1996 AIME Problems/Problem 11
Problem
Let be the product of the roots of that have a positive imaginary part, and suppose that , where and . Find .
Contents
Solution 1
Thus ,
or
(see cis).
Discarding the roots with negative imaginary parts (leaving us with ), we are left with ; their product is .
Solution 2
Let the fifth roots of unity, except for . Then , and since both sides have the fifth roots of unity as roots, we have . Long division quickly gives the other factor to be . The solution follows as above.
Solution 3
Divide through by . We get the equation . Let . Then . Our equation is then , with solutions . For , we get . For , we get (using exponential form of ). For , we get . The ones with positive imaginary parts are ones where , so we have .
Solution 4
This is just a slight variation of Solution 1.
We start off by adding to both sides, to get a neat geometric sequence with and , which gives us . From here, multiply by to both sides, noting that then since, then we are multiplying by which makes it undefined. We now note that . (This is the part that it becomes almost identical to Solution 1). Factor from the LHS, to get . Call the set of roots from as , and set of roots from as . We are to find . (Essentially, we are to find the roots that are not common to both equations/sets, or else we are overcounting a root two times, rather than once. Try out some equation to see where this might apply) Thankfully in this case, none of the roots are overcounted. From here, proceed with Solution 1.
Solution 5
We recognize that and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting and as shown below.
Now, knowing that is not a root, we multiply by to obtain Now, we see the and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. .
Now, it is clear that we have two cases to consider.
Case : We obtain that or Obviously, the answers to this case are
Case : Completing the square and then algebra allows us to find that which has
Hence, the answer is
Solution 6
Add 1 to both sides of the equation to get . We can rearrange to find that . Then, using sum of a geometric series, .
Combining the two terms of the LHS, we get that , so , and simplifying, we see that , so by SFFT, . Then, the roots of our polynomial are the fifth and sixth roots of unity. However, looking back at our expression when it had fractions, we realize that if is a second or third root of unity, it would cause a denominator to be zero, so the roots of our polynomial are the fifth and sixth roots of unity that are not the second or third roots of unity. The only roots in this category are , so our desired sum is , and we are done.
-coolak
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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