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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?

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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: enter image description here

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:

  1. 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.

  2. 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.

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    Very nice answer! +1 – 2017-02-03
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    I understand the second part, the first one not so much. – 2017-02-03
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    He is counting the triangles you don't want to include, then subtracting those "bad" triangles from the total number of triangles, to get the number of "good" triangles. He first has to count the number of "bad" triangles that have exactly one side with the 12-gon. He does this by first choosing the side that is shared, which determines two vertices of the triangle, then choosing the third vertex. There are 12 choices for the side that's shared, and 8 for the last vertex (because 2 vertices are on the side already, and another 2 are adjacent to the side). This gives 12*8=96. – 2017-02-03
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Another method, choosing the points of the triangle one by one then dividing by $6$ because each triangle has been chosen $6$ times (by choosing the vertices in different orders):

Choose the first vertex. $12$ options.

Choose the second vertex.
- case (a) a next-but-one vertex: $2$ options
- case (b) any other vertex: $7$ options

Choose the last vertex.
- in case (a): $7$ options
- in case (b): $6$ options

This gives $12 \times (2\times 7 + 7\times 6) = 672$ options, then the number of triangles is $ 672/6 = \fbox{112}$.

The reason case (a) gives more selection options for the third vertex is that the next-but-one point shares a neighbour with the first point.