2003 AIME II problem 15 (part 1)
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.