$12 \cdot 52$ Given an equilateral triangle with a side length of 10, divide it into several equilateral triangles with a side length of 1 using lines parallel to its sides. There are $m$ equilateral triangles with a side length of 2 and $25-m$ parallelograms (as shown in figure (a)).
(1) If $m=10$, can they be used to form the original triangle?
(2) Find all $m$ for which they can be used to form the original triangle.
[Solution] Divide the large equilateral triangle into 100 smaller equilateral triangles and alternately color them white or black, ensuring that any two triangles sharing a common side are of different colors (see figure (b)). Thus, there are 55 white triangles and 45 black triangles. The given parallelogram blocks always cover 2 white and 2 black triangles, while the equilateral triangle blocks with a side length of 2 can cover 1 white and 3 black or 1 black and 3 white triangles. Since there are 10 more white triangles than black triangles, the number of blocks covering 3 white and 1 black must be 5 more than the number of blocks covering 1 white and 3 black. Therefore, $m \geqslant 5$ and $m$ must be odd. Hence, when $m=10$, it is impossible to form the original triangle using them.

On the other hand, divide the original triangle into 5 equilateral triangles with a side length of 2 and 10 rhombus blocks with a side length of 2 as shown in figure (c). Since each rhombus block can be divided into two equilateral triangles with a side length of 2, it follows that for all odd numbers $m$ satisfying the inequality $5 \leqslant m \leqslant 25$, the original triangle can be formed using the given 25 shapes.