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

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The actual function:
 
The actual function:
 
<math>f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}</math>
 
<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}\text{ triple root.}</math>
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<math>\text{Complete the cube (!)}</math>
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<math>f(x)=0\rightarrow x=\sqrt{3}</math> triple root! "Complete the cube."
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:31, 5 February 2016

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

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. 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 $x^3-ax^2+bx-a$,

\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\] \[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(C) }9}\]

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."

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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