Difference between revisions of "2011 AIME II Problems/Problem 4"

Revision as of 01:30, 10 March 2014

Problem 4

In triangle $ABC$ , $AB=\frac{20}{11} AC$ . The angle bisector of $\ang A$ (Error compiling LaTeX. Unknown error_msg) intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solutions

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 $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$ . It follows that $\triangle BPC \sim \triangle DD'C$ , so $\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}$ by the Angle Bisector Theorem. Similarly, we see by the midline theorem that $AP = PD'$ . Thus, $\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},$ and $m+n = \boxed{051}$ .

Solution 2

Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$ , so we assign $m(B) = 11, m(C) = 20, m(D) = 31$ . Since $AM = MD$ , then $m(A) = 31$ , and $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$ .

Solution 3

By Menelaus' Theorem on $\triangle ACD$ with transversal $PB$ , $1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.$

Solution 4

We will use barycentric coordinates. Let $A = (1, 0, 0)$ , $B = (0, 1, 0)$ , $C = (0, 0, 1)$ . By the angle bisector theorem, $D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)$ . Since $M$ is the midpoint of $AD$ , $M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)$ . Therefore, the equation for line BM is $20x = 31z$ . Let $P = (x, 0, 1-x)$ . Using the equation for $BM$ , we get $20x = 31(1-x)$ $x = \frac{31}{51}$ Therefore, $\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}$ so the answer is $\boxed{051}$ .

@@ Line 20: / Line 20: @@
 === 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>
+===Solution 4===
+We will use barycentric coordinates. Let <math>A = (1, 0, 0)</math>, <math>B = (0, 1, 0)</math>, <math>C = (0, 0, 1)</math>. By the angle bisector theorem, <math>D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)</math>. Since <math>M</math> is the midpoint of <math>AD</math>, <math>M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)</math>. Therefore, the equation for line BM is <math>20x = 31z</math>. Let <math>P = (x, 0, 1-x)</math>. Using the equation for <math>BM</math>, we get <cmath>20x = 31(1-x)</cmath>
+<cmath>x = \frac{31}{51}</cmath> Therefore, <math>\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}</math> so the answer is <math>\boxed{051}</math>.
 == See also ==

2011 AIME II (Problems • Answer Key • Resources)
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