Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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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> | 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> | ||
+ | == Solution 3 == | ||
+ | We denote CD by <math>y</math> and DB by <math>x</math>. Then, with the Angle Bisector Theorem in triangle ACB with angle bisector AD, we have | ||
+ | <math>\frac{x}{6}=\frac{y}{8}</math> or <math>y=\frac{4x}{3}.</math> However, <math>x+y=7,</math> so <math>x+\frac{4x}{3}=7</math> or <math>x=3.</math> | ||
+ | Now, we use the Angle Bisector Theorem again in triangle ADB with angle bisector BF. We get | ||
+ | <math>\frac{AF}{6}=\frac{FD}{3}</math> or | ||
+ | <math>\frac{AF}{FD}=\frac{2}{1},</math> which gives us the answer <math>\frac{AF}{AD} =\boxed{\textbf{(C)}\; 2 : 1}</math> | ||
==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 21:56, 9 February 2016
Problem 12
In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?
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,
Solution 3
We denote CD by and DB by . Then, with the Angle Bisector Theorem in triangle ACB with angle bisector AD, we have or However, so or Now, we use the Angle Bisector Theorem again in triangle ADB with angle bisector BF. We get or which gives us the answer
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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