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

1986 AMC 12 Problem 16

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16.

\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

7

8

9

10

11

Use the vertex order in

\triangle PAB\sim\triangle PCA

to match corresponding sides

Set

PC=t

and use both

PB=PC+7

and the repeated similarity ratio

The stated order of similarity gives

\frac{PA}{PC}=\frac{PB}{PA}=\frac{AB}{CA}=\frac43.

Let

PC=t.

Then

PA=\frac{4t}{3}

and

PB=\frac{16t}{9}.

Since

PB=PC+CB=t+7,

\frac{16t}{9}=t+7,

which gives

t=9.

Thus the correct answer is C.