Difference between revisions of "2018 AIME II Problems/Problem 5"
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Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>. | Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>. | ||
− | ==Solution 1== | + | ==Solution 1 (Euler's formula and Substitution)== |
− | First we evaluate the magnitudes. <math>|xy|=80\sqrt{17}</math>, <math>|yz|=60</math>, and <math>|zx|=24\sqrt{17}</math>. Therefore, <math>|x^2y^2z^2|=17\cdot80\cdot60\cdot24</math>, or <math>|xyz|=240\sqrt{34}</math>. Divide to find that <math>|z|=3\sqrt{2}</math>, <math>|x|= | + | The First (pun intended) thing to notice is that <math>xy</math> and <math>zx</math> have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is <math>e^{\frac{3\pi i}{2}}</math>. Anyway this results in getting that <math>\left(\frac{-3i}{10}\right)y=z</math>. Then substitute this into <math>yz</math> to get, after some calculation, that <math>y=10e^{\frac{5\pi i}{4}}\sqrt{2}</math> and <math>z=-3e^{\frac{7\pi i}{4}}\sqrt{2}</math>. Then plug <math>z</math> into <math>zx</math>, you could do the same thing with <math>xy</math> but <math>zx</math> looks like it's easier due to it being smaller. Anyway you get <math>x=20+12i</math>. Then add all three up, it turns out easier than it seems because for <math>z</math> and <math>y</math> the <math>\sqrt{2}</math> disappears after you expand the root of unity (e raised to a specific power). Long story short, you get <math>x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}</math>. |
+ | |||
+ | ~First | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | First we evaluate the magnitudes. <math>|xy|=80\sqrt{17}</math>, <math>|yz|=60</math>, and <math>|zx|=24\sqrt{17}</math>. Therefore, <math>|x^2y^2z^2|=17\cdot80\cdot60\cdot24</math>, or <math>|xyz|=240\sqrt{34}</math>. Divide to find that <math>|z|=3\sqrt{2}</math>, <math>|x|=4\sqrt{34}</math>, and <math>|y|=10\sqrt{2}</math>. | ||
<asy> | <asy> | ||
draw((0,0)--(4,0)); | draw((0,0)--(4,0)); | ||
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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{ | + | 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== | + | |
− | 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> | + | == Solution 3 (Pretty easy, no hard stuff, just watch ur arithmetic) :) == |
− | 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{ | + | Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out <math>60</math> to find <math>x^2</math>, then use standard techniques that are used to evaluate square roots of irrationals. let <cmath>x = c+di</cmath>, then you get <cmath>c^2 - d^2 = 256</cmath> and <cmath>2cd = 480</cmath> Solve to get <math>x</math> as <math>20+12i</math> and <math>-20-12i</math>. Both will give us the same answer, so use the positive one. Divide <math>-80-320i</math> by <math>x</math>, and you get <math>10+10i</math> as <math>y</math>. This means that <math>z</math> is a multiple of <math>1-i</math> to get a real product, so you find <math>z</math> is <math>3-3i</math>. Now, add the real and imaginary parts separately to get <math>-7-5i</math>, and calculate <math>a^2 + b^2</math> to get <math>\boxed{74}</math>. |
+ | ~minor latex improvements done by jske25 and jdong2006 | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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{74}</math> | ||
+ | |||
Solution by ktong | Solution by ktong | ||
− | |||
− | |||
+ | ==Solution 5 == | ||
+ | |||
+ | 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 6== | ||
+ | |||
+ | We observe that by multiplying <math>xy,</math> <math>yz,</math> and <math>zx,</math> we get <math>(xyz)^2=(-80-320i)(60)(-96+24i).</math> Next, we divide <math>(xyz)^2</math> by <math>(yz)^2</math> to | ||
+ | get <math>x^2.</math> We have <math>x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.</math> We can write <math>x</math> in the form of <math>a+bi,</math> so we get | ||
+ | <math>(a+bi)^2=256+480i.</math> Then, <math>a^2-b^2+2abi=256+480i,</math> <math>a^2-b^2=256,</math> and <math>2ab=480.</math> Solving this system of equations is relatively | ||
+ | simple. We have two cases, <math>a=20, b=12,</math> and <math>a=-20, b=-12.</math> | ||
+ | Case 1: <math>a=20, b=12,</math> so <math>x=20+12i.</math> We solve for <math>y</math> and <math>z</math> by plugging in <math>x</math> to the two equations. We see | ||
+ | <math>y=\frac{-80-320i}{20+12i}=-10-10i</math> and <math>z=\frac{-96+24i}{20+12i}=-3+3i.</math> <math>x+y+z=7+5i,</math> so <math>a=7</math> and <math>b=5.</math> Solving, we end up with | ||
+ | <math>7^2+5^2=\boxed{074}</math> as our answer. | ||
+ | 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> | ||
+ | Solution by Airplane50 | ||
+ | |||
+ | ==Solution 7 (Based on advanced mathematical knowledge)== | ||
+ | According to the Euler's Theory, we can rewrite <math>x</math>, <math>y</math> and <math>z</math> as <cmath>x=r_{1}e^{i{\theta}_1}</cmath> <cmath>y=r_{2}e^{i{\theta}_2}</cmath> <cmath>x=r_{3}e^{i{\theta}_3}</cmath> As a result, <cmath>|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}</cmath> <cmath>|yz|=r_{2}r_{3}=60</cmath> <cmath>|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}</cmath> Also, it is clear that <cmath>yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60</cmath> So <math>{\theta}_2+{\theta}_3=0</math>, or <cmath>{\theta}_2=-{\theta}_3</cmath> Also, we have <cmath>xy=-80\sqrt{17}e^{i\arctan{4}}</cmath> <cmath>yz=60</cmath> <cmath>xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}</cmath> So now we have <math>r_{1}r_{2}=80\sqrt{17}</math>, <math>r_{2}r_{3}=60</math>, <math>r_{1}r_{3}=24\sqrt{17}</math>, <math>{\theta}_1+{\theta}_2=\arctan{4}</math> and <math>{\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}</math>. Solve these above, we get <cmath>r_{1}=4\sqrt{34}</cmath> <cmath>r_{2}=10\sqrt{2}</cmath> <cmath>r_{3}=3\sqrt{2}</cmath> <cmath>{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}</cmath> So we can get <cmath>y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i</cmath> <cmath>z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i</cmath> Use <math>xy=-80-320i</math> we can find that <cmath>x=-20-12i</cmath> So <cmath>x+y+z=-20-12i+10+10i+3-3i=-7-5i</cmath> So we have <math>a=-7</math> and <math>b=-5</math>. | ||
+ | |||
+ | As a result, we finally get <cmath>a^2+b^2=(-7)^2+(-5)^2=\boxed{074}</cmath> | ||
+ | |||
+ | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | We can turn the expression <math>x+y+z</math> into <math>\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}</math>, and this would allow us to plug in the values after some computations. | ||
+ | |||
+ | Based off of the given products, we have | ||
+ | |||
+ | <math>xy^2z=60(-80-320i)</math> | ||
+ | |||
+ | <math>xyz^2=60(-96+24i)</math> | ||
+ | |||
+ | <math>x^2yz=(-96+24i)(-80-320i)</math>. | ||
+ | |||
+ | Dividing by the given products, we have | ||
+ | |||
+ | <math>y^2=\frac{60(-80-320i)}{-96+24i}</math> | ||
+ | |||
+ | <math>z^2=\frac{60(-96+24i)}{-80-320i}</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{74}</math>. | ||
+ | |||
+ | <math>\textbf{-RootThreeOverTwo}</math> | ||
+ | |||
+ | ==Solution 9 (A Little Rigorous, but Straightforward and Easy)== | ||
+ | Multiplying <math>xy \cdot yz \cdot zx = (xyz)^2</math> we obtain <math>60 \cdot 960(16+30i)</math> (too lazy to do <math>60 \cdot 960</math>, you don't need to). Taking the square root, we get <math>240\sqrt{16+30i}</math>. Letting <math>(a+bi)^2=16+30i,</math> we have <math>a^2+2abi-b^2=16+30i.</math> Thus, <math>(a+b)(a-b)=16,</math> and <math>2ab=30.</math> Guessing and checking, we get <math>a+bi=5+3i</math>. Therefore, <math>xyz=240(5+3i).</math> Dividing this by each of the equations provided in the original problem, we get <math>x=20+12i,y=-10-10i,</math> and <math>z=-3+3i</math>. <math>20+12i-10-10i-3+3i=7+5i</math>. Finally, <math>7^2+5^2=\boxed{074}.</math> | ||
− | + | ~SirAppel | |
− | + | ==See Also== | |
+ | {{AIME box|year=2018|n=II|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:15, 12 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Euler's formula and Substitution)
- 3 Solution 2
- 4 Solution 3 (Pretty easy, no hard stuff, just watch ur arithmetic) :)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Solution 7 (Based on advanced mathematical knowledge)
- 9 Solution 8
- 10 Solution 9 (A Little Rigorous, but Straightforward and Easy)
- 11 See Also
Problem
Suppose that ,
, and
are complex numbers such that
,
, and
, where
. Then there are real numbers
and
such that
. Find
.
Solution 1 (Euler's formula and Substitution)
The First (pun intended) thing to notice is that and
have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is
. Anyway this results in getting that
. Then substitute this into
to get, after some calculation, that
and
. Then plug
into
, you could do the same thing with
but
looks like it's easier due to it being smaller. Anyway you get
. Then add all three up, it turns out easier than it seems because for
and
the
disappears after you expand the root of unity (e raised to a specific power). Long story short, you get
.
~First
Solution 2
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 3 (Pretty easy, no hard stuff, just watch ur arithmetic) :)
Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out to find
, then use standard techniques that are used to evaluate square roots of irrationals. let
, then you get
and
Solve to get
as
and
. Both will give us the same answer, so use the positive one. Divide
by
, and you get
as
. This means that
is a multiple of
to get a real product, so you find
is
. Now, add the real and imaginary parts separately to get
, and calculate
to get
.
~minor latex improvements done by jske25 and jdong2006
Solution 4
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 5
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 6
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 Airplane50
Solution 7 (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 8
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
.
Solution 9 (A Little Rigorous, but Straightforward and Easy)
Multiplying we obtain
(too lazy to do
, you don't need to). Taking the square root, we get
. Letting
we have
Thus,
and
Guessing and checking, we get
. Therefore,
Dividing this by each of the equations provided in the original problem, we get
and
.
. Finally,
~SirAppel
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
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.