Difference between revisions of "2020 AIME I Problems/Problem 14"
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== Solution 1 == | == Solution 1 == | ||
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | ||
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+ | <b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>. | ||
== Solution 2 == | == Solution 2 == |
Revision as of 15:37, 29 March 2020
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 to 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
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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