math> so <math>x=-20-12i.</math> Again, we solve for <math>y</math> and <math>z.</math> We find <math>y=\frac{-80-320i}{-20-12i}=10+10i,</math>  
 
Case 2: <math>a=-20, b=-12,</math> so <math>x=-20-12i.</math> Again, we solve for <math>y</math> and <math>z.</math> We find <math>y=\frac{-80-320i}{-20-12i}=10+10i,</math>  
  
<math>z=\frac{-96+24i}{-20-12i}=3-3i,</math> so <math>x+y+z=-7-5i.</math> We again have <math>(-7)^2+(-5)^2=\boxed{074}.</math>
+
<math>z=\frac{-96+24i}{-20-12i}=3-3i,</math> so <math>x+y+z=-7-5i.</math> We again have <math>(-7)^2+(-5)^2=\boxed{74}.</math>
  
 
Solution by <math>Airplane50</math>
 
Solution by <math>Airplane50</math>
Line 80: Line 80:
 
<math>x^2=\frac{(-96+24i)(-80-320i)}{60}</math>.
 
<math>x^2=\frac{(-96+24i)(-80-320i)}{60}</math>.
  
Simplifying, we get that this expression becomes <math>\sqrt{24+70i}</math>. This equals <math>\pm{(7+5i)}</math>, so the answer is <math>7^2+5^2=\boxed{074}</math>.
+
Simplifying, we get that this expression becomes <math>\sqrt{24+70i}</math>. This equals <math>\pm{(7+5i)}</math>, so the answer is <math>7^2+5^2=\boxed{74}</math>.
  
 
<math>\textbf{-RootThreeOverTwo}</math>
 
<math>\textbf{-RootThreeOverTwo}</math>

Latest revision as of 12:00, 12 October 2019

Problem

Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$.

Solution 1

First we evaluate the magnitudes. $|xy|=80\sqrt{17}$, $|yz|=60$, and $|zx|=24\sqrt{17}$. Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$, or $|xyz|=240\sqrt{34}$. Divide to find that $|z|=3\sqrt{2}$, $|x|=40\sqrt{34}$, and $|y|=10\sqrt{2}$. [asy] draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); [/asy] This allows us to see that the argument of $y$ is $\frac{\pi}{4}$, and the argument of $z$ is $-\frac{\pi}{4}$. We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$. More division is needed to find what $x$ is. \[x=-20-12i\] \[x+y+z=-7-5i\] \[(-7)^2+(-5)^2=\boxed{74}\] \[QED\blacksquare\] Written by a1b2

Solution 2

Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$. Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$, and their magnitudes multiply to $60$. Thus, we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$. To find $x$, we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$. Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}$

Solution by ktong

Solution 3

We are given that $xy=-80-320i$. Thus $y=\frac{-80-320i}{x}$. We are also given that $xz= -96+24i$. Thus $z=\frac{-96+24i}{x}$. We are also given that $yz$ = $60$. Substitute $y=\frac{-80-320i}{x}$ and $z=\frac{-96+24i}{x}$ into $yz$ = $60$. We have $\frac{(-80-320i)(-96+24i)}{x^2}=60$. Multiplying out $(-80-320i)(-96+24i)$ we get $(1920)(8+15i)$. Thus $\frac{1920(8+15i)}{x^2} =60$. Simplifying this fraction we get $\frac{32(8+15i)}{x^2}=1$. Cross-multiplying the fractions we get $x^2=32(8+15i)$ or $x^2= 256+480i$. Now we can rewrite this as $x^2-256=480i$. Let $x= (a+bi)$.Thus $x^2=(a+bi)^2$ or $a^2+2abi-b^2$. We can see that $a^2+2abi-b^2-256=480i$ and thus $2abi=480i$ or $ab=240$.We also can see that $a^2-b^2-256=0$ because there is no real term in $480i$. Thus $a^2-b^2=256$ or $(a+b)(a-b)=256$. Using the two equations $ab=240$ and $(a+b)(a-b)=256$ we solve by doing system of equations that $a=-20$ and $b=-12$. And $x=a+bi$ so $x=-20-12i$. Because $y=\frac{-80-320i}{x}$, then $y=\frac{-80-320i}{-20-12i}$. Simplifying this fraction we get $y=\frac{-80(1+4i)}{-4(5+3i)}$ or $y=\frac{20(1+4i)}{(5+3i)}$. Multiplying by the conjugate of the denominator ($5-3i$) in the numerator and the denominator and we get $y=\frac{20(17-17i)}{34}$. Simplifying this fraction we get $y=10-10i$. Given that $yz$ = $60$ we can substitute $(10-10i)(z)=60$ We can solve for z and get $z=3+3i$. Now we know what $x$, $y$, and $z$ are, so all we have to do is plug and chug. $x+y+z= (-20-12i)+(10+10i)+(3-3i)$ or $x+y+z= -7-5i$ Now $a^2 +b^2=(-7)^2+(-5)^2$ or $a^2 +b^2 = 74$. Thus $74$ is our final answer.(David Camacho)

Solution 4

We observe that by multiplying $xy,$ $yz,$ and $zx,$ we get $(xyz)^2=(-80-320i)(60)(-96+24i).$ Next, we divide $(xyz)^2$ by $(yz)^2$ to

get $x^2.$ We have $x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.$ We can write $x$ in the form of $a+bi,$ so we get

$(a+bi)^2=256+480i.$ Then, $a^2-b^2+2abi=256+480i,$ $a^2-b^2=256,$ and $2ab=480.$ Solving this system of equations is relatively

simple. We have two cases, $a=20, b=12,$ and $a=-20, b=-12.$

Case 1: $a=20, b=12,$ so $x=20+12i.$ We solve for $y$ and $z$ by plugging in $x$ to the two equations. We see

$y=\frac{-80-320i}{20+12i}=-10-10i$ and $z=\frac{-96+24i}{20+12i}=-3+3i.$ $x+y+z=7+5i,$ so $a=7$ and $b=5.$ Solving, we end up with

$7^2+5^2=\boxed{074}$ as our answer.

Case 2: $a=-20, b=-12,$ so $x=-20-12i.$ Again, we solve for $y$ and $z.$ We find $y=\frac{-80-320i}{-20-12i}=10+10i,$

$z=\frac{-96+24i}{-20-12i}=3-3i,$ so $x+y+z=-7-5i.$ We again have $(-7)^2+(-5)^2=\boxed{74}.$

Solution by $Airplane50$

Solution 5 (Based on advanced mathematical knowledge)

According to the Euler's Theory, we can rewrite $x$, $y$ and $z$ as \[x=r_{1}e^{i{\theta}_1}\] \[y=r_{2}e^{i{\theta}_2}\] \[x=r_{3}e^{i{\theta}_3}\] As a result, \[|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}\] \[|yz|=r_{2}r_{3}=60\] \[|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}\] Also, it is clear that \[yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60\] So ${\theta}_2+{\theta}_3=0$, or \[{\theta}_2=-{\theta}_3\] Also, we have \[xy=-80\sqrt{17}e^{i\arctan{4}}\] \[yz=60\] \[xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}\] So now we have $r_{1}r_{2}=80\sqrt{17}$, $r_{2}r_{3}=60$, $r_{1}r_{3}=24\sqrt{17}$, ${\theta}_1+{\theta}_2=\arctan{4}$ and ${\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}$. Solve these above, we get \[r_{1}=4\sqrt{34}\] \[r_{2}=10\sqrt{2}\] \[r_{3}=3\sqrt{2}\] \[{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\] So we can get \[y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i\] \[z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i\] Use $xy=-80-320i$ we can find that \[x=-20-12i\] So \[x+y+z=-20-12i+10+10i+3-3i=-7-5i\] So we have $a=-7$ and $b=-5$.

As a result, we finally get \[a^2+b^2=(-7)^2+(-5)^2=\boxed{074}\]

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 6

We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$, and this would allow us to plug in the values after some computations.

Based off of the given products, we have

$xy^2z=60(-80-320i)$

$xyz^2=60(-96+24i)$

$x^2yz=(-96+24i)(-80-320i)$.

Dividing by the given products, we have

$y^2=\frac{60(-80-320i)}{-96+24i}$

$z^2=\frac{60(-96+24i)}{-80-320i}$

$x^2=\frac{(-96+24i)(-80-320i)}{60}$.

Simplifying, we get that this expression becomes $\sqrt{24+70i}$. This equals $\pm{(7+5i)}$, so the answer is $7^2+5^2=\boxed{74}$.

$\textbf{-RootThreeOverTwo}$

2018 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS