If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:11:54

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 know you you kind of don't even know where to start here so let P of X equal 24 X to the 24th plus the sum from J equals 1 to 23 if 24 minus J times X to the 24th minus J plus X to the 24th plus J let Z 1 Z 2 all the way to Zr be the distinct zeros the of P of X fair enough let Z sub K squared equal a sub k plus B sub K I for K equals 1 through R so there's Z sub K or I guess Z sub they're saying just look a sub k plus B sub K I is going to be the square of each of these roots so let's see for K is equal to 1 through R where I is equal to square root of negative 1 we know that and a and a sub K and V sub K are real numbers let the sum from k equal 1 are 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 radical 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 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 let's just try to visualize this polynomial a little better and 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 number is the first term as the way it's written first term could be anything 24 X to the 24th and now let's write out let's not write out this over here so if we set J equal to 1 what do we get we get 23 23 times X to the 23 x to the 23 plus X to the 25 all right this is 24 plus J plus X to the 25 and then when J is equal to 2 I'll just go down in a big column here when J is equal to two we have plus 22x to the 22 plus X to the 26 and then we could keep going plus I'll just keep you get the general idea this term over here the 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 - lets go to J equal 22 so when J is equal to 20 to 24 - 22 is 2 it's going to be 2x squared 22 plus 24 is 46 so plus X to the X to the 46 I did it right 22 right 46 and then this last term over here when J is equal to 23 because that's where we stopped 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 traditional we're used to seeing polynomials and that's you know I start with the highest degree term and then you go down from there so we can write P of X we can rewrite 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 a coefficient here is 1 if you distribute it so it's going to be it's going to be X to the 47th X to the 47th Plus this term over here 2 X to the 46 I'm just distributing this multiplying factor or this I guess it's coefficient it's going to be 2x squared + 2 X to the 46 and then we're just going to keep going we're going to keep going all the way down all the way up to 22 X to the 26 22 X to the 26 so you see what's happening the coefficient is increasing by 1 each time and the exponent is decreasing by 1 + 23 + 23 X to the 25th and then the next degree we don't have a 20 X to the 24th year but we have the X to the 24th right there it so it seems almost by design for the problem that they stuck it there so plus 24 X to the 24th and now we can do all of these terms here so we have plus 23 plus 23 plus 23 x to the 23rd plus 22x to this 22nd and then we go all the way down to plus 2x squared plus X so this is just another way so 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 up to the 47th power and then the coefficients started 1 and they keep increasing up to 24 and 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 P of X is equal to x times it's equal to x times X to the 46 plus 2x to the 45th well I'm just dividing all of these by X or factoring out an X plus all the way to 22 X to the 25th plus 23 X 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 23 X to the 22nd plus 22x to the 21 all the way to 2x plus 1 now x equals 0 is clearly a root if you set P of x equals 0 x equals 0 as 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 the sum of the roots of everything else all the stuff in parenthesis and this part I guess the next step it might not be obvious to you but now that once you see it once you see the pattern I guess if you ever see it 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 to recognize it's just not something you normally see in your everyday curriculum and so what do to see the pattern here to see the pattern is think about what happens when we square different polynomials so if we square if we square where all of the coefficients are one they don't all have to be one 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 one so X plus one X plus one squared is x squared plus two x plus one we've seen that multiple times now what is x squared what is x squared plus X plus one squared so all I'm doing is I'm just taking all of the powers of X and summing them up well that's going to be let me just well we've done enough practice with 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 one which is going to be x squared plus x times x squared which is going to be is going to be X to the third plus x times X which is going to be x squared plus x times one which is going to be X and then finally it's going to be one times all of this stuff which is going to be plus x squared plus X plus one and so if you take the sum you're going to get X to the fourth plus two x to the third plus three x squared plus 2x plus one and so you might already see a pattern emerging what happened over here when you just took X plus one squared you started with the one you went up to a two and then you went back to one 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 one two three the middle term the three peaked out at the hot at D I guess the middle term and then you start going down in coefficient again one two three and then two one and you can actually prove that this is the case for any problem 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 to the third plus x squared plus X plus one if I were to square this I just based on the pattern and you could do it on paper if you like you hopefully know how to multiply this 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 one X to the sixth plus two X to the fifth plus three X to the fourth plus and we have to be careful the middle term so this is going to have this right here is going to have you're going to have each of these terms x also the last term is going to be so you're going to have X to the fourth let me write not write the coefficients you're gonna have X to the fourth plus something times X to the third plus something times x squared plus something times X plus one so you're gonna have one two three four five six seven 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 three two oh sorry it should be three whoops 4x and three then two and then one again and you 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 were picking 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 of the X term and this was a first degree polynomial we're squaring this is the exact same pattern but we peak out we peak out at X to the 23rd at X to the 23rd so using the same pattern using that and you can actually prove it for the general case it just gets messy and I think you 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 re-written 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 plus X to the xxi all the way down to plus X to the third plus x squared plus X plus one squared squared and we know that don't we and you know the only way you'd really know that is if you've seen this pattern before if you've seen this pattern before but that's the way you do it you look at this is a pattern the coefficient is just keep increasing the 24 then keep decreasing that's going to be whatever the coefficient the highest degree over here so it's extra 23 it's going to be extra 23 plus all 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 gonna leave you in this video this simplifies it a good bit but there's going to be a few more kind of aha moments to be able to that we're gonna have to have to be able to solve this problem