If the perpendiculars from the vertices to the opposite faces of a tetrahedron be concurrent, then a sphere passes through the four feet of the perpendiculars, and consequently through the centre of gravity of each of the four faces, and through the mid-points of the segments of the perpendiculars between the vertices and their common point of intersection.
This theorem has been generalized for any tetrahedron; a sphere can be drawn through the four feet of the perpendiculars, and consequently through the mid-points of the lines from the vertices to the centre of the hyperboloid having these perpendiculars as generators, and through the orthogonal projections of these points on the opposite faces.
Also the auxiliarly circle is the locus of the feet of the perpendiculars from the foci on any tangent.
Since the area of the triangle FPP' is one half the product of FP into the perpendicular p from P on FP', it follows that if these perpendiculars were equal all round the orbit, the areas described during the infinitesimal time would be smallest at the pericentre and continually increase during the passage of the body to B.
Inside an equilateral triangle, for instance, of height h, - 2Ra/3y/h, (8) where a, 13, y are the perpendiculars on the sides of the triangle.
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