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

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Problem 4

In triangle $ABC$ , $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ 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 $m+n = \boxed{51}$ .

Solution 2 (mass points)

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 $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$ , so $m+n = \boxed{51}$ .

Solution 3

By Menelaus' Theorem on $\triangle ACD$ with transversal $PB$ , $1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.$ So $m+n = \boxed{051}$ .

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 $\boxed{051}$ .

Solution 5

Let $DC=x$ . Then by the Angle Bisector Theorem, $BD=\frac{20}{11}x$ . By the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.$ Notice that $[\triangle BAM]=[\triangle BMD]$ since their bases have the same length and they share a height. By the sin area formula, we have that $\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.$ Simplifying, we get that $\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.$ Plugging this into what we got from the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}$

Solution 6 (quick Menelaus)

First, we will find $\frac{MP}{BP}$ . By Menelaus on $\triangle BDM$ and the line $AC$ , we have $\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.$ This implies that $\frac{MB}{BP}=1-\frac{MP}{BP}=\frac{51}{62}$ . Then, by Menelaus on $\triangle AMP$ and line $BC$ , we have $\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.$ Therefore, $\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.$ The answer is $\boxed{051}$ . -brainiacmaniac31

Solution 7 (Visual)

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 34: / Line 34: @@
 <cmath>\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.</cmath>
 Therefore, <math>\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.</math> The answer is <math>\boxed{051}</math>. -brainiacmaniac31
+==Solution 7 (Visual)==
+[[File:2011 AIME II 4.png|400px]]
+'''vladimir.shelomovskii@gmail.com, vvsss'''
 == See also ==

2011 AIME II (Problems • Answer Key • Resources)
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