Outline of Key Results in
Chapter V of Algebra
- Between two given lines it is required to find two other lines such that all four will form a continuous proportion.
Given the rectangular parallepiped ABCDE, whose base is the square AD, and the square MH, construct on MH a rectangular parallelpiped equal to ABCDE.
- Kasir's Notes:
- AB : x = x : y = y : BC
- In the parabola BDE, HD^2 = BT^2 = BH * BC, Hence BC : BT = BT :HT.
- In the parabola BDZ, DT^2 = HB^2 = BA * BT, Hence BT : HB = HB :BA.
- Consequently, BC:BT = BT : HB = HB : BA or AB: BH = BH : BT = BT: BC.
Given the solid ABCD whose base AC is a square, it is required to construct a solid whose base is a square, whose height is equivalent to a given line ET, and which is equal to the solid ABCD.
After these preliminary proofs we shall be able to give the solution of the third species of the simple equations, a cube is equal to a number.
Eves*: Solution of the cubic equation :x^3 + b^2x + a^3 = cx^2.
"First of all, by the basic construction, find the segment z such that b:a = a : z. Then, again by the basic construction, find line segment m such that b:z=a:m. we easily find that m = a^3/b^2. Now,...,
construct AB = m = = a^3/b^2 and BC = c. Draw a semicircle on AC as diameter and let the perpendicular to AC at B cut it in D. On BD mark off BE = b and through E draw EF parallel to AC.
By the basic construction, find G on BC such that ED:BE=AB:BG and
complete the rectangle DBGH. Through H draw thwe rectangular hyperbola
having EF and ED for asymptotes (that is, the hyperbola through H whose
equation with respect to EF and ED as x and y axes is of the form xy
= a constant). Let the hyperbola cut the semicircle in J , and
let the parallel to DE through J cut EF in K and BC in L. Let GH cut EF
in M. Now:
Eves, Howard. "Omar Khayyam's Solution of Cubic Equations", Mathematics Teacher 51 (Apr., 1958): 285-86.
- Since J and H are on the Hyperbola,
(EK)(KJ) = (EM)(MH).
- Since ED:BE = AB:BG, we have
(BG)(ED) = (BE)(AB).
- Therefore, from (1) and (2),
(EK)(KJ) = (EM)(MH) = (BG)(ED) = (BE)(AB).
- Now (BL)(LJ) = (EK)(BE + KJ)= (EK)(BE) + (EK)(KJ) = (EK)(BE) + (AB)(BE) ( by ) = (BE)(EK + AB)=(BE) (AL),
whence, (BL)^2 (LJ)^2 = (BE)^2(AL)^2
- But, from elementary geometry,
(LJ)^2 = (AL)(LC)
- Therefore, from (4) and (5), (BE)^2(AL)= (BL)^2(LC), or
(BE)^2(BL + AB) = (BL)^2(BC-BL).
- Setting BE= b, AB = a^3/b^2, BC = c in (6), we obtain b^2(BL + a^3/b^2) = (BL)^2(c - BL).
- Expanding the last equation in (7), and arranging terms, we find
(BL)^3 + b^2(BL) + a^3 = c (BL)^2, and it follows that BL= x, a root of the given equation."