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

Latest revision as of 17:36, 16 October 2024

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

-Sorry, but the number $3$ doesn't actually come up very often as you said, so this solution might not actually be a good idea. ~get-rickrolled

-It's also probably better to get 1.5 points for not answering than using this method. Sorry! ~slamgirls~

Video Solution by the Art of Problem Solving

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

- AMBRIGGS

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math>
-==Solution==
-===Solution 1===
+==Solution 1 (calculus) ==
 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>:
+Derive <math>f(x)</math> so that the acceleration <math>f''(x)=0</math>. Using the power rule,
-<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).
+<cmath>\begin{align*}
+f(x) &= x^3-ax^2+bx-a \\
+f’(x) &= 3x^2-2ax+b \\
+f’’(x) &= 6x-2a
+\end{align*}</cmath>
+So <math>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>:
@@ Line 16: / Line 21: @@
 <cmath>f(x)=\left(x-\frac{a}{3}\right)^3</cmath>
 <cmath>x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}</cmath>
-Since this is equal to the original equation <math>x^3-ax^2+bx-a</math>,
+Since this is equal to the original equation <math>x^3-ax^2+bx-a</math>, equating the coefficients, we get that
 <cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath>
@@ Line 26: / Line 31: @@
 <math>f(x)=0\rightarrow x=\sqrt{3}</math> triple root. "Complete the cube."
+==Solution 2==
-===Solution 2===
+Let the roots of the polynomial be <math>r, s, t</math>. By Vieta's formulas we have <math>r+s+t=a</math>, <math>rs+st+rt=b</math>, and <math>rst=a</math>. Since both <math>a</math> and <math>b</math> are positive, it follows that all 3 roots <math>r, s, t</math> are positive as well, and so we can apply AM-GM to get <cmath>\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.</cmath> 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 <cmath>a^2 \ge 27 \quad \Rightarrow \quad a \ge  3\sqrt{3}.</cmath>
+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>.
-Note that since both <math>a</math> and <math>b</math> are positive, all 3 roots of the polynomial are positive as well.
+==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>.
-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>.
+-Sorry, but the number <math>3</math> doesn't actually come up very often as you said, so this solution might not actually be a good idea. ~get-rickrolled
-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>.
+-It's also probably better to get 1.5 points for not answering than using this method. Sorry! ~slamgirls~
+==Video Solution by the Art of Problem Solving==
+https://www.youtube.com/watch?v=OkI1HDEj2B8&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=4
-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>.
+- AMBRIGGS
-==Solution 3==
-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>.
 ==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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