The Complete Works of Archimedes

And let $A$ be greater than $B$. I say that it is possible to find two unequal straight lines, having performed the stated requirement. For taking the double of the first of the circles, and $BG$ [equal] to the equal $B$, and taking the straight line $ZH$. And $HA$, being added to itself, will exceed $A$. Let it be multiplied then, and let it be $AΘ$, and as many times as $AΘ$ is of $AΓ$, so many times let $ZH$ be of $H$. Therefore, as $AΘ$ is to $AΓ$, so is $ZH$ to $H$. And inversely, as $H$ is to $HZ$, so is $AΓ$ to $AΘ$. And since $AΘ$ is greater than $D$, that is, than $GB$, therefore $GA$ to $AΘ$ has a lesser ratio than $GA$ to $GB$. And by composition, $ZH$ therefore to $ZH$ has a lesser ratio than $AB$ to $BG$. (Lemma) And $BG$ is equal to $D$. Therefore $ZH$ to $ZH$ has a lesser ratio than $AB$ to $D$. Therefore two straight lines $HZ$ have been found, performing the requirement. That is, the greater has to the lesser a lesser ratio than the greater magnitude to the lesser.

A vertical geometric diagram consisting of a main line segment labeled η and ζ, and two smaller segments to the right labeled θ and ικ.

III

Two unequal magnitudes being given, and a circle, it is possible to inscribe a polygon in the circle, and to circumscribe another, so that the side of the circumscribed polygon to the side of the inscribed polygon has a ratio less than the greater magnitude to the lesser. Let the two given magnitudes be $A, B$. And let the given circle lying below be $Θ$. I say indeed that it is possible to perform the requirement. Let two straight lines $D, KL$ be found, of which let $Θ$ be the greater, so that $Θ$ to $KL$ has a ratio less than the greater magnitude to the lesser. And let $LM$ be drawn from $L$ at right angles to $KL$, and from $K$ let $KM$ be drawn down equal to $Θ$; for this is possible. And let two diameters of the circle be drawn at right angles to one another, $GE, DH$, bisecting the angle under $DHG$, and bisecting the half of it, and always doing this, we shall leave some angle less than double the angle under $LKM$. Let it be left, and let it be the angle under $VHG$. And let $VG$ be joined. Therefore $VG$ is a side...

A geometric diagram showing a circle with center labeled η. Within the circle is an inscribed polygon, and outside is a circumscribed polygon. Various radii and tangent lines are drawn, labeled with Greek letters: α, β, γ, δ, ε, ζ, θ, κ, λ, μ, ν, ξ, π, ρ, σ. The diagram illustrates the construction of polygons to bound the circumference or area of a circle.

...of an equilateral polygon. Since the angle under $VHG$ is a measure of the angle under $DHG$ which is a right angle, then the arc $VG$ also measures $GD$, being a fourth part of the circle; so that it also measures the circle. Therefore $GV$ is a side of an equilateral polygon; for it is clear that it is. And let the angle under $VHG$ be bisected by the straight line $HX$. And from $X$ let $OXP$ be tangent to the circle. And let $HVP, HGO$ be produced. So that $PO$ is also a side of a circumscribed polygon about the circle, and equilateral, and it is clear that it is also similar to the inscribed one, whose side is $VG$. And since the angle under $VHG$ is less than double the angle under $LKM$, and it is double the angle under $XHG$, therefore the angle under $XHG$ is less than the angle under $LKM$. And the angles at $L$ and $G$ are right. Therefore $MK$ to $LK$ has a greater ratio than $GP$ to $PH$, and $GP$ is equal...