Proposition VI

[The center of gravity of a hemisphere is so divided on its axis] that the portion near the surface of the hemisphere is in the ratio of 5: 3 to the remaining portion.

Let a sphere be cut by a plane through its center intersecting the surface in the circle αβγδ [Fig.6], αγ and βδ being two diameters of the circle perpendicular to each other. Let a plane be constructed on βδ perpendicular to αγ. Then imagine a cone whose base is the circle with the diameter βδ, whose vertex is at α and its lateral boundaries are βα and αδ; let γα be produced so that αθ = γα, imagine the straight line θγ to be a scale-beam with its center at α and in the semi-circle βαδ draw a straight line ξo βδ; let it cut the circumference of the semicircle in ξ and o, the lateral boundaries of the cone in π and ρ, and αγ in. On ξo construct a plane perpendicular to α; it will intersect the hemisphere in a circle with the diameter ξo, and the cone in a circle with the diameter πρ. Now because αγ: α = ξα²: α² and ξα² = α² + ξ² and α = π, therefore αγ: α = ξ² + π²: π². But ξ² + π²: π² = the circle with the diameter ξo + the circle with the diameter πρ: the circle with the diameter πρ, and γα = αθ, hence θα: α = the circle with the diameter ξo + the circle with the diameter πρ: circle with the diameter πρ. Fig. 6. Therefore the two circles whose diameters are ξo and πρ in their present position are in equilibrium at the point α with the circle whose diameter is πρ if it is transferred and so arranged at θ that θ is its center of gravity. Now since the center of gravity of the two circles whose diameters are ξo and πρ in their present position [is the point, but of the circle whose diameter is πρ when its position is changed is the point θ, then θα: α = the circles whose diameters are] ξo [,πρ: the circle whose diameter is πρ. In the same way if another straight line in the] hemisphere βαδ [is drawn βδ and a plane is constructed] perpendicular to [αγ the] two [circles produced in the cone and in the hemisphere are in their position] in equilibrium at α [with the circle which is produced in the cone] if it is transferred and arranged on the scale at θ. [Now if] the hemisphere and the cone [are filled up with circles then all circles in the] hemisphere and those [in the cone] will in their present position be in equilibrium [with all circles] in the cone, if these are transferred and so arranged on the scale-beam at θ that θ is their center of gravity; [therefore the hemisphere and cone also] are in their position [in equilibrium at the point α] with the cone if it is transferred and so arranged [on the scale-beam at θ] that θ is its center of gravity.