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Revision as of 02:30, 28 December 2021
Contents
Problem 10
Let and
where
Let
be the unique complex number with the properties that
is a real number and the imaginary part of
is the greatest possible. Find the real part of
.
Solution 1
This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable.
Instead of thinking of complex numbers as purely a real plus a constant times , let’s graph them and hope that the geometric visualization adds insight to the problem.
Note that when we subtract two vectors, the geometric result is the line segment between the two endpoints of the vectors. Thus we can fill in and
as so;
looks similar to
, so let’s try to prove that they are congruent. We can show this in two ways;
Solution 1.1
Let’s look back at the information given to us. The problem states that is a real number. Let the real number be some constant,
. Rearranging yields
.
But how do we relate this expression to our angles? Well, let’s take a look at the divisions themselves.
The subtraction of two vectors yields a vector, and we can write any vector division as where
is a complex number, as the division of two vectors also yields a vector. We can rewrite this as
. We can think of this expression as transforming
directly on to
, and
is the transformation function. However, this transformation must be some kind of rotation, which means that the degree measure of
is equal to the angle between the two vectors since we need to rotate by that angle to lay
flat on
.
Thus we can rewrite our previous equation as , where the angle of
equals the angle between
and
and likewise for
. More precisely, we can write
as
and
, respectively by Euler’s formula. Then
is the claim we wish to prove.
We can now do some simple algebra to prove this;
is obviously real, so
must be real as well. But the only way that can happen is if
.
Solution 1.2
Let us write as some complex number with form
Similarly, we can write
as some
The product must be real, so we have that is real.
is real by definition, so dividing the real number above by
will still yield a real number. (Note that we can see that
from the definitions of
and
). Thus we have
is real. The imaginary part of this is which we recognize as
This is only
when
for some integer
. Here
represents the major angle
, and the angle we want is the minor angle, so we can rewrite the equation as
. We can see from the diagram that both
and
are obtuse, so therefore
.
Solution 1 Rejoined
Now that we’ve proven this fact, we know that all four points lie on a circle, so let’s draw that in;
</asy>
Note that when we subtract two vectors, the geometric result is the line segment between the two endpoints of the vectors. Thus we can fill in and
as so;
Bashy Solution :)
We know thatHence,
where
.
Let
. Then,
The numerator is:
The ratio of the imaginary part to the real part must be
because
Hence,
Evidently,
is maximized when
is minimized or when
~AopsUser101
Solution 3
Algebra Bash
First we calculate , which becomes
.
Next, we define to be
for some real numbers
and
. Then,
can be written as
Multiplying both the numerator and denominator by the conjugate of the denominator, we get:
In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of be a multiple of the conjugate of
, namely
So, we have
and
for some real number
.
Then, we get:
Expanding both sides and combining like terms, we get:
which can be rewritten as:
Now, common sense tells us that to maximize , we would need to maximize
. Therefore, we must set
to its lowest value, namely 0. Therefore,
must be
You can also notice that the ab terms cancel out so all you need is the x-coordinate of the center and only expand the a parts of the equation.
~stronto
Solution 4 (algebra but much cleaner)
We see that . Now, let
, in which case
and
. We now have that
is real, meaning that
is real.
We see that , so therefore
is real.
This means that , so we now have that
, so
, which can be rewritten as
. In order to maximize
we want to maximize
, and in order to maximize
we want
and
, so
.
(Note:
is the imaginary part of
, and
is the real part of
)
~Stormersyle
Solution 5
We will just bash. Let where
. We see that
after doing some calculations. We also see that
We note that
is a multiple of
because the numerator has to be real. Thus, expanding it out, we see that
Hence,
To maximize the imaginary part,
must equal
so hence,
.
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.