## Tetradivision

Divide the perimeter of a triangle through four points, so that the distances between successive points are equal.
This is a particular case of a more general problem initiated in the file DivisionProblem.html .

The two cases above are the classical, of inscribed square and rhombus, correspondingly. The second case is the unique solution, where the dividing, through its vertices, quadrilateral, has a vertex coinciding with a vertex of the triangle.

Both cases are particular cases of the preceding figure, constructed as follows. From a point D on a side, AB say, draw parallel DE to another side, BC say. Then complete the figure to a rhombus as shown. When D coincides with the foot of the bisector from C, we obtain a solution like the (II). When DF becomes orthogonal to BC, we become (I). It is plain that there are no other solutions. The equilateral (rhombus) inscribed of minimum perimeter is one of the three, in general, squares inscribed, of the case (I).

Notice that for N=3, the problem is equivalent to inscribing an equilateral in a triangle. This, like the present problem, admits of infinite solutions. For a discussion of the case N=5, look at the Pentadivision.html .

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