Mathematical Collections (Commandino 1588)

PAPPI MATH. COLL.

the ratio of KH to HR, and KH is given (for this must be posited), then HR will be given, and the remainder RK. But SR, which is half of RK, and RH were already given; therefore, the whole HS will be given, and thus the ratio of KH to HS. And as KH is to HS, so is SH to HT. And SH is given, as was shown; therefore, HT will also be given. For the same reason, HΦ will be given, and thus the difference between the straight lines HR and HΦ; and it has been found that Φ falls between HR, as has already been demonstrated by numbers. And since the difference between ΦR and the straight line connecting RQ, which is equal to XK, is given, the right-angled triangle ΦQR will be given in species and size. Therefore, the angle RΦQ is given, which is equal to the exterior angle KSψ. Therefore, when ωψ is extended to F, the right-angled triangle SFψ will be given both in species and in size. For since both RK and RQ are given, QK will also be given, and the ratio of QK to Kψ is given; because it is the same as the given ratio of ΦK to KS. Therefore, ψK is also given. But ψS is also given, since as ΦK is to KS, so is ΦQ to ψS. And ΦQ was shown to be given; therefore, ψS is also given. The angle ψSK was also given. Therefore, the right-angled triangle ψSF will be given in species and size. Hence, ψF is parallel to XK and is in a straight line with ψω. Thus, ωL is also given, equal to FK. And since HK is equal to ML, but ωL is smaller than SK. Indeed, ωL is equal to KF, and as KH is to HS, so is SH to HT, and TH to HΦ; but as LM is to Mω, so is ωM to Mα, and αM to Mβ; Mβ will be greater than HΦ. Indeed, this will also be demonstrated subsequently. Therefore, the remainder βL is smaller than ΦK. Again, since ωL is given, being shown equal to the given FK, and LM is also given, because KH is given: the ratio of LM to Mω will be given. And as LM is to Mω, so is ωM to Mα; and ωM is given. Therefore, Mα will be given. For the same reason, Mβ will also be given. Therefore, the point β is given; let it be placed wherever it wants, either between VM, as it now is, or between Vα, provided the straight line VL is set equal to each of KR, AB, QX, τN. If he says that β falls at V, he still assumes the requirement as if it were conceded. For it appears again that in the straight line ML, given in position, and with some point V given on it, to take between LV two points ω and A, and to make it so that as LM is to Mω, so is ωM to Mα, and αM to MV; which none of them concedes. For even the ancients, when seeking this, doubted whether to find it by means of plane geometry, as he himself has nothing to say that refutes it when we say, putting forward their own words in the demonstration. If V is necessarily the point of section of the third proportion, show that it cannot fall between Vα nor between MV. For we showed in the beginning that point Φ falls above R and below it. For it falls according to the position of the proposition. Similarly, therefore, as the resolution proceeds from the fact that the triangle V, β, Γ is given in species and size, even if the section point β falls between Vα. But with the triangle A, ω, L given as above, and A, ε given, the ratio of DG to Gε will be given. This is the ratio of ζG to Gθ, and in no way again DG to GE, since they are posited as equal, and now KR, that is AB, is posited as equal to D, τ, although he wants θ to fall between γ, τ; for he has nothing to say that refutes it; hearing from us, show that it does not fall between τ, G, nor between τ, ζ. If, however, he simply wants the point of section of this kind to be at τ by concession, he is still assuming the requirement as conceded. But if it is not conceded to him that the section is at point τ, since we do not even concede that R makes the demonstration on the straight line KH. And if he wishes to take any other point between ε, G, such as ζ, he does not know how θ, deceived, takes it. But as he wishes, let it be posited to be separable according to τ, and connecting θ, C, and drawing parallels to C, θ, namely ζ, κ, ε, λ, and through κ, λ parallels to AC, namely κμ, λν; he shows that he did not understand the problem. For the straight line θC not being made parallel to EG, the angle CθG is indeed obtuse, with point θ falling between G, τ, but acute when θ falls between τ, ζ, for the angle at τ is a right angle, according to which alone the problem is effected. But if anyone concedes, as we said above, in the straight line DG given in position, and with a given point τ, to take two points, such as ε, ζ, such that the ratio which DG has to Gε, εG has to Gζ and ζG to Gτ.