12. (10 points) Cut a pentagon along a straight line into two polygons, then cut one of the polygons along a straight line into two parts, resulting in three polygons, and then cut one of the polygons along a straight line into two parts, $\cdots$, and so on. To have 20 pentagons among the resulting polygons, what is the minimum number of cuts needed?
【Analysis】According to the problem, a polygon can be divided into two parts, and the sum of its interior angles can increase by at most $360^{\circ}$. Cutting $K$ times can increase the degree by at most $K \times 360^{\circ}$, so the sum of the angles of these $(K+1)$ polygons is at most $K \times 360^{\circ} + 540^{\circ}$. On the other hand, the sum of the angles of 20 pentagons is $20 \times 540^{\circ}$, and the sum of the angles of the remaining $(K-19)$ polygons is at least $(K-19) \times 180^{\circ}$. This gives us: $(K-19) \times 180^{\circ} + 20 \times 540^{\circ} \leqslant K \times 360^{\circ} + 540^{\circ}$, solving this yields the result.

【Solution】According to the analysis, when a polygon is divided into two parts, the sum of its interior angles can increase by at most $360^{\circ}$. Cutting $K$ times can increase the degree by at most $K \times 360^{\circ}$, so the sum of the angles of these $(K+1)$ polygons is at most $K \times 360^{\circ} + 540^{\circ}$,

On the other hand, the sum of the angles of 20 pentagons is $20 \times 540^{\circ}$, and the sum of the angles of the remaining $(K-19)$ polygons is at least $(K-19) \times 180^{\circ}$,

This gives us: $(K-19) \times 180^{\circ} + 20 \times 540^{\circ} \leqslant K \times 360^{\circ} + 540^{\circ}$;
Simplifying, we get: $K \geqslant 38$,
When $K=38$, we can first cut a pentagon into one pentagon and one quadrilateral, then use 18 cuts to divide the quadrilateral into 19 quadrilaterals,

Then use 19 cuts to cut each quadrilateral into a pentagon, thus using 38 cuts to divide it into 20 pentagons.
In summary, the minimum number of cuts required is 38.
Therefore, the answer is: 38.