Difference between revisions of "2011 AIME II Problems/Problem 4"
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</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 === | + | === 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>. | ||
Revision as of 11:14, 9 August 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 (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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