2017 AIME I Problems/Problem 10
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
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
is some multiple of
In this problem, this implies
and
must form a cyclic quadrilateral, so the possibilities of
lie on the circumcircle of
and
To maximize the imaginary part of it must lie at the top of the circumcircle, which means the real part of
is the same as the real part of the circumcenter. The center of the circumcircle can be found in various ways, (such as computing the intersection of the perpendicular bisectors of the sides) and when computed gives us that the real part of the circumcenter is
so the real part of
is
and thus our answer is
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 | ||
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