Difference between revisions of "2016 AMC 12A Problems/Problem 12"
Matt44206068 (talk | contribs) (Added another solution, using only the Angle Bisector Theorem) |
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== Solution 3 == | == 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 | + | We denote <math>CD</math> by <math>y</math> and <math>DB</math> by <math>x</math>. Then, with the Angle Bisector Theorem in triangle <math>ACB</math> with angle bisector <math>AD</math>, 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> | <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 | + | Now, we use the Angle Bisector Theorem again in triangle <math>ADB</math> with angle bisector <math>BF.</math> We get |
<math>\frac{AF}{6}=\frac{FD}{3}</math> or | <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> | <math>\frac{AF}{FD}=\frac{2}{1},</math> which gives us the answer <math>\frac{AF}{AD} =\boxed{\textbf{(C)}\; 2 : 1}</math> |
Revision as of 20:57, 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 by and by . Then, with the Angle Bisector Theorem in triangle with angle bisector , we have or However, so or Now, we use the Angle Bisector Theorem again in triangle with angle bisector 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.