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

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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

Solution 1

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 $f''(x)=0$ : $x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}$ 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

Note that since both $a$ and $b$ are positive, all 3 roots of the polynomial are positive as well.

Let the roots of the polynomial be $r, s, t$ . By Vieta's $a=r+s+t$ and $a=rst$ .

Since $r, s, t$ are positive we can apply AM-GM to get $\frac{r+s+t}{3} \ge \sqrt[3]{rst} \rightarrow \frac{a}{3} \ge \sqrt[3]{a}$ . 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 $\frac{a^2}{27} \ge 1 \rightarrow a \ge 3\sqrt{3}$ .

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 $a$ , $b=\boxed{\textbf{(B) }9}$ .

@@ Line 6: / Line 6: @@
 ==Solution==
+===Solution 1===
 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>:
@@ Line 23: / Line 25: @@
 <math>f(x)=0\rightarrow x=\sqrt{3}</math> triple root. "Complete the cube."
+===Solution 2===
+Note that since both <math>a</math> and <math>b</math> are positive, all 3 roots of the polynomial are positive as well.
+Let the roots of the polynomial be <math>r, s, t</math>. By Vieta's <math>a=r+s+t</math> and <math>a=rst</math>.
+Since <math>r, s, t</math> are positive we can apply AM-GM to get <math>\frac{r+s+t}{3} \ge \sqrt[3]{rst} \rightarrow \frac{a}{3} \ge \sqrt[3]{a}</math>. Cubing both sides and then dividing by <math>a</math> (since <math>a</math> is positive we can divide by <math>a</math> and not change the sign of the inequality) yields <math>\frac{a^2}{27} \ge 1 \rightarrow a \ge 3\sqrt{3}</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>, <math>b=\boxed{\textbf{(B) }9}</math>.
 ==See Also==
 {{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}}
 {{MAA Notice}}

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