2020 AIME I Problems/Problem 14
Contents
Problem
Let be a quadratic polynomial with complex coefficients whose coefficient is Suppose the equation has four distinct solutions, Find the sum of all possible values of
Solution 1
Either or not. We first see that if it's easy to obtain by Vieta's that . Now, take and WLOG . Now, consider the parabola formed by the graph of . It has vertex . Now, say that . We note . Now, we note by plugging in again. Now, it's easy to find that , yielding a value of . Finally, we add . ~awang11, charmander3333
Remark: We know that from .
Solution 2
Let the roots of be and , then we can write . The fact that has solutions implies that some combination of of these are the solution to , and the other are the solution to . It's fairly easy to see there are only possible such groupings: and , or and (Note that are interchangeable, and so are and ). We now casework: If , then so this gives . Next, if , then Subtracting the first part of the first equation from the first part of the second equation gives Hence, , and so . Therefore, the solution is ~ktong
Solution 3
Write . Split the problem into two cases: and .
Case 1: We have . We must have Rearrange and divide through by to obtain Now, note that Now, rearrange to get and thus Substituting this into our equation for yields . Then, it is clear that does not have a double root at , so we must have and or vice versa. This gives and or vice versa, implying that and .
Case 2: We have . Then, we must have . It is clear that (we would otherwise get implying or vice versa), so and .
Thus, our final answer is . ~GeronimoStilton
Solution 4
Let . There are two cases: in the first case, equals (without loss of generality), and thus . By Vieta's formulas .
In the second case, say without loss of generality and . Subtracting gives , so . From this, we have .
Note , so by Vieta's, we have . In this case, .
The requested sum is .~TheUltimate123
Solution 5 (Official MAA)
Note that because , and are roots of . There are two cases. CASE 1: . Then is symmetric about ; that is to say, for all . Thus the remaining two roots must sum to . Indeed, the polynomials satisfy the conditions. CASE 2: . Then and are the two distinct roots of , sofor all . Note that any solution to must satisfy either or . Because is quadratic, the polynomials and each have the same sum of roots as the polynomial , which is . Thus the answer in this case is , and so it suffices to compute the value of .
Let and . Substituting and into the above quadratic polynomial yields the system of equations Subtracting the first equation from the second gives , yielding Substituting this value into the second equation givesyielding The sum of the two solutions is . In this case, .
The requested sum of squares is .
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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