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=\frac{20}{11} AC</math>. The angle bisector of <math>\ | + | In triangle <math>ABC</math>, <math>AB=\frac{20}{11} AC</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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Revision as of 20:23, 13 March 2015
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 .
Solution 3
By Menelaus' Theorem on with transversal ,
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 .
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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