Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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(Much easier solution (solution 1)) |
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<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math> | <math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math> | ||
− | == Solution == | + | == Solution 1== |
+ | |||
+ | Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have | ||
+ | |||
+ | <cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath> | ||
+ | |||
+ | Thus, we have | ||
+ | |||
+ | <cmath>\frac{CD}{8}=\frac{BD}{6},</cmath> | ||
+ | |||
+ | and cross multiplying and dividing by <math>2</math> gives us | ||
+ | |||
+ | <cmath>3\cdot CD=4\cdot BD.</cmath> | ||
+ | |||
+ | Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>. | ||
+ | |||
+ | |||
+ | Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected. This gives us | ||
+ | |||
+ | <cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath> | ||
+ | |||
+ | and since <math>AB=6</math> and <math>BD=3</math>, we have | ||
+ | |||
+ | <cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath> | ||
+ | |||
+ | Cross multiplying and dividing by <math>3</math> gives us | ||
+ | |||
+ | <cmath>AF=2\cdot FD.</cmath> | ||
+ | |||
+ | Dividing by <math>FD</math> gives us | ||
+ | |||
+ | <cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath> | ||
+ | |||
+ | == Solution 2== | ||
+ | |||
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | ||
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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 | + | == Solution 3== |
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 | + | == Solution 4== |
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 | 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> |
Revision as of 17:45, 17 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 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
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 . Note that is the incenter. Then,
Solution 4
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 |
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