2020 AIME I Problems/Problem 14
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
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 .
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
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
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 .
Let for some , .
Then, we can write . Setting the expression equal to and solving for gives:
Therefore, we have that takes on the four values , , , and . Two of these values are and , and the other two are and .
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.
: Both the 3 and 4 values are from the same group.
In this case, the and values are both from the other group. The sum of this is just because the radical cancels out. Because of this, we can see that is just the average of and , so we have , so .
: The 3 and 4 values come from different groups.
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for . Without loss of generality, assume that and . Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.
We can write .
Moving most terms to the left side and simplifying gives .
We can square both sides and simplify:
Squaring both sides again gives the following:
Nearly all terms cancel out, yielding .
By substituting this back in, we obtain and .
The sum of and is equal to , so .
Adding up both values gives as our final answer.
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