Proposition XI

When in a perpendicular prism with square bases a cylinder is inscribed whose bases lie in opposite squares and whose curved surface touches the four other parallelograms, and when a plane is passed through the center of the circle which is the base of the cylinder and one side of the opposite square, then the body which is cut off by this plane [from the cylinder] will be 1/6 of the entire prism. This can be perceived through the present method and when it is so warranted we will pass over to the geometrical proof of it.

Fig. 9.

Fig. 10.

Imagine a perpendicular prism with square bases and a cylinder inscribed in the prism in the way we have described. Let the prism be cut through the axis by a plane perpendicular to the plane which cuts off the section of the cylinder; this will intersect the prism containing the cylinder in the parallelogram αβ [Fig. 9] and the common intersecting line of the plane which cuts off the section of the cylinder and the plane lying through the axis perpendicular to the one cutting off the section of the cylinder will be βγ; let the axis of the cylinder and the prism be γδ which is bisected at right angles by ζ and on ζ let a plane be constructed perpendicular to γδ. This will intersect the prism in a square and the cylinder in a circle. Now let the intersection of the prism be the square μν [Fig. 10], that of the cylinder, the circle ξoπρ and let the circle touch the sides of the square at the points ξ, o, π and ρ; let the common line of intersection of the plane cutting off the cylinder-section and that passing through ζ perpendicular to the axis of the cylinder, be κλ; this line is bisected by πθξ. In the semicircle oπρ draw a straight line στ perpendicular to πχ, on στ construct a plane perpendicular to ξπ and produce it to both sides of the plane enclosing the circle ξoπρ; this will intersect the half-cylinder whose base is the semicircle oπρ and whose altitude is the axis of the prism, in a parallelogram one side of which = στ and the other = the vertical boundary of the cylinder, and it will intersect the cylinder-section likewise in a parallelogram of which one side is στ and the other μν [Fig. 9]; and accordingly μν will be drawn in the parallelogram δ βω and will cut off ι = πχ. Now because γ is a parallelogram and νι θγ, and θ and βγ cut the parallels, therefore θ: θι = ωγ: γν = βω: υν. But βω: υν = parallelogram in the half-cylinder: parallelogram in the cylinder-section, therefore both parallelograms have the same side στ; and θ = θπ, ιθ = χθ; and since πθ = θξ therefore θξ: θχ = parallelogram in half-cylinder: parallelogram in the cylinder-section. lmagine the parallelogram in the cylinder-section transferred and so brought to ξ that ξ is its center of gravity, and further imagine πξ to be a scale-beam with its center at θ; then the parallelogram in the half-cylinder in its present position is in equilibrium at the point θ with the parallelogram in the cylinder-section when it is transferred and so arranged on the scale-beam at ξ that ξ is its center of gravity. And since χ is the center of gravity in the parallelogram in the half-cylinder, and ξ that of the parallelogram in the cylinder-section when its position is changed, and ξθ: θχ = the parallelogram whose center of gravity is χ: the parallelogram whose center of gravity is ξ, then the parallelogram whose center of gravity is χ will be in equilibrium at θ with the parallelogram whose center of gravity is ξ. In this way it can be proved that if another straight line is drawn in the semicircle oπρ perpendicular to πθ and on this straight line a plane is constructed perpendicular to πθ and is produced towards both sides of the plane in which the circle ξoπρ lies, then the parallelogram formed in the half-cylinder in its present position will be in equilibrium at the point θ with the parallelogram formed in the cylinder-section if this is transferred and so arranged on the scale-beam at ξ that ξ is its center of-gravity; therefore also all parallelograms in the half-cylinder in their present positions will be in equilibrium at the point θ with all parallelograms of the cylinder-section if they are transferred and attached to the scale-beam at the point ξ; consequently also the half-cylinder in its present position will be in equilibrium at the point θ with the cylinder-section if it is transferred and so arranged on the scale-beam at ξ that ξ is its center of gravity.