Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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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== |
+ | |||
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> | ||
Line 14: | Line 15: | ||
Similarly, <math>CD = 4</math>. | Similarly, <math>CD = 4</math>. | ||
− | Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>. | + | There are two ways to solve from here. |
+ | First way: | ||
+ | |||
+ | Note that <math>DB = 7 - 4 = 3.</math> By the angle bisector theorem on <math>\triangle ADB,</math> <math>\frac{AF}{FD} = \frac{AB}{DB} = \frac{6}{3}.</math> Thus the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math> | ||
+ | |||
+ | Second way: | ||
+ | |||
+ | Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>. | ||
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math> | <math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math> | ||
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<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math> | <math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math> | ||
− | So <math>\frac{AF}{ | + | So <math>\frac{AF}{FD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math> |
== Solution 2== | == 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>. There are two ways to continue from here: | ||
+ | |||
+ | <math>1.</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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | Draw the third angle bisector, and denote the point where this bisector intersects <math>AB</math> as <math>P</math>. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13, AP=16/5, PB=14/5</math>. Applying [https://artofproblemsolving.com/wiki/index.php/Van_Aubel%27s_Theorem Van Aubel's Theorem], <math>AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1</math>, and so the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math>. | ||
+ | |||
+ | == Solution 4== | ||
+ | One only needs the angle bisector theorem to solve this question. | ||
+ | |||
+ | The question asks for <math>AF:FD = \frac{AF}{FD}</math>. Apply the angle bisector theorem to <math>\triangle ABD</math> to get | ||
+ | <cmath>\frac{AF}{FD} = \frac{AB}{BD}.</cmath> | ||
+ | |||
+ | <math>AB = 6</math> is given. To find <math>BD</math>, apply the angle bisector theorem to <math>\triangle BAC</math> to get | ||
+ | <cmath>\frac{BD}{DC} = \frac{BA}{AC} = \frac{6}{8} = \frac{3}{4}.</cmath> | ||
+ | |||
+ | Since | ||
+ | <cmath>BD + DC = BC = 7,</cmath> | ||
+ | it is immediately obvious that <math>BD = 3</math>, <math>DC = 4</math> satisfies both equations. | ||
+ | |||
+ | Thus, | ||
+ | <cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath> | ||
+ | ~revision by [[User:emerald_block|emerald_block]] | ||
==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}} |
Latest revision as of 11:09, 3 November 2021
Problem 12
In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?
Solution 1
By the angle bisector theorem,
so
Similarly, .
There are two ways to solve from here. First way:
Note that By the angle bisector theorem on Thus the answer is
Second way:
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 . There are two ways to continue from here:
Note that is the incenter. Then,
Apply the angle bisector theorem on to get
Solution 3
Draw the third angle bisector, and denote the point where this bisector intersects as . Using angle bisector theorem, we see . Applying Van Aubel's Theorem, , and so the answer is .
Solution 4
One only needs the angle bisector theorem to solve this question.
The question asks for . Apply the angle bisector theorem to to get
is given. To find , apply the angle bisector theorem to to get
Since it is immediately obvious that , satisfies both equations.
Thus, ~revision by emerald_block
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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