Current time:0:00Total duration:9:31

0 energy points

Studying for a test? Prepare with this lesson on AIME.

See lesson

# 2003 AIME II problem 15 (part 3)

Video transcript

Where we left off
in the last video, we were beginning
to try to figure out what the 24 roots of 1 were. And then we were going
to square those 24 roots, take the absolute value
of their imaginary part, and then sum them up. Once again, let's just think
about the roots that matter. So this over here, we figured
out this is pi over 12. So let me-- this is pi over 12. That's the angle, I
guess I should say. That's e to the pi over 12i. Let me just focus
on the angle, here. This over here is 2 pi
over 12, or pi over 6. This over here is 3 pi
over 12, or pi over 4. This over here is 4 pi
over 12, or pi over 3. And this right over
here is 5 pi over 12. And then this, obviously, is
6 pi over 12, or pi over 2. We could keep going, but
before we keep going-- and this will
really just simplify the math a little
bit-- remember, we're going to have
to square these roots. So let's just think
about if we have some complex number,
a plus bi, let's think about what happens
when we square it. This is going to be
equal to a squared plus 2abi minus b squared. Or we could rewrite it as
a squared minus b squared-- will be the real part--
and then, plus 2abi. 2ab is the imaginary part. And the reason why I
even took the trouble of writing it this
way is to realize that what we're
going to do is we're going to square all of
these complex numbers, and then take the absolute
value of the imaginary part. So really, all that matters
is the absolute value of 2ab, or really, 2 times the
absolute value of ab. Now these guys all
have analogs where either a or b would be negative. So if this guy is--
this right over here-- let's call this, a plus bi. If you had a minus bi,
you would be down here. So if this guy is a
plus bi, a minus bi is going to be over here. Or you could have
negative a minus bi is going to be
over here, or you could have negative
a plus bi over here. But the reason why
I did this is we showed you, just over
here, all of these guys when you square
them and you take the absolute value
of the imaginary part of that squared value, it's
all going to be the same. Because when you take--
it's going to be ab, absolute value is
going to be ab. Negative ab absolute value
just still going to be ab. So they're all going
to be the same. So what we can do is--
and each of these guys are going to have
the same analog. So what we could is just find
the values for these guys one, two, three, four, for
these five guys over here. And then just multiply whatever
we get by 4, because they each have-- I guess there's
four of these around the unit circle, and that'll
save a lot of our work. Now, the other thing we
want to think about-- we already said that
we have to ignore 1, because we added that root. But even if we didn't
remember to ignore it, it wouldn't matter, because 1
doesn't have an imaginary part. 1 squared is still 1-- does
not have an imaginary part. We can also ignore an angle
of pi over 2, or 90 degrees, because this has no real part. And you see over here,
that when you square it, it's actually going
to take you over here, and it won't have
any imaginary part. When you square it it's
2 times the real times the imaginary part. This has no real part,
so this is going to be 0. So this guy will
also not contribute. So we really just care about
these angles right over here. And then we'll square them,
find the absolute value of their imaginary
parts, and then we'll multiply everything
by 4, because that'll correspond to these
other-- if you took the imaginary or the real
parts and made them negative. And it would take us all the
way around the unit circle. So let's think
about a little bit. Let me just write
these over here. So the z's that we'll
think about right now are going to be e to the pi
over 12, e to the pi over 6, e to the pi over 4-- oh,
e to the pi over 12i, I should say-- pi over 6i, pi
over 4i, e to the pi over 3i, and then we have e
to the 5 pi over 12i. Now we're going to
square each of them, and it's nice to leave it
in this exponential form when you square it. It's much easier to
square these values. If you square this,
you're just, essentially, multiplying the
exponent times 2. So this is going to be
e to the pi over 6i. This will be e to the pi over
3i-- we're just squaring them. So we're just taking
the square of each of these values,
each of these roots. So e to the pi over 3i,
and then here you're going to have e
to the pi over 2i. And over here, you'll have
e to the 2 pi over 3i. And then finally over here,
you'll have e to the 5 pi over 6i. Now these are the squared
values of these roots here. Now, let's just think about
their imaginary parts. So this guy right
over here can be rewritten as cosine of pi over
6 plus i sine of pi over 6. So his imaginary part
is sine of pi over 6. This guy's imaginary part--
if you expand it out, Euler's identity,
Euler's formula. Cosine of pi over 3 plus
i sine of pi over 3. So this imaginary part is just
going to be sine of pi over 3. Here is going to be
sine of pi over 2. Here is going to be
sine of 2 pi over 3. And here is going to be
a sine of 5 pi over 6. Now we just have to
evaluate these guys, and take their absolute
value, and then take the sum, and then multiply
everything times 4. And we're essentially
in the home stretch. So pi over 6 is 180--
if we were in degrees, and my brain has an easier
time processing that. So, actually, let me draw
another unit circle over here, just so we can
visualize these angles. So now we have
sine of pi over 6. Pi over 6 is the same
thing as 30 degrees. So it looks like this. And we know that the sine
of 30 degrees is 1/2. This is one, this is 1/2,
this the cosine square root of 3 over 2. But this right
over here, is 1/2. Sine of pi over 3. pi over 3 is the
equivalent of 60 degrees. The sin over here is
square root of 3 over 2. You can figure that out
from 30, 60, 90 triangles. Then you have sine of pi over 2. Pi over 2 is that
right over there. Well, the sine of
that is just 1. So this is the
imaginary part of this, or I guess, this would
be-- essentially, this evaluates just to i,
but the imaginary part, which is viewed as the
coefficient on the i, which is kind of non-intuitive. You would kind of think
it's the whole thing, but when people say
the imaginary part. So this will just be 1. And then sine of 2 pi over 3. Let's see, that is-- so
pi over 3 is 60 degrees. So you could view
that is 120 degrees. So it's 120 degrees. So it's right over here. So it's going to have the
same sin value as pi over 3, so it's going to be
square root of 3 over 2. And then, sine of 5 pi over 6. Pi over 6 is 30 degrees. So this is sine of 150 degrees. So it's going to
be just like that. And it's going to have the
same sine value as pi over 6. So it's going to be
1/2, and lucky for us, these are all positive values. So let's just take the sum. And so we have 1/2 plus
1/2 is 1, plus another 1 over here is going to be
two, plus the square root of 3 over 2 plus square root of
3 over 2 is square root of 3. Now remember, we just did it
for this quadrant over here. We have to do it for
all of the quadrants, so we just need to multiply
everything times 4. So the sum of the absolute
value of the imaginary part of the square of the roots is
8 plus 4 square roots of 3. And going back to
the original problem, we got the answer here. This is 8 plus 4
square roots of 3. I want to make sure
I didn't mess it up. Right. 8 plus 4 square roots of 3. So if we want find m plus
n plus p, it is 8 plus 4 plus 3, which is equal to 15. And we're done.