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2025 AMC 12A Problems/Problem 9: Difference between revisions

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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.  
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 <imath>(2,1)</imath> and <imath>(3,4)</imath>. The slope of the line passing through these points is <imath>\frac{4-1}{3-2} = 3</imath>, so the equation of this line is
Translating onto the Cartesian plane, we have the points <imath>(2,1)</imath> and <imath>(3,4)</imath>. The slope of the line passing through these points is <imath>\frac{4-1}{3-2} = 3</imath>, so the equation of this line is
<cmath>y = 3(x-2)+1</cmath>
<cmath>\begin{align*}
<cmath>y = 3x-5.</cmath>
y &= 3(x-2)+1 \\
We want the real number that passes through this line, which is equivalent to the <imath>x-</imath>intercept. This occurs when <imath>y=0</imath>, so the <imath>x</imath>-intercept of this line is <imath>x = \boxed{\frac{5}{3}}.</imath>
y &= 3x-5.
\end{align*}</cmath>
We want the real number that passes through this line, which is equivalent to the <imath>x-</imath>intercept. This occurs when <imath>y=0</imath>, so the <imath>x</imath>-intercept of this line is <imath>x=\boxed{\textbf{(E)}~\frac53}.</imath>


~lprado (minor edits ~Logibyte)
~lprado (minor edits ~Logibyte)
Line 22: Line 24:
By De Moivre's Theorem, we have <imath>w^2=5\operatorname{cis}(2\theta),</imath> from which <cmath>\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{4}{3}.</cmath>
By De Moivre's Theorem, we have <imath>w^2=5\operatorname{cis}(2\theta),</imath> from which <cmath>\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{4}{3}.</cmath>


We obtain the following diagram: [Will add it later].
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 <imath>\left|w^2\right|=5,</imath> we have <imath>w^2=3+4i</imath> by a <imath>3</imath>-<imath>4</imath>-<imath>5</imath> triangle. The complex numbers <imath>w</imath> and <imath>w^2</imath> correspond to the points <imath>(2,1)</imath> and <imath>(3,4),</imath> respectively, and <imath>r</imath> corresponds to the <imath>x</imath>-intercept of this line. In slope-intercept form, the line containing these two points is <imath>y=3x-5.</imath> Therefore, the <imath>x</imath>-intercept is <imath>r=\boxed{\textbf{(E)}~\frac53}.</imath>
Since <imath>\left|w^2\right|=5,</imath> we have <imath>w^2=3+4i</imath> by a <imath>3</imath>-<imath>4</imath>-<imath>5</imath> triangle. The complex numbers <imath>w</imath> and <imath>w^2</imath> correspond to the points <imath>(2,1)</imath> and <imath>(3,4),</imath> respectively, and <imath>r</imath> corresponds to the <imath>x</imath>-intercept of this line. In slope-intercept form, the line containing these two points is <imath>y=3x-5.</imath> Therefore, the <imath>x</imath>-intercept is <imath>r=\boxed{\textbf{(E)}~\frac53}.</imath>


~MRENTHUSIASM
~MRENTHUSIASM
==Video Solution by Power Solve==
https://www.youtube.com/watch?v=Oy5hEE1fmJU
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==See Also==
==See Also==

Latest revision as of 11:40, 11 November 2025

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

See Also

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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