Solution: Fix one vertex and look at how many isosceles triangles exist where the legs are adjacent to this vertex. There will be a total of $3 k-1$ such triangles. Summing over all possible fixed vertices, we get $6 k \cdot(3 k-1)$. In this sum, all equilateral triangles were counted 3 times. There are a total of $2 k$ equilateral triangles. Therefore, the number of isosceles triangles is

$$
6 k \cdot(3 k-1)-2 \cdot 2 k=18 k^{2}-10 k
$$

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 53900 | 64200 | 75400 | 87500 | 11000 | 15900 | 21700 | 28400 | 36000 | 44500 |
4. (7 points) Find the number of isosceles triangles formed by the sides and diagonals of a regular $6k$-gon?