19.3 $ABCDE$ is a regular pentagon, $AP$, $AQ$, and $AR$ are the perpendiculars from $A$ to the lines $CD$, $CB$ extended, and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP=1$, then $AO + AQ + AR$ equals
(A) 3.
(B) $1+\sqrt{5}$.
(C) 4.
(D) $2+\sqrt{5}$.
(E) 5.
(37th American High School Mathematics Examination, 1986)
[Solution] Connect $A D, A C$, and connect $O D, O C$, and let the side length of the regular pentagon be $a$.
$$
\begin{aligned}
\because S_{\triangle B C D E} & =5 S_{\triangle O C D}=5 \cdot \frac{1}{2} \cdot 1 \cdot a \\
& =\frac{5}{2} a,
\end{aligned}
$$

and
$$
\begin{aligned}
S_{A B C D E} & =S_{\triangle A E D}+S_{\triangle A D C}+S_{\triangle A C B} \\
& =\frac{1}{2} a \cdot A R+\frac{1}{2} a \cdot A P+\frac{1}{2} a \cdot A Q \\
& =\frac{1}{2} a(A R+A P+A Q), \\
\therefore \quad \frac{5}{2} a & =\frac{1}{2} a(A R+A P+A Q) .
\end{aligned}
$$

Thus, $A R+A P+A Q=5$,
and $O P=1, A O=A P-1$.
$$
\therefore \quad A R+A O+A Q=4 \text {. }
$$

Therefore, the answer is $(C)$.