The Complete Works of Archimedes

...in square, that is, [the ratio of] γ to δ; but in length, the ratio which γ has to δ in length; [the same ratio] which the polygon circumscribed about circle α has to the surface of the pyramid circumscribed about the cone. For γ is the radius of the base of the cone, and δ is one side of the polygon. And the [line] to the side [is] from the [elements] of α. The perimeter of the polygon is a common height, [related] to half of the surface. Therefore, the rectilinear figure associated with circle α has the same ratio to the rectilinear figure associated with circle β. And that rectilinear figure [has the same ratio] to the surface of the pyramid circumscribed about the cone. So that the surface of the pyramid is [equal] to the rectilinear figure circumscribed about circle β. Since, therefore, the rectilinear figure circumscribed about circle β has a lesser ratio to the inscribed one than the surface of the cone has to circle β, the surface of the pyramid circumscribed about the cone will have a lesser ratio to the rectilinear figure inscribed in circle β than the surface of the cone has to circle β. Which is impossible. For the surface of the pyramid is shown to be greater than the surface of the cone. But the rectilinear figure inscribed in circle β will be less than circle β. Therefore, circle β is not less than the surface of the cone.

I say then that it is not greater either. For if possible, let it be greater. Again, let there be conceived a polygon inscribed in circle β, and another circumscribed, so that the circumscribed figure has to the inscribed a lesser ratio than circle β has to the surface of the cone. Let there be conceived a polygon inscribed in circle α similar to the one inscribed in circle β, and let there be erected from it a pyramid having the same vertex as the cone. Since, therefore, the figures inscribed in α and β are similar, they have the same ratio to one another as the squares of their radii [lines from the centers] have to one another; therefore, the ratio which polygon has to polygon, γ also has to δ in length. But γ has a greater ratio to δ than the polygon inscribed in circle α has to the surface of the pyramid inscribed in the cone. For the radius of circle α has a greater ratio to the side of the cone than the perpendicular drawn from the center to one side of the polygon has to the perpendicular drawn from the vertex of the cone to the side of the polygon. Therefore, the polygon inscribed in circle α has a greater ratio to the polygon inscribed in β than the polygon itself has to the surface of the pyramid. Therefore, the surface of the pyramid is greater than the polygon inscribed in β. But the polygon circumscribed about circle β has a lesser ratio to the inscribed one than circle β has to the surface of the cone. By much more, therefore, the polygon circumscribed about circle β has a lesser ratio to the surface of the pyramid inscribed in the cone than circle β has to the surface of the cone; which is impossible. For the circumscribed polygon is greater than circle β, while the surface of the pyramid [inscribed] in the cone is less than the surface of the cone. Since, therefore, the circle is neither greater than the surface of the cone, and it was shown that it is not less, it is therefore equal.

16 The surface of every isosceles cone has the same ratio to the base as the side of the cone has to the radius of the base of the cone. Let there be an isosceles cone, the base of which is circle α. Let the radius of α be equal to β, and the side of the cone be γ. It is to be shown that the surface of the cone has the same ratio to circle α as γ has to β. For let there be taken a mean proportional δ between β and γ, and let there be set out circle δ, having its radius equal to δ. Therefore, circle δ is equal to the surface of the cone; for this was shown in the [proposition] before this. Since, therefore, circle δ has the same ratio to circle α as the length of γ has to β. For each of them is in the [ratio] of the square of γ to β. This is because circles are to one another as the squares on their diameters; and likewise the squares on the radii of the circles. For the diameters and their halves, that is the radii, are [proportional], and the radii are equal to β and γ. It is manifest, then, that the surface of the cone has the same ratio to circle α as γ has to β in length.

Geometric diagram accompanying Proposition 16. On the left, a circle 'α' is shown with an inscribed hexagon. Below it are four horizontal line segments labeled 'β', 'γ', 'δ', and 'ε'. To the right of the circle is a diagram representing a cone, shown as a right-angled triangle with a vertical altitude, a slanted side (πλευρὰ), and a base radius along the horizontal.