Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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The angle bisector of the acute angle formed at the origin by the graphs of the lines <math>y = x</math> and <math>y=3x</math> has equation <math>y=kx.</math> What is <math>k?</math> | The angle bisector of the acute angle formed at the origin by the graphs of the lines <math>y = x</math> and <math>y=3x</math> has equation <math>y=kx.</math> What is <math>k?</math> | ||
| − | <math>\textbf{(A)} \ | + | <math>\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}</math> |
==Solution== | ==Solution== | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2021 Fall|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:55, 23 November 2021
Problem 13
The angle bisector of the acute angle formed at the origin by the graphs of the lines 
and 
has equation 
What is 
Solution
IN PROGRESS. APPRECIATE IT IF NO EDITS. THANKS.
~MRENTHUSIASM
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. 
