AMC12 2016 A

AMC12 2016 A · Q24

AMC12 2016 A · Q24. It mainly tests Quadratic equations, Inequalities (AM-GM etc. basic).

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 $b$?

存在一个最小的正实数 $a$，使得存在一个正实数 $b$，从而多项式 $x^3-ax^2+bx-a$ 的所有根都是实数。事实上，对于这个 $a$ 的取值，$b$ 的取值是唯一的。求这个 $b$ 的值。

(A) 8 8

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because $a$ and $b$ are positive, all the roots must be positive. Let the roots be $r$, $s$, and $t$. Then $x^3-ax^2+bx-a=(x-r)(x-s)(x-t)=x^3-(r+s+t)x^2+(rs+st+tr)x-rst.$ Therefore $r+s+t=a=rst$. The Arithmetic Mean–Geometric Mean Inequality implies that $27rst\le (r+s+t)^3=(rst)^3$, from which $a=rst\ge 3\sqrt{3}$. Furthermore, equality is achieved if and only if $r=s=t=\sqrt{3}$. In this case $b=rs+st+tr=9$.

答案（B）：因为 $a$ 和 $b$ 为正，所以所有根都必须为正。设根为 $r$、$s$、$t$。则 $x^3-ax^2+bx-a=(x-r)(x-s)(x-t)=x^3-(r+s+t)x^2+(rs+st+tr)x-rst.$ 因此 $r+s+t=a=rst$。算术-几何平均不等式给出 $27rst\le (r+s+t)^3=(rst)^3$，从而 $a=rst\ge 3\sqrt{3}$。并且当且仅当 $r=s=t=\sqrt{3}$ 时取等号。此时 $b=rs+st+tr=9$。

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