Three. (25 points) A rectangular glass tabletop has shattered into several small glass fragments, each of which is a convex polygon. After reassembling the fragments into the original rectangle, there are 30 intersection points, 20 of which are on the boundary of the original rectangle (including the four vertices of the original rectangle), and the remaining 10 points are inside the rectangle. Inside the rectangle,

there are 45 seams (a line segment between two nodes is counted as one seam, as shown in Figure 3).

Determine the number of pieces of each type (referring to triangles, quadrilaterals, pentagons, etc.) that the rectangular tabletop has shattered into.

Note: If a convex polygon has $n$ points on its boundary, it is considered an $n$-sided polygon. For example, the polygon $A B C D E$ in Figure 3 is considered a pentagon.
Three, suppose among all the fragments, there are $a_{3}$ triangles, $a_{4}$ quadrilaterals, $\cdots \cdots, k$-sided polygons $a_{k}$ $\left(a_{3}, a_{4}, \cdots, a_{k}\right.$ are non-negative integers
$$
S_{\text {angle }}=a_{3} \times \pi+a_{4} \times 2 \pi+\cdots+a_{k}(k-2) \pi \text {. }
$$

On the other hand, there are 10 nodes inside the rectangle. For each node, the sum of the angles at the vertices of the surrounding polygons is $2 \pi$, and 10 internal nodes together obtain 10 $\times 2 \pi$ radians; there are 16 nodes on the boundary of the rectangle (excluding the 4 vertices), and at each of these nodes, the angles of the polygons meet to form a straight angle, and 16 such nodes together obtain $16 \pi$ radians; at the 4 vertices of the original rectangle, a total of $2 \pi$ radians are obtained from the polygon fragments. Therefore,
$$
S_{\text {angle }}=20 \pi+16 \pi+2 \pi=38 \pi \text {. }
$$

Thus, $a_{3}+2 a_{4}+\cdots+(k-2) a_{k}=38$.
Let the sum of the sides of these polygons be $S_{\text {side }}$.
Since each $n$-sided polygon has $n$ sides, then
$$
S_{\text {side }}=3 a_{3}+4 a_{4}+\cdots+k a_{k} \text {. }
$$

On the other hand, there are 45 seams inside the rectangle, each of which is a common side of two polygons, so they are counted twice; the 20 line segments on the boundary of the rectangle are each counted once, thus,
$$
S_{\text {side }}=2 \times 45+20=110 \text {. }
$$

Thus, $3 a_{3}+4 a_{4}+\cdots+k a_{k}=110$.
(2) - (1) gives $2\left(a_{3}+a_{4}+\cdots+a_{k}\right)=72$.

Hence $a_{3}+a_{4}+\cdots+a_{k}=36$.
(1) - (3) gives
$$
a_{4}+2 a_{5}+3 a_{6}+\cdots+(k-3) a_{k}=2 \text {. }
$$

Since all $a_{i} \in \mathbf{N}$, therefore
$$
a_{6}=a_{7}=\cdots=a_{k}=0, a_{4}+2 a_{5}=2 \text {. }
$$

Thus, either $a_{4}=2, a_{5}=0$; or $a_{4}=0, a_{5}=1$.
In summary, there are two solutions to this problem, i.e., there are a total of 36 fragments, among which, either 34 triangles and 2 quadrilaterals; or 35 triangles and 1 pentagon.
(Provided by Tao Pingsheng)