# 2021 Fall AMC 12A Problems/Problem 13

## Problem

The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y=3x$ has equation $y=kx.$ What is $k?$

$\textbf{(A)} \ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)} \ \frac{1+\sqrt{7}}{2} \qquad \textbf{(C)} \ \frac{2+\sqrt{3}}{2} \qquad \textbf{(D)} \ 2\qquad \textbf{(E)} \ \frac{2+\sqrt{5}}{2}$

## Diagram

$[asy] /* Made by MRENTHUSIASM */ size(250); real xMin = -1; real xMax = 4; real yMin = -1; real yMax = 4; real k = (1+sqrt(5))/2; pair O; O = origin; draw(anglemark(dir((1,1)),O,dir((1,k)),20), red); draw(anglemark(dir((1,k)),O,dir((1,3)),20), red); add(pathticks(anglemark(dir((1,1)),O,dir((1,k)),20), n = 1, r = 0.05, s = 5, red)); add(pathticks(anglemark(dir((1,k)),O,dir((1,3)),20), n = 1, r = 0.05, s = 5, red)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); label("y=x",4*dir((1,1))); label("y=3x",4*dir((1,3))); label("y=kx",4*dir((1,k))); draw(O--3.75*dir((1,1))^^O--3.75*dir((1,3))^^O--3.75*dir((1,k))); [/asy]$ ~MRENTHUSIASM

## Solution 1 (Angle Bisector Theorem)

This solution refers to the Diagram section.

Let $O=(0,0), A=(3,3), B=(1,3),$ and $C=\left(\frac3k,3\right).$ As shown below, note that $\overline{OA}, \overline{OB},$ and $\overline{OC}$ are on the lines $y=x, y=3x,$ and $y=kx,$ respectively. By the Distance Formula, we have $OA=3\sqrt2, OB=\sqrt{10}, AC=3-\frac3k,$ and $BC=\frac3k-1.$ $[asy] /* Made by MRENTHUSIASM */ size(250); real xMin = -1; real xMax = 4; real yMin = -1; real yMax = 4; real k = (1+sqrt(5))/2; pair O, A, B, C; O = origin; A = (3,3); B = (1,3); C = (3/k,3); draw(anglemark(dir((1,1)),O,dir((1,k)),20), red); draw(anglemark(dir((1,k)),O,dir((1,3)),20), red); dot("O",O,1.5*SW,linewidth(4.5)); dot("A",A,1.5*N,linewidth(4.5)); dot("B",B,1.5*N,linewidth(4.5)); dot("C",C,1.5*N,linewidth(4.5)); add(pathticks(anglemark(dir((1,1)),O,dir((1,k)),20), n = 1, r = 0.05, s = 5, red)); add(pathticks(anglemark(dir((1,k)),O,dir((1,3)),20), n = 1, r = 0.05, s = 5, red)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); draw(A--B--O--cycle^^O--C); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); label("3\sqrt{2}",midpoint(O--A),1.5*E,red+fontsize(10)); label("\sqrt{10}",midpoint(O--B),W,red+fontsize(10)); label("3-\frac3k",midpoint(A--C),N,red+fontsize(10)); label("\frac3k-1",midpoint(B--C),N,red+fontsize(10)); [/asy]$ By the Angle Bisector Theorem, we get $\frac{OA}{OB}=\frac{AC}{BC},$ or \begin{align*} \frac{3\sqrt2}{\sqrt{10}}&=\frac{3-\frac3k}{\frac3k-1} \\ \frac{3\sqrt2}{\sqrt{10}}&=\frac{3k-3}{3-k} \\ \frac{\sqrt2}{\sqrt{10}}&=\frac{k-1}{3-k} \\ \frac15&=\frac{(k-1)^2}{(3-k)^2} \\ 5(k-1)^2&=(3-k)^2 \\ 4k^2-4k-4&=0 \\ k^2-k-1&=0 \\ k&=\frac{1\pm\sqrt5}{2}. \end{align*} Since $k>0,$ the answer is $k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$

Remark

The value of $k$ is known as the Golden Ratio: $\phi=\frac{1+\sqrt{5}}{2}\approx 1.61803398875.$

~MRENTHUSIASM

## Solution 2 (Analytic and Plane Geometry)

$[asy] size(180); real xMin = -0.5; real xMax = 2; real yMin = -0.5; real yMax = 4.5; real k = (1+sqrt(5))/2; real m = sqrt(2); real n = sqrt(10); real q = sqrt((5+sqrt(5))/2); pair O; O = origin; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("O",(-0.2,-0.2),(0,0)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); label("A",(1,0.95),(1,1)); label("B",(1,2.80),(1,3)); label("C",(1.06,k-0.05),(1,k)); draw(O--m*dir((1,1))^^O--n*dir((1,3))^^O--q*dir((1,k))); draw((1,1)--(1,3)); [/asy]$ Consider the graphs of $f(x)=x$ and $g(x)=3x$. Since it will be easier to consider at unity, let $x=1$, then we have $f(1)=1$ and $g(1)=3$.

Now, let $O$ be $(0,0)$, $A$ be $(1,1)$, and $B$ be $(1,3)$. Cutting through side $AB$ of triangle $OAB$ is the angle bisector $OC$ where $C$ is on side $AB$.

Hence, by the Angle Bisector Theorem, we get $\frac{OB}{OA}=\frac{BC}{AC}$.

By the Pythagorean Theorem, $OA=\sqrt{2}$ and $OB=\sqrt{10}$. Therefore, $\frac{BC}{AC}=\sqrt{5} \implies BC=\sqrt{5}AC$.

Since $AB=AC+BC=2$, it is easy derive $AC+\sqrt{5}AC=2 \implies AC=\frac{2}{1+\sqrt{5}}=\frac{-1+\sqrt{5}}{2}$.

The vertical distance between the $x$-axis and $C$ is $\frac{-1+\sqrt{5}}{2}+1=\frac{1+\sqrt{5}}{2}$. Because the $x$-coordinate of point $C$ is $1$, the slope we need to find is just the $y$-coordinate $\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$

~Wilhelm Z

## Solution 3 (Analytic and Plane Geometry)

Let's begin by drawing a triangle that starts at the origin. Assume that the base of the triangle goes to the point $x = 1$. The line $x = y$ is the hypotenuse of a right triangle with side length $1$. The hypotenuse' length is $\sqrt 2$. Then, let's draw the line $x = 3y$. We extend it to when $x = 1$. The length of the hypotenuse of the larger triangle is $\sqrt {10}$ with legs $1, 3$. We then draw the angle bisector. We should label the triangle, so here we go. $AC$ is $1$. $BC$ is $3$. $AB$ is $\sqrt {10}$. When the line with angle $45 ^\circ$ intersects the line $x = 1$, call the point $D$. When the angle bisector intersects the line $x = 1$, call the point $E$. By Angle Bisector Theorem, $\frac {DE}{DB} = \frac {\sqrt {2}}{\sqrt{10}}$. Since $BC$ is $3$ and $DC$ is $1$, we have that $BD$ is $2$. Solving for $DE$, we get that $DE$ is $\frac {\sqrt 5 - 1}{2}$.

Since $DE$ is $\frac {\sqrt 5 - 1}{2}$, we have that $CE$ is just one more than that. Therefore, $CE$ is $\frac {1+\sqrt 5}{2}$. Since $AC$ is $1$, we get that $k$ is $\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}$.

Remark

The answer turns out to be the golden ratio or phi ($\phi$). Phi has many properties and is related to the Fibonacci sequence. See Phi.

~Arcticturn $\blacksquare$

## Solution 4 (Distance Between a Point and a Line)

Note that the distance between the point $(m,n)$ to line $Ax + By + C = 0,$ is $\frac{|Am + Bn +C|}{\sqrt{A^2 +B^2}}.$ Because line $y=kx$ is a perpendicular bisector, a point on the line $y=kx$ must be equidistant from the two lines($y=x$ and $y=3x$), call this point $P(z,w).$ Because, the line $y=kx$ passes through the origin, our requested value of $k,$ which is the slope of the angle bisector line, can be found when evaluating the value of $\frac{w}{z}.$ By the Distance from Point to Line formula we get the equation, $$\frac{|3z-w|}{\sqrt{10}} = \frac{|z-w|}{\sqrt{2}}.$$ Note that $|3z-w|\ge 0,$ because $y=3x$ is higher than $P$ and $|z-w|\le 0,$ because $y=x$ is lower to $P.$ Thus, we solve the equation, $$(3z-w)\sqrt{2} = (w-z)\sqrt{10} \Rightarrow 3z-w = \sqrt{5} \cdot(w-z)\Rightarrow (\sqrt{5} +1)w = (3+\sqrt{5})z.$$ Thus, the value of $\frac{w}{z} = \frac{3+\sqrt{5}}{1+\sqrt{5}} = \frac{1+\sqrt{5}}{2}.$ Thus, the answer is $\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$

(Fun Fact: The value $\frac{1+\sqrt{5}}{2}$ is the golden ratio $\phi.$)

~NH14

## Solution 5 (Trigonometry)

This problem can be trivialized using basic trig identities. Let the angle made by $y=x$ and the $x$-axis be $\theta_{1}$ and the angle made by $y=3x$ and the $x$-axis be $\theta_{3}$. Note that $\tan(\theta_{1})=1$ and $\tan(\theta_{3})=3$, and this is why we named them as such. Let the angle made by $y=kx$ be denoted as $\theta_{k}$. Since $y=kx$ bisects the two lines, notice that $$\theta_k-\theta_1=\theta_3-\theta_k.$$

Now, we can take the tangent and apply the tangent subtraction formula: \begin{align*} \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\ \frac{\tan(\theta_k)-\tan(\theta_1)}{1+\tan(\theta_k)\tan(\theta_1)}&=\frac{\tan(\theta_3)-\tan(\theta_k)}{1+\tan(\theta_3)\tan(\theta_k)}\\ \frac{k-1}{1+k}&=\frac{3-k}{1+3k}\\ (k-1)(1+3k)&=(1+k)(3-k)\\ 3k^2-2k-1&=-k^2+2k+3\\ 4k^2-4k-4&=0\\ k^2-k-1&=0\\ \implies k&=\frac{1\pm \sqrt{5}}{2} \end{align*} Since $y=kx$ is increasing, $k>0$; thus, $k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$

~Indiiiigo

## Solution 6 (Trigonometry)

Denote by $\alpha_1$, $\alpha_2$, $\alpha_3$ the acute angles formed between the $x$-axis and lines $y = x$, $y = 3 x$, $y = k x$, respectively. Hence, $\tan \alpha_1 = 1$, $\tan \alpha_2 = 3$, $\tan \alpha_3 = k$.

Denote by $\theta$ the acute angle formed by lines $y = x$ and $y = 3 x$.

Hence, \begin{align*} \tan \theta & = \tan \left( \alpha_2 - \alpha_1 \right) \\ & = \frac{\tan \alpha_2 - \tan \alpha_1}{1 + \tan \alpha_1 \tan \alpha_2} \\ &= \frac{3 - 1}{1 + 1 \cdot 3} \\ & = \frac{1}{2} . \end{align*}

Following from the double-angle identity, we have $$\tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} .$$

Hence, $\tan \frac{\theta}{2} = - 2 \pm \sqrt{5}$.

Because $\theta$ is acute, $\frac{\theta}{2}$ is acute. Hence, $\tan \frac{\theta}{2} > 0$. Hence, $\tan \frac{\theta}{2} = - 2 + \sqrt{5}$.

Because line $y = kx$ is the angle bisector of $\theta$, the angle between lines $y = x$ and $y = k x$ is $\frac{\theta}{2}$.

Hence, \begin{align*} \tan \alpha_3 & = \tan \left( \alpha_1 + \frac{\theta}{2} \right) \\ & = \frac{\tan \alpha_1 + \tan \frac{\theta}{2}}{1 - \tan \alpha_1 \tan \frac{\theta}{2}} \\ &= \frac{1 + \left( - 2 + \sqrt{5} \right)}{1 - 1 \cdot \left( - 2 + \sqrt{5} \right) } \\ & = \frac{\sqrt{5} - 1}{3 - \sqrt{5}} \\ & = \frac{1 + \sqrt{5}}{2} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(A) }\frac{1 + \sqrt{5}}{2}}$.

~Steven Chen (www.professorchenedu.com)

## Solution 7 (Vectors)

When drawing the lines $y=x$ and $y=3x$, it is natural to choose points $A(1,1)$ and $B(1,3)$ together with origin $O$. See the figure attached. We utilize the fact that if $\mathbf{u}$ and $\mathbf{v}$ are vectors of same length, then $\mathbf{u} + \mathbf{v}$ bisects the angle between $\mathbf{u}$ and $\mathbf{v}$.

In particular, we scale the vector $\overrightarrow{OA} = (1,1)$ by the factor of $\frac{OB}{OA} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5}$ to get $\overrightarrow{OA'} \coloneqq \sqrt{5}\,\overrightarrow{OA} = \left(\sqrt{5}, \sqrt{5}\right)$. So by adding vectors $\overrightarrow{OA'}$ and $\overrightarrow{OB} = (1,3)$ we get $${\color[rgb]{0.666667,0,0}\overrightarrow{OC}} \coloneqq {\color[rgb]{0,0.4,0.65}\overrightarrow{OA'}} + {\color[rgb]{0,0.4,0.65}\overrightarrow{OB}} = \left( 1 + \sqrt{5}, 3 + \sqrt{5} \right)$$ which bisects the acute angle formed by lines $OA: y = x$ and $OB: y = 3x$. (In other words, quadrilateral $OBCA'$ is a rhombus.) Finally, observe that $C\!\left(1+\sqrt{5}, 3+\sqrt{5}\right)$ lies on the line $y = kx$ whose slope is $$k = \frac{3+\sqrt{5}}{1+\sqrt{5}} = \frac{1+\sqrt{5}}{2}.$$ Thus, the answer is $\boxed{\textbf{(A)}\;\frac{1+\sqrt{5}}{2}}$. $\blacksquare$

~VensL.

## Solution 8 (Dot Product)

We notice that the line $y = x$ can be represented as the vector $\vec{A} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $y = 3x$ as $\vec{B} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ as the "slope" of both vectors represent the coefficient of $x$.

Then, we can represent $y = kx$ as $\vec{C} = \begin{bmatrix} 1 \\ k \end{bmatrix}$ and notice that since $C$ is in essence an angle bisector, $$\theta = \angle \vec{A}\vec{C} = \angle \vec{B}\vec{C}$$

$$A \cdot C = |A|\cdot|C|\cos(\theta_1)$$ where $\theta_1 = \angle \vec{A}\vec{C} = \theta$ $$B \cdot C = |B|\cdot|C|\cos(\theta_2)$$ where $\theta_2 = \angle \vec{B}\vec{C} = \theta$

Since both $\theta_i$'s are equivalent, we may simply represent them with $\theta$.

Simplifying both equations by performing the necessary operations, we get

$$1+k = \sqrt{2} \cdot \sqrt{k^2 + 1} \cdot \cos(\theta) \implies \frac{1+k}{\sqrt{2}} = \sqrt{k^2 + 1} \cos(\theta)$$

$$1+3k = \sqrt{10} \cdot \sqrt{k^2 + 1} \cdot \cos(\theta)$$

Substituting the first into the second, we get $$1 + 3k = \sqrt{10} \cdot \frac{1+k}{\sqrt{2}} \implies k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}$$

~ $\color{magenta} zoomanTV$

## Solution 9

$[asy] size(250); real xMin = -1; real xMax = 4; real yMin = -1; real yMax = 4; real k = (1+sqrt(5))/2; pair O; O = origin; draw(anglemark(dir((1,1)),O,dir((1,k)),20), red); draw(anglemark(dir((1,k)),O,dir((1,3)),20), red); add(pathticks(anglemark(dir((1,1)),O,dir((1,k)),20), n = 1, r = 0.05, s = 5, red)); add(pathticks(anglemark(dir((1,k)),O,dir((1,3)),20), n = 1, r = 0.05, s = 5, red)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); label("y=x",4*dir((1,1))); label("y=3x",4*dir((1,3))); label("y=kx",4*dir((1,k))); draw((1, 3)--(2, 2)); draw((2, 2)--(2, 0)); draw((2, 0)--(2, 0.2)--(1.8, 0.2)--(1.8, 0)); draw((2, 2)--(1.9, 2.1)--(1.8, 2)--(1.9, 1.9)); label("A", (2, 2), E); label("(a, a)", (2.3, 2), E); label("B", (1.5, 2.5), E); label("(b, 3b)", (0.9, 3), SW); label("C", (1, 3), NW); label("D", (0, 0), NW); label("E", (2, 0), NE); draw(O--3.75*dir((1,1))^^O--3.75*dir((1,3))^^O--3.75*dir((1,k))); [/asy]$

Refer to the modified version of Solution 1's diagram, seen above.

Label $A = (a, a)$ and $B = (b, 3b)$, by definition of a point on a line.

The slope of the line through $A$ and $C$ is $-1$, as it is perpendicular to $y=x.$ Therefore, we have $\frac{3b-a}{b-a} = -1,$ so $2b=a.$

We label $A = (2b, 2b)$ to simplify. By $45$-$45$-$90$ triangle relations (or distance formula), $AD = 2b\sqrt{2}.$

By the distance formula, $AC = b\sqrt{2},$ and by the Pythagorean Theorem, $CD = b\sqrt{10}.$

From the angle bisector theorem,

$$\frac{AD}{CD} = \frac{AB}{BC}$$ $$\frac{2b\sqrt{2}}{b\sqrt{10}} = \frac{AB}{BC}.$$ $$\frac{2b\sqrt{5}}{5} BC = AB.$$

Note that $AB + BC = b\sqrt{10},$ so:

$$BC + \frac{2b\sqrt{5}}{5} BC = b\sqrt{10}$$

(WIP, will finish by 5/8/23)

~IceMatrix