Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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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} | + | 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}}{1+\sqrt{5}} = \frac{\sqrt{5}+1}{2}.</math> Thus, our answer is <math>\boxed{\textbf{(A.)}}.</math> |
(Fun Fact: The value <math>\frac{\sqrt{5}+1}{2}</math> is the golden ratio <math>\phi.</math>) | (Fun Fact: The value <math>\frac{\sqrt{5}+1}{2}</math> is the golden ratio <math>\phi.</math>) |
Revision as of 20:02, 23 November 2021
Contents
Problem 13
The angle bisector of the acute angle formed at the origin by the graphs of the lines and has equation What is
Solution
IN PROGRESS. APPRECIATE IT IF NO EDITS. THANKS.
~MRENTHUSIASM
Solution 2
Note that the distance between the point to line is Because line is a perpendicular bisector, a point on the line must be equidistant from the two lines( and ), call this point Because, the line passes through the origin, our requested value of which is the slope of the angle bisector line, can be found when evaluating the value of By the Distance from Point to Line formula we get the equation, Note that because is higher than and because is lower to Thus, we solve the equation, Thus, the value of Thus, our answer is
(Fun Fact: The value is the golden ratio )
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.