2016 AMC 12A Problems/Problem 24

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 $f''(x)=0$ . Using the power rule, $\begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*}$ So $x=\frac{a}{3}$ 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 $x^3-ax^2+bx-a$ , equating the coefficients, we get that

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

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 $r+s+t=a$ , $rs+st+rt=b$ , and $rst=a$ . 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 $b=\boxed{\textbf{(B) }9}$ .

Solution 3 (SUPER RISKY) do if you have no time

We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{\textbf{(B) }9}$ .

Video Solution by the Art of Problem Solving

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

- AMBRIGGS

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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