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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.