We can start by considering the case of n=3, which is just a triangle. There is only one possible triangle that can be formed by joining the vertices of a triangle, so T3=1.

For n=4, we can draw a square and see that there are two possible triangles that can be formed by joining any three vertices. However, we need to be careful not to count the same triangle twice. For example, if we choose vertices A, B, and C, we get one triangle, but if we choose vertices A, C, and B (in a different order), we get the same triangle. So we need to divide by 3, which is the number of ways we can arrange the vertices of a triangle. This gives us T4=2/3.

For n=5, we can draw a pentagon and see that there are three possible triangles that can be formed by joining any three vertices. Again, we need to be careful not to count the same triangle twice, so we divide by 3 to get T5=3/3=1.

For n=6, we can draw a hexagon and see that there are four possible triangles that can be formed by joining any three vertices. This time, we need to divide by 6, which is the number of ways we can arrange the vertices of a triangle. This gives us T6=4/6=2/3.

We can continue this pattern and find that:

T7=5/3
T8=7/3
T9=8/3
T10=11/3

It seems that Tn is always a fraction with numerator n-2 and denominator 3. To prove this, we can use combinatorics. There are n choose 3 ways to choose three vertices from an n-sided polygon. However, we need to be careful not to count degenerate triangles (i.e. triangles with zero area). A triangle has zero area if and only if its vertices are collinear. There are n ways to choose three collinear vertices, so we need to subtract this from the total. This gives us:

Tn = (n choose 3) - n = (n(n-1)(n-2)/6) - n = (n^3 - 3n^2 + 2n)/6

Simplifying this expression gives:

Tn = (n-2)(n-1)/3

which confirms our observation. Therefore, the answer to the problem is T_2022 = (2022-2)(2022-1)/3 = 3417078.