Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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Pi is 3.14 (talk | contribs) (→Solution 4) |
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<cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath> | <cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath> | ||
~revision by [[User:emerald_block|emerald_block]] | ~revision by [[User:emerald_block|emerald_block]] | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/Gjt25jRiFns?t=43 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:30, 23 January 2023
Contents
Problem 12
In ,
,
, and
. Point
lies on
, and
bisects
. Point
lies on
, and
bisects
. The bisectors intersect at
. What is the ratio
:
?
Solution 1
By the angle bisector theorem,
so
Similarly, .
There are two ways to solve from here. First way:
Note that By the angle bisector theorem on
Thus the answer is
Second way:
Now, we use mass points. Assign point a mass of
.
, so
Similarly, will have a mass of
So
Solution 2
Denote as the area of triangle ABC and let
be the inradius. Also, as above, use the angle bisector theorem to find that
. There are two ways to continue from here:
Note that
is the incenter. Then,
Apply the angle bisector theorem on
to get
Solution 3
Draw the third angle bisector, and denote the point where this bisector intersects as
. Using angle bisector theorem, we see
. Applying Van Aubel's Theorem,
, and so the answer is
.
Solution 4
One only needs the angle bisector theorem to solve this question.
The question asks for . Apply the angle bisector theorem to
to get
is given. To find
, apply the angle bisector theorem to
to get
Since
it is immediately obvious that
,
satisfies both equations.
Thus,
~revision by emerald_block
Video Solution by OmegaLearn
https://youtu.be/Gjt25jRiFns?t=43
~ pi_is_3.14
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.