Difference between revisions of "2020 AIME I Problems/Problem 14"
m |
(→Solution 3) |
||
Line 39: | Line 39: | ||
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton | Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>. | ||
+ | |||
+ | In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>. | ||
+ | |||
+ | Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>. | ||
+ | |||
+ | The requested sum is <math>36+49=85</math>.~TheUltimate123 | ||
==See Also== | ==See Also== |
Revision as of 16:48, 13 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.