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## 2003 AIME

Current time:0:00Total duration:11:33

# 2003 AIME II problem 15 (part 2)

## Video transcript

Where we left off
in the last video, we had somewhat simplified our
polynomial p of x from the way that it was originally
described in the problem, but it still doesn't
seem like we're any closer to finding the roots. At least we were able to reduce
the degree from 47th degree polynomial to the product of a
first degree and a 23rd degree polynomial. But it still
doesn't seem like we are any closer to finding
the roots, especially the roots that matter. And here, we're going to have
to-- there's another little bit of an aha moment here-- that
this part in a blue, which is really just the sum of all
of the degrees of x from x to the 0 all the way to x to 23. This actually looks a lot. It is actually, if you
view the common ratio as x as a finite
geometric series. And I've done it
in other videos, there is a way to find the sum
of a finite geometric series. And we'll use a similar
technique right here to simplify this a little
bit-- to help us find the root. So let's just call
this whole thing S. So let's just call this S is
equal to x to the 23 plus x to the 22nd plus all the
way down to x plus 1. So that's this thing over here. Now let's multiply S times x. So x times S is going
to be equal to-- and actually, let
me do it like this, because this will
help us a little bit. I'm going to write
it up here, and it's going to look a little
bit messy. x times S is going to be-- actually, let
me-- I'll just write it over here. x times S is equal
to-- let me save some space, actually. x times S is equal to-- x times
x to 23 is x to the 24th power, x times x to the 22nd is going
to be x to the 23rd power, and then there's going
to be an x to the 21st. When you multiply by x, you're
going to get x to the 22nd. All the way down to when
you multiply x times x, you're going to
get plus x squared. And then when you multiply the 1
times x, you're going to get x. Now, what I want to do, let's
subtract this from that. Let's subtract it. I won't put the
negative sign out there. So on the left hand
side of the equation you have S minus xS
is equal to-- what happens on the right hand
side of the equation? Well, you're going to have
this guy minus this guy. These are going to cancel out. These are going to cancel out. Everything is going
to cancel out, except you're going to have
a 1 minus x to the 24th. So you're going to have
1 minus x to the 24th. Now let's just multiply
both sides by a negative 1, because this will just get it in
a slightly easier form for us. So this is the same thing. If we multiply both
sides times negative 1 it's xS minus S is equal
to x to the 24th minus 1. We can factor out an S here. I'll do it over here. So if we factor out an S
from the left hand side, we get S times x minus 1 is
equal to x to the 24th minus 1. Or we could say that the sum is
equal to x to the 24th minus 1 over x minus 1. So this whole thing
in the blue over here can be rewritten as
this right over here-- as x to the 24th minus
1 over x minus 1. Now let's be clear here. What we really care
about for this problem is to figure out the
roots that have-- the roots whose squares
have complex parts. We already know
that one of the root is 0, because if you
set this equal to 0, x could be equal to 0. 0 squared is still 0-- it
does not have a complex part. So it really doesn't matter. So for our problem--
and the other thing that we want to keep
in mind is that we don't want to
double count roots. Let's re-read the
problem a little bit. Because what we're going
to have is a bunch of-- I guess you could view
them as double roots. You're only going to
have 23 distinct roots from this part of
the polynomial. And they say, let
z1, z2, and zr be the distinct zeroes of p sub x. So what we only care about
right now are these 23 zeroes. We don't have to think
about those 23 zeroes twice. We're not going to add
their imaginary parts twice. So we can kind of substitute
p of x with-- let's call it q of x. We really just care
about the zeroes of q of x, which is just x to
the 23 plus x to the 22nd plus all the way down to x
squared plus x plus 1. That's this inside part. The fact that it's squared,
that just doubles the roots, but we just care about
the distinct root. So we really just care about
the sum of the imaginary part-- of the absolute value of the
imaginary parts of the squares of the roots of this
guy, right over here. Now we just figured
out that this guy can be rewritten
as this over here. Or even better, let's define--
instead of our defining q of x like this, let's define
q of x as this times x-- let me write it a little
bit-- as this times x minus 1. Let's just define q of x here. So this q of x is going to have
all of the same distinct roots, that matter as this
p of x, but it's also going to have the root x is
equal to 1, which is clearly not a root here. If you put x equals
1 here, you're actually going to get 24. So it's not going
to be equal to 0. But what makes this better,
easier to find the roots of, is that this thing
right over here is equivalent to-- this part in
yellow is x to the 24th minus 1 over x minus 1 times
x minus 1, which is equal to x to
the 24th minus 1. So we've simplified the
problem a huge amount. What we've said is the roots
that we care about are actually the roots of this
polynomial right here, except for
x is equal to 1-- we just kind of added that
there to simplify it-- which is equal to the roots
of this polynomial here, except for 1, because
we added 1 as a root. So we just care about the
roots of this polynomial, except for 1. And the roots of this polynomial
are-- well, I won't say easy, but it's much more
straightforward to calculate. If we set this
polynomial to equal to 0, to figure out the roots,
we just get x to the 24th add 1 to both sides,
is equal to 1. So we're going to
have-- remember, this is going to have 24 roots. One of them is going to be 1. We want to ignore the 1, so we
care about the 23 non-1 roots. And we care about squaring
them, and then taking the absolute value of
the imaginary parts. So we essentially need
to think about all of the 24th root of
1, and then ignore 1. So let's think about how
to do that a little bit. And this is going to
delve a little bit into the complex plane. So just as a refresher, how do
we find all of the roots of 1? The 24th roots of 1? And just remember,
we can rewrite 1. So let's draw a little
complex plane over here. So this is the real axis, and
this is the imaginary axis. 1 is sitting right over here. It only has a real
part-- it is 1. But we know that we
can rewrite this. We can rewrite 1 as--
let me write it this way. It could be e to the i times 0. Actually, so this is e to
the 0, obviously equal to 1. Or we can add 2 pi to it, if
we're dealing with radians. So 1 is equal to e to i0, which
is also equal to e to the 2 pi i, but you could keep
adding, because if you add 2 pi goes around
the complex plane once. If you add another 2 pi,
it goes around again. It also equals e to the 4 pi i. It also equals e to the 6 pi i. It also equals e to the 8 pi i. I think you get
the general idea. So if you want to
find the 24th roots, you want to just raise all of
these things to the 1 over 24. So essentially, you're going
to take this divide by 24, divide by 24, divide by 24,
divide by 24, divided by 24, and then keep going. You can actually go all the way. Well, I'll depict it over here. So what is this first root? The first root is
e to the 0 over 24. That's just e to the 0. That's just 1. That's the root that
we can ignore, right? Because we added that root
to q of x to simplify it a little bit. Now what's this root over here? This is e to the pi over
12i, which remember, is the same thing as
cosine of pi over 12-- that's the real part--
plus i sine of pi over 12. And if we were to
draw that, we would have an angle of pi over 12. And if you wanted
to think in degrees, that would be 15 degrees. So it would look something
like this right over here. So we do care about that
root right over here. Now what would this root be? Well this root is
going to be pi over 6. It's going to have
twice the angles. So it's going to look like this. So this root right here is
going to look like this. This root over here-- and
notice, they're pi over 12 apart. This root over here,
this is pi over 4. So it's going to look
like this-- pi over 4. That's 45 degrees. So it's going to look like this. So notice, each of the
roots we're just adding are just multiples
of this-- I shouldn't say multiples-- are powers
of this first root over here. We're just increasing
the degree by pi over 12. And remember, we just care
about the imaginary parts of their squares. So let's just first of all,
just think about the roots that matter. So one way to think
about it is we could literally just
draw a unit circle here and divide it into 24 sections. So let me draw that. Let me draw it a little bit. Let me redraw it
a little bit, so I don't overload this
graph over here. Actually, I'm
running out of time in this video-- I might
continue it in the next. So we could draw a
unit circle here. They're all going to
have magnitude one. We could draw the
unit circle over here. One of the roots is
1, but we're going to have 24 roots, so we can
divide it into 24 sections. So we have four
sections right here. We're going to have six
roots in each section. So we're going to have 1,
pi over 12, 2pi over 12, pi over 4, and then you have
the one symmetric to that, then you have the one
symmetric to that-- let me make sure I'm
doing this right. And then you're going
to have pi over 2. So these are the roots that
are-- actually, let me just continue this in the next video. I don't want to-- when
I start rushing this-- this is kind of a new
concept to everybody. I'll continue this
in the next video.