Difference between revisions of "2011 AIME II Problems/Problem 4"
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− | Problem | + | == Problem == |
+ | In triangle <math>ABC</math>, <math>AB=\frac{20}{11} AC</math>. The angle bisector of <math>\ang 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>. | ||
− | + | __TOC__ | |
+ | == Solution == | ||
+ | === Solution 1 === | ||
+ | <asy> | ||
+ | pointpen = black; pathpen = linewidth(0.7); | ||
− | ---- | + | pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); |
− | |||
− | + | 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 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{051}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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>. | ||
+ | |||
+ | === 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> | ||
+ | [/hide] | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year = 2011|n=II|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 12:57, 31 March 2011
Problem
In triangle , . The angle bisector of $\ang A$ (Error compiling LaTeX. ! Undefined control sequence.) 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
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 , [/hide]
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 |