Difference between revisions of "2021 Fall AMC 12A Problems/Problem 13"
(→Solution 3 (Easy Test)) |
(→Solution 3 (Easy Test)) |
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Line 110: | Line 110: | ||
real yMin = 0; | real yMin = 0; | ||
real yMax = 5; | real yMax = 5; | ||
− | real | + | real k = (1+sqrt(5))/2; |
pair O; | pair O; | ||
O = origin; | O = origin; | ||
− | draw(anglemark(dir((1,1)),O,dir((1, | + | draw(anglemark(dir((1,1)),O,dir((1,k)),20), blue); |
− | draw(anglemark(dir((1, | + | draw(anglemark(dir((1,k)),O,dir((1,3)),20), blue); |
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); |
Revision as of 06:27, 24 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
~MRENTHUSIASM
Solution 1
This solution refers to the Diagram section.
Let and
Note that
is on the line
is on the line
and
is on the line
as shown below.
By the Angle Bisector Theorem, we have
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 (Easy Test)
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 1, 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.