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 $\begin{align*} y &= 3(x-2)+1 \\ y &= 3x-5. \end{align*}$ 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 $x=\boxed{\textbf{(E)}~\frac53}.$

~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: $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 5; int yMin = -2; int yMax = 5; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); pair W1, W2, R; W1 = (2,1); W2 = (3,4); R = (5/3,0); label("$w=2+i$", W1, 2*dir(W1)); label("$w^2$", W2, 2*dir(W2)); label("$r$", R, 2*dir(R-W2)); label("$\theta$", origin, 6*dir(14)); label("$\theta$", origin, 6*dir(40)); draw(W1--origin--W2); draw(W2--R,dashed); dot(W1^^W2^^R, linewidth(4)); [/asy]$ 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 $r=\boxed{\textbf{(E)}~\frac53}.$

~MRENTHUSIASM

Video Solution by Power Solve

https://www.youtube.com/watch?v=Oy5hEE1fmJU

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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