Hexadivision

Construct inscribed equilateral hexagons LMNOPQ in a triangle, given the angles (alpha), (beta), (psi), as shown.

Essentially the angles of triangle ABC are given and triangle AKL, determining the slope of side MN to AB, is also given. Taking E arbitrary on AC and drawing EF parallel to KL, we find F. Then from E and F we lay correspondingly segments equal to EF, finding points D and G. Draw from G parallel to BC and take GJ = GF. Then construct the isosceles with basis on DJ and sides DI=IJ=JG. Find H as the vertex of rhombus GJIH. This completes the construction of an equilateral hexagon DEFGHI. The only shortcoming of the hexagon is that IH is not on BC but parallel to it, in general. The remedy lies in the fact that all hexagons constructed that way, starting with an arbitrary point E on AC, are similar to each other and their vertices I, J lie respectively on two fixed lines [AQ] and [AP]. This determines side QP on BC and this, in turn, determines the hexagon LMNOPQ. Later e.g. by taking the similarity with center at A and ratio k = QP/IH and applying it to the hexagon DEFGHI.

Problem: Find the inscribed hexagon of minimum perimeter/area. Show that it coincides with the one defined by a segment EF parallel to BC (psi+beta=pi). The hexagon then is symmetric, with sides parallel to the sides of the triangle ABC. The center of the hexagon is the triangle-center X(37). A picture of this particular inscribed hexagon is contained in HexadivisionSymmetric.html .

This is a particular case of the more general problem initiated in the file DivisionProblem.html . There are other special cases of inscribed equilateral hexagons corresponding to the partition (3,2,1) i.e. 3 vertices on one side, 2 on another and 1 on the third. These can be handled in a similar way. Finally there are cases where one of the vertices of the hexagon coincides with a vertex of the triangle.

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