AMC12 2025 A

AMC12 2025 A · Q9

AMC12 2025 A · Q9. It mainly tests Complex numbers (rare), Coordinate geometry.

Let $w$ be the complex number $2+i$, where $i=\sqrt{-1}$. What real number $r$ has the property that $r$, $w$, and $w^2$ are three collinear points in the complex plane?

设复数$w=2+i$，其中$i=\sqrt{-1}$。求实数$r$，使得$r$、$w$和$w^2$在复平面中三点共线。

(A) \frac34 \frac34

(B) 1 1

(D) \frac32 \frac32

(E) \frac53 \frac53

Answer

Correct choice: (E)

正确答案：（E）

Solution

We begin by calculating $w^2$: \[w^2 = (2+i)^2 = 4+4i-1 = 3+4i.\] Values on the complex plane can easily be represented as points on the Cartesian plane, so we go ahead and do that so we are in a more familiar place. Translating onto the Cartesian plane, we have the points $(2,1)$ and $(3,4)$. The slope of the line passing through these points is $\frac{4-1}{3-2} = 3$, so the equation of this line is \begin{align*} y &= 3(x-2)+1 \\ y &= 3x-5. \end{align*} We want the real number that passes through this line, which is equivalent to the $x-$intercept. This occurs when $y=0$, so the $x$-intercept of this line is $x=\boxed{\textbf{(E)}~\frac53}.$

先计算$w^2$： \[w^2 = (2+i)^2 = 4+4i-1 = 3+4i.\] 复平面上的值可表示为笛卡尔平面上的点，因此我们这样表示。将点$(2,1)$和$(3,4)$放在笛卡尔平面上，通过这两点的直线斜率为$\frac{4-1}{3-2} = 3$，故直线方程为 \begin{align*} y &= 3(x-2)+1 \\ y &= 3x-5. \end{align*} 我们要求通过这条直线的实数，即$x$轴截距。当$y=0$时，$x$轴截距为$x=\boxed{\textbf{(E)}~\frac{5}{3}}$。

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