QuadDivision

Quandrangle division

Here is the answer to a problem, proposed by Michael Metaxas: Given a convex quadrangle, divide it through two intersecting lines in four equal parts (in area). Following an idea of Michael Papadimitrakis, find first the envelope of the lines [EF] dividing the quadrangle in two equal parts. As pointed out by Antreas Varverakis, this envelope is a hyperbola, with asymptotic lines [AB] and [CD] (drawn in red), or [AD] and [BC] (not drawn). Then draw a tangent [EF] to this hyperbola at a point H of it. This creates (under certain restrictions) two other quadrangles: AEID and BEIC, having equal areas. Repeat the construction of the envelope for these two quadrangles. This determines two other hyperbolas (green and blue respectively). Draw a common tangent [FG] to these two hyperbolas. This completes the construction of the two lines, intersecting at F. Obviously there are infinite many solutions.

The fact that the envelope of the lines, dividing in two equal parts the quadrangle, is a hyperbola, is due to the fact that asymptotic triangles of a hyperbola have a constant area. Look at the file AsymptoticTriangle.html for a picture.

Look at the file ParaDivision.html for the solution of the special problem of a four-division of a parallelogram. In that case point F is always the center of the parallelogram. Which is the locus of F in the general case?

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