Difference between revisions of "2016 AMC 12A Problems/Problem 24"
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All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is <math>\sqrt{3}</math>. | All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is <math>\sqrt{3}</math>. | ||
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 <math>b=\boxed{\textbf{(B) }9}</math>. | 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 <math>b=\boxed{\textbf{(B) }9}</math>. | ||
+ | (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== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:50, 23 December 2020
Contents
[hide]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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.