Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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Consider the graphs of <math>f(x)=x</math> and <math>g(x)=3x</math>. Since it will be easier to consider at unity, let <math>x=1</math>, then we have <math>f(1)=1</math> and <math>g(1)=3</math>. | Consider the graphs of <math>f(x)=x</math> and <math>g(x)=3x</math>. Since it will be easier to consider at unity, let <math>x=1</math>, then we have <math>f(1)=1</math> and <math>g(1)=3</math>. | ||
Revision as of 13:12, 24 November 2021
Contents
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
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
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 3 (Angle Bisector Theorem)
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 .
As 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
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.