2016 AMC 12A Problems/Problem 24
Contents
Problem
There is a smallest positive real number such that there exists a positive real number such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is this value of ?
Solution 1 (calculus)
The acceleration must be zero at the -intercept; this intercept must be an inflection point for the minimum value. Derive so that the acceleration : for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum :
Since this is equal to the original equation ,
The actual function:
triple root. "Complete the cube."
Solution 2
Let the roots of the polynomial be . By Vieta's formulas we have and . Since both and are positive, it follows that all 3 roots are positive as well, and so we can apply AM-GM to get Cubing both sides and then dividing by (since is positive we can divide by and not change the sign of the inequality) yields Thus, the smallest possible value of is which is achieved when all the roots are equal to . For this value of , we can use Vieta's to get .
Video Solution by the Art of Problem Solving
https://www.youtube.com/watch?v=OkI1HDEj2B8&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=4
- AMBRIGGS
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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