Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math> | So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math> | ||
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+ | == Solution 2== | ||
+ | Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>. Note that <math>F</math> is the incenter. Then, <math>\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}</math> |
Revision as of 18:07, 4 February 2016
Problem 12
In ,
,
, and
. Point
lies on
, and
bisects
. Point
lies on
, and
bisects
. The bisectors intersect at
. What is the ratio
:
?
TODO: Diagram
Solution
By the angle bisector theorem,
so
Similarly, .
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
. Note that
is the incenter. Then,