Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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== Solution 1== | == Solution 1== | ||
− | + | By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | |
− | < | + | <math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math> |
− | + | Similarly, <math>CD = 4</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> | |
+ | Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math> | ||
− | + | <math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math> | |
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− | < | + | 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>\frac{ | + | <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> | + | <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]] | ||
− | + | ==Solution 5 (Luck-Based)== | |
− | + | Note that <cmath>AF</cmath> and <cmath>BD</cmath> look like medians. Assuming they are medians, we mark the answer <cmath>\boxed{\textbf{(C)}\ 2:1}</cmath> as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a <cmath>2:1</cmath> ratio, with the shorter part being closer to the side it bisects. | |
− | + | ~[[User:scthecool|scthecool]] | |
− | + | Note: This is heavily luck based, and if the figure had been not drawn to scale, for example, this answer would have easily been wrong. It is thus advised to not use this in a real competition unless absolutely necessary. | |
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− | + | == Video Solution by OmegaLearn == | |
+ | https://youtu.be/Gjt25jRiFns?t=43 | ||
− | + | ~ pi_is_3.14 | |
==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 18:19, 20 March 2024
Contents
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
Solution 5 (Luck-Based)
Note that and
look like medians. Assuming they are medians, we mark the answer
as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a
ratio, with the shorter part being closer to the side it bisects.
~scthecool
Note: This is heavily luck based, and if the figure had been not drawn to scale, for example, this answer would have easily been wrong. It is thus advised to not use this in a real competition unless absolutely necessary.
Video Solution by OmegaLearn
https://youtu.be/Gjt25jRiFns?t=43
~ pi_is_3.14
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.