Archimedis Syracusani Opera (1544 Basel Editio Princeps)

and let $A$ be greater than $B$. I say that it is possible to find two such straight lines to satisfy the stated requirement. Take the second proposition of the first book of Euclid referring to Euclid's Elements, Book I, Prop 2, regarding placing a line equal to a given line, setting $BΓ$ equal to $ΔΓ$, and let there be a straight line $ZH$. Let $Γ A$ added to itself exceed $ΔΓ$. Let it be multiplied, and let it be $AΘ$. And let $ZK$ be the same multiple of $HE$ as $AΘ$ is of $AΓ$. Therefore, as $A$ is to $AΓ$, so is $ZK$ to $HE$. And inversely, as $EH$ is to $KZ$, so is $AΓ$ to $AΘ$. And since $AΘ$ is greater than $ΔΓ$, that is to say $BΓ$, therefore $Γ A$ to $AΘ$ has a smaller ratio than $Γ A$ to $Γ B$. By composition, $EZ$ to $ZK$ has a smaller ratio than $AB$ to $BΓ$. (By lemma) But $BΓ$ is equal to $Δ H$. Therefore, $EZ$ to $ZK$ has a smaller ratio than $AB$ to $Δ H$. Thus, two straight lines $ZH$ and $HE$ have been found, satisfying the given requirement. That is to say, the ratio of the greater to the smaller is less than the ratio of the greater magnitude to the smaller.

A diagram on the upper right margin consists of several vertical line segments of different lengths labeled with Greek letters: $θ$, $β$, $γ$, $α$, and $δ$. Below them is a right-angled construction with labels $θ$, $β$, $γ$, $α$, and $δ$.

3

Given two unequal magnitudes and a circle, it is possible to inscribe a polygon in the circle, and to circumscribe another, such that the side of the circumscribed polygon to the side of the inscribed polygon has a ratio less than the ratio of the greater magnitude to the smaller. Let the two given magnitudes be $A$ and $B$, and the given circle be the one here. I say that it is possible to satisfy the requirement. Let two straight lines be found, $Θ$ and $KΛ$, of which $Θ$ is the greater, such that $Θ$ to $KΛ$ has a smaller ratio than the greater magnitude to the smaller. And from $Λ$, draw $Λ M$ at right angles to $KΛ$, and from $K$, let $KM$ be drawn equal to $Θ$. This is possible. And let two diameters of the circle be drawn at right angles to each other, $Γ E$ and $ΔΠ$. Bisecting the angle $Π HΓ$, and bisecting its half, and always doing this, we will leave some angle smaller than double the angle $Λ KM$. Let it be left, and let it be $\angle NHΓ$. And join $NΓ$. Therefore $NΓ$

A large circular geometric diagram centered at H with horizontal diameter $εγ$ and vertical diameter $δπ$ (labeled with lowercase Greek letters $ε, η, γ$ and $δ, η, π$). In the lower-right quadrant, a chord $νγ$ ($NΓ$) is drawn. Radius $ην$ ($HN$) is shown, and the angle $γην$ is bisected by the line $ηξ$ ($HΞ$). A tangent line $oξπ$ touches the circle at $ξ$, with points $o$ and $π$ formed by the extensions of the radii. To the right, a separate right-angled triangle labeled $κλμ$ ($KLM$) is shown.

is the side of an equilateral polygon. Since the angle $NHΓ$ measures the angle $Δ HΓ$, which is a right angle, then the arc $NΓ$ measures the arc $ΓΔ$, which is a quarter of the circle; so it also measures the circle. Therefore $Γ N$ is a side of an equilateral polygon. This is clear. And let the angle $Γ HN$ be bisected by the straight line $HΞ$. And from $Ξ$, let $OΞΠ$ be tangent to the circle. And let $HNΠ$ and $HΓ O$ be extended. Thus, $Π O$ is a side of the polygon circumscribed about the circle, and it is equilateral, and it is clear that it is similar to the inscribed one, whose side is $NΓ$. Since the double of the angle $NHΓ$ is less than $LK M$, and it is double of $\angle THΓ$, then $\angle THΓ$ is less than $\angle LKM$. And the angles at $L$ and $T$ are right angles. Therefore, $MK$ to $LK$ has a greater ratio than $HT$ to $HΞ$. But $HΓ$ is equal to...