Archimedis Syracusani Opera (1544 Basel Editio Princeps)

16

If an isosceles cone is cut by a plane parallel to its base, the surface of the cone between the parallel planes is equal to a circle whose radius is the 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 triangle through the axis is equal to α β γ. And let it be cut by a plane parallel to the base, forming the section δ ε. Let the axis of the cone be β η. And let a circle θ be set out whose radius is the mean proportional between α δ and the sum of δ ζ and α κ. I say that the circle θ is equal to the surface of the cone between δ ε and α γ. For let circles λ and π be set out. And let the square of the radius of circle π be equal to the rectangle contained by β δ and δ ζ. And the square of the radius of λ be equal to the rectangle contained by β α and κ α. Therefore, circle λ is equal to the surface of the cone α β γ. And circle π is equal to the surface of the cone δ ε β. And since the rectangle contained by β α and α η is equal to the rectangle contained by β δ and δ ζ plus the rectangle contained by δ α and the sum of δ ζ and α η, due to the parallelogram δ ζ π κ α κ. But the square of the radius of circle λ is equal to the rectangle α β by α η. And the square of the radius of circle π is equal to the rectangle β δ by δ ζ. And the square of the radius of θ is equal to the rectangle of δ α and the sum of δ ζ and α η. Therefore, the square of the radius of circle λ is equal to the sum of the squares of the radii of circles π and θ. So that circle λ is equal to the sum of circles π and θ. But π is equal to the surface of the cone β δ ε. Therefore, the remainder, the surface of the cone between the parallel planes δ ε and α γ, is equal to circle θ.

A geometric diagram showing a triangle αβγ representing the cross-section of a cone through its axis βη. A horizontal line δε is drawn parallel to the base αγ. Within the lower part of the diagram, there are several concentric circular arcs and lines illustrating geometric proportions, labeled with various Greek letters including π, κ, θ, λ, and ζ.

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

A rectangular geometric diagram representing a parallelogram ΒΑΚ subdivided into several smaller rectangles and sections. A diagonal line βκ runs across the figure. Various internal lines create sub-rectangles and areas labeled with Greek letters (α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ν, ξ, π). Curved arcs are drawn within some of the sections to illustrate mathematical relationships or the gnomon described in the text.