Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 6(Angle Bisector Theorem 2))
(Rearranged the solutions based on titles. Let me know if you disagree ...)
Line 136: Line 136:
 
~Wilhelm Z
 
~Wilhelm Z
  
−
==Solution 3 (Distance Between a Point and a Line)==
+
==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 <math>x = 1</math>. The line <math>x = y</math> is the hypotenuse of a right triangle with side length <math>1</math>. The hypotenuse' length is <math>\sqrt 2</math>. Then, let's draw the line <math>x = 3y</math>. We extend it to when <math>x = 1</math>. The length of the hypotenuse of the larger triangle is <math>\sqrt {10}</math> with legs <math>1, 3</math>. We then draw the angle bisector. We should label the triangle, so here we go. <math>AC</math> is <math>1</math>. <math>BC</math> is <math>3</math>. <math>AB</math> is <math>\sqrt {10}</math>. When the line with angle <math>45 ^\circ</math> intersects the line <math>x = 1</math>, call the point <math>D</math>. When the angle bisector intersects the line <math>x = 1</math>, call the point <math>E</math>. By Angle Bisector Theorem, <math>\frac {DE}{DB} = \frac {\sqrt {2}}{\sqrt{10}}</math>. Since <math>BC</math> is <math>3</math> and <math>DC</math> is <math>1</math>, we have that <math>BD</math> is <math>2</math>. Solving for <math>DE</math>, we get that <math>DE</math> is <math>\frac {\sqrt 5 - 1}{2}</math>.
 +
 
 +
Since <math>DE</math> is <math>\frac {\sqrt 5 - 1}{2}</math>, we have that <math>CE</math> is just one more than that. Therefore, <math>CE</math> is <math>\frac {1+\sqrt 5}{2}</math>. Since <math>AC</math> is <math>1</math>, we get that <math>k</math> is <math>\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}</math>.
 +
 
 +
Remark: The answer turns out to be the golden ratio or phi (<math>\phi</math>). Phi has many properties and is related to the [[Fibonacci sequence]]. See [[Phi]].
 +
 
 +
~Arcticturn <math>\blacksquare</math>
 +
 
 +
==Solution 4 (Distance Between a Point and a Line)==
 
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{1+\sqrt{5}}{2}.</math> Thus, the answer is <math>\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.</math>
 
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{1+\sqrt{5}}{2}.</math> Thus, the answer is <math>\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.</math>
  
Line 143: Line 152:
 
~NH14
 
~NH14
  
−
== Solution 4 (Trigonometry) ==
+
== Solution 5 (Trigonometry) ==
 
Denote by <math>\alpha_1</math>, <math>\alpha_2</math>, <math>\alpha_3</math> the acute angles formed between the <math>x</math>-axis and lines <math>y = x</math>, <math>y = 3 x</math>, <math>y = k x</math>, respectively.
 
Denote by <math>\alpha_1</math>, <math>\alpha_2</math>, <math>\alpha_3</math> the acute angles formed between the <math>x</math>-axis and lines <math>y = x</math>, <math>y = 3 x</math>, <math>y = k x</math>, respectively.
 
Hence, <math>\tan \alpha_1 = 1</math>, <math>\tan \alpha_2 = 3</math>, <math>\tan \alpha_3 = k</math>.
 
Hence, <math>\tan \alpha_1 = 1</math>, <math>\tan \alpha_2 = 3</math>, <math>\tan \alpha_3 = k</math>.
Line 189: Line 198:
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  
−
== Solution 5 (Vectors) ==
+
== Solution 6 (Vectors) ==
  
 
[[Image:2021FallAMC12AProblem13.png|center|500px]]
 
[[Image:2021FallAMC12AProblem13.png|center|500px]]
Line 211: Line 220:
  
 
~VensL.
 
~VensL.
−
 
−
==Solution 6 (Angle Bisector Theorem 2)==
 
−
Let's begin by drawing a triangle that starts at the origin. Assume that the base of the triangle goes to the point <math>x = 1</math>. The line <math>x = y</math> is the hypotenuse of a right triangle with side length <math>1</math>. The hypotenuse' length is <math>\sqrt 2</math>. Then, let's draw the line <math>x = 3y</math>. We extend it to when <math>x = 1</math>. The length of the hypotenuse of the larger triangle is <math>\sqrt {10}</math> with legs <math>1, 3</math>. We then draw the angle bisector. We should label the triangle, so here we go. <math>AC</math> is <math>1</math>. <math>BC</math> is <math>3</math>. <math>AB</math> is <math>\sqrt {10}</math>. When the line with angle <math>45 ^\circ</math> intersects the line <math>x = 1</math>, call the point <math>D</math>. When the angle bisector intersects the line <math>x = 1</math>, call the point <math>E</math>. By Angle Bisector Theorem, <math>\frac {DE}{DB} = \frac {\sqrt {2}}{\sqrt{10}}</math>. Since <math>BC</math> is <math>3</math> and <math>DC</math> is <math>1</math>, we have that <math>BD</math> is <math>2</math>. Solving for <math>DE</math>, we get that <math>DE</math> is <math>\frac {\sqrt 5 - 1}{2}</math>.
 
−
 
−
Since <math>DE</math> is <math>\frac {\sqrt 5 - 1}{2}</math>, we have that <math>CE</math> is just one more than that. Therefore, <math>CE</math> is <math>\frac {1+\sqrt 5}{2}</math>. Since <math>AC</math> is <math>1</math>, we get that <math>k</math> is <math>\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}</math>.
 
−
 
−
Remark: The answer turns out to be the golden ratio or phi (<math>\phi</math>). Phi has many properties and is related to the [[Fibonacci sequence]]. See [[Phi]].
 
−
 
−
~Arcticturn <math>\blacksquare</math>
 
  
 
==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 00:27, 29 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

[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(5)); dot("$A$",A,1.5*N,linewidth(5)); dot("$B$",B,1.5*N,linewidth(5)); dot("$C$",C,1.5*N,linewidth(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)

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 6 (Vectors)

2021FallAMC12AProblem13.png

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.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_13&oldid=166962"