Geometrical Solutions Derived from Mechanics

 Table of Contents

 Introduction

 Geometrical Solutions Derived from Mechanics.

 Proposition I

 Proposition II

 Proposition III

 Proposition IV

 Proposition V

 Proposition VI

 Proposition VII

 Proposition VIIa

 Proposition VIII

 Proposition IX

 Proposition X

 Proposition XI

 Proposition XII

 Proposition XIII

 Proposition XIV

Proposition XIV

[Inscribe a cylinder in] a perpendicular prism with square bases [and let it be cut by a plane passed through the center of the base of the cylinder and one side of the opposite square.] Then this plane will cut off a prism from the whole prism and a portion of the cylinder from the cylinder. It may be proved that the portion cut off from the cylinder by the plane is one-sixth of the whole prism. But first we will prove that it is possible to inscribe a solid figure in the cylinder-section and to circumscribe another composed of prisms of equal altitude and with similar triangles as bases, so that the circumscribed figure exceeds the inscribed less than any given magnitude. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

But it has been shown that the prism cut off by the inclined plane < 3/2 the body inscribed in the cylinder-section. Now the prism cut off by the inclined plane: the body inscribed in the cylinder-section = parallelogram δη: the parallelograms which are inscribed in the segment bounded by the parabola and the straight line η. Hence the parallelogram δη < 3/2 the parallelograms in the segment bounded by the parabola and the straight line η. But this is impossible because we have shown elsewhere that the parallelogram δη is one and one half times the segment bounded by the parabola and the straight line η, consequently is. . . . . . . . . . . . . . . . . . . . not greater. . . . . . . . . . . . . . . . .. ..

And all prisms in the prism cut off by the inclined plane: all prisms in the figure described around the cylinder-section = all parallelograms in the parallelogram δη: all parallelograms in the figure which is described around the segment bounded by the parabola and the straight line η, i. e., the prism cut off by the inclined plane: the figure described around the cylinder-section = parallelogram δη: the figure bounded by the parabola and the straight line η. But the prism cut off by the inclined plane is greater than one and one half times the solid figure circumscribed around the cylinder-section. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..