The Complete Works of Archimedes

...rectilinear figure, so that the surface of the prism circumscribed about cylinder A is also equal to the rectilinear figure about circle B. And since the rectilinear figure about circle A has a lesser ratio to the one inscribed in circle A than the surface of cylinder A has to circle B, the surface of the prism circumscribed about cylinder A will also have a lesser ratio to the rectilinear figure inscribed in circle B than the surface of the cylinder has to circle B; and alternately. Which is impossible. For the surface of the prism circumscribed about the cylinder is greater, as has been shown, than the surface of the [inscribed] prism. But the inscribed rectilinear figure in circle B is less than circle B. Therefore, circle B is not less than the surface of the cylinder.

Let it be, then, if possible, greater. Again, let there be conceived a rectilinear figure inscribed in circle B, and another circumscribed, so that the circumscribed figure has to the inscribed a lesser ratio than circle B has to the surface of the cylinder. And let a polygon be inscribed in circle A similar to the one inscribed in circle B. And let a prism be erected from the polygon inscribed in circle A. And again, let line $KΔ$ be equal to the perimeter of the rectilinear figure inscribed in circle A. And let $HΛ$ be equal to it. Then the triangle from $KΔ$ will be greater than the rectilinear figure inscribed in circle A, because it has its perimeter as a base, and its height is greater than the radius—that is, the perpendicular drawn to one side of the polygon. And the [surface] composed of the parallelograms is equal to the surface of the prism, being composed of the parallelograms because it is contained by the side of the cylinder and the line equal to the perimeter of the rectilinear figure which is the base of the prism. So that the triangle $RHL$ is also equal to the surface of the prism.

Since the rectilinear figures inscribed in circles A and B are similar, they have the same ratio to each other as the squares on their radii. And triangles $KΛ T$ and $ZRL$ also have the same ratio to each other as the squares on the radii of the circles. Therefore, the rectilinear figure inscribed in circle A has the same ratio to the rectilinear figure inscribed in circle B as triangle $KΛ T$ has to triangle $LZR$. But the rectilinear figure inscribed in circle A is less than triangle $KΛ T$; therefore, the rectilinear figure inscribed in circle B is also less than triangle $ZRL$, and consequently, it is less than the surface of the prism inscribed in the cylinder. Which is impossible. For since the rectilinear figure circumscribed about circle B has a lesser ratio to the inscribed figure than circle B has to the surface of the cylinder; and alternately. But the figure circumscribed about circle B is greater than circle B. Therefore, the inscribed figure in circle B is greater than the surface of the cylinder; so also is the surface of the prism. If, then, circle B is greater than the surface of the cylinder [it leads to a contradiction]. But it was shown to be not less. Therefore, it is equal.

15

The surface of any isosceles cone, excluding the base, is equal to a circle whose radius is a mean proportional between the side of the cone and the radius of the circle which is the base of the cone.

Let the base of an isosceles cone be circle A. Let its radius be $Γ$. Let the side of the cone be equal to $KT$. And let $E$ be a mean proportional between $Γ$ and $LΔ$. Let circle B have its radius equal to $E$. I say that circle B is equal to the surface of the cone, excluding the base.

For if it is not equal, it is either greater or less. First, let it be less. There are then two unequal magnitudes, the surface of the cone and circle B, and the surface of the cone is greater. It is therefore possible to inscribe an equilateral polygon in circle B, and to circumscribe another similar to the inscribed one, such that the circumscribed has to the inscribed a lesser ratio than the surface of the cone has to circle B. Let there also be conceived a polygon circumscribed about circle A, similar to the one circumscribed about circle B. And let a pyramid be erected from the polygon about circle A, having the same vertex as the cone. Since, then, the polygons circumscribed about circles A and B are similar, they have the same ratio to each other as the squares on their radii; that is, as the ratio which $Γ$ has to...

Geometric diagram showing several vertical lines and line segments. A long vertical line on the left is marked with points κ and τ. To its right, a shorter vertical line is marked with ζ and ρ. At the bottom right, four horizontal or floating line segments are labeled with the Greek letters γ, λ, δ, and ε respectively.