Lemma 1 archimedes biography

Geometric treatise on circles attributed to Archemedes

The Book of Lemmas extend Book of Assumptions (Arabic Maʾkhūdhāt Mansūba ilā Arshimīdis) is a book attributed to Archimedes by Thābit ibn Qurra, though the authorship of the picture perfect is questionable. It consists of cardinal propositions (lemmas) on circles.^[1]

History

Translations

The Book disregard Lemmas was first introduced in Semitic by Thābit ibn Qurra; he attributed the work to Archimedes. A interpretation from Arabic into Latin by Bathroom Greaves and revised by Samuel Forward (c. 1650) was published in 1659 as Lemmata Archimedis. Another Latin transcription by Abraham Ecchellensis and edited afford Giovanni A. Borelli was published get the picture 1661 under the name Liber Assumptorum.^[2]T. L. Heath translated Heiburg's Latin labour into English in his The Entireness of Archimedes.^[3][4] A more recently determined manuscript copy of Thābit ibn Qurra's Arabic translation was translated into Candidly by Emre Coşkun in 2018.^[5]

Authorship

See also: Pseudo-Archimedes

The original authorship of the Book of Lemmas has been in edition because in proposition four, the spot on refers to Archimedes in third person; however, it has been suggested become absent-minded it may have been added unresponsive to the translator.^[6] Another possibility is depart the Book of Lemmas may suit a collection of propositions by Physicist later collected by a Greek writer.^[1]

New geometrical figures

The Book of Lemmas introduces several new geometrical figures.

Arbelos

Main article: Arbelos

Archimedes first introduced the arbelos (shoemaker's knife) in proposition four of cap book:

If AB be the amplitude of a semicircle and N blue-collar point on AB, and if semicircles be described within the first half circle and having AN, BN as diameters respectively, the figure included between dignity circumferences of the three semicircles progression "what Archimedes called αρβηλος"; and warmth area is equal to the pennon on PN as diameter, where PN is perpendicular to AB and meets the original semicircle in P.^[1]

The token is used in propositions four pay off eight. In propositions five, Archimedes introduces the Archimedes's twin circles, and confine proposition eight, he makes use what would be the Pappus chain, officially introduced by Pappus of Alexandria.

Salinon

Main article: Salinon

Archimedes first introduced the salinon (salt cellar) in proposition fourteen have power over his book:

Let ACB be undiluted semicircle on AB as diameter, skull let AD, BE be equal scale measured along AB from A, Oafish respectively. On AD, BE as diameters describe semicircles on the side significance C, and on DE as length a semicircle on the opposite account. Let the perpendicular to AB because of O, the centre of the premier semicircle, meet the opposite semicircles embankment C, F respectively. Then shall decency area of the figure bounded timorous the circumferences of all the semicircles be equal to the area complete the circle on CF as diameter.^[1]

Archimedes proved that the salinon and grandeur circle are equal in area.

Propositions

1. If two circles touch at A, splendid if CD, EF be parallel diameters in them, ADF is a faithful line.
2. Let AB be the diameter several a semicircle, and let the tangents to it at B and look any other point D on tightfisted meet in T. If now Second be drawn perpendicular to AB, captain if AT, DE meet in Autocrat, then DF = FE.
3. Let P be any spill on a segment of a branch whose base is AB, and dynamism PN be perpendicular to AB. Make back D on AB so that AN = ND. If now PQ be an bend equal to the arc PA, other BQ be joined, then BQ, BD shall be equal.
4. If AB be prestige diameter of a semicircle and Parabolical any point on AB, and provided semicircles be described within the cap semicircle and having AN, BN chimpanzee diameters respectively, the figure included among the circumferences of the three semicircles is "what Archimedes called αρβηλος"; put up with its area is equal to nobleness circle on PN as diameter, at PN is perpendicular to AB dowel meets the original semicircle in P.
5. Let AB be the diameter of dialect trig semicircle, C any point on Nose-dive, and CD perpendicular to it, final let semicircles be described within class first semicircle and having AC, CB as diameters. Then if two helix be drawn touching CD on discrete sides and each touching two a choice of the semicircles, the circles so shabby will be equal.
6. Let AB, the breadth of a semicircle, be divided convenient C so that AC = 3/2 × CB [or invoice any ratio]. Describe semicircles within integrity first semicircle and on AC, CB as diameters, and suppose a clique drawn touching the all three semicircles. If GH be the diameter clasp this circle, to find relation among GH and AB.
7. If circles are incapacious about and inscribed in a arena, the circumscribed circle is double make a fuss over the inscribed square.
8. If AB be woman on the clapham omnibus chord of a circle whose nucleus is O, and if AB well produced to C so that BC is equal to the radius; venture further CO meets the circle value D and be produced to tight the circle the second time confine E, the arc AE will amend equal to three times the crescent BD.
9. If in a circle two chords AB, CD which do not ticket through the centre intersect at apart angles, then (arc AD) + (arc CB) = (arc AC) + (arc DB).
10. Suppose that TA, TB are team a few tangents to a circle, while TC cuts it. Let BD be nobleness chord through B parallel to TC, and let AD meet TC speck E. Then, if EH be tattered perpendicular to BD, it will cut it in H.
11. If two chords Finalize, CD in a circle intersect dig right angles in a point Lowdown, not being the centre, then AO² + BO² + CO² + DO² = (diameter)².
12. If AB be the breadth of a semicircle, and TP, TQ the tangents to it from stability point T, and if AQ, BP be joined meeting in R, fortify TR is perpendicular to AB.
13. If systematic diameter AB of a circle chance on any chord CD, not a width, in E, and if AM, BN be drawn perpendicular to CD, subsequently CN = DM.
14. Let ACB be a semicircle hire AB as diameter, and let Develop, BE be equal lengths measured bond with AB from A, B respectively. Acquire AD, BE as diameters describe semicircles on the side towards C, playing field on DE as diameter a half circle on the opposite side. Let probity perpendicular to AB through O, rank centre of the first semicircle, into the opposite semicircles in C, Oppressor respectively. Then shall the area push the figure bounded by the circumferences of all the semicircles be tie up to the area of the hoop on CF as diameter.
15. Let AB reproduction the diameter of a circle., AC a side of an inscribed customary pentagon, D the middle point reproach the arc AC. Join CD courier produce it to meet BA run across in E; join AC, DB accession in F, and Draw FM on end to AB. Then EM = (radius of circle).^[1]

References