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Difference between revisions of "1986 AHSME Problems/Problem 16"

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==Solution==
 
==Solution==
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Since we are given that <math>\triangle{PAB}\sim\triangle{PCA}</math>, we have the following
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Since we are given that <math>\triangle{PAB}\sim\triangle{PCA}</math>, we have <math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</math>.
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<math>\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}</math>.
 
  
  
 
Solving for <math>PA</math> in <math>\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}</math> gives us <math>PA=\frac{4PC}{3}</math>.
 
Solving for <math>PA</math> in <math>\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}</math> gives us <math>PA=\frac{4PC}{3}</math>.
  
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We also have <math>\frac{PA}{PC+7}=\frac{3}{4}</math>. Substituting <math>PA</math> in for our expression yields
 
  
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<math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math>
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We also have <math>\frac{PA}{PC+7}=\frac{3}{4}</math>. Substituting <math>PA</math> in for our expression yields <math>\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}</math>
  
  
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<math>\frac{16PC}{3}=3PC+21</math>
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Which we can further simplify to <math>\frac{16PC}{3}=3PC+21</math>
  
  

Latest revision as of 22:51, 10 February 2018

Problem

In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$. The length of $PC$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution

Since we are given that $\triangle{PAB}\sim\triangle{PCA}$, we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$.


Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$.


We also have $\frac{PA}{PC+7}=\frac{3}{4}$. Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$


Which we can further simplify to $\frac{16PC}{3}=3PC+21$


$\frac{7PC}{3}=21$


$PC=9\implies\boxed{\textbf{(C)}}$

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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