The Works of Archimedes (Dover Books on Mathematics)

The Works of Archimedes (Dover Books on Mathematics)

Archimedes

Language: English

Pages: 326

ISBN: 0486420841

Format: PDF / Kindle (mobi) / ePub


The complete works of antiquity's great geometer appear here in a highly accessible English translation by a distinguished scholar. Remarkable for his range of thought and his mastery of treatment, Archimedes addressed such topics as the famous problems of the ratio of the areas of a cylinder and an inscribed sphere; the measurement of a circle; the properties of conoids, spheroids, and spirals; and the quadrature of the parabola. This edition offers an informative introduction with many valuable insights into the ancient mathematician's life and thought as well as the views of his contemporaries. Modern mathematicians, physicists, science historians, and logicians will find this volume a source of timeless fascination. Unabridged reprint of the classic 1897 edition, with supplement of 1912.

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Measurement

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

HenceKG = 3GF ButKG = 3LK, from (1) above. Therefore And, from (2), LF = (AO−AL−OF)=AO = OF. ThereforeOF = 5GF, andOG = 6GF. ButAO = 3OF = 15GF. Therefore, by subtraction, Proposition 9 (Lemma). If a, b, c, d be four lines in continued proportion and in descending order of magnitude, and if d : (a−d) = x : (a −c), and (2a + 4b + 6c + 3d) : (5a +10b + 10c + 5d) = y : (a − c), is required to prove that x+y = a. [The following is the proof given by Archimedes, with the only difference

by the use of this same lemma that they have shown that circles are to one another in the duplicate ratio of their diameters, and that spheres are to one another in the triplicate ratio of their diameters, and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height; also, that every cone is one third part of the cylinder having the same base as the cone and equal height they proved by assuming a certain lemma similar to that aforesaid.

is a vertical straight line. “For this is proved †.” Therefore, as before, there will be equilibrium. Thus orP = ΔBCD. Proposition 8, 9. Suppose a lever AOB placed horizontally and supported at its middle point O. Let a triangle BCD, right-angled or obtuseangled at C, be suspended from the points B, E on OB, the angular point O being so attached to E that the side CD is in the same vertical line with E. Let Q be an area such that SO : OE = ΔBCD : Q. Then, if ctu area P suspended from A

is known. Therefore BO2, or OE2, can be found, and therefore O. * To prove this, suppose that, in the figure on the opposite page, BR1 is produced to meet the outer parabola in R2. We have, as before, whenceER1 : ER = BQ2 : BQ1. And, since R1 is a point within the outer parabola, ER : ER1 = BR1 : BR2, in like manner. HenceBQ1 : BQ2 = BR1 : BR2. BOOK OF LEMMAS. Proposition 1. If two circles touch at A, and if BD, EF be parallel diameters in them, ADF is a straight line [The proof in the

is no special technical term, but we find such phrases as the following: άν εις τòτ κύκλον εθεîα γραµµή �πéση if in a circle a straight line be placed, and the chord is then the straight line so placed ή �πεσοσα , or quite commonly ή ν τ κύκλω (εθεîα) simply. For the chord subtending one 656th part of the circumference of a circle we have the following interesting phrase, ποτεìνονσα éν τµµα διαιρεθείσας τς το ΑΒΓ κύκλου περιφερείας éς χνς’. A segment of a circle is τµµα κύκλου ; sometimes, to

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