The Complete Works of Archimedes

16

If an isosceles cone is cut by a plane parallel to the base, there is a circle equal to the surface of the cone between the parallel planes, the radius of which circle is a mean proportional between the side of the cone between the parallel planes and the sum of the radii of the circles in the parallel planes. Let there be a cone whose axial triangle is equal to $α β γ$. And let it be cut by a plane parallel to the base, and let it make the section $δ ε$. Let the axis of the cone be $β η$. Let a certain circle $θ$ be set out, the radius of which is a mean proportional between $α δ$ and the sum of $α ζ, α η$. Let the circle be $θ$. I say that circle $θ$ is equal to the surface of the cone between $δ ε$ and $α γ$. For let circles $λ, κ$ be set out; and let the square on the radius of circle $λ$ be equal to the rectangle contained by $β α, α ζ$, and let the square on the radius of $κ$ be equal to the rectangle contained by $β δ, δ ζ$. Therefore, circle $λ$ is equal to the surface of the cone $α β γ$, and circle $κ$ is equal to the surface of the cone $δ β ε$. And since the rectangle $β α, α ζ$ is equal both to the rectangle $β δ, α ζ$ and the rectangle $δ α, α ζ$, and to the sum of $δ ζ, α η$, because $ζ η$ is parallel to $α κ$; but the square on the radius of circle $λ$ is equal to the rectangle $β α, α κ$. And the square on the radius of circle $κ$ is equal to the rectangle $β δ, δ ζ$. And the square on the radius of $θ$ is equal to the rectangle contained by $δ α$ and the sum of $α ζ, α η$. Therefore, the square on the radius of circle $λ$ is equal to the squares on the radii of circles $κ, θ$. Consequently, circle $λ$ is also equal to circles $κ, θ$. But $λ$ is equal to the surface of cone $α β γ$, and $κ$ is equal to the surface of cone $δ β ε$. Therefore, the remaining surface of the cone between the parallel planes $δ ε, α γ$ is equal to circle $θ$.

A geometric diagram showing a large triangle with vertex β at the top and base α-γ. A line segment δ-ε is drawn parallel to the base, cutting the triangle. Below the triangle is a construction involving circles and radial lines. Labels include α, β, γ, δ, ε, ζ, η, θ, κ, λ.

Let there be a parallelogram $β α κ η$, and let its diagonal be $β η$. Let side $β α$ be cut at random at $δ$, and through $δ$ let $δ θ$ be drawn parallel to $α η$, and through $ζ$ let $κ λ$ [be drawn parallel] to $β α$. I say that the rectangle $β α, α κ$ is equal both to the rectangle $β δ, δ ζ$ and the rectangle contained by $δ α$ and the sum of $δ ζ, α η$. For since the rectangle $β α, α κ$ is the whole $β η$, and the rectangle $β δ, δ ζ$ is $β ζ$, and the rectangle contained by $δ α$ and the sum of $δ ζ, α η$ is the gnomon $μ ν ξ$. For the rectangle $δ α, α κ$ is equal to $κ η$, because the complement $κ θ$ is equal to the complement $δ λ$. And the rectangle $δ α, δ ζ$ is $δ λ$. Therefore the whole $β η$, which is the rectangle $β α, α κ$, is equal to the rectangle $β δ, δ ζ$ and to the gnomon $μ ν ξ$, which is indeed the rectangle contained by $δ α$ and the sum of $α ζ, α η$.

A rectangular geometric construction (parallelogram β-α-κ-η) with a diagonal β-η. It is subdivided into smaller rectangles and squares by horizontal and vertical lines (δ-θ, κ-λ). Circular arcs are inscribed within parts of the figure. The labels μ, ν, ξ denote sections of a gnomon shape within the diagram.