A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921
MACEDONIA is GREECE and will always be GREECE- (if they are desperate to steal a name, Monkeydonkeys suits them just fine) ΚΑΤΩ Η ΣΥΓΚΥΒΕΡΝΗΣΗ ΤΩΝ ΠΡΟΔΟΤΩΝ!!! ΦΕΚ,ΚΚΕ,ΚΝΕ,ΚΟΜΜΟΥΝΙΣΜΟΣ,ΣΥΡΙΖΑ,ΠΑΣΟΚ,ΝΕΑ ΔΗΜΟΚΡΑΤΙΑ,ΕΓΚΛΗΜΑΤΑ,ΔΑΠ-ΝΔΦΚ, MACEDONIA,ΣΥΜΜΟΡΙΤΟΠΟΛΕΜΟΣ,ΠΡΟΣΦΟΡΕΣ,ΥΠΟΥΡΓΕΙΟ,ΕΝΟΠΛΕΣ ΔΥΝΑΜΕΙΣ,ΣΤΡΑΤΟΣ, ΑΕΡΟΠΟΡΙΑ,ΑΣΤΥΝΟΜΙΑ,ΔΗΜΑΡΧΕΙΟ,ΝΟΜΑΡΧΙΑ,ΠΑΝΕΠΙΣΤΗΜΙΟ,ΛΟΓΟΤΕΧΝΙΑ,ΔΗΜΟΣ,LIFO,ΛΑΡΙΣΑ, ΠΕΡΙΦΕΡΕΙΑ,ΕΚΚΛΗΣΙΑ,ΟΝΝΕΔ,ΜΟΝΗ,ΠΑΤΡΙΑΡΧΕΙΟ,ΜΕΣΗ ΕΚΠΑΙΔΕΥΣΗ,ΙΑΤΡΙΚΗ,ΟΛΜΕ,ΑΕΚ,ΠΑΟΚ,ΦΙΛΟΛΟΓΙΚΑ,ΝΟΜΟΘΕΣΙΑ,ΔΙΚΗΓΟΡΙΚΟΣ,ΕΠΙΠΛΟ, ΣΥΜΒΟΛΑΙΟΓΡΑΦΙΚΟΣ,ΕΛΛΗΝΙΚΑ,ΜΑΘΗΜΑΤΙΚΑ,ΝΕΟΛΑΙΑ,ΟΙΚΟΝΟΜΙΚΑ,ΙΣΤΟΡΙΑ,ΙΣΤΟΡΙΚΑ,ΑΥΓΗ,ΤΑ ΝΕΑ,ΕΘΝΟΣ,ΣΟΣΙΑΛΙΣΜΟΣ,LEFT,ΕΦΗΜΕΡΙΔΑ,ΚΟΚΚΙΝΟ,ATHENS VOICE,ΧΡΗΜΑ,ΟΙΚΟΝΟΜΙΑ,ΕΝΕΡΓΕΙΑ, ΡΑΤΣΙΣΜΟΣ,ΠΡΟΣΦΥΓΕΣ,GREECE,ΚΟΣΜΟΣ,ΜΑΓΕΙΡΙΚΗ,ΣΥΝΤΑΓΕΣ,ΕΛΛΗΝΙΣΜΟΣ,ΕΛΛΑΔΑ, ΕΜΦΥΛΙΟΣ,ΤΗΛΕΟΡΑΣΗ,ΕΓΚΥΚΛΙΟΣ,ΡΑΔΙΟΦΩΝΟ,ΓΥΜΝΑΣΤΙΚΗ,ΑΓΡΟΤΙΚΗ,ΟΛΥΜΠΙΑΚΟΣ, ΜΥΤΙΛΗΝΗ,ΧΙΟΣ,ΣΑΜΟΣ,ΠΑΤΡΙΔΑ,ΒΙΒΛΙΟ,ΕΡΕΥΝΑ,ΠΟΛΙΤΙΚΗ,ΚΥΝΗΓΕΤΙΚΑ,ΚΥΝΗΓΙ,ΘΡΙΛΕΡ, ΠΕΡΙΟΔΙΚΟ,ΤΕΥΧΟΣ,ΜΥΘΙΣΤΟΡΗΜΑ,ΑΔΩΝΙΣ ΓΕΩΡΓΙΑΔΗΣ,GEORGIADIS,ΦΑΝΤΑΣΤΙΚΕΣ ΙΣΤΟΡΙΕΣ, ΑΣΤΥΝΟΜΙΚΑ,ΦΙΛΟΣΟΦΙΚΗ,ΦΙΛΟΣΟΦΙΚΑ,ΙΚΕΑ,ΜΑΚΕΔΟΝΙΑ,ΑΤΤΙΚΗ,ΘΡΑΚΗ,ΘΕΣΣΑΛΟΝΙΚΗ,ΠΑΤΡΑ, ΙΟΝΙΟ,ΚΕΡΚΥΡΑ,ΚΩΣ,ΡΟΔΟΣ,ΚΑΒΑΛΑ,ΜΟΔΑ,ΔΡΑΜΑ,ΣΕΡΡΕΣ,ΕΥΡΥΤΑΝΙΑ,ΠΑΡΓΑ,ΚΕΦΑΛΟΝΙΑ, ΙΩΑΝΝΙΝΑ,ΛΕΥΚΑΔΑ,ΣΠΑΡΤΗ,ΠΑΞΟΙ
MACEDONIA is GREECE and will always be GREECE- (if they are desperate to steal a name, Monkeydonkeys suits them just fine)
ΚΑΤΩ Η ΣΥΓΚΥΒΕΡΝΗΣΗ ΤΩΝ ΠΡΟΔΟΤΩΝ!!!
ΦΕΚ,ΚΚΕ,ΚΝΕ,ΚΟΜΜΟΥΝΙΣΜΟΣ,ΣΥΡΙΖΑ,ΠΑΣΟΚ,ΝΕΑ ΔΗΜΟΚΡΑΤΙΑ,ΕΓΚΛΗΜΑΤΑ,ΔΑΠ-ΝΔΦΚ, MACEDONIA,ΣΥΜΜΟΡΙΤΟΠΟΛΕΜΟΣ,ΠΡΟΣΦΟΡΕΣ,ΥΠΟΥΡΓΕΙΟ,ΕΝΟΠΛΕΣ ΔΥΝΑΜΕΙΣ,ΣΤΡΑΤΟΣ, ΑΕΡΟΠΟΡΙΑ,ΑΣΤΥΝΟΜΙΑ,ΔΗΜΑΡΧΕΙΟ,ΝΟΜΑΡΧΙΑ,ΠΑΝΕΠΙΣΤΗΜΙΟ,ΛΟΓΟΤΕΧΝΙΑ,ΔΗΜΟΣ,LIFO,ΛΑΡΙΣΑ, ΠΕΡΙΦΕΡΕΙΑ,ΕΚΚΛΗΣΙΑ,ΟΝΝΕΔ,ΜΟΝΗ,ΠΑΤΡΙΑΡΧΕΙΟ,ΜΕΣΗ ΕΚΠΑΙΔΕΥΣΗ,ΙΑΤΡΙΚΗ,ΟΛΜΕ,ΑΕΚ,ΠΑΟΚ,ΦΙΛΟΛΟΓΙΚΑ,ΝΟΜΟΘΕΣΙΑ,ΔΙΚΗΓΟΡΙΚΟΣ,ΕΠΙΠΛΟ, ΣΥΜΒΟΛΑΙΟΓΡΑΦΙΚΟΣ,ΕΛΛΗΝΙΚΑ,ΜΑΘΗΜΑΤΙΚΑ,ΝΕΟΛΑΙΑ,ΟΙΚΟΝΟΜΙΚΑ,ΙΣΤΟΡΙΑ,ΙΣΤΟΡΙΚΑ,ΑΥΓΗ,ΤΑ ΝΕΑ,ΕΘΝΟΣ,ΣΟΣΙΑΛΙΣΜΟΣ,LEFT,ΕΦΗΜΕΡΙΔΑ,ΚΟΚΚΙΝΟ,ATHENS VOICE,ΧΡΗΜΑ,ΟΙΚΟΝΟΜΙΑ,ΕΝΕΡΓΕΙΑ, ΡΑΤΣΙΣΜΟΣ,ΠΡΟΣΦΥΓΕΣ,GREECE,ΚΟΣΜΟΣ,ΜΑΓΕΙΡΙΚΗ,ΣΥΝΤΑΓΕΣ,ΕΛΛΗΝΙΣΜΟΣ,ΕΛΛΑΔΑ, ΕΜΦΥΛΙΟΣ,ΤΗΛΕΟΡΑΣΗ,ΕΓΚΥΚΛΙΟΣ,ΡΑΔΙΟΦΩΝΟ,ΓΥΜΝΑΣΤΙΚΗ,ΑΓΡΟΤΙΚΗ,ΟΛΥΜΠΙΑΚΟΣ, ΜΥΤΙΛΗΝΗ,ΧΙΟΣ,ΣΑΜΟΣ,ΠΑΤΡΙΔΑ,ΒΙΒΛΙΟ,ΕΡΕΥΝΑ,ΠΟΛΙΤΙΚΗ,ΚΥΝΗΓΕΤΙΚΑ,ΚΥΝΗΓΙ,ΘΡΙΛΕΡ, ΠΕΡΙΟΔΙΚΟ,ΤΕΥΧΟΣ,ΜΥΘΙΣΤΟΡΗΜΑ,ΑΔΩΝΙΣ ΓΕΩΡΓΙΑΔΗΣ,GEORGIADIS,ΦΑΝΤΑΣΤΙΚΕΣ ΙΣΤΟΡΙΕΣ, ΑΣΤΥΝΟΜΙΚΑ,ΦΙΛΟΣΟΦΙΚΗ,ΦΙΛΟΣΟΦΙΚΑ,ΙΚΕΑ,ΜΑΚΕΔΟΝΙΑ,ΑΤΤΙΚΗ,ΘΡΑΚΗ,ΘΕΣΣΑΛΟΝΙΚΗ,ΠΑΤΡΑ, ΙΟΝΙΟ,ΚΕΡΚΥΡΑ,ΚΩΣ,ΡΟΔΟΣ,ΚΑΒΑΛΑ,ΜΟΔΑ,ΔΡΑΜΑ,ΣΕΡΡΕΣ,ΕΥΡΥΤΑΝΙΑ,ΠΑΡΓΑ,ΚΕΦΑΛΟΝΙΑ, ΙΩΑΝΝΙΝΑ,ΛΕΥΚΑΔΑ,ΣΠΑΡΤΗ,ΠΑΞΟΙ
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MENAECHMUS'S PROCEDURE 113<br />
particular hyperbola which we call rectangular or equilateral,<br />
and also <strong>to</strong> obtain its property with reference <strong>to</strong> its asymp<strong>to</strong>tes,<br />
a considerable advance on what was necessary in the<br />
case <strong>of</strong> the parabola. Two methods <strong>of</strong> obtaining the particular<br />
hyperbola were possible, namely (1) <strong>to</strong> obtain the hyperbola<br />
arising <strong>from</strong> the section <strong>of</strong> any obtuse-angled cone <strong>by</strong> a plane<br />
at right angles <strong>to</strong> a genera<strong>to</strong>r, and then <strong>to</strong> show how a<br />
rectangular hyperbola can be obtained as a particular case<br />
<strong>by</strong> finding the vertical angle which the cone must have <strong>to</strong><br />
give a rectangular hyperbola when cut in the particular way,<br />
or (2) <strong>to</strong> obtain the rectangular hyperbola direct <strong>by</strong> cutting<br />
another kind <strong>of</strong> cone <strong>by</strong> a section not necessarily perpendicular<br />
<strong>to</strong><br />
a genera<strong>to</strong>r.<br />
(1) Taking the first method, we draw (Fig. 2) a cone with its<br />
vertical angle BOG obtuse. Imagine a section perpendicular<br />
<strong>to</strong> the plane <strong>of</strong> the paper and passing through AG which is<br />
perpendicular <strong>to</strong> OB. Let GA produced meet GO produced in<br />
A\ and complete the same construction as in the case <strong>of</strong><br />
the<br />
parabola.<br />
In this case we have<br />
1523.2<br />
PN 2<br />
Fig. 2.<br />
= BN.NG = AN.NG.<br />
i<br />
114 CONIC SECTIONS<br />
But, <strong>by</strong> similar triangles,<br />
NG:AF=NC:AD<br />
= A'N:AA'.<br />
Hence PlY 2 = AN. A'N<br />
AA f<br />
.<br />
^-,<br />
AA<br />
AN.A'N.<br />
which is the property <strong>of</strong> the hyperbola, A A' being what we<br />
call the transverse axis, and 2 AL the parameter <strong>of</strong> the principal<br />
ordinates.<br />
Now, in order that the hyperbola may be rectangular, we<br />
must have 2 AL :<br />
AA<br />
f equal <strong>to</strong> 1. The problem therefore now<br />
is: given a straight line AA\ and AL along A''A produced<br />
equal <strong>to</strong> \ A A\ <strong>to</strong> find a cone such that L is on its axis and<br />
the section through AL perpendicular <strong>to</strong> the genera<strong>to</strong>r through<br />
A is a rectangular hyperbola with A'A as transverse axis.<br />
other words, we have <strong>to</strong> find a point on the straight line<br />
through A perpendicular <strong>to</strong> AA f such that OX bisects the<br />
angle which is the supplement <strong>of</strong> the angle A'OA.<br />
This is the case if A'O : OA = A'L : LA<br />
= 3:1;<br />
therefore is on the circle which is the locus <strong>of</strong> all points<br />
such that their distances <strong>from</strong> the two fixed points A' , A<br />
are in the ratio 3:1. This circle is the circle on KL as<br />
diameter, where A'K-.KA = A'L: LA = 3:1. Draw this<br />
circle, and is then determined as the point in which AO<br />
drawn perpendicular <strong>to</strong> AA' intersects the circle.<br />
It is <strong>to</strong> be observed, however, that this deduction <strong>of</strong> a<br />
particular <strong>from</strong> a more general case is not usual in early<br />
<strong>Greek</strong> <strong>mathematics</strong> ; on the contrary, the particular usually<br />
led <strong>to</strong> the more general. Notwithstanding, therefore, that the<br />
orthodox method <strong>of</strong> producing conic sections is said <strong>to</strong> have<br />
been <strong>by</strong> cutting the genera<strong>to</strong>r <strong>of</strong> each cone perpendicularly,<br />
I am inclined <strong>to</strong> think that Menaechmus would get his rectangular<br />
hyperbola directly, and in an easier way, <strong>by</strong> means <strong>of</strong><br />
a different cone differently cut. Taking the right-angled cone,<br />
already used for obtaining a parabola, we have only <strong>to</strong> make<br />
a section parallel <strong>to</strong> the axis (instead <strong>of</strong> perpendicular <strong>to</strong> a<br />
genera<strong>to</strong>r) <strong>to</strong> get a rectangular hyperbola.<br />
In