I would approach the issue from a completely different direction. Consider a triangle in 3D with vertices at $\vec {v}_0$, $\vec {v}_1$, and $\vec {v}_2$. It has a directed normal $\vec {n}$, $$\vec {n} = \left (\vec {v}_1 - \vec {v}_0\right)\times\left (\vec {v}_2 - \vec {v}_0\right) \tag {1}\label {1}$$ If we look along $\vec {n}$ in one direction, the vertices are clockwise; in the opposite ...
geometry - Orientation of a triangle's vertices in 3D space: Clockwise ...
Distance between triangle incenter and vertices Ask Question Asked 2 years ago Modified 2 years ago
Anyone know of an online tool available for making graphs (as in graph theory - consisting of edges and vertices)? I have about 36 vertices and even more edges that I wish to draw. (why do I have so
Think about the vertices of the polygon as potential candidates for vertices of the triangle. Using that, you get (n choose 3) as the number of possible triangles that can be formed by the vertices of a regular polygon of n sides.
combinatorics - How many triangles can be formed by the vertices of a ...
To get a formula where the vertices can be anywhere, just subtract the coordinates of the third vertex from the coordinates of the other two (translating the triangle) and then use the above formula.
Suppose that we had a set of vertices labelled $1,2,\ldots,n$. There will several ways to connect vertices using edges. Assume that the graph is simple and connected. In what efficient (or if the...
The sum of all the degrees is equal to twice the number of edges. Since the sum of the degrees is even and the sum of the degrees of vertices with even degree is even, the sum of the degrees of vertices with odd degree must be even. If the sum of the degrees of vertices with odd degree is even, there must be an even number of those vertices.