Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math> | <math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math> | ||
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==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 16:10, 30 December 2016
Problem 12
In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?
Solution 1
Applying the angle bisector theorem to with being bisected by , we have
Thus, we have
and cross multiplying and dividing by gives us
Since , we can substitute into the former equation. Therefore, we get , so .
Apply the angle bisector theorem again to with being bisected. This gives us
and since and , we have
Cross multiplying and dividing by gives us
and dividing by gives us
Therefore,
Solution 2
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 3
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
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 |
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