1986 AHSME Problems/Problem 16

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 $\frac{16PC}{3}=3PC+21$

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

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

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1986 AHSME Problems/Problem 16

Problem

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