Difference between revisions of "2011 AIME II Problems/Problem 4"
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== Problem 4 == | == Problem 4 == | ||
− | In triangle <math>ABC</math>, <math>AB= | + | 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>. |
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D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); | D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); | ||
− | </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 | + | </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 === | + | === 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>. | ||
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=== Solution 5 === | === 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> | ||
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+ | === 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 | ||
+ | <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> | ||
+ | This implies that <math>\frac{MB}{BP}=1-\frac{MP}{BP}=\frac{51}{62}</math>. Then, by Menelaus on <math>\triangle AMP</math> and line <math>BC</math>, we have | ||
+ | <cmath>\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.</cmath> | ||
+ | Therefore, <math>\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.</math> The answer is <math>\boxed{051}</math>. -brainiacmaniac31 | ||
== See also == | == See also == |
Latest revision as of 17:26, 21 November 2020
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 .
Contents
Solutions
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
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 |
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