In a regular 12-sided polygon, how many triangles can be formed using vertices of the polygon that do not share a side with the polygon? I think is is 12 choose 3 minus 12, because that is the number of triangles that share a side with the polygon. Is that correct?

You're not quite right.  For each side of the 12-gon, there are 12-4=8 non-adjacent vertices which you can use to form a triangle, so there are 12*8=96 such triangles.  There are also those triangles which share two sides with the 12-gon, of which there are 12 (one for each vertex of the 12-gon--choose the two sides adjacent to the vertex).  So there are 108 total excluded triangles, and thus the answer is $\binom{12}{3}-108 = 220-108 = 112$.
This figure shows two of the 8 possible triangles sharing the right edge of the 12-gon:


Here's another way to get the answer:
For a triangle to not share an edge with the 12-gon, between each pair of triangle vertices there must be another vertex of the 12-gon.  Number the vertices of the 12-gon 1 through 12.  Consider two cases for a valid triangle:


*

*Vertex 1 is not on the triangle.  In this case, consider getting the 12-gon by starting with a 9-gon and putting 3 triangle vertices between the existing 9 vertices.  There are $\binom{9}{3} = 84$ ways to do this.

*Vertex 1 is on the triangle.  In this case, consider getting the 12-gon by starting with a 10-gon consisting of vertex 1 and the 9 other vertices which are not triangle vertices.  Then the remaining 2 triangle vertices can go between vertices 2 & 3, 3 & 4, ..., 9 & 10, so there are 8 places to put them.  There are $\binom{8}{2} = 28$ ways to do this.
Any valid triangle falls into case 1 or 2, so there are $84+28 = 112$ possible triangles.
