2021 Fall AMC 12A Problems/Problem 13
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 Video Solution by TheBeautyofMath
- 12 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
~
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 |
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