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2021 Fall AMC 12A Problems/Problem 13

Revision as of 13:18, 24 November 2021 by MRENTHUSIASM (talk | contribs) (Rearranged the solutions and prioritize ANGLE BISECTOR THEOREM. Let me know if you are unhappy with this edit.)

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("$y=x$",4.5*dir((1,1))); label("$y=3x$",3.75*dir((1,3))); label("$y=kx$",4*dir((1,k))); draw(O--4.25*dir((1,1))^^O--3.5*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: $\phi=\frac{1+\sqrt{5}}{2}\approx1.618.$

~MRENTHUSIASM

Solution 2 (Angle Bisector Theorem)

[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 $AC+\sqrt{5}AC=2 \implies AC=\frac{2}{1+\sqrt{5}}=\frac{-1+\sqrt{5}}{2}$.

As 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 (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

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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