Mathematical Collections (Commandino 1588)

PAPPI MATH. COLL.

the same ratio to LX, and the rest will be constructed in a similar manner. From which it clearly appears that it is impossible for the proposition to be solved through planes.

COMMENTARY.

A "And let triangle AEH remain fixed, but triangle MFK move on the rulers AB, CD": From the letter of Eratosthenes, which is read in the commentaries of Eutocius on the second book of Archimedes' On the Sphere and Cylinder, it appears that he intended the middle parallelogram or triangle to be fixed, not the first. But the matter comes to the same result; for even if the last remains and the other two move, the same thing will clearly happen. The Greek codex, however, is corrupt and defective, which perhaps may be restored thus: τὸ δὲ μὲν τῶν κινήσιν ἐχέτω ἐπὶ τῶν α β, γ δ κανόνων let one of them have motion on the AB, CD rulers; let the rest be added from the manuscript codex.

B "For since it is as AC to PH, so is AR to RP, and as AH to PK, and as HR to RK": From the fourth proposition of the sixth book of the Elements; for the triangles ARC and PRH are similar to each other, as are the triangles ARH and PRK. Wherefore, as AC is to PH, so is AR to RP; and as AR is to RP, so is AH to PK, and HR to RK.

C "And as PH to OK, and PR to RO, and PK to OL, and KR to RL, and as OK to LX": For again, the triangles PRH and ORK become similar, as do PRK and ORL. Therefore, as HR is to RK, so is PH to OK, and PR to RO. But as PR is to RO, so is PK to OL, and KR to RL. And for the same reason, as KR is to RL, so is OK to XL. From which it follows, by the eleventh of the fifth of the Elements, that as AC is to PH, so is PH to OK, and OK to XL.

For two given straight lines CD, DA; two mean proportionals in continuous analogy are assumed in this way:

A geometric diagram illustrates the construction of a conchoid to find mean proportionals between two lines, showing intersecting lines and triangles labeled A through M.

A Let parallelogram ABCD be completed, and let each of AB and BC be bisected at points L and E; and having joined LD, let it be produced and meet CB produced at G. To this BC, at right angles, let EF be drawn, and let CF be joined, which is equal to AL. Let FG be joined as well, and let CH be parallel to it. Since the angle is contained by KCH, from the given point F, let FHK be drawn, which makes the line HK equal to AL or CF. For it has been shown that this can be done via the conchoid line. And having joined KD, let it be produced and meet BA produced at point M. I say that as DC is to CK, so is CK to MA, and MA to AD. For since BC is bisected at E, and CK is added to it, the rectangle BKC 6, Book II together with the square from CE is equal to the square from EK. Let the square from EF be added in common.