## Outline of Key Results in Chapter V of Algebra (Kasir translation)

1. Between two given lines it is required to find two other lines such that all four will form a continuous proportion.
• 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.

2. Given the rectangular parallepiped ABCDE, whose base is the square AD, and the square MH, construct on MH a rectangular parallelpiped equal to ABCDE.

3. 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.

4. 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:
1. Since J and H are on the Hyperbola,
(EK)(KJ) = (EM)(MH).
2. Since ED:BE = AB:BG, we have
(BG)(ED) = (BE)(AB).
3. Therefore, from (1) and (2),
(EK)(KJ) = (EM)(MH) = (BG)(ED) = (BE)(AB).
4. Now (BL)(LJ) = (EK)(BE + KJ)= (EK)(BE) + (EK)(KJ) = (EK)(BE) + (AB)(BE) ( by [3]) = (BE)(EK + AB)=(BE) (AL),
whence, (BL)^2 (LJ)^2 = (BE)^2(AL)^2
5. But, from elementary geometry,
(LJ)^2 = (AL)(LC)
6. Therefore, from (4) and (5), (BE)^2(AL)= (BL)^2(LC), or
(BE)^2(BL + AB) = (BL)^2(BC-BL).
7. 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).
8. 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."
Eves, Howard. "Omar Khayyam's Solution of Cubic Equations", Mathematics Teacher 51 (Apr., 1958): 285-86.