Difference between revisions of "2016 AMC 12A Problems/Problem 24"
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Thus, the smallest possible value of <math>a</math> is <math>3\sqrt{3}</math> which is achieved when all the roots are equal to <math>\sqrt{3}</math>. For this value of <math>a</math>, we can use Vieta's to get <math>b=\boxed{\textbf{(B) }9}</math>. | Thus, the smallest possible value of <math>a</math> is <math>3\sqrt{3}</math> which is achieved when all the roots are equal to <math>\sqrt{3}</math>. For this value of <math>a</math>, we can use Vieta's to get <math>b=\boxed{\textbf{(B) }9}</math>. | ||
− | ==Solution 3 (SUPER RISKY)== | + | ==Solution 3 (SUPER RISKY) do if you have no time == |
We see that with cubics, the number <math>3</math> comes up a lot, and as <math>9=3\cdot3</math> has the most relation to <math>3</math>, we can assume <math>b=\boxed{\textbf{(B) }9}</math>. | We see that with cubics, the number <math>3</math> comes up a lot, and as <math>9=3\cdot3</math> has the most relation to <math>3</math>, we can assume <math>b=\boxed{\textbf{(B) }9}</math>. | ||
Revision as of 12:14, 2 November 2023
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
. Using the power rule,
So
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
, equating the coefficients, we get that
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
.
Solution 3 (SUPER RISKY) do if you have no time
We see that with cubics, the number comes up a lot, and as
has the most relation to
, we can assume
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.