[ Law of Cosines $\quad]$ [ Properties and Criteria of Parallelograms ]

Given a parallelogram with sides 1 and 2 and an acute angle of $60^{\circ}$. On two of its opposite sides, isosceles triangles with angles of $120^{\circ}$ at the vertices are constructed outside the parallelogram. Find the distance between these vertices.
Let the angle at vertex $A$ of parallelogram $A B C D$ be $60^{\circ} ; \mathrm{M}$ and $>\mathrm{AB}=1, B C=2$. In the isosceles triangles $A D E$ and $B C F$, it is known that

$\angle \mathrm{AED}=\angle \mathrm{BFC}=120^{\circ}, A E=D E=B F=F C=\frac{B C}{\sqrt{3}}=\frac{2}{\sqrt{3}}$.

Note that quadrilateral $B E D F$ is a parallelogram, since its opposite sides $B F$ and $D E$ are equal and parallel. Therefore, its diagonal $E F$ passes through the midpoint $O$ of diagonal $B D$, i.e., through the center of parallelogram $A B C D$.

In triangle $A B D$, side $A B$ is half the length of side $A D$, and $\angle B A D=60^{\circ}$, so $\angle A B D=90^{\circ}$. Then

$$
\begin{aligned}
& B D=A B \cdot \operatorname{tg} 60^{\circ}=1 \cdot \sqrt{3}=\sqrt{3}, B O=\frac{1}{2} B D=\frac{\sqrt{3}}{2} \\
& \angle C B D=\angle A B C-\angle A B D=120^{\circ}-90^{\circ}=30^{\circ} \\
& \angle D B F=\angle C B D+\angle C B F=30^{\circ}+30^{\circ}=60^{\circ}
\end{aligned}
$$

From triangle $O B F$ using the cosine rule, we find that

$$
O F=\sqrt{B O^{2}+B F^{2}-2 B O \cdot B F \cos 60^{\circ}}=\sqrt{\frac{3}{1}+\frac{1}{3}-2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{2}}=\sqrt{\frac{13}{12}} \text {. }
$$

Therefore,

$$
E F=2 O F=2 \sqrt{\frac{13}{12}}=\sqrt{\frac{13}{3}}
$$

Now consider the isosceles triangles $A L B$ and $C K D$ with angles $120^{\circ}$ at vertices $L$ and $K$. Reasoning similarly, we will prove that $O$ is the midpoint of $K L$. From triangle $O K D$ using the cosine rule, we find that

$$
O K=\sqrt{D O^{2}+D K^{2}-2 D O \cdot D K \operatorname{ccs} 120^{\circ}}=\sqrt{\frac{3}{1}+\frac{1}{3}+2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{2}}=\sqrt{\frac{19}{12}} .
$$

Therefore,

$$
K L=2 O K=2 \sqrt{\frac{19}{12}}=\sqrt{\frac{19}{3}}
$$