Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
− | < | + | Let <math>O=(0,0), A=(3,3),</math> and <math>B=(1,3).</math> Note that <math>\overline{OA}</math> is on the line <math>y=x,</math> and <math>\overline{OB}</math> is on the line <math>y=3x.</math> |
+ | |||
+ | Suppose that the line <math>y=kx</math> intersects <math>\overline{AB}</math> at <math>C.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 21:09, 23 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
Solution 1
Let and
Note that
is on the line
and
is on the line
Suppose that the line intersects
at
~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
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.