Thus, there is only one point (B') whose successive rotations come back to close to a triangle.

One has to represent the rotations as products of reflections, using the bisectors of the triangle.

The cyclic permutations f2*f1*f3 and f1*f3*f2 give the corresponding symmetries w.r. to the other feet of I on the sides of ABC. The other (3) remaining permutations give the inverse transformations of the preceding ones.

Free movable is the triangle ABC and the point X. Switch to the selection-tool (Ctrl+1) to catch and modify the figure.

Produced with EucliDraw© |