Difference between revisions of "2016 AMC 12A Problems/Problem 12"
Matt44206068 (talk | contribs) |
(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 18:45, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.