Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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~NH14 | ~NH14 | ||
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+ | ==Solution 3 (Easy Test) == | ||
+ | Consider the graphs of <math>f(x)=x</math> and <math>g(x)=3x</math>. Set <math>x=1</math>, we have <math>f(1)=1</math> and <math>f(1)=3</math>. | ||
+ | |||
+ | Now, let <math>O</math> be <math>(0,0)</math>, <math>A</math> be <math>(1,1)</math>, and <math>B</math> be <math>(1,3)</math>. Cutting through side <math>AB</math> of the triangle <math>OAB</math> is the angle bisector <math>OC</math> where <math>C</math> is on side <math>AB</math>. | ||
+ | |||
+ | Hence, by Angle Bisector Theorem, we have <math>OB/OA=BC/AC</math>. | ||
+ | |||
+ | By Pythagorean Theorem, <math>OA=\sqrt{2}</math> and <math>OB=\sqrt{10}</math>. Therefore, <math>BC/AC=\sqrt{5} \Rightarrow BC=\sqrt{5}AC</math> | ||
+ | |||
+ | Since <math>AB=AC+BC=2</math>, it is easy derive <math>AC+\sqrt{5}AC=2 \Rightarrow AC=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}</math> | ||
+ | |||
+ | As the vertical distance between the x-axis and <math>C</math> is <math>\frac{\sqrt{5}-1}{2}+1=</math>\frac{\sqrt{5}+1}{2}<math>. Thus, the answer is | ||
+ | </math>\boxed{\textbf{(A)}}$. | ||
+ | |||
+ | ~Wilhelm Z | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=A|num-b=12|num-a=14}} | {{AMC12 box|year=2021 Fall|ab=A|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:37, 23 November 2021
Problem
The angle bisector of the acute angle formed at the origin by the graphs of the lines and has equation What is
Diagram
Solution 1
Let and Note that is on the line is on the line and is on the line as shown below.
DIAGRAM WILL BE READY VERY VERY SOON. NO EDIT PLEASE
By the Angle Bisector Theorem, we have or Since the answer is
~MRENTHUSIASM
Solution 2
Note that the distance between the point to line is Because line is a perpendicular bisector, a point on the line must be equidistant from the two lines( and ), call this point Because, the line passes through the origin, our requested value of which is the slope of the angle bisector line, can be found when evaluating the value of By the Distance from Point to Line formula we get the equation, Note that because is higher than and because is lower to Thus, we solve the equation, Thus, the value of Thus, the answer is
(Fun Fact: The value is the golden ratio )
~NH14
Solution 3 (Easy Test)
Consider the graphs of and . Set , we have and .
Now, let be , be , and be . Cutting through side of the triangle is the angle bisector where is on side .
Hence, by Angle Bisector Theorem, we have .
By Pythagorean Theorem, and . Therefore,
Since , it is easy derive
As the vertical distance between the x-axis and is \frac{\sqrt{5}+1}{2}\boxed{\textbf{(A)}}$.
~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.