Archimedis Syracusani Opera (1544 Basel Editio Princeps)

to $KΞ$. Thus $KΞ$ to $HΘ$ has a smaller ratio, that is, $η$ original notation to $NΓ$. And $MK$ to $KL$. Further, $KM$ to $KL$ has a smaller ratio than $A$ to $B$. The side $η$ is that of the circumscribed polygon, and $Γ N$ is that of the inscribed; which was what was proposed to be found.

Again, there being two unequal magnitudes and a sector, it is possible to circumscribe a polygon about the sector, and to inscribe another, so that the side of the circumscribed polygon to the side of the inscribed polygon has a smaller ratio than the greater magnitude to the smaller. For let there be again two

A geometric diagram illustrates a proposition from Archimedes. On the left is a sector of a circle with arc $αβγ$ and center $δ$. Radial lines connect the center to points $α, ν, μ, β$ on the arc. A tangent line and various auxiliary lines are drawn, with points labeled $κ, λ, θ, μ, ν$. To the right of the sector are three vertical line segments of decreasing length, labeled $η, ε$, and $ζ$ from left to right.

unequal magnitudes $ε$ and $ζ$. Let $ε$ be the greater, and let $αβγ$ be the sector of the circle, having center $δ$. And let the sector $αδβ$ be established at $δ$. It is necessary to circumscribe and inscribe a polygon about the sector $αβδ$, having equal sides apart from $βδα$, so that the requirement is fulfilled. Let two unequal straight lines $κ, θ, η$ be found, and let $κ$ be the greater. So that $κ$ to $θκ$ has a smaller ratio than the greater magnitude to the smaller. This is possible. And from $θ$ similarly, having drawn $θλ$ at right angles to $θκ$, let $κλ$ be set equal to $η$. This is possible. Since $ζθκ$ is greater, the angle $αδβ$ being bisected, and the half being bisected, and this always happening, an angle will be taken smaller than double the angle $λκθ$.
Let the angle $αδμ$ be taken. The line $αμ$ is a side of a polygon inscribed in the circle. And if we bisect the angle $αδμ$ by the line $δν$, and from $ν$ we draw the tangent to the circle $ξ o$, this will be a side of the polygon circumscribed about the same circle, similar to the one inscribed. And similar to the previously stated, $ξ o$ to $αμ$ has a smaller ratio than the magnitude $ε$ to $ζ$.

A circle being given and two unequal magnitudes, to circumscribe about the circle a polygon, and to inscribe another, so that the circumscribed to the inscribed has a smaller ratio than the greater magnitude to the smaller. Let a circle $α$ be set out, and two unequal magnitudes $ε$ and $ζ$, and $ε$ is the greater. It is necessary to inscribe a polygon in the circle, and to circumscribe another, so that the required condition is met. For I take two unequal straight lines $η, θ$, of which $η$ is the greater, so that $η$ to $θ$ has a smaller ratio than $ε$ to $ζ$. And having taken the mean proportional $κ$ of $η$ and $θ$, $η$ is therefore greater than $κ$. Let a polygon be circumscribed about the circle, and another inscribed, so that the circumscribed...

A second geometric diagram showing a full circle labeled $α$. An inscribed triangle and a circumscribed polygon segment are shown. Radial lines connect to points $γ, ν, μ$ on the circumference, and a tangent segment is labeled $ξ o$. To the right are three vertical line segments labeled $η, κ, θ$. At the bottom are two horizontal line segments labeled $ε$ and $ζ$.