AMC12 2022 A

AMC12 2022 A · Q22

AMC12 2022 A · Q22. It mainly tests Inequalities (AM-GM etc. basic), Vieta / quadratic relationships (basic).

Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$. Points $z_1$, $z_2$, $\frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?

设 $c$ 是一个实数，$z_1$ 和 $z_2$ 是满足方程 $z^2 - c z + 10 = 0$ 的两个复数。点 $z_1$、$z_2$、$\frac{1}{z_1}$ 和 $\frac{1}{z_2}$ 是复平面中（凸）四边形 $\mathcal{Q}$ 的顶点。当 $\mathcal{Q}$ 的面积取得最大可能值时，$c$ 最接近以下哪个值？

(A) 4.5 4.5

(B) 5 5

(D) 6 6

(E) 6.5 6.5

Answer

Correct choice: (A)

正确答案：（A）

Solution

Because $c$ is real, $z_2 = \bar z_1$. We have \begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*} where the first equality follows from Vieta's formula. Thus, $|z_1| = \sqrt{10}$. We have \begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*} where the first equality follows from Vieta's formula. Thus, ${\rm Re}(z_1) = \frac{c}{2}$. We have \begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*} where the second equality follows from Vieta's formula. We have \begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*} where the second equality follows from Vieta's formula. Therefore, \begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*} where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$. Thus, $|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}$.

因为 $c$ 是实数，$z_2 = \bar z_1$。我们有 \begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*} 其中第一个等式由 Vieta 公式得。因此，$|z_1| = \sqrt{10}$。我们有 \begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*} 其中第一个等式由 Vieta 公式得。因此，${\rm Re}(z_1) = \frac{c}{2}$。我们有 \begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*} 其中第二个等式由 Vieta 公式得。我们有 \begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*} 其中第二个等式由 Vieta 公式得。因此， \begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*} 其中不等式由 AM-GM 不等式得，且等号当且仅当 $c^2 = 40 - c^2$ 时成立。因此，$|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}$。

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