ARCHIMEDES ON THE SPHERE

side of the circumscribed polygon to that of the inscribed polygon has a smaller ratio than $Γ$ to $Δ$, as we have learned. Therefore, the ratio of the doubled [polygon] to the doubled is also smaller. And the ratio of the side to the side is doubled, as is the ratio of the polygons to the polygons, for they are similar. $Γ$ to $Δ$ is the whole of $Γ$ to $Δ$. Thus, the circumscribed polygon has a smaller ratio to the inscribed polygon than $Γ$ to $Δ$. Therefore, the circumscribed polygon also has a much smaller ratio to the inscribed polygon than $E$ to $Z$. Similarly, we shall show that, given two unequal magnitudes and a sector, it is possible to circumscribe a polygon about the sector and to inscribe another similar to it, so that the circumscribed polygon has a smaller ratio to the inscribed polygon than the greater magnitude to the smaller. It is also evident that if a circle or sector and a certain area are given, it is possible to inscribe equilateral polygons in the circle or sector. And furthermore, by always leaving out segments of the circle or sector in the remaining parts, these will be smaller than the proposed area. These things have been handed down in the elements.

6

It must also be shown that, given a circle or sector and an area, it is possible to circumscribe a polygon about the circle or sector such that the segments left over by the circumscription are smaller than the given area. For let it be as it remains of the circle, to transfer the similar ratio to the sector as well. Let circle $A$ be given, and a certain area $B$. It is possible to circumscribe a polygon about the circle such that the segments intercepted between the circle and the polygon are smaller than area $B$. For indeed, there being two unequal magnitudes—the greater being the sum of the area and the circle, and the lesser being the circle—let a polygon be circumscribed about the circle and another inscribed, such that the circumscribed has a smaller ratio to the inscribed than the stated greater magnitude to the lesser. This circumscribed polygon is such that the remainders will be smaller than the proposed area $B$. For if the circumscribed has a smaller ratio to the inscribed than the sum of the circle and area $B$ to the circle itself; and since the circle is greater than the inscribed, much more does the circumscribed have a smaller ratio to the circle than the sum of the circle and area $B$ to the circle itself. And by dividing, the remainders of the circumscribed polygon have a smaller ratio to the circle than area $B$ to the circle. Thus, the remainders of the circumscribed polygon are smaller than area $B$, or not. Since the circumscribed has a smaller ratio to the circle than the sum of the circle and area $B$ to the circle, for this reason the circumscribed will be less than the sum. So that the remainders will be smaller than area $B$. And similarly for the sector.

A geometric diagram shows a circle inscribed within a square. A regular hexagon is inscribed within the circle. Radial lines connect the center (α) to the vertices of the hexagon. A corner of the square is labeled (β).

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If a pyramid is inscribed in an isosceles cone, having an equilateral base, its surface, excluding the base, is equal to a triangle having a base equal to the perimeter of the base, and a height equal to the perpendicular drawn from the apex to one side of the base. Let there be an isosceles cone whose base is the circle $ABΓ$, and let an equilateral pyramid be inscribed in it having triangle $ABΓ$ as its base. I say that its surface, excluding the base, is equal to the stated triangle. For since the cone is isosceles and the base of the pyramid is equilateral, the heights of the triangles containing the pyramid are equal to one another. And the triangles have bases $AB, BΓ, Γ A$, and the stated height. Thus, the triangles are equal to a triangle having a base equal to the sum of $AB, BΓ, Γ A$, and height equal to the stated straight line; that is to say, the surface of the pyramid, excluding the triangle $ABΓ$.

A geometric diagram depicts a triangle with vertices labeled α at the top, and β and γ at the base. Internal lines meet at a central point, with a vertical altitude drawn from vertex α to the base line βγ.

THE PROOF MORE CLEARLY.

Let there be an isosceles cone whose base is the circle $ABΓ$ and whose apex is the point $D$.