2020 AIME II Problems/Problem 11
Contents
Problem
Let , and let and be two quadratic polynomials also with the coefficient of equal to . David computes each of the three sums , , and and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If , then , where and are relatively prime positive integers. Find .
Solution 1
Let and . We can write the following: Let the common root of be ; be ; and be . We then have that the roots of are , the roots of are , and the roots of are .
By Vieta's, we have:
Subtracting from , we get . Adding this to , we get . This gives us that from . Substituting these values into and , we get and . Equating these values, we get . Thus, our answer is . ~ TopNotchMath
Solution 2
We know that .
Since , the constant term in is . Let .
Finally, let .
. Let its roots be and .
Let its roots be and .
. Let its roots be and .
By vietas,
We could work out the system of equations, but it's pretty easy to see that .
~quacker88
Solution 3 (Official MAA)
Let the common root of and be , the common root of and be , and the common root of and be . Because and are both roots of and has leading coefficient , it follows that Similarly, and . Adding these three equations together and dividing by yieldsso Similarly, Comparing the coefficients yields , and comparing the constant coefficients yields . The fact that implies that . Adding these two equations yields , and so substituting back in to solve for gives . Finally,The requested sum is . Note that and .
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM
Another one:
https://www.youtube.com/watch?v=AXN9x51KzNI
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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