ARCHIMEDES ON THE SPHERE AND CYLINDER

are greater than the surface of the cylinder excluding the base.

13 The surface of every right cylinder, excluding the base, is equal to a circle whose radius has a mean proportional relationship between the side of the cylinder and the diameter of the base of the cylinder. Let the circle $α$ be the base of a right cylinder. Let $γδ$ be equal to the diameter of circle $α$, and $εζ$ be equal to the side of the cylinder. Let $θλ$ be the mean proportional between $γδ$ and $εζ$. Let circle $β$ be constructed such that its radius is equal to $θλ$. I must show that circle $β$ is equal to the surface of the cylinder, excluding the base. For if it is not equal, it is either greater or smaller. Let it first be smaller, if possible. Since there are two unequal magnitudes, the surface of the cylinder and circle $β$, it is possible to inscribe an isopleuron equilateral polygon in circle $β$ and circumscribe another such that the ratio of the circumscribed to the inscribed is less than the ratio of the surface of the cylinder to circle $β$. Let us assume this has been done. Also, let a polygon be circumscribed about circle $α$, similar to the one circumscribed about circle $β$. And let a prism be constructed from the polygon. This will be circumscribed about the cylinder. Let $κθ$ be equal to the perimeter of the polygon circumscribed about circle $α$. And let $λθ$ be a right angle likely referring to the construction of a height. And since $θλ$ is half of $γδ$, the triangle $κθλ$ will be equal to the polygon circumscribed about circle $α$, because it has a base equal to the perimeter and a height equal to the radius of circle $α$. And the parallelogram $ελ$ is equal to the surface of the prism circumscribed about the cylinder, because it is contained by the side of the cylinder and the line equal to the perimeter of the base of the prism. Let $ερ$ be equal to $εζ$. The triangle $ζελ$ is a right triangle, equal to the parallelogram $ελ$, and thus equal to the surface of the prism. And since the polygons circumscribed about circles $α$ and $β$ are similar, the polygons will have the same ratio as the squares of their radii. Therefore, the ratio of triangle $κθλ$ to the polygon circumscribed about circle $β$ is the ratio of the square of $θλ$ to the square of $ηκ$. For $θλ$ and $ηκ$ are equal to the radii. But the ratio of the square of $θλ$ to the square of $ηκ$ is the same as the ratio of $γδ$ to $εζ$ in length. For $ηκ$ is the mean proportional between $θλ$ and $εζ$, because $γδ$ and $εζ$ are also proportional. How so? Since $δγ$ is to $γθ$ This geometric derivation follows the standard Euclidean mean proportional construction... A geometric diagram shows two circles with centers labeled alpha (α) and beta (β) at the top. A large rectangle occupies the central and lower part of the space. To the left, inside the rectangle's boundary, is a triangle with vertices kappa (κ), theta (θ), and lambda (λ). To the right are two vertical line segments: the left one is labeled with points gamma (γ), theta (θ), and delta (δ); the right one is labeled epsilon (ε), rho (ρ), and zeta (ζ).