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

Solution 1

Let $O=(0,0), A=(3,3), B=(1,3),$ and $C=\left(\frac3k,3\right).$ Note that $\overline{OA}$ is on the line $y=x,$ $\overline{OB}$ is on the line $y=3x,$ and $\overline{OC}$ is on the line $y=kx,$ as shown below.

DIAGRAM WILL BE READY VERY VERY SOON. NO EDIT PLEASE

By the Angle Bisector Theorem, we have $\frac{OA}{OB}=\frac{AC}{BC},$ 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 $k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.$

~MRENTHUSIASM

Solution 2

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 3 (Easy Test)

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, $BC/AC=\sqrt{5} \Rightarrow BC=\sqrt{5}AC$

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

As the vertical distance between the x-axis and $C$ is $\frac{\sqrt{5}-1}{2}+1=\frac{\sqrt{5}+1}{2}$ . 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

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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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