The Complete Works of Archimedes

...to $KΞ$. So that the [line] $KΞ$ to the whole $KΞ$ has a lesser ratio, that is, $Π$ to $ΥΓ$, or $MK$ to $KΛ$. Furthermore $M$ to $KΛ$ has a lesser ratio than $A$ to $B$. And $Π$ is the side of the circumscribed polygon, and $ΓΥ$ that of the inscribed; which was the thing proposed to be found.

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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 lesser ratio than the greater magnitude to the lesser. For let there be again two

A geometric diagram showing a sector of a circle with center Α and arc ΒΓ. A radius ΑΔ bisects the angle ΒΑΓ. A tangent line ΚΘ is drawn through point Δ. To the right of the sector, there is a separate construction involving a vertical line segment ΚΛ bisected at Θ, with various horizontal magnitudes labeled Ε, Ζ, Η below it. Points are labeled with Greek letters: Α, Β, Γ, Δ, Ε, Ζ, Η, Θ, Ι, Κ, Λ, Μ.

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unequal magnitudes $E, Z$, of which let $E$ be the greater; and a circle $ABΓ$ having center $A$; and at $A$ let the sector $ABΓ$ be constituted. It is necessary indeed to circumscribe and to inscribe a polygon about the sector $BΔ$, having equal sides apart from $BA, Δ A$, so that the requirement is fulfilled. For let two unequal straight lines $HK$ and $Θ K$ be found, and let $HK$ be the greater, so that $HK$ to $Θ K$ has a lesser ratio than the greater magnitude to the lesser. For this is possible. And at $Θ$, let $ΘΛ$ be drawn likewise at right angles to $Θ K$, and let $KΛ$ be joined. This is possible. Since the [line] $H$ is greater than $Θ K$, if the angle at $A$ is bisected, and the half again bisected, and this being done continually, there will be left a certain angle that is less than double the angle $LKΘ$. Let the angle $AΔ M$ be left. For this is the side of the polygon inscribed in the circle. And if we bisect the angle $AΔ M$ by $Δ N$, and from $N$ we draw the tangent to the circle $Ξ O$, this will be the side of the polygon circumscribed about the same circle, similar to the one mentioned. And in a manner similar to the things said before, $Ξ O$ to $AM$ has a lesser ratio than the magnitude $E$ to $Z$.

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A circle being given and two unequal magnitudes, to circumscribe a polygon about the circle, and to inscribe another, so that the circumscribed [polygon] to the inscribed [polygon] has a lesser ratio than the greater magnitude to the lesser. Let the circle $A$ be set forth, and two unequal magnitudes $E, Z$, and let $E$ be the greater. It is necessary indeed to inscribe a polygon in the circle, and to circumscribe another, so that the thing commanded is fulfilled. For I take two unequal straight lines $H, Θ$, of which let $H$ be the greater, so that $H$ to $Θ$ has a lesser ratio than $E$ to $Z$. And a mean proportional $K$ being taken between $H$ and $Θ$, then $H$ is also greater than $K$. Let a polygon be circumscribed about the circle, and another be inscribed, so that the [side] of the circumscribed...

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