AMC12 2013 B

AMC12 2013 B · Q17

AMC12 2013 B · Q17. It mainly tests Linear equations, Inequalities (AM-GM etc. basic).

Let $a$, $b$, and $c$ be real numbers such that $$ \begin{cases} a+b+c=2, \text{ and}\\ a^2+b^2+c^2=12. \end{cases} $$ What is the difference between the maximum and minimum possible values of $c$?

设 $a$、$b$、$c$ 为实数，满足 $$ \begin{cases} a+b+c=2，\text{且}\\ a^2+b^2+c^2=12。 \end{cases} $$ 求 $c$ 的最大可能值与最小可能值之差是多少？

(A) 2 2

(B) \frac{10}{3} \frac{10}{3}

(D) \frac{16}{3} \frac{16}{3}

(E) \frac{20}{3} \frac{20}{3}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): From the equations, $a+b=2-c$ and $a^2+b^2=12-c^2$. Let $x$ be an arbitrary real number, then $(x-a)^2+(x-b)^2\ge 0$; that is, $2x^2-2(a+b)x+(a^2+b^2)\ge 0$. Thus $2x^2-2(2-c)x+(12-c^2)\ge 0$ for all real values $x$. That means the discriminant $4(2-c)^2-4\cdot 2(12-c^2)\le 0$. Simplifying and factoring gives $(3c-10)(c+2)\le 0$. So the range of values of $c$ is $-2\le c\le \frac{10}{3}$. Both maximum and minimum are attainable by letting $(a,b,c)=(2,2,-2)$ and $(a,b,c)=(-\frac{2}{3},-\frac{2}{3},\frac{10}{3})$. Therefore the difference between the maximum and minimum possible values of $c$ is $\frac{10}{3}-(-2)=\frac{16}{3}$.

答案（D）：由方程可得，$a+b=2-c$ 且 $a^2+b^2=12-c^2$。令 $x$ 为任意实数，则 $(x-a)^2+(x-b)^2\ge 0$；即 $2x^2-2(a+b)x+(a^2+b^2)\ge 0$。因此 $2x^2-2(2-c)x+(12-c^2)\ge 0$ 对所有实数 $x$ 都成立。这意味着判别式 $4(2-c)^2-4\cdot 2(12-c^2)\le 0$。化简并因式分解得 $(3c-10)(c+2)\le 0$。所以 $c$ 的取值范围为 $-2\le c\le \frac{10}{3}$。取 $(a,b,c)=(2,2,-2)$ 与 $(a,b,c)=(-\frac{2}{3},-\frac{2}{3},\frac{10}{3})$ 时分别可达到最小值与最大值。因此 $c$ 的最大可能值与最小可能值之差为 $\frac{10}{3}-(-2)=\frac{16}{3}$。

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