Example 4.2.12 Use various regular polygons as tiles to cover the ground, without allowing the vertices of the regular polygons to be placed on the edges of another regular polygon, and requiring that the placement of regular polygons at each vertex be the same. How many choices are there?
Solution: (1) If six regular polygons surround each vertex, then these six regular polygons can only be equilateral triangles, and the scheme is $(3,3,3,3,3,3)$.
(2) If five regular polygons surround each vertex, since $60^{\circ} \times 2 + 90^{\circ} \times 3 > 360^{\circ}$, at least three equilateral triangles are required. Also, $360^{\circ} - 60^{\circ} \times 3 = 180^{\circ}$, so the remaining two regular polygons can only be two squares or one equilateral triangle and one regular hexagon. The schemes are $(3,3,3,4,4)$ and $(3,3,3,3,6)$.
(3) If four regular polygons surround each vertex, let the number of sides of these four regular polygons be $n_{1}, n_{2}, n_{3}, n_{4}$, with $n_{1} \leqslant n_{2} \leqslant n_{3} \leqslant n_{4}$. Since $\sum_{i=1}^{4} \frac{n_{i}-2}{n_{i}} \times 180^{\circ} = 360^{\circ}$, we have $\sum_{i=1}^{4} \frac{1}{n_{i}} = 1$,
$$
\frac{1}{n_{1}} \geqslant \frac{1}{n_{i}}, \text{ hence } \frac{1}{n_{1}} \geqslant \frac{1}{4}, n_{1} \leqslant 4.
$$
A. $n_{1}=4$, then $\frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}} = \frac{3}{4}$, which can only be $(4,4,4,4)$.
B. $n_{1}=3$, then $\frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}} = \frac{2}{3}$. Since $\frac{1}{n_{2}} \geqslant \frac{1}{n_{3}} \geqslant \frac{1}{n_{4}}$, we have $\frac{1}{n_{2}} \geqslant \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9} > \frac{1}{5}$, so $n_{2} \leqslant 4$.
(i) $n_{2}=4$, then $\frac{1}{n_{3}} + \frac{1}{n_{4}} = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$.
$\frac{1}{n_{3}} \geqslant \frac{1}{2} \cdot \frac{5}{12} > \frac{1}{5}$, so $n_{3} = 4$, and the scheme is $(3,4,4,6)$.
(ii) $n_{2}=3$, then $\frac{1}{n_{3}} + \frac{1}{n_{4}} = \frac{1}{3}$, $\frac{1}{n_{3}} \geqslant \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$, so $n_{3} \leqslant 6$.
$n_{3}=6, n_{4}=6$, the scheme is $(3,3,6,6)$.
$n_{3}=5$, no solution for $n_{4}$.
$n_{3}=4, n_{4}=12$, the scheme is $(3,3,4,12)$.
$n_{3}=3$, no solution for $n_{4}$.
$(3,3,4,12)$ satisfies the equation $\sum_{i=1}^{4} \frac{1}{n_{i}} = 1$, but the tiling scheme $(3,3,4,12)$ does not exist.

In fact, if the tiling scheme $(3,3,4,12)$ exists, then each vertex of the equilateral triangle $\triangle ABC$ should be paired with one equilateral triangle, one square, and one regular dodecagon.

Let the polygon adjacent to $AB$ be a regular $n_{1}$-gon, and the polygon adjacent to $AC$ be a regular $n_{2}$-gon, then the polygon opposite $\angle A$ must be a regular $n_{3}$-gon (where $n_{1}, n_{2}, n_{3}$ are different from 3, 4, 12), and the polygon adjacent to $BC$ can only be a regular $n_{3}$-gon, the polygon opposite $\angle B$ can only be a regular $n_{2}$-gon, and the polygon opposite $\angle C$ can only be a regular $n_{3}$-gon.
The three polygons opposite the three vertices of $\triangle ABC$ have side numbers $n_{1}, n_{2}, n_{3}$, respectively. Without loss of generality, assume the polygon opposite $\angle A$ is an equilateral triangle. Suppose the polygon adjacent to $AB$ is a square $ABDE$ (the proof is the same if the polygon adjacent to $AC$ is a square), then the polygon adjacent to $BC$ is an equilateral triangle, the polygon adjacent to $BD$ is a regular dodecagon, the polygon adjacent to $DE$ is an equilateral triangle, and the polygons adjacent to $AF, EF$ should both be regular dodecagons. The configuration around point $F$ is different, so the scheme $(3,3,4,12)$ does not exist.
(4) If three regular polygons surround each vertex, let the number of sides be $n_{1}, n_{2}, n_{3}, n_{1} \leqslant n_{2} \leqslant n_{3}$.
$\frac{1}{n_{1}} \geqslant \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$, so $n_{1} \leqslant 6$.
A. $n_{1}=3$, consider the equilateral $\triangle ABC$, let the polygon adjacent to $AB$ be a regular $n_{2}$-gon, and the polygon adjacent to $AC$ be a regular $n_{3}$-gon. If $n_{2} \neq n_{3}$, the polygon adjacent to $BC$ cannot be chosen, so $n_{2} = n_{3}$, and the scheme is $(3,12,12)$.
B. $n_{1}=4$, $\frac{1}{n_{2}} + \frac{1}{n_{3}} = \frac{1}{4}$.
$$
\frac{1}{n_{2}} \geqslant \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}, n_{2} \leqslant 8.
$$
(i) $n_{2}=4$, no solution;
(ii) $n_{2}=5, n_{3}=20$, the scheme $(4,5,20)$ satisfies the equation $\sum_{i=1}^{3} \frac{1}{n_{i}} = \frac{1}{2}$.

However, the scheme $(4,5,20)$ does not exist. In fact, consider the regular pentagon. If $AB$ is adjacent to a square, then $BC, AE$ should be adjacent to a regular 20-gon, and $CD, DE$ should be adjacent to a square, but the configuration around point $D$ is different.
(iii) $n_{2}=6, n_{3}=12$, the scheme is $(4,6,12)$.
(iv) $n_{2}=7$, no solution for $n_{3}$.
(v) $n_{2}=8, n_{3}=8$, the scheme is $(4,8,8)$.
C. $n_{1}=5$,
For the scheme $(5, n_{2}, n_{3})$ to exist, $n_{2} = n_{3}$ must hold,
$\frac{2}{n_{2}} = \frac{1}{2} - \frac{1}{5} = \frac{3}{10}$, no solution.
D. $n_{1}=6$,
$$
\frac{1}{n_{2}} + \frac{1}{n_{3}} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}.
$$

The scheme is $(6,6,6)$.
The ten tiling schemes discussed above can all be realized through drawing. The ten realizable tiling schemes are listed in the table below:
\begin{tabular}{|c|c|c|c|c|}
\hline Scheme \\
Regular polygons \\
surrounding each vertex \\
number
\end{tabular}