All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?

Options are

- 49
- 65
- 70
- 96

Option 1 which is **49** is the correct answer.

Let the number of intersections be x.

We know that x <= ^{8}C_{4 }= 70, as every vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract ^{4}C_{2} - 1 = 5. (70 - 5 = 65)

There are 8 type of three diagonals intersecting at the same point, so we need to subtract 8 x 2 = 16 (as we will count it once)

**65 - 16 = 49
Another simple method using option**

As we know maximum number of intersection is 70. there will be four diagonals intersect in the center, so we need to subtract

^{4}C

_{2}- 1 = 5. (70 - 5 = 65)Probably there will be more.

We have only 1 option which is less than 65.

i.e. 49