Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"

Revision as of 20:09, 23 November 2021

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 $O=(0,0), A=(3,3),$ and $B=(1,3).$ Note that $\overline{OA}$ is on the line $y=x,$ and $\overline{OB}$ is on the line $y=3x.$

Suppose that the line $y=kx$ intersects $\overline{AB}$ at $C.$

~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

@@ Line 7: / Line 7: @@
 ==Solution 1==
-<b>IN PROGRESS. APPRECIATE IT IF NO EDITS. THANKS.</b>
+Let <math>O=(0,0), A=(3,3),</math> and <math>B=(1,3).</math> Note that <math>\overline{OA}</math> is on the line <math>y=x,</math> and <math>\overline{OB}</math> is on the line <math>y=3x.</math>
+Suppose that the line <math>y=kx</math> intersects <math>\overline{AB}</math> at <math>C.</math>
 ~MRENTHUSIASM

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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