Mathematical Collections (Commandino 1588)

BOOK THREE.

COMMENTARY.

A

"Whatever problems are solved by assuming in the construction one or more conic sections are called solid": The Greek codex reads thus: παραλαμβανομένης εἰς τὴν γένεσιν μιᾶς τῶν τοῦ κώνου τομῶν a conic section being taken into the genesis. I would prefer to read παραλαμβανομένης εἰς τὴν κατασκευὴν μιᾶς τῶν τοῦ κώνου τομῶν a conic section being taken into the construction.

B

"Such as spirals and those which the Greeks call τετραγωνιζούσας quadratrices": The Greek codex is defective, reading thus: ὁποῖαι τυγχάνουσιν ... ... καὶ τετραγωνίζουσαι. Perhaps it should be read: ὁποῖαι τυγχάνουσιν αἱ ἕλικες, καὶ τετραγωνίζουσαι such as happen to be the spirals, and the quadratrices.

C

"Since it is not easy to draw conic sections on a plane": In the Greek codex, it is read thus: ἐπεὶ μὴ δὲ τὰς τοῦ κώνου τομὰς ῥᾴδιον ἐν ἐπιπέδῳ γράφειν ἦν since it was not easy to draw the sections of a cone on a plane. The parts that follow are redundant and I believe should be abolished, as they were inserted by a copyist's error: ὡς δεῖ δύο δοθεισῶν εὐθειῶν ἀνίσων δύο μέσας ἀνάλογον λαβεῖν ἐν συνεχῇ ἀναλογίᾳ as it is necessary to take two mean proportionals in continuous analogy between two given unequal straight lines.

D

"For with any solid given, another solid similar to the given one is constructed to a given ratio": We shall correct the Greek codex thus: στερεὸν γὰρ παντὸς δεδομένου, ἕτερον στερεὸν ὅμοιον τῷ δοθέντι κατασκευάζεται πρὸς τὸν δοθέντα λόγον for with any solid given, another solid similar to the given one is constructed to the given ratio.

PROBLEM I. PROPOSITION V.

To find two mean proportionals in continuous analogy for two given straight lines.

A geometric diagram shows a rectangular frame (ABCD) containing three right-angled triangles (AEH, MFK, NGL) positioned along parallel tracks (AB, CD). A diagonal line (APOX) intersects the triangles and the frame, with points labeled A through X.

Let there be a slab τόπα frame/slab ABCD, and within it equal right-angled triangles AEH, MFK, NGL, having right angles at points E, F, G. Let triangle AEH remain fixed, while triangle MFK moves on the rulers AB and CD, such that MF moves on ruler AB—which has a channel throughout—and vertex K moves on CD—which also has a channel hollowed along its entire length. Similarly, let triangle NGL move on the rulers AB and CD through said channels. These having been prepared in this way, if anyone wishes to make a cube double the cube, taking AC as the double of LX, and moving the triangles MFK and NGL until points A, P, Q, X are established on the same straight line, where the intersections of the triangles P, Q occur with the straight line APOX meeting CD at R; for this must necessarily happen. And thus, that which is proposed will be achieved. For since it is as AC to PH, so is AR to RP, and as AH to PK, and as HR to RK, and as PH to OK, and as PR to RO, and as PK to OL, and as KR to RL, and as OK to LX, the two means PH, OK will be in continuous analogy for the lines AC, LX; and AC is double LX. Therefore, the cube made from AC will be double the one made from PH. But if the cube has some other ratio to a cube, AC must have 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.