Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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== Solution 5 (Trigonometry) == | == Solution 5 (Trigonometry) == | ||
+ | |||
+ | This problem can be trivialized using basic trig identities. Let the angle made by <math>y=x</math> and the <math>x</math>-axis be <math>\theta_{1}</math> and the angle made by <math>y=3x</math> and the <math>x</math>-axis be <math>\theta_{3}</math>. Note that <math>\tan(\theta_{1})=1</math> and <math>\tan(\theta_{3})=3</math>, and this is why we named them as such. Let the angle made by <math>y=kx</math> be denoted as <math>\theta_{k}</math>. Since <math>y=kx</math> bisects the two lines, notice that | ||
+ | <cmath>\theta_k-\theta_1=\theta_3-\theta_k.</cmath> | ||
+ | |||
+ | Now, we can take the tangent and apply the tangent subtraction formula: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\ | ||
+ | \frac{\tan(\theta_k)-\tan(\theta_1)}{1+\tan(\theta_k)\tan(\theta_1)}&=\frac{\tan(\theta_3)-\tan(\theta_k)}{1+\tan(\theta_3)\tan(\theta_k)}\\ | ||
+ | \frac{k-1}{1+k}&=\frac{3-k}{1+3k}\\ | ||
+ | (k-1)(1+3k)&=(1+k)(3-k)\\ | ||
+ | 3k^2-2k-1&=-k^2+2k+3\\ | ||
+ | 4k^2-4k-4&=0\\ | ||
+ | k^2-k-1&=0\\ | ||
+ | \implies k&=\frac{1\pm \sqrt{5}}{2} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Since <math>y=kx</math> is increasing, <math>k>0</math>; thus, <math>k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}.</math> | ||
+ | |||
+ | ~Indiiiigo | ||
+ | |||
+ | == Solution 6 (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>. | ||
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
− | == Solution | + | == Solution 7 (Vectors) == |
[[Image:2021FallAMC12AProblem13.png|center|500px]] | [[Image:2021FallAMC12AProblem13.png|center|500px]] | ||
Line 223: | Line 245: | ||
~VensL. | ~VensL. | ||
− | == Solution | + | == Solution 8 (Dot Product) == |
+ | |||
+ | We notice that the line <math>y = x</math> can be represented as the vector <math>\vec{A} = \begin{bmatrix} | ||
+ | 1 \\ 1 | ||
+ | \end{bmatrix}</math> | ||
+ | and <math>y = 3x</math> as <math>\vec{B} = \begin{bmatrix} | ||
+ | 1 \\ 3 | ||
+ | \end{bmatrix}</math> | ||
+ | as the "slope" of both vectors represent the coefficient of <math>x</math>. | ||
+ | |||
+ | Then, we can represent <math>y = kx</math> as <math>\vec{C} = \begin{bmatrix} | ||
+ | 1 \\ k | ||
+ | \end{bmatrix}</math> and notice that since <math>C</math> is in essence an angle bisector, <cmath>\theta = \angle \vec{A}\vec{C} = \angle \vec{B}\vec{C}</cmath> | ||
+ | |||
+ | <cmath>A \cdot C = |A|\cdot|C|\cos(\theta_1)</cmath> where <math>\theta_1 = \angle \vec{A}\vec{C} = \theta</math> | ||
+ | <cmath>B \cdot C = |B|\cdot|C|\cos(\theta_2)</cmath> where <math>\theta_2 = \angle \vec{B}\vec{C} = \theta</math> | ||
+ | |||
+ | Since both <math>\theta_i</math>'s are equivalent, we may simply represent them with <math>\theta</math>. | ||
+ | |||
+ | Simplifying both equations by performing the necessary operations, we get | ||
+ | |||
+ | <cmath>1+k = \sqrt{2} \cdot \sqrt{k^2 + 1} \cdot \cos(\theta) \implies \frac{1+k}{\sqrt{2}} = \sqrt{k^2 + 1} \cos(\theta)</cmath> | ||
+ | |||
+ | <cmath>1+3k = \sqrt{10} \cdot \sqrt{k^2 + 1} \cdot \cos(\theta)</cmath> | ||
+ | |||
+ | Substituting the first into the second, we get | ||
+ | <cmath>1 + 3k = \sqrt{10} \cdot \frac{1+k}{\sqrt{2}} \implies k=\boxed{\textbf{(A)} \ \frac{1+\sqrt{5}}{2}}</cmath> | ||
+ | |||
+ | ~ <math>\color{magenta} zoomanTV</math> | ||
+ | |||
+ | ==Solution 9== | ||
+ | |||
+ | <asy> | ||
+ | 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((1, 3)--(2, 2)); | ||
+ | draw((2, 2)--(2, 0)); | ||
+ | draw((2, 0)--(2, 0.2)--(1.8, 0.2)--(1.8, 0)); | ||
+ | draw((2, 2)--(1.9, 2.1)--(1.8, 2)--(1.9, 1.9)); | ||
+ | label("$A$", (2, 2), E); | ||
+ | label("$(a, a)$", (2.3, 2), E); | ||
+ | label("$B$", (1.5, 2.5), E); | ||
+ | label("$(b, 3b)$", (0.9, 3), SW); | ||
+ | label("$C$", (1, 3), NW); | ||
+ | label("$D$", (0, 0), NW); | ||
+ | label("$E$", (2, 0), NE); | ||
+ | |||
+ | |||
+ | draw(O--3.75*dir((1,1))^^O--3.75*dir((1,3))^^O--3.75*dir((1,k))); | ||
+ | </asy> | ||
+ | |||
+ | Refer to the modified version of Solution 1's diagram, seen above. | ||
− | + | Label <math>A = (a, a)</math> and <math>B = (b, 3b)</math>, by definition of a point on a line. | |
− | <cmath>\ | + | |
+ | The slope of the line through <math>A</math> and <math>C</math> is <math>-1</math>, as it is perpendicular to <math>y=x.</math> Therefore, we have <math>\frac{3b-a}{b-a} = -1,</math> so <math>2b=a.</math> | ||
+ | |||
+ | We label <math>A = (2b, 2b)</math> to simplify. By <math>45</math>-<math>45</math>-<math>90</math> triangle relations (or distance formula), <math>AD = 2b\sqrt{2}.</math> | ||
+ | |||
+ | By the distance formula, <math>AC = b\sqrt{2},</math> and by the Pythagorean Theorem, <math>CD = b\sqrt{10}.</math> | ||
+ | |||
+ | From the angle bisector theorem, | ||
+ | |||
+ | <cmath>\frac{AD}{CD} = \frac{AB}{BC}</cmath> | ||
+ | <cmath>\frac{2b\sqrt{2}}{b\sqrt{10}} = \frac{AB}{BC}.</cmath> | ||
+ | <cmath>\frac{2b\sqrt{5}}{5} BC = AB.</cmath> | ||
+ | |||
+ | Note that <math>AB + BC = b\sqrt{10},</math> so: | ||
+ | |||
+ | <cmath>BC + \frac{2b\sqrt{5}}{5} BC = b\sqrt{10}</cmath> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | + | (WIP, will finish by 5/8/23) | |
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Latest revision as of 00:53, 6 May 2023
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Angle Bisector Theorem)
- 4 Solution 2 (Analytic and Plane Geometry)
- 5 Solution 3 (Analytic and Plane Geometry)
- 6 Solution 4 (Distance Between a Point and a Line)
- 7 Solution 5 (Trigonometry)
- 8 Solution 6 (Trigonometry)
- 9 Solution 7 (Vectors)
- 10 Solution 8 (Dot Product)
- 11 Solution 9
- 12 Video Solution by TheBeautyofMath
- 13 See Also
Problem
The angle bisector of the acute angle formed at the origin by the graphs of the lines and
has equation
What is
Diagram
~MRENTHUSIASM
Solution 1 (Angle Bisector Theorem)
This solution refers to the Diagram section.
Let and
As shown below, note that
and
are on the lines
and
respectively. By the Distance Formula, we have
and
By the Angle Bisector Theorem, we get
or
Since
the answer is
Remark
The value of is known as the Golden Ratio:
~MRENTHUSIASM
Solution 2 (Analytic and Plane Geometry)
Consider the graphs of
and
. Since it will be easier to consider at unity, let
, then we have
and
.
Now, let be
,
be
, and
be
. Cutting through side
of triangle
is the angle bisector
where
is on side
.
Hence, by the Angle Bisector Theorem, we get .
By the Pythagorean Theorem, and
. Therefore,
.
Since , it is easy derive
.
The vertical distance between the -axis and
is
. Because the
-coordinate of point
is
, the slope we need to find is just the
-coordinate
~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 . The line
is the hypotenuse of a right triangle with side length
. The hypotenuse' length is
. Then, let's draw the line
. We extend it to when
. The length of the hypotenuse of the larger triangle is
with legs
. We then draw the angle bisector. We should label the triangle, so here we go.
is
.
is
.
is
. When the line with angle
intersects the line
, call the point
. When the angle bisector intersects the line
, call the point
. By Angle Bisector Theorem,
. Since
is
and
is
, we have that
is
. Solving for
, we get that
is
.
Since is
, we have that
is just one more than that. Therefore,
is
. Since
is
, we get that
is
.
Remark
The answer turns out to be the golden ratio or phi (). Phi has many properties and is related to the Fibonacci sequence. See Phi.
~Arcticturn
Solution 4 (Distance Between a Point and a Line)
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, the answer is
(Fun Fact: The value is the golden ratio
)
~NH14
Solution 5 (Trigonometry)
This problem can be trivialized using basic trig identities. Let the angle made by and the
-axis be
and the angle made by
and the
-axis be
. Note that
and
, and this is why we named them as such. Let the angle made by
be denoted as
. Since
bisects the two lines, notice that
Now, we can take the tangent and apply the tangent subtraction formula:
Since
is increasing,
; thus,
~Indiiiigo
Solution 6 (Trigonometry)
Denote by ,
,
the acute angles formed between the
-axis and lines
,
,
, respectively.
Hence,
,
,
.
Denote by the acute angle formed by lines
and
.
Hence,
Following from the double-angle identity, we have
Hence, .
Because is acute,
is acute.
Hence,
.
Hence,
.
Because line is the angle bisector of
, the angle between lines
and
is
.
Hence,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 7 (Vectors)
When drawing the lines and
, it is natural to choose points
and
together with origin
. See the figure attached.
We utilize the fact that if
and
are vectors of same length, then
bisects the angle between
and
.
In particular, we scale the vector by the factor of
to get
.
So by adding vectors
and
we get
which bisects the acute angle formed by lines
and
. (In other words, quadrilateral
is a rhombus.)
Finally, observe that
lies on the line
whose slope is
Thus, the answer is
.
~VensL.
Solution 8 (Dot Product)
We notice that the line can be represented as the vector
and
as
as the "slope" of both vectors represent the coefficient of
.
Then, we can represent as
and notice that since
is in essence an angle bisector,
where
where
Since both 's are equivalent, we may simply represent them with
.
Simplifying both equations by performing the necessary operations, we get
Substituting the first into the second, we get
~
Solution 9
Refer to the modified version of Solution 1's diagram, seen above.
Label and
, by definition of a point on a line.
The slope of the line through and
is
, as it is perpendicular to
Therefore, we have
so
We label to simplify. By
-
-
triangle relations (or distance formula),
By the distance formula, and by the Pythagorean Theorem,
From the angle bisector theorem,
Note that so:
(WIP, will finish by 5/8/23)
Video Solution by TheBeautyofMath
https://youtu.be/ToiOlqWz3LY?t=504
~IceMatrix
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.