# Download PDF by Roger Herz-Fischler: A Mathematical History of the Golden Number (Dover Books on

By Roger Herz-Fischler

ISBN-10: 0486400077

ISBN-13: 9780486400075

A entire research of the historical improvement of department in severe and suggest ratio ("the golden number"), this article strains the concept's improvement from its first visual appeal in Euclid's Elements in the course of the 18th century. The coherent yet rigorous presentation bargains transparent causes of DEMR's historic transmission and lines a number of illustrations.

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Extra info for A Mathematical History of the Golden Number (Dover Books on Mathematics)

Sample text

We are given a line AB, a parallelogram D and a rectilineal figure C. On AB we are to construct a parallelogram ABRT so that the subparallelogram ASQT has the same area as C, while the subparallelogram SBRQ (II) is similar to the given parallelogram D. In order for this construction to be possible, a limitation must be put on the area of C. This condition is as follows: let E be the midpoint of AB and let parallelogram EBFG be similar to parallelogram D. We assume that the area of parallelogram EBFG is greater than or equal to that of C.

Note: Heiberg [IV,254] considers this lemma to be not genuine. THEOREM SCAD) IV o Proof: Draw the squares and rectangles as indicated. We are given that 5· S(AC) = S(AB) = I + II + III + S(AC) == gnomon + S(AC). Subtracting S(AC) from each side we obtain (a) 4· S(AC) == gnomon = I + II + III. Also, since DC = 2· AC, we have S(CD) = 4· S(AC) and thus from (a) we have (b) I + II + III == gnomon = 4· S(AC) = S(CD) = IV + V. Again, since KC = CD = 2· AC = CH and II = III (1,43), we have (c) II + III = 2· II = V (by VI,l).

A THEOREM 111,37 (the converse of 111,36). Let D be a point outside a circle with centre F and let DA be a secant line which cuts the circle at C . Now suppose that B is a point on the circle such that the relationship R(AD,DC) = S(DB) holds, then the line DB is in fact tangent to the circle at B. E FIGURE 1-15. IV,10 111,36. IV,lO. Discussed in: Section 2. Proof· Take any line AB and draw a circle with A as centre and AB as radius. Now use 11,11 to obtain point C with R(AB,BC) = S(CA). BAD isosceles with AB = AD and 1-ABD = ~BDA -such that BD = AC.