11.16. (England, 66). Find the number of sides of a regular polygon if for four of its consecutive vertices \( A, B, C, D \) the equality

\[
\frac{1}{A B}=\frac{1}{A C}+\frac{1}{A D}
\]

is satisfied.
11.16. Let a circle with center 0 and radius $R$ be circumscribed around a polygon (Fig. 58). Denote $\alpha=\angle A O B$, then $0<\alpha<$ $<120^{\circ}$ and

$A B=2 R \sin (\alpha / 2), A C=2 R \sin \alpha$,

$$
A D=2 R \sin (3 \alpha / 2)
$$

from which we have

$$
\frac{1}{\sin (\alpha / 2)}=\frac{1}{\sin \alpha}+\frac{1}{\sin (3 \alpha / 2)}
$$

Therefore, we obtain

$$
\begin{array}{r}
0=\sin \alpha \sin \frac{3 \alpha}{2}-\left(\sin \alpha+\sin \frac{3 \alpha}{2}\right) \times \\
\times \sin \frac{\alpha}{2}=\frac{1}{2}\left(\cos \frac{\alpha}{2}-\cos \frac{5 \alpha}{2}\right)-\frac{1}{2}\left(\cos \frac{\alpha}{2}-\cos \frac{3 \alpha}{2}\right)- \\
-\frac{1}{2}(\cos \alpha-\cos 2 \alpha)=\frac{1}{2}\left(\left(\cos \frac{3 \alpha}{2}+\cos 2 \alpha\right)-\left(\cos \alpha+\cos \frac{5 \alpha}{2}\right)\right)= \\
=\cos \frac{7 \alpha}{4}\left(\cos \frac{\alpha}{4}-\cos \frac{3 \alpha}{4}\right)=2 \cos \frac{7 \alpha}{4} \sin \frac{\alpha}{4} \sin \frac{\alpha}{2}
\end{array}
$$

from which

$$
7 \alpha / 4=90^{\circ} \text { and } \alpha=360^{\circ} / 7
$$

i.e., the original polygon has seven sides.