Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Problem == | == Problem == | ||
− | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\ | + | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\overline{AC}</math> at <math>N.</math> What is the perimeter of <math>\triangle AMN?</math> |
<math> | <math> | ||
Line 9: | Line 9: | ||
\textbf{(E)}\ 42 </math> | \textbf{(E)}\ 42 </math> | ||
− | == Solution == | + | ==Solution 1== |
− | Let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have | + | Let <math>O</math> be the incenter of <math>\triangle{ABC}</math>. Because <math>\overline{MO} \parallel \overline{BC}</math> and <math>\overline{BO}</math> is the angle bisector of <math>\angle{ABC}</math>, we have |
<cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath> | <cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath> | ||
Line 23: | Line 23: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math> O </math> be the incenter. <math> AO </math> is the angle bisector of <math> \angle MAN </math>. Let the angle bisector of <math> \angle BAC </math> meets <math> BC </math> at <math> P </math> and the angle bisector of <math> \angle ABC </math> meets <math> AC </math> at <math> Q </math>. By applying both angle bisector theorem and Menelaus' theorem, | ||
+ | |||
+ | |||
+ | |||
+ | <math>\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1 </math> | ||
+ | |||
+ | |||
+ | <math>\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1 </math> | ||
+ | |||
+ | |||
+ | <math>\frac{AO}{OP}=\frac{5}{4} </math> | ||
+ | |||
+ | |||
+ | <math>\frac{AO}{AP}=\frac{5}{9} </math> | ||
+ | |||
+ | |||
+ | Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 \rightarrow \boxed{(B)}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | Like in other solutions, let <math>O</math> be the incenter of <math>\triangle ABC</math>. Let <math>AO</math> intersect <math>BC</math> at <math>D</math>. By the angle bisector theorem, <math>\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}</math>. Since <math>BD+DC = 24</math>, we have <math>\frac{BD}{24-BD} = \frac{2}{3}</math>, so <math>3BD = 48 - 2BD</math>, so <math>BD = \frac{48}{5} = 9.6</math>. By the angle bisector theorem on <math>\triangle ABD</math>, we have <math>\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8</math>, so <math>\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8</math>, so <math>\frac{AO}{AD} = \frac{5}{9}</math>. Because <math>\triangle AMN \sim \triangle ABC</math>, the perimeter of <math>\triangle AMN</math> must be <math>\frac{5}{9} (12 + 18 + 24) = 30</math>, so our answer is <math>\boxed{\textbf{(B)}\ 30}</math>. | ||
+ | |||
+ | Another way to find <math>\frac{AO}{OD}</math> is to use mass points. Assign a mass of 24 to <math>A</math>, a mass of 18 to <math>B</math>, and a mass of 12 to <math>C</math>. Then <math>D</math> has mass 30, so <math>\frac{AO}{OD} = \frac{30}{24} = \frac{5}{4} = 1.25</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | We know that the ratio of the perimeter of <math>\triangle AMN</math> and <math>\triangle ABC</math> is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from <math>A</math> to <math>BC</math> is <math>\frac{9\sqrt{15}}{4}</math> from Herons and then <math>A=\frac{bh}{2}</math> and also that the height from <math>A</math> to <math>MN</math> is simply the height from <math>A</math> to <math>BC</math> minus the inradius. We know the area and the semiperimeter so <math>r=\frac{A}{s}</math> which gives us <math>r=\sqrt{15}</math>. Now we know that the altitude from <math>A</math> to <math>MN</math> is <math>\frac{5\sqrt{15}}{4}</math> so the ratios of the heights from <math>A</math> for <math>\triangle AMN</math> and <math>\triangle ABC</math> is <math>\frac{5}{9}</math>. Thus the perimeter of <math>\triangle AMN</math> is <math>\frac{5}{9} \times 54 = 30</math> so our answer is <math>\boxed{B}</math> | ||
+ | |||
+ | -srisainandan6 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=u23iWcqbJlE | ||
+ | ~Shreyas S | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/5jwD5UViZO8?t=1013 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}} | {{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:12, 23 January 2023
Contents
Problem
Triangle has side-lengths
and
The line through the incenter of
parallel to
intersects
at
and
at
What is the perimeter of
Solution 1
Let be the incenter of
. Because
and
is the angle bisector of
, we have
It then follows due to alternate interior angles and base angles of isosceles triangles that . Similarly,
. The perimeter of
then becomes
Solution 2
Let be the incenter.
is the angle bisector of
. Let the angle bisector of
meets
at
and the angle bisector of
meets
at
. By applying both angle bisector theorem and Menelaus' theorem,
Perimeter of
Solution 3
Like in other solutions, let be the incenter of
. Let
intersect
at
. By the angle bisector theorem,
. Since
, we have
, so
, so
. By the angle bisector theorem on
, we have
, so
, so
. Because
, the perimeter of
must be
, so our answer is
.
Another way to find is to use mass points. Assign a mass of 24 to
, a mass of 18 to
, and a mass of 12 to
. Then
has mass 30, so
.
Solution 4
We know that the ratio of the perimeter of and
is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from
to
is
from Herons and then
and also that the height from
to
is simply the height from
to
minus the inradius. We know the area and the semiperimeter so
which gives us
. Now we know that the altitude from
to
is
so the ratios of the heights from
for
and
is
. Thus the perimeter of
is
so our answer is
-srisainandan6
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
Video Solution by OmegaLearn
https://youtu.be/5jwD5UViZO8?t=1013
~ pi_is_3.14
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.