Question

设 $x$ 、 $y$ 、 $z$ 为复数，满足 $xy = -80 - 320i$ 、 $yz = 60$ 、 $zx = -96 + 24i$ ，其中 $i = \sqrt{-1}$ 。那么存在实数 $a$ 与 $b$ ，使得 $x + y + z = a + bi$ 。求 $a^2 + b^2$ 。

Suppose that $x,$ $y,$ and $z$ are complex numbers such that $xy = -80 - 320i,$ $yz = 60,$ and $zx = -96 + 24i,$ where $i = \sqrt{-1}.$ Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi.$ Find $a^2 + b^2.$

答案

答案：74

知识点：复数方程组代数变形

难度评级：2450

解答：

将三个方程相乘，得到 $\begin{aligned} (xyz)^2 &= (-80 - 320i)(60) \\ &\quad {}\cdot (-96 + 24i) \\ &= 80 \cdot 60 \cdot 24 \\ &\quad {}\cdot \,(-1 - 4i)(-4 + i) \\ &= 240^2 (16 + 30i). \end{aligned}$ 因为 $16 + 30i = (5 + 3i)^2$ ，所以 $xyz = \pm 240(5 + 3i)$ 。

用 $xyz$ 分别除以给定的乘积，得 $\begin{aligned} x &= \frac{xyz}{yz} = \pm(20 + 12i), \\ y &= \frac{xyz}{zx} = \pm(-10 - 10i), \\ z &= \frac{xyz}{xy} = \pm(-3 + 3i), \end{aligned}$ 且符号一致。因此 $x + y + z = \pm(7 + 5i)$ ，所以 $(a, b) = \pm(7, 5)$ ，从而 $a^2 + b^2 = 49 + 25 = 74$ 。

Multiplying the three equations gives $\begin{aligned} (xyz)^2 &= (-80 - 320i)(60) \\ &\quad {}\cdot (-96 + 24i) \\ &= 80 \cdot 60 \cdot 24 \\ &\quad {}\cdot \,(-1 - 4i)(-4 + i) \\ &= 240^2 (16 + 30i). \end{aligned}$ Since $16 + 30i = (5 + 3i)^2,$ we get $xyz = \pm 240(5 + 3i).$

Dividing $xyz$ by each given product yields $\begin{aligned} x &= \frac{xyz}{yz} = \pm(20 + 12i), \\ y &= \frac{xyz}{zx} = \pm(-10 - 10i), \\ z &= \frac{xyz}{xy} = \pm(-3 + 3i), \end{aligned}$ with matching signs. Hence $x + y + z = \pm(7 + 5i),$ so $(a, b) = \pm(7, 5)$ and $a^2 + b^2 = 49 + 25 = 74.$

2018 AIME II 第 5 题

答案：74

难度评级：2450

解答：

其他年份的第 5 题