(Song Qing, Nanchang University Affiliated High School, 330029)
In a cyclic convex quadrilateral $ABCD$, $\angle A = 60^{\circ}$, $BC = CD = 1$. Extend $AB$ and $DC$ to intersect at point $E$, and extend $BC$ and $AD$ to intersect at point $F$. It is known that the perimeters of $\triangle BCE$ and $\triangle DCF$ are both integers. Find the perimeter of quadrilateral $ABCD$.
Solution: As shown in Figure 3, it is easy to see that
$\angle E B C + \angle F D C = 180^{\circ}$.
It is also easy to see that $\angle E B C \neq 90^{\circ}$. Otherwise, if $\angle E B C = 90^{\circ}$, then the perimeter of $\triangle B C E$ would be $3 + \sqrt{3}$, which contradicts the given information.

Since $\angle E B C$ and $\angle F D C$ are one acute and one obtuse, let's assume $\angle E B C < 90^{\circ}$.
$$

It can also be proven that $B E + E C < B K + K C = 2 + \sqrt{3}$. Therefore,
$$
2 < B E + E C < 2 + \sqrt{3}.
$$

Since the perimeter of $\triangle B C E$ is an integer and $B C = 1$, $B E + E C$ must be an integer. Thus, $B E + E C = 3$.
Let $E C = m$, then $B E = 3 - m$.
Noting that $\angle B C E = 60^{\circ}$, by the cosine rule we have
$$
(3 - m)^{2} = 1^{2} + m^{2} - 1 \times m.
$$

Solving this, we get $m = \frac{8}{5}, 3 - m = \frac{7}{5}$.
Thus, $E C = \frac{8}{5}, B E = \frac{7}{5}$.
Since $\frac{C F}{D F} = \frac{\sin \angle C D F}{\sin \angle D C F} = \frac{\sin \angle C B E}{\sin \angle B C E} = \frac{E C}{B E} = \frac{8}{7}$,
when $C F = 8n$, $D F = 7n$.
Noting that $\angle D C F = 60^{\circ}$, by the cosine rule we have
$$
(7n)^{2} = 1^{2} + (8n)^{2} - 1 \times (8n).
$$

Solving this, we get $n = \frac{1}{5}$ or $\frac{1}{3}$.
Thus, $C F = 8n = \frac{8}{5}$ or $\frac{8}{3}$.
If $C F = \frac{8}{5} = C E$, it would lead to $\triangle D C F \cong \triangle B C E$, thus, $\angle F D C = \angle E B C = 90^{\circ}$. This contradicts the given information.
Therefore, $C F = \frac{8}{3}, D F = \frac{7}{3}$.
Let $A D = x, A B = y$.
By $\triangle E C B \sim \triangle E A D$, we have $\frac{E C}{E A} = \frac{C B}{A D}$. Thus,
$$
\frac{8}{5} x = \frac{7}{5} + y.
$$

By $\triangle F C D \sim \triangle F A B$, similarly we get
$$
\frac{8}{3} y = \frac{7}{3} + x.
$$

From equations (1) and (2), we have $\frac{8}{5} x = \frac{7}{5} + \frac{3}{8} \left( \frac{7}{3} + x \right)$.
Solving this, we get $x = \frac{13}{7}$.
Substituting into equation (1), we get $y = \frac{11}{7}$.
Therefore, the perimeter of quadrilateral $A B C D$ is $\frac{38}{7}$.
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)