$36 \cdot 44 \triangle ABC$ has side lengths of $6, 8, 10$, with the center at $P$, and a radius of 1. A circle rolls inside $\triangle ABC$, always remaining tangent to at least one side. When $P$ first returns to its initial position, the distance traveled by $P$ is
(A) 10.
(B) 12.
(C) 14.
(D) 15.
(E) 17.
(44th American High School Mathematics Examination, 1993)
[Solution] As shown in the figure, the path traced by point $P$ is $\triangle A^{\prime} B^{\prime} C^{\prime}$, which is similar to $\triangle A B C$.
$$
\begin{aligned}
\because \quad A D & =A E=\operatorname{ctg} \frac{A}{2}=\frac{1+\cos A}{\sin A} \\
& =\frac{1+\frac{4}{5}}{\frac{3}{5}}=3 . \\
\therefore \quad A^{\prime} B^{\prime} & =A B-\operatorname{ctg} \frac{A}{2}-\operatorname{ctg} \frac{B}{2}=4=\frac{1}{2} A B,
\end{aligned}
$$

That is, the similarity ratio of $\triangle A^{\prime} B^{\prime} C^{\prime}$ to $\triangle A B C$ is $\frac{1}{2}$.
Therefore, the perimeter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is
$$
\frac{8+6+10}{2}=12 \text {. }
$$

Hence, the answer is $(B)$.