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

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==Problem==
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There is a smallest positive real number <math>a</math> such that there exists a positive real number <math>b</math> such that all the roots of the polynomial <math>x^3-ax^2+bx-a</math> are real. In fact, for this value of <math>a</math> the value of <math>b</math> is unique. What is this value of <math>b</math>?
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<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math>
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==Solution 1 (calculus) ==
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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.
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Derive <math>f(x)</math> so that the acceleration <math>f''(x)=0</math>. Using the power rule,
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<cmath>\begin{align*}
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f(x) &= x^3-ax^2+bx-a \\
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f’(x) &= 3x^2-2ax+b \\
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f’’(x) &= 6x-2a
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\end{align*}</cmath>
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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).
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The function with the minimum <math>a</math>:
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<cmath>f(x)=\left(x-\frac{a}{3}\right)^3</cmath>
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<cmath>x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}</cmath>
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Since this is equal to the original equation <math>x^3-ax^2+bx-a</math>, equating the coefficients, we get that
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<cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath>
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<cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}</cmath>
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The actual function:
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<math>f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}</math>
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<math>f(x)=0\rightarrow x=\sqrt{3}</math> triple root. "Complete the cube."
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==Solution 2==
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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>
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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>.
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==Solution 3 (SUPER RISKY) do if you have no time ==
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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>.
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-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
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-It's also probably better to get 1.5 points for not answering than using this method. Sorry! ~slamgirls~
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==Video Solution by the Art of Problem Solving==
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https://www.youtube.com/watch?v=OkI1HDEj2B8&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=4
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- AMBRIGGS
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==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}}
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{{MAA Notice}}

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

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

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