Step 1: Understanding the Concept:
We are forming triangles using the vertices of a 20-sided polygon.
We need to exclude any triangles that share one or two sides with the polygon itself.
We can solve this by finding the total number of triangles and subtracting those with shared sides, or by using a direct formula.
Step 2: Key Formula or Approach:
For a polygon with $n$ sides (and $n$ vertices):
1. Total number of triangles formed by vertices = $\binom{n}{3}$.
2. Number of triangles sharing exactly two sides with the polygon = $n$ (these are formed by any three consecutive vertices).
3. Number of triangles sharing exactly one side with the polygon = $n(n - 4)$ (choose one side in $n$ ways, the third vertex must not be adjacent to this side's endpoints, leaving $n-4$ choices).
4. Number of triangles sharing NO sides = Total - (sharing 2 sides) - (sharing 1 side).
Alternative direct formula: $\frac{n(n-4)(n-5)}{6}$.
Step 3: Detailed Explanation:
Here, $n = 20$.
Let's use the subtraction method first.
Total number of triangles = $\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 20 \times 19 \times 3 = 1140$.
Triangles sharing 2 sides = $n = 20$.
Triangles sharing 1 side = $n(n - 4) = 20(20 - 4) = 20 \times 16 = 320$.
Triangles sharing zero sides = Total - (sharing 2 sides) - (sharing 1 side)
$= 1140 - 20 - 320 = 1140 - 340 = 800$.
Alternatively, using the direct formula:
Number of such triangles $= \frac{n(n-4)(n-5)}{6}$
$= \frac{20 \times (20 - 4) \times (20 - 5)}{6}$
$= \frac{20 \times 16 \times 15}{6}$
$= \frac{20 \times 8 \times 30}{6}$ (simplifying first) or just cancel: $\frac{15}{6} = \frac{5}{2}$, then $\frac{16}{2} = 8$.
$= 20 \times 8 \times 5 = 160 \times 5 = 800$.
Step 4: Final Answer:
The number of such triangles is $800$.