8,9 A heptagon, three angles of which are equal to $120^{\circ}$, is inscribed in a circle. Can all its sides be of different lengths

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Let in this heptagon there are two adjacent angles of $120^{\circ}$, for example, the angles at vertices $A$ and $B$ (Fig.1). Then

$$
G A B=A B C=360^{\circ}-240^{\circ}=120^{\circ},
$$

therefore, $\angle B O G=\angle A O C=120^{\circ}$, where $O-$ is the center of the circle. Hence, $\angle A O G=\angle B O C$. Consequently, $A G=B C$. If the angles of $120^{\circ}$ are not adjacent, then without loss of generality, we can assume that these are the angles at vertices $A, C$ and $E$ (Fig.2). Then

$$
\angle B O G=\angle B O D=\angle D O F=120^{\circ}, \angle F O G=360^{\circ}-(\angle B O G+\angle B O D+\angle D O F)=360^{\circ}-3 \cdot 120^{\circ}=0^{\circ},
$$

## Answer

No.