2025 AMC 12A Problems/Problem 9

Problem

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?

$\textbf{(A)}~\frac34\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\frac75\qquad\textbf{(D)}~\frac32\qquad\textbf{(E)}~\frac53$

Solution 1 (Rectangular Form)

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 $y = 3(x-2)+1$ $y = 3x-5.$ 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 $x = \boxed{\frac{5}{3}}.$

~lprado (minor edits ~Logibyte)

Solution 2 (Polar Form)

Recall that the slope of a line is $m=\tan\phi,$ where $\phi$ is the angle formed by the line and the positive $x$ -axis.

Note that $|w|=\sqrt{2^2+1^2}=\sqrt{5}.$ In polar coordinates, let $w=\sqrt{5}\operatorname{cis}\theta.$ It follows that $\tan\theta=\frac12.$

By De Moivre's Theorem, we have $w^2=5\operatorname{cis}(2\theta),$ from which $\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{4}{3}.$

We obtain the following diagram: [Will add it later].

Since $\left|w^2\right|=5,$ we have $w^2=3+4i$ by a $3$ - $4$ - $5$ triangle. The complex numbers $w$ and $w^2$ correspond to the points $(2,1)$ and $(3,4),$ respectively, and $r$ corresponds to the $x$ -intercept of this line. In slope-intercept form, the line containing these two points is $y=3x-5.$ Therefore, the $x$ -intercept is $\boxed{\textbf{(E)}~\frac53}.$

~MRENTHUSIASM

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2025 AMC 12A Problems/Problem 9

Problem

Solution 1 (Rectangular Form)

Solution 2 (Polar Form)