The Complete Works of Archimedes

...by the straight line subtending. And let a certain circle $Ξ$ be set out, of which the square on the radius is equal to the rectangle contained by $AE$ and a line equal to [the sum of] $EZ, HΘ, ΓΛ, MN$. I say that this circle is equal to the surface of the figure inscribed in the sphere. For let there be set out circles $O, Π, P, Σ, T, Υ$. And let the square on the radius of $O$ be equal to the rectangle contained by $AE$ and half of $EZ$. And let the square on the radius $Π$ of the circle $Π$ be equal to the rectangle contained by $AE$ and half of [the sum of] $EZ, HΘ$. And let the square on the radius of $P$ be equal to the rectangle contained by $AE$ and half of [the sum of] $HΘ, ΓΛ$. And let the square on the radius of $Σ$ be equal to the rectangle contained by $AE$ and half of [the sum of] $ΓΛ, KΛ$. And let the square on the radius of $T$ be equal to the rectangle contained by $AE$ and half of [the sum of] $KΛ, MN$. And let the square on the radius of $Υ$ be equal to the rectangle contained by $AE$ and half of $MN$. On account of these things, circle $O$ is equal to the surface of the cone $AEZ$, and $Π$ to the surface of the cone between $EZ, HΘ$, and $P$ to the surface between $HΘ, ΓΛ$, and $Σ$ to the surface between $ΓΛ, KΛ$. Moreover, circle $T$ is equal to the surface of the cone between $KΛ, MN$, and $Υ$ is equal to the surface of the cone $MBN$. Therefore, all the circles together are equal to the surface of the inscribed figure. And it is manifest that the squares on the radii of the circles $O, Π, P, Σ, T, Υ$ are equal to the rectangle contained by $AE$ and half of [the sum of] $EZ, HΘ, ΓΛ, KΛ, MN$. The whole lines are $EZ, HΘ, ΓΛ, MN$. Therefore, the squares on the radii of the circles $O, Π, P, Σ, T, Υ$ would be equal to the rectangle contained by $AE$ and all of $EZ, HΘ, ΓΛ, KΛ, MN$. But also the square on the radius of circle $Ξ$ is equal to the rectangle contained by $AE$ and the sum of all $EZ, HΘ, ΓΛ, KΛ, MN$. Therefore, the square on the radius of circle $Ξ$ is equal to [the sum of] the squares on the radii of circles $O, Π, P, Σ, T, Υ$. And thus circle $Ξ$ is equal to circles $O, Π, P, Σ, T, Υ$. But circles $O, Π, P, Σ, T, Υ$ were demonstrated to be equal to the surface of the figure. Therefore circle $Ξ$ will also be equal to the surface of the figure.

21 Of the figure inscribed in the sphere, the surface contained by the conical surfaces is less than four times the greatest circle of those in the sphere. For let there be a sphere, and let $ABCD$ be its greatest circle. And let there be inscribed in it an equilateral polygon of an even number of sides, of which the sides are measured by four. And let a surface be conceived upon it contained by conical surfaces. I say that the surface of the inscribed figure is less than four times the greatest circle of those in the sphere. For let the lines subtending two sides of the polygon be joined, $EZ, HΘ, MK$, and let $ZK, HB, ΛΠ$ be parallel to these. And let a certain circle $P$ be set out, of which the square on the radius is equal to the rectangle contained by $AE$ and a line equal to all of $EZ, ZK, HB, ΛΠ, BM$. On account of what was said before, circle $P$ is equal to the aforementioned surface of the figure. And it was shown that this [line] equal to all $EZ, ZK, HB, ΔΛ, KΛ, BM$ [stands] to the diameter of the circle $AG$ as $GE$ is to $EA$. Therefore, the rectangle contained by the line equal to all those mentioned and $EA$, that is, the square on the radius of circle $P$, is equal to the rectangle contained by $AG, GE$.

But the rectangle contained by $AG, GE$ is less than the square on $AG$. Therefore, the square on the radius of $P$ is less than the square on $AG$. Therefore, the radius of $P$ is less than $AG$. So that the diameter of circle $P$ is less than double the diameter of circle $ABCD$. And therefore two diameters of circle $ABCD$ are greater than the diameter of circle $P$. And four times the square on the diameter of circle $ABCD$, that is, of $AG$, is greater than the square on the diameter of circle $P$. As four times the square on $AG$ is to the square on the diameter of circle $P$, so are four circles $ABCD$ to circle $P$. Therefore, four circles $ABCD$ are greater than circle $P$. Thus, circle $P$ is less than four times the greatest circle. And the circle was shown to be equal [to the surface].

A geometric diagram showing a large circle (representing a sphere) with points labeled with Greek letters (α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ν). A horizontal diameter (αγ) and a vertical diameter (βδ) intersect at the center. An inscribed polygon is represented by chords. Parallel horizontal chords and vertical lines extending from the vertices to the horizontal diameter are shown, illustrating the construction of conical surfaces within the sphere.