Part 4

Let the square $\alpha\beta\gamma\delta$ [Fig.~12] be the base of a perpendicular prism with square bases and let a cylinder be inscribed in the prism whose base is the circle $\epsilon\zeta\eta\theta$ which touches the sides of the parallelogram $\alpha\beta\gamma\delta$ at $\epsilon$, $\zeta$, $\eta$, and $\theta$. Pass a plane through its center and the side in the square opposite the square $\alpha\beta\gamma\delta$ corresponding to the side $\gamma\delta$; this will cut off from the whole prism a second prism which is $\frac{1}{4}$ the size of the whole prism and which will be bounded by three parallelograms and two opposite triangles. In the semicircle $\epsilon\zeta\eta$ describe a parabola whose origin is $\eta\epsilon$ and whose axis is $\zeta\kappa$, and in the parallelogram $\delta\eta$ draw $\mu\nu \| \kappa\zeta$; this will cut the circumference of the semicircle at $\xi$, the parabola at $\lambda$, and $\mu\nu \times \nu\lambda = \nu\zeta^2$ (for this is evident [Apollonios, \emph{Con.} I, 11]). Therefore $\mu\nu : \nu\lambda = \kappa\eta^2 : \lambda\sigma^2$. Upon $\mu\nu$ construct a plane parallel to $\epsilon\eta$; this will intersect the prism cut off from the whole prism in a right-angled triangle one side of which is $\mu\nu$ and the other a straight line in the plane upon $\gamma\delta$ perpendicular to $\gamma\delta$ at $\nu$ and equal to the axis of the cylinder, but whose hypotenuse is in the intersecting plane. It will intersect the portion which is cut off from the cylinder by the % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig12.png} \begin{center} {\small Fig.~12.} \end{center} \end{wrapfigure} % plane passed through $\epsilon\eta$ and the side of the square opposite the side $\gamma\delta$ in a right-angled triangle one side of which is $\mu\xi$ and the other a straight line drawn in the surface of the cylinder perpendicular to the plane $\kappa\nu$, \linebreak and the hypotenuse\dotfill\linebreak and all the triangles in the prism : all the triangles in the cylinder-section = all the straight lines in the parallelogram $\delta\eta$ : all the straight lines between the parabola and the straight line $\epsilon\eta$. And the prism consists of the triangles in the prism, the cylinder-section of those in the cylinder-section, the parallelogram $\delta\eta$ of the straight lines in the parallelogram $\delta\eta \| \kappa\zeta$ and the segment of the parabola of the straight lines cut off by the parabola and the straight line $\epsilon\eta$; hence prism : cylinder-section = parallelogram $\eta\delta$ : segment $\epsilon\zeta\eta$ that is bounded by the parabola and the straight line $\epsilon\eta$. But the parallelogram $\delta\eta = \frac{3}{2}$ the segment bounded by the parabola and the straight line $\epsilon\eta$ as indeed has been shown in the previously published work, hence also the prism is equal to one and one half times the cylinder-section. Therefore when the cylinder-section = 2, the prism = 3 and the whole prism containing the cylinder equals 12, because it is four times the size of the other prism; hence the cylinder-section is equal to $\frac{1}{6}$ of the prism, Q. E. D.

\section*{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 magni-\-\linebreak tude.\dotfill

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

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 $\delta\eta$ : all parallelograms in the figure which is described around the segment bounded by the parabola and the straight line $\epsilon\eta$, i. e., the prism cut off by the inclined plane : the figure described around the cylinder-section = parallelogram $\delta\eta$ : the figure bounded by the parabola and the straight line $\epsilon\eta$. 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\dotfill\linebreak.\dotfill\linebreak

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End of the Project Gutenberg EBook of Geometrical Solutions Derived from Mechanics, by Archimedes