AMC12 2022 A

AMC12 2022 A · Q13

AMC12 2022 A · Q13. It mainly tests Complex numbers (rare), Geometry misc.

Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$, where $z_1$ lies on the segment with endpoints $3$ and $4i$, and $z_2$ has magnitude at most $1$. What integer is closest to the area of $\mathcal{R}$?

在复平面上，区域 $\mathcal{R}$ 由所有复数 $z$ 组成，这些 $z$ 可以写成复数 $z_1$ 和 $z_2$ 的和，其中 $z_1$ 位于端点为 $3$ 和 $4i$ 的线段上，且 $z_2$ 的模不超过 $1$。$\mathcal{R}$ 的面积最接近的整数是多少？

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (A)

正确答案：（A）

Solution

If $z$ is a complex number and $z = a + bi$, then the magnitude (length) of $z$ is $\sqrt{a^2 + b^2}$. Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$, the bounds of the region $\mathcal{R}$ could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of $\pi \approx 3$. Therefore, the total area is $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$.

若 $z$ 是复数且 $z = a + bi$，则 $z$ 的模（长度）为 $\sqrt{a^2 + b^2}$。因此，$z_1$ 的模为 $5$。若 $z_2$ 的模至多为 $1$，则对于 $z_1$ 给定的线段上的每个点，区域 $\mathcal{R}$ 的边界至多距离 $1$。沿线段（不含端点），形成一个宽度为 $2$、长度为 $5$（模长）的矩形。在端点处，形成两个半径为 $1$ 的半圆，总面积为 $\pi \approx 3$。因此，总面积为 $5\times2 + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$。

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