# 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> | ||

+ | |||

+ | == 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,