2016 AMC 12A Problem 24

Below is the professionally curated solution for Problem 24 of the 2016 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12A solutions, or check the answer key.

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Concepts:Vieta’s Formulas AM-GM Inequality

Difficulty rating: 2380

24.

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 $b?$

$8$

$9$

$10$

$11$

$12$

Solution:

Since $a$ and $b$ are positive, all roots $r,s,t$ must be positive. By Vieta's formulas, $r+s+t=a,$ $rs+st+tr=b,$ and $rst=a,$ so $r+s+t=rst.$

By the AM-GM inequality, $27rst\le(r+s+t)^3=(rst)^3,$ so $a=rst\ge 3\sqrt3,$ with equality if and only if $r=s=t=\sqrt3.$ At this smallest $a,$ $b=rs+st+tr=3r^2=3\cdot 3=9.$

Thus, the correct answer is B.

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