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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Solution 2==
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Note that the distance between the point <math>(m,n)</math> to line <math>Ax + By + C = 0,</math> is <math>\frac{|Am + Bn +C|}{\sqrt{A^2 +B^2}}.</math> Because line <math>y=kx</math> is a perpendicular bisector, a point on the line <math>y=kx</math> must be equidistant from the two lines(<math>y=x</math> and <math>y=3x</math>), call this point <math>P(z,w).</math> Because, the line <math>y=kx</math> passes through the origin, our requested value of <math>k,</math> which is the slope of the angle bisector line, can be found when evaluating the value of <math>\frac{w}{z}.</math> By the Distance from Point to Line formula we get the equation, <cmath>\frac{|3z-w|}{\sqrt{10}} = \frac{|z-w|}{\sqrt{2}}.</cmath> Note that <math>|3z-w|\ge 0,</math> because <math>y=3x</math> is higher than <math>P</math> and <math>|z-w|\le 0,</math> because <math>y=x</math> is lower to <math>P.</math> Thus, we solve the equation, <cmath>(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.</cmath> Thus, the value of <math>\frac{w}{z} = \frac{3+\sqrt{5}}{\sqrt{5}+1} = \frac{\sqrt{5}+1}{2}.</math> Thus, our answer is <math>\boxed{\textbf{(A.)}}.</math>
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(Fun Fact: The value <math>\frac{\sqrt{5}+1}{2}</math> is the golden ratio <math>\phi.</math>)
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~NH14
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:00, 23 November 2021

Problem 13

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 $k?$

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

Solution

IN PROGRESS. APPRECIATE IT IF NO EDITS. THANKS.

~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 $\frac{w}{z} = \frac{3+\sqrt{5}}{\sqrt{5}+1} = \frac{\sqrt{5}+1}{2}.$ Thus, our answer is $\boxed{\textbf{(A.)}}.$

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

~NH14

See Also

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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