An Icosagon Riddle

The total number of quadrilaterals formed by the 4 vertices of the icosagon is $C(20,4) = 4845$ .

Case 1: Quadrilaterals share exactly 1 side with the icosagon:

Choosing 1 side (2 vertices) out of 20 sides of the icosagon, we have $20$ options.

Choosing the other 2 vertices out of remaining 16 vertices (not neighbors of the 2 chosen vertices) of the icosagon, we have $C(16,2) = 120$ options but the 2 vertices mustn’t be neighbors. Because 2 neighboring vertices will form a shared side with the icosagon, and there are 16 such vertices so there are 15 sides formed by them. So we need to subtract $120$ with these $15$ unconditioned sides. Therefore, choosing the other 2 non-neighboring vertices will have $105$ options.

Overall, the number of options in this case is $20*105=2100$

Case 2: Quadrilaterals share exactly 2 sides with the icosagon:

1. The two sides are adjacent:

Choosing 3 points: $20$ options. Choosing the other point, we have $20$ subtract 3 chosen points and 2 points neighboring them so $15$ options. Combining, we have $20*15 = 300$ options.

1. The two sides are non-adjacent:

Choosing 1 side out of 20 sides of the icosagon, we have $20$ options.

The remaining 16 vertices (excluding the 2 chosen points and 2 points adjacent to them) form 15 sides so there are $options$ . The combining number of options is $20*15 = 300$ . However, this number needs to be divided by 2 because half of the quadrilaterals are the same as the other half.

Overall, the number of options in this case is $300 + 150 = 450$ .

Case 3: Quadrilaterals share exactly 3 sides with the icosagon: $20$ options.

The number is quadrilaterals who don’t share any side with the icosagon is the total number of quadrilaterals in general subtracted by the specific cases combined. Therefore, $4845 - 2100 - 450 - 20 = \boxed{2275}$ .