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

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 one were and then we're going to square those 24 roots take the absolute value of their imaginary part and then sum them up so let's 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 as the angle I guess I should say that's e to the PI over 12 I let me just focus on the angle here this over here is two PI over 12 or PI over six this over here is three PI over 12 or PI over four this over here is four PI over 12 4 PI over 12 or PI over three and this right over here is five PI over 12 this right over here is five PI over 12 and then this obviously is six PI over 12 or PI over two we could keep going but before we keep going and this is this will really just simplify the math a little bit remember we're going to square these roots so let's just think about if we have some complex number A plus bi 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 two a bi minus B squared or we could rewrite it as a squared minus B squared will be the real part and then plus 2 a B I to a B is the imaginary part and the reason why I even took the trouble of writing it this way is to realize that the comp what we're going to do is 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 all that matters is the absolute value of 2 a B or really two times the absolute value two times the absolute value of a B now these guys all have analogs where either A or B would be negative so if this guy is if this guy this right over here is let's call this a plus bi if you had a minus bi you would be down here 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 negative a minus bi is going to be over there or you could have negative a plus bi over here negative a plus bi over here but the reason why I did this as 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 a be AB absolute value it's going to be a be negative a be absolute values are still going to be a B so they're all going to be the same so what we can do is and each of these guys we're going to have the same analog so what we can do is just find the values for these guys 1 2 3 4 for these five guys over here and then just multiply whatever we get by 4 because they each have I guess there's 4 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 one square one doesn't have an imaginary part 1 squared is not still 1 does not have an imaginary part we can also ignore we can also ignore 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 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 and then we'll square them find the absolute value of the imaginary parts and then we'll multiply everything by 4 because that'll correspond to these other two there too what if you took the 8 the imaginary of the real parts and made them negative and it would take us all the way around the unit circle so let's think about it a little bit let me just write these over here so the Z's the Z's that we'll think about right now are going to be e to the PI over e to the PI over 12 e to the PI over e to the PI over 6 e to the PI over 4 of e to the PI over 12 I should say PI over 6 I PI over 4 I eetu the PI over 3 I and then we have e to the e to the 5 PI over 12 I five PI over 12 I now we're going to square each of them and it's nice to leave it in this exponential form when you square it much easier to square these values if you square this you're just essentially multiplying the exponent times two so this is going to be e to the PI over six I this will be e to the PI over three I 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 three I and then here you're gonna have eetu the e to the PI over two I eat ooh the PI over two I then over here you'll have e to the 2 PI over 3 2 PI over 3 I and then finally over here you'll have e to the 5 PI 5 PI over 6 I now these are the squared of the squared values of these roots here now let's just think about their imaginary parts let's just think about their imaginary parts so this guy right over here can be rewritten this guy 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 is imaginary part same is going 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 it's going to be sine of here it is going to be sine of 5 PI 5 PI over 6 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 homestretch so PI over 6 PI over 6 is 180 if we are in degrees and my brain has 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 PI over 6 is the same thing as 30 degrees so it looks like this and we know that the sine of 30 degrees the sine of 30 degrees is 1/2 this is 1 this is 1/2 this the cosine is 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 is the equivalent of 60 degrees the sine over here is square root of 3 over 2 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 is just 1 so this is the imagine it's if it's the imaginary part of this or I guess this would be this is essentially this evaluates just 2i 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 this will just be one and then sine of 2 pi over 3 2 PI over 3 let's see that is so pi over 3 is 60 degrees so you can view that as 120 degrees so it's 120 degrees so it's right over here so it's going to have the same sine value it's going to have the same sine value as PI over 3 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 so we have we have 1/2 plus 1/2 is 1 plus another one over here is going to be 2 plus the square root of 3 over 2 plus square root of 3 over 2 is square root of 3 now remember this was we just we just did for this quadrant over here we have to do it for all of the quadrants so we just need to multiply we just need to multiply everything times four so the sum of the absolute value of the imaginary part of the square of the square of the roots is eight plus four square roots of three eight plus four square roots of three and going back to the original problem we got the answer here this is eight plus four square roots of three I don't want to make I want to make sure I didn't mess it up right eight plus four square roots of three so if you want to find M plus n plus P it is eight plus four plus three which is equal to fifteen and we're done