2012 AMC 12B Problems/Problem 23
Contents
Problem
Consider all polynomials of a complex variable, , where and are integers, , and the polynomial has a zero with What is the sum of all values over all the polynomials with these properties?
Solutions
Solution 1
Since is a root of , and has integer coefficients, must be algebraic. Since is algebraic and lies on the unit circle, must be a root of unity (Comment: this is not true. See this link: [1]). Since has degree 4, it seems reasonable (and we will assume this only temporarily) that must be a 2nd, 3rd, or 4th root of unity. These are among the set . Since complex roots of polynomials come in conjugate pairs, we have that has one (or more) of the following factors: , , , or . If then ; a contradiction since are non-negative. On the other hand, suppose . Then . This implies while correspondingly. After listing cases, the only such valid are , , , , and .
Now suppose . Then whereupon and . But then and . This gives only the cases equals , which we have already counted in a previous case.
Suppose . Then so that and . This only gives rise to equal which we have previously counted.
Finally suppose divides . Using polynomial division ((or that to make the same deductions) we ultimately obtain that . This can only happen if is .
Hence we've the polynomials However, by inspection has roots on the unit circle, because which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that in an th root of unity where , and is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If is prime, then \textit{every} th root of unity except 1 must satisfy our polynomial, but since and the degree of our polynomial is 4, this is impossible. Suppose is composite. If it has a prime factor greater than 5 then again every th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose is divisible only by 2,3,or 5. Since by hypothesis is not a 2nd or 3rd root of unity, must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy . But has exactly all 5th roots of unity excluding 1, and . Thus this must divide which implies . This completes the proof.
Solution 2
First, assume that , so or . does not work because . Assume that . Then , we have , so . Also, has to be true since . Now gives , therefore the only possible choices for are . In these cases, . The sum of over these cases is .
Second, assume that , so for some real , . By conjugate roots theorem we have that , therefore is a factor of , and we may assume that
for some real . Expanding this polynomial and comparing the coefficients, we have the following equations:
From the first and the third we may deduce that and that , if (we will consider by the end). Let . From the second equation, we know that is non-negative.
Consider the following cases:
Case 1: . Then , , so , . However, this has already been found (i.e. the form of ).
Case 2: . Then since , we have . However, , therefore . This is true only when . Also, we get again. In this case, , so , , . has a root . .
Last case: . We have and that has a root . .
Therefore the desired sum is .
Solution 3
First, notice that cannot be a root of the polynomial because . Multiplying the polynomial by yields , so for to be a root of , . Now we consider the root with . , so the right hand side must have absolute value 4. By the triangle inequality, , with equality if and only if each of , , , , , and is either zero or in the same direction as all the others when looked at as vectors in the complex plane.
We can now divide into two cases: and . If , then must be real by the previous argument, so is a fifth root of unity. Also, , , , and must all be zero because if is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, , , , and , leading to the solution . Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.
If , then each of , , , , and must either be zero or in the same direction as all the others, so each of , , , , and must either be zero or in the same direction as all the others. We can divide this into two cases: and . If , then must be real. Then, is a fourth root of unity. If is not a second root of unity, , , and must be zero, implying that , , and , leading to the solution . If is also a second root of unity, and must be zero but can be anything. This implies and with no other restrictions, leading to the new solutions .
If , then we can similarly show that each of , , , and must be zero or in the same direction as all the others. If , then must be a third root of unity, so and must be zero, implying , leading to the new solution .
If , then we can similarly show that each of , , and must be zero or in the same direction as the others. For , , but we have already counted the solution .
Then, the complete list of solutions is , leading to a sum of .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12b/278
~dolphin7
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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