Difference between revisions of "2016 AMC 12A Problems/Problem 24"

Revision as of 14:33, 19 July 2023

Problem

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$ ?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution 1 (calculus)

The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value. Derive $f(x)$ so that the acceleration $f''(x)=0$ : $f(x)=x^3-ax^2+bx-a\rightarrow f’(x)=3x^2-2ax+b\rightarrow f’’(x)=6x-2a\rightarrow x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).

The function with the minimum $a$ :

$f(x)=\left(x-\frac{a}{3}\right)^3$ $x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}$ Since this is equal to the original equation $x^3-ax^2+bx-a$ ,

$\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}$ $b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}$

The actual function: $f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$

$f(x)=0\rightarrow x=\sqrt{3}$ triple root. "Complete the cube."

Solution 2

Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a=rst$ and $rs+st+rt=b$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get $\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.$ Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields $a^2 \ge 27 \quad \Rightarrow \quad a \ge 3\sqrt{3}.$ Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{\textbf{(B) }9}$ .

Video Solution by the Art of Problem Solving

https://www.youtube.com/watch?v=OkI1HDEj2B8&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=4

- AMBRIGGS

@@ Line 10: / Line 10: @@
 The acceleration must be zero at the <math>x</math>-intercept; this intercept must be an inflection point for the minimum <math>a</math> value.
 Derive <math>f(x)</math> so that the acceleration <math>f''(x)=0</math>:
-<math>x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}</math> for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at <math>x=a/3</math> (if the slope is greater than zero, there will be two complex roots and we do not want that).
+<math>f(x)=x^3-ax^2+bx-a\rightarrow f’(x)=3x^2-2ax+b\rightarrow f’’(x)=6x-2a\rightarrow x=\frac{a}{3}</math> for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at <math>x=a/3</math> (if the slope is greater than zero, there will be two complex roots and we do not want that).
 The function with the minimum <math>a</math>:

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

Page

Toolbox

Search