Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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Similarly, <math>CD = 4</math>. | Similarly, <math>CD = 4</math>. | ||
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+ | There are two ways to solve from here. | ||
+ | First way: | ||
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+ | Note that <math>DB = 7 - 4 = 3.</math> By the angle bisector theorem on <math>\triangle ADB,</math> <math>\frac{AF}{FD} = \frac{AB}{DB} = \frac{6}{3}.</math> Thus the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math> | ||
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+ | Second way: | ||
Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>. | Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>. |
Revision as of 03:26, 5 August 2021
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, the segment addition postulate, and some simple algebra to solve this question.
The question asks for AF:DF. Apply the angle bisector theorem to to get the ratio
:
=
or, equivalently,
=
.
AB is given. To find BD apply the angle bisector theorem to to get:
=
---> =
since, by the segment addition postulate, BD + CD = BC
---> BD = .
Substituting this expression for BD into the proportion =
yields:
=
=
=
= 2.
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.