Difference between revisions of "2011 AIME II Problems/Problem 4"
Pi is 3.14 (talk | contribs) (→Solution 8 (Cheese)) |
|||
(5 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
+ | __TOC__ | ||
+ | |||
== Problem 4 == | == Problem 4 == | ||
In triangle <math>ABC</math>, <math>AB=20</math> and <math>AC=11</math>. The angle bisector of <math>\angle A</math> intersects <math>BC</math> at point <math>D</math>, and point <math>M</math> is the midpoint of <math>AD</math>. Let <math>P</math> be the point of the intersection of <math>AC</math> and <math>BM</math>. The ratio of <math>CP</math> to <math>PA</math> can be expressed in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | In triangle <math>ABC</math>, <math>AB=20</math> and <math>AC=11</math>. The angle bisector of <math>\angle A</math> intersects <math>BC</math> at point <math>D</math>, and point <math>M</math> is the midpoint of <math>AD</math>. Let <math>P</math> be the point of the intersection of <math>AC</math> and <math>BM</math>. The ratio of <math>CP</math> to <math>PA</math> can be expressed in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | + | == Solution 1 == | |
− | |||
− | |||
− | |||
<asy> | <asy> | ||
pointpen = black; pathpen = linewidth(0.7); | pointpen = black; pathpen = linewidth(0.7); | ||
Line 15: | Line 14: | ||
</asy> Let <math>D'</math> be on <math>\overline{AC}</math> such that <math>BP \parallel DD'</math>. It follows that <math>\triangle BPC \sim \triangle DD'C</math>, so <cmath>\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}</cmath> by the [[Angle Bisector Theorem]]. Similarly, we see by the Midline Theorem that <math>AP = PD'</math>. Thus, <cmath>\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},</cmath> and <math>m+n = \boxed{51}</math>. | </asy> Let <math>D'</math> be on <math>\overline{AC}</math> such that <math>BP \parallel DD'</math>. It follows that <math>\triangle BPC \sim \triangle DD'C</math>, so <cmath>\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}</cmath> by the [[Angle Bisector Theorem]]. Similarly, we see by the Midline Theorem that <math>AP = PD'</math>. Thus, <cmath>\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},</cmath> and <math>m+n = \boxed{51}</math>. | ||
− | + | == Solution 2 (mass points) == | |
Assign [[mass points]] as follows: by Angle-Bisector Theorem, <math>BD / DC = 20/11</math>, so we assign <math>m(B) = 11, m(C) = 20, m(D) = 31</math>. Since <math>AM = MD</math>, then <math>m(A) = 31</math>, and <math>\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}</math>, so <math>m+n = \boxed{51}</math>. | Assign [[mass points]] as follows: by Angle-Bisector Theorem, <math>BD / DC = 20/11</math>, so we assign <math>m(B) = 11, m(C) = 20, m(D) = 31</math>. Since <math>AM = MD</math>, then <math>m(A) = 31</math>, and <math>\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}</math>, so <math>m+n = \boxed{51}</math>. | ||
− | + | == Solution 3 == | |
By [[Menelaus' Theorem]] on <math>\triangle ACD</math> with [[transversal]] <math>PB</math>, <cmath>1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}. </cmath> So <math>m+n = \boxed{051}</math>. | By [[Menelaus' Theorem]] on <math>\triangle ACD</math> with [[transversal]] <math>PB</math>, <cmath>1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}. </cmath> So <math>m+n = \boxed{051}</math>. | ||
− | + | ==Solution 4== | |
We will use barycentric coordinates. Let <math>A = (1, 0, 0)</math>, <math>B = (0, 1, 0)</math>, <math>C = (0, 0, 1)</math>. By the [[Angle Bisector Theorem]], <math>D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)</math>. Since <math>M</math> is the midpoint of <math>AD</math>, <math>M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)</math>. Therefore, the equation for line BM is <math>20x = 31z</math>. Let <math>P = (x, 0, 1-x)</math>. Using the equation for <math>BM</math>, we get <cmath>20x = 31(1-x)</cmath> | We will use barycentric coordinates. Let <math>A = (1, 0, 0)</math>, <math>B = (0, 1, 0)</math>, <math>C = (0, 0, 1)</math>. By the [[Angle Bisector Theorem]], <math>D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)</math>. Since <math>M</math> is the midpoint of <math>AD</math>, <math>M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)</math>. Therefore, the equation for line BM is <math>20x = 31z</math>. Let <math>P = (x, 0, 1-x)</math>. Using the equation for <math>BM</math>, we get <cmath>20x = 31(1-x)</cmath> | ||
<cmath>x = \frac{31}{51}</cmath> Therefore, <math>\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}</math> so the answer is <math>\boxed{051}</math>. | <cmath>x = \frac{31}{51}</cmath> Therefore, <math>\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}</math> so the answer is <math>\boxed{051}</math>. | ||
− | + | == Solution 5 == | |
Let <math>DC=x</math>. Then by the Angle Bisector Theorem, <math>BD=\frac{20}{11}x</math>. By the Ratio Lemma, we have that <math>\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.</math> Notice that <math>[\triangle BAM]=[\triangle BMD]</math> since their bases have the same length and they share a height. By the sin area formula, we have that <cmath>\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.</cmath> Simplifying, we get that <math>\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.</math> Plugging this into what we got from the Ratio Lemma, we have that <math>\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}</math> | Let <math>DC=x</math>. Then by the Angle Bisector Theorem, <math>BD=\frac{20}{11}x</math>. By the Ratio Lemma, we have that <math>\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.</math> Notice that <math>[\triangle BAM]=[\triangle BMD]</math> since their bases have the same length and they share a height. By the sin area formula, we have that <cmath>\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.</cmath> Simplifying, we get that <math>\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.</math> Plugging this into what we got from the Ratio Lemma, we have that <math>\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}</math> | ||
− | + | == Solution 6 (quick Menelaus) == | |
First, we will find <math>\frac{MP}{BP}</math>. By Menelaus on <math>\triangle BDM</math> and the line <math>AC</math>, we have | First, we will find <math>\frac{MP}{BP}</math>. By Menelaus on <math>\triangle BDM</math> and the line <math>AC</math>, we have | ||
<cmath>\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.</cmath> | <cmath>\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.</cmath> | ||
Line 35: | Line 34: | ||
Therefore, <math>\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.</math> The answer is <math>\boxed{051}</math>. -brainiacmaniac31 | Therefore, <math>\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.</math> The answer is <math>\boxed{051}</math>. -brainiacmaniac31 | ||
− | ==Solution 7 (Visual)== | + | == Solution 7 (Visual) == |
[[File:2011 AIME II 4.png|400px]] | [[File:2011 AIME II 4.png|400px]] | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==Solution 8 (Cheese)== | + | == Solution 8 (Cheese) == |
− | Assume <math>ABC</math> is a right triangle at <math>A</math>. Line <math>AD = x</math> and <math>BC = \tfrac{-11}{20}x + 11</math>. These two lines intersect at <math>D</math> which have coordinates <math>(\frac{220}{31},\frac{220}{31})</math> and thus <math>M</math> has coordinates <math>(\frac{110}{31},\frac{110}{31})</math>. Thus, the line <math>BM = \tfrac{11}{51} \cdot (20-x)</math>. When <math>x = 0</math>, <math>P</math> has <math>y</math> coordinate equal to <math>\frac{11\cdot20}{51} | + | Assume <math>ABC</math> is a right triangle at <math>A</math>. Line <math>AD = x</math> and <math>BC = \tfrac{-11}{20}x + 11</math>. These two lines intersect at <math>D</math> which have coordinates <math>(\frac{220}{31},\frac{220}{31})</math> and thus <math>M</math> has coordinates <math>(\frac{110}{31},\frac{110}{31})</math>. Thus, the line <math>BM = \tfrac{11}{51} \cdot (20-x)</math>. When <math>x = 0</math>, <math>P</math> has <math>y</math> coordinate equal to <math>\frac{11\cdot20}{51} \frac{AP + CP}{AP} = 1 + \frac{CP}{AP}</math> = <math>\tfrac{51}{20} = 1 + \frac{CP}{AP},</math> which equals <math>{\tfrac{31}{20}},</math> giving an answer of <math>\boxed{51}.</math> |
+ | == Solution 9 (Menelaus + Ceva's + Angle Bisector Theorem) == | ||
+ | We start by using Menelaus' theorem on <math>\triangle ABD</math> and <math>EC</math>. | ||
+ | So, we see that <math>\frac{BC}{DC}\cdot\frac{DM}{AM}\cdot\frac{AE}{EB}=1</math>. | ||
+ | By Angle Bisector theorem, <math>\frac{BC}{DC}=\frac{31}{11}</math>, and therefore after plugging in our values we get <math>\frac{AE}{EB}=\frac{11}{31}</math>. | ||
+ | Then, by Ceva's on the whole figure, we have <math>\frac{CP}{PA}\cdot\frac{AE}{EB}\cdot\frac{BD}{DC}=1</math>. | ||
+ | Plugging in our values, we get <math>\frac{CP}{PA}=\frac{31}{20}</math>, giving an answer of <math>\boxed{51}</math>. | ||
+ | ~ESAOPS | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Latest revision as of 21:31, 14 January 2024
Contents
Problem 4
In triangle ,
and
. The angle bisector of
intersects
at point
, and point
is the midpoint of
. Let
be the point of the intersection of
and
. The ratio of
to
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let
be on
such that
. It follows that
, so
by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that
. Thus,
and
.
Solution 2 (mass points)
Assign mass points as follows: by Angle-Bisector Theorem, , so we assign
. Since
, then
, and
, so
.
Solution 3
By Menelaus' Theorem on with transversal
,
So
.
Solution 4
We will use barycentric coordinates. Let ,
,
. By the Angle Bisector Theorem,
. Since
is the midpoint of
,
. Therefore, the equation for line BM is
. Let
. Using the equation for
, we get
Therefore,
so the answer is
.
Solution 5
Let . Then by the Angle Bisector Theorem,
. By the Ratio Lemma, we have that
Notice that
since their bases have the same length and they share a height. By the sin area formula, we have that
Simplifying, we get that
Plugging this into what we got from the Ratio Lemma, we have that
Solution 6 (quick Menelaus)
First, we will find . By Menelaus on
and the line
, we have
This implies that
. Then, by Menelaus on
and line
, we have
Therefore,
The answer is
. -brainiacmaniac31
Solution 7 (Visual)
vladimir.shelomovskii@gmail.com, vvsss
Solution 8 (Cheese)
Assume is a right triangle at
. Line
and
. These two lines intersect at
which have coordinates
and thus
has coordinates
. Thus, the line
. When
,
has
coordinate equal to
=
which equals
giving an answer of
Solution 9 (Menelaus + Ceva's + Angle Bisector Theorem)
We start by using Menelaus' theorem on and
.
So, we see that
.
By Angle Bisector theorem,
, and therefore after plugging in our values we get
.
Then, by Ceva's on the whole figure, we have
.
Plugging in our values, we get
, giving an answer of
.
~ESAOPS
Video Solution by OmegaLearn
https://youtu.be/Gjt25jRiFns?t=314
~ pi_is_3.14
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.