Difference between revisions of "2011 AIME II Problems/Problem 4"
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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 | + | === 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> |
Revision as of 21:08, 4 March 2020
Problem 4
In triangle , . 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
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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