2018 AIME II Problems/Problem 5
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.