## Rotation-product

Consider a convex pentagon and the clockwise rotations about its vertices by the angles defined there. What is their product? As shown below, the product is a point-symmetry. The center of the symmetry lies on the last side of the pentagon, starting counting from the vertex of the first rotation.

The figure speaks from itself. The rotations are divided in two parts. Those that turn about the vertices of the quadrangle, composed by the first four vertices (GFJI) and the rotations of the remaining triangle IHG. More precisely, denote by g2, f, j, i1 the rotations clockwise about the vertices of the quadrangle (GFJI) by its angles there. Denote also by s the translation by the vector 2*XY. Denote then by g1, i2, h the rotation about the vertices of the triangle GIH. The requested product is the composition of rotations w = h*i2*i1*j*f*g2*g1. The composition is applied from right to left ( w(x) = h(....(g1(x)...) ).

In the file RotationsOnQuadrangleVertices.html it is proved that s = (g2*f*j*i1). Thus, replacing in the long product, w becomes w = h*i2*s*g1. s being a translation parallel to side GI of the triangle GIH.

In the file Isometries_product.html it is proved that such a product w = h*i2*s*g1 is always a symmetry with center on the side GA and at distance |ZA| equal to half the length of the translation vector of s, as shown in the figure.
In the above figure freely movable are the vertice of the pentagon (GFJIH) and the point N. S is simply w(N) and A denotes the center of the point-symmetry w.

The extension to more general odd sided polygons can be carried easily, taking into acount the result discussed in the file Isometries_product2.html .