2016 AMC 12A Problems/Problem 24
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
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
Note that since both and are positive, all 3 roots of the polynomial are positive as well.
Let the roots of the polynomial be . By Vieta's and .
Since are positive 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 .
Solution 3
All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is . Using Vieta's, the coefficient we desire is the sum of the pairwise products of the roots. Because our root is unique, the answer is simply . (note that this is only true since for the min value of a, applying AM-GM to the sums and products of roots, equality condition produced min value of a )
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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