Difference between revisions of "2018 AIME II Problems/Problem 5"
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− | 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{ | + | 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{ | + | 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 | ||
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==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> | + | 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 == | ||
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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{ | + | <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> | ||
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<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{ | + | 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> |
Revision as of 12:00, 12 October 2019
Contents
Problem
Suppose that , , and are complex numbers such that , , and , where . Then there are real numbers and such that . Find .
Solution 1
First we evaluate the magnitudes. , , and . Therefore, , or . Divide to find that , , and . This allows us to see that the argument of is , and the argument of is . We need to convert the polar form to a standard form. Simple trig identities show and . More division is needed to find what is. Written by a1b2
Solution 2
Dividing the first equation by the second equation given, we find that . Substituting this into the third equation, we get . 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 is the negative of that of , and their magnitudes multiply to . Thus, we have and . To find , we can use the previous substitution we made to find that . Therefore,
Solution by ktong
Solution 3
We are given that . Thus . We are also given that . Thus . We are also given that = . Substitute and into = . We have . Multiplying out we get . Thus . Simplifying this fraction we get . Cross-multiplying the fractions we get or . Now we can rewrite this as . Let .Thus or . We can see that and thus or .We also can see that because there is no real term in . Thus or . Using the two equations and we solve by doing system of equations that and . And so . Because , then . Simplifying this fraction we get or . Multiplying by the conjugate of the denominator () in the numerator and the denominator and we get . Simplifying this fraction we get . Given that = we can substitute We can solve for z and get . Now we know what , , and are, so all we have to do is plug and chug. or Now or . Thus is our final answer.(David Camacho)
Solution 4
We observe that by multiplying and we get Next, we divide by to
get We have We can write in the form of so we get
Then, and Solving this system of equations is relatively
simple. We have two cases, and
Case 1: so We solve for and by plugging in to the two equations. We see
and so and Solving, we end up with
as our answer.
Case 2: so Again, we solve for and We find
so We again have
Solution by
Solution 5 (Based on advanced mathematical knowledge)
According to the Euler's Theory, we can rewrite , and as As a result, Also, it is clear that So , or Also, we have So now we have , , , and . Solve these above, we get So we can get Use we can find that So So we have and .
As a result, we finally get
~Solution by (Frank FYC)
Solution 6
We can turn the expression into , and this would allow us to plug in the values after some computations.
Based off of the given products, we have
.
Dividing by the given products, we have
.
Simplifying, we get that this expression becomes . This equals , so the answer is .
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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All AIME Problems and Solutions |
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