2018 AIME II Problems/Problem 5
Contents
[hide]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
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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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.