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## 2003 AIME

# 2003 AIME II problem 15 (part 1)

## Video transcript

So I'll tell you ahead of
time, this problem is no joke. But if we do it step by step,
it's actually not too bad. But it's not an easy problem,
so don't get discouraged if you kind of don't
even know where to start on this
one right over here. So let p of x equal
24x to the 24th plus the sum from j equals 1 to
23 of 24 minus j times x to the 24th minus j plus
x to the 24th plus j. Let z1, z2, all the way
to zr be the distinct 0's of p of x-- fair enough. Let z sub k squared
equal a sub k plus b sub ki for k equals 1 through r. So there's z sub k
or-- I guess, they're saying a sub k plus
b sub ki is going to be the square of
each of these roots. So let's see, for k
is equal 1 through r where i is equal to
square root of negative 1. We know that. And a sub k and b sub
k are real numbers. Let the sum from k equal 1 to
r of the absolute value of b sub k equal this
number over here where m, n, and p
are integers and p is not divisible by the
square of any prime. So they're saying
that we've just simplified the radicals
as much as possible. Find m plus n plus p. So this sum right
here, this is the sum of the absolute value
of the imaginary parts of the square of the roots. Right? b sub k is the imaginary
part of the square of root k, I guess you could call it. And so what we're going to
do is take the absolute value of the imaginary part of
the square of the roots and sum them up. So before we even think about
how to do that, let's just, I don't know, let's
just try to visualize this polynomial a little better. In fact, I have
a feeling this is going to take us
multiple videos to do. So we have p of x
is equal to-- well, the first term here is just 24. Well, it's not the
first term, it's the first term as
the way it's written. The first term
could be anything. 24x to the 24th. And now, let's write
out this over here. So if we set j equal
to 1, what do we get? We get 23 times x to
the 23 plus x to the 25. This is 24 plus j. And then when j is
equal to 2-- I'll just go down in a
big column here. When j is equal to 2, we
have plus 22, x to the 22 plus x to the 26th. And then we could keep going. Plus-- well, you get
the general idea. This term over here,
the coefficients are going to go down. So is this exponent, but
this exponent over here is going to go up. So then we can keep
adding all the way-- let's go to j equal 22. So when j is equal to
22, 24 minus 22 is 2. It's going to be 2, x
squared-- 22 plus 24 is 46-- so plus x to the 46th. Did I do that right? 22, 46. And then, this last term over
here, when j is equal to 23, because that's
where we stop, we're going to get 1 times
x plus x to the 47th. So that's our polynomial. What I want to do
is I'm just going to rewrite this in
kind of the way we're used to seeing polynomials, and
that's start with the highest degree term and then
you go down from there. So we can write p of x as
being equal to-- what's the highest degree term here? Well, we have an x to the 47th
over here, right over here-- x to the 47th. And then the coefficient here
is 1, if you distribute it. So it's going to be x to the
47th plus this term over here, 2x to the 46th. I'm just distributing
this multiplying factor, or this, I guess,
its coefficient-- it's going to be 2x squared
plus 2x to the 46th. And then we're just going
to keep going all the way up to 22x to the 26th. So you see what's happening? The coefficient is
increasing by 1 each time, and the exponent
is decreasing by 1. Plus 23x to the 25th. And then the next degree, we
don't have a x to the 24th here, but we have the x
to the 24th right there. So it seems almost by
design for the problem that they stuck it there. So plus 24x to the 24th. And now we can do all
of these terms here. So we have plus 23x to the
23rd, plus 22x to the 22nd, and then we go all the way
down to plus 2x squared plus x. So this is just another way
that, at least in my brain, simplifies this
polynomial a good bit. We now have a good sense
of what's happening here. It's really all of the powers
of x up to the 47th power. And then the
coefficients start at 1 and they keep
increasing up to 24, then they start
decreasing again. Now, we're interested in the
roots of this polynomial. And there's one root
that's pretty easy to find. We can just factor
an x out of this. Every term here
is divisible by x. So we can rewrite p of x is
equal to x times x to the 46th plus 2x to the 45th-- I'm
just dividing all of these by x or factoring out an
x-- plus all the way to 22x to the 25th plus 23x
to the 24-- and then we have the one in magenta--
plus 24x to the 23rd-- and then we have all
of these characters down here-- so then
we have, plus 23x to the 22nd plus 22x to the
21 all the way to 2x plus 1. Now x equals 0 is
clearly a root-- or if you set p of x equals
0, x equals 0 is a root. It's actually not
going to matter much, because it has no
imaginary part. So if you take 0-- one of the
roots here is 0, you square 0, it's not going to have
an imaginary part. So the absolute value
of its imaginary part is not going to contribute
to this sum over here, but maybe it simplifies this. So we really care about the
sum of the roots other than 0. So we care about sum of the
roots of everything else, all the stuff in parentheses. And this part, I
guess the next step, it might not be obvious to
you, but once you see it-- once you see the pattern, I
guess if you ever see again in your life, maybe in some
of these AMI type problems or competition problems-- you
should be able to recognize it, because it's an easy
pattern recognize. It's just not
something you normally see in your everyday curriculum. And so what I want to do
to see the pattern here is think about what happens when
we square different polynomial. So if we square where all
of the coefficients are 1. They don't all have
to be 1, but it'll help us see the pattern there. The pattern will obviously
be slightly different if the coefficients
are different. Now, if I just take x
plus 1-- x plus 1 squared is x squared plus 2x plus 1. We've seen that multiple times. Now what is x squared
plus x plus 1 squared? So all I'm doing
is I'm just taking all of the powers of
x and summing them up. Well, it's going to be-- well,
we've done enough practice for this, and if you're
watching this problem, you know how to do this. It's going to be x
squared times x squared. So it's going to be x to the
fourth plus x squared times x, which is going to be x to the
third, plus x squared times 1, which is going to be x squared,
plus x times x squared, which is going to
be x to the third, plus x times x, which is
going to be x squared, plus x times 1, which
is going to be x. And then finally, it's
going to be 1 times all of this stuff,
which is going to be plus x squared
plus x plus 1. And so if you take
the sum, you're going to get x to
the fourth plus 2x to the third plus 3x
squared plus 2x plus 1. And so you might already
see a pattern emerging. What happened over here when
you just took x plus 1 squared? You started with a 1,
you went up to a 2, and then you went back to a 1. So not clear that
you have a pattern, but it seems to have this-- the
coefficients increase and then decrease. What happens when you
took this thing squared? You started with a 1,
then the coefficients went 1, 2, 3, the
middle term, the 3, peaked out at the
middle term, and then you start going down
in coefficient again. 1, 2, 3, and then 2, 1. And you can actually
prove that this is the case for any polynomial. So I could just write out--
I won't multiply it out. If you have time, you can do it. But if I were to take
x to the third plus x squared plus x plus 1. If I were to square this,
just based on the pattern-- you could do it on
paper if you like. You hopefully know
how to multiply this type of polynomial. There are videos on
it, if you don't. This is going to be equal
to-- the highest degree term is going to
be x to the sixth. So it's going to be 1x to the
sixth plus 2x to the fifth plus 3x to the fourth
plus-- and we have to be careful-- the middle term. So this right here
is going to have-- you're going to have
each of these terms times all-- So the last term is going
to be-- so you're going to have x to fourth-- Let me not
write the coefficients. You're going to have x to the
fourth plus something times x to the third plus something
times x squared plus something times x plus 1. So you're going to have
1, 2, 3, 4, 5, 6, 7 terms. The middle term is going
to be this right over here. So that's where we peak
out our coefficients, and then they go
back down again. 3, 2, oh, sorry. It should be 3, whoops, 4x, then
3, then 2, and then 1 again. And you can multiply
this out for yourself, but I think you now
see the pattern. So when you look
at this thing here, where the coefficients
are just going from 1, 2, all the way to 24
is where they peak out. And notice they peaked
out-- the middle term where you're peaking
out-- in this case is, really, the highest. So we're peaking out here
at the x to the third term, and this was a third
degree polynomial that we were squaring. Over here, we peaked out
at the x squared term, and this was a second degree
polynomial we were squaring. Here we peak out at the x term,
and this was a first degree polynomial we're squaring. This is the exact same
pattern, but we peak out at x to the 23rd. So using the same pattern--
and you can actually prove it for the general case. It just gets messy, and I think
you get the general gist of it. This thing over here-- and
this is what I talked about, it's not the easiest
thing in the world to recognize-- p of x can
be rewritten as equal to x times-- this thing
over here is x to the 23rd-- x to the 23rd
plus x to the 22nd plus x to the 21st all the way down
to plus x to the third plus x squared plus x plus 1 squared. And we know that-- the only
way you'd really know that is if you've seen
this pattern before. But that's the way you do it. You look at-- this is a pattern. The coefficients just
keep increasing to 24, then keep decreasing. That's going to be whatever
the coefficient of the highest degree over here. So it's x to 23. And it's going to
be x to 23 plus all of the powers of x less than
that all the way down to x to 0 squared. And we saw the pattern
here a couple of times. I'm going to leave
you in this video. This simplifies it a
good bit, but there's going to be a few more
kind of aha moments that we're going to
have to have to be able to solve this problem.