## al- Khayyami

## Outline of Key Results in

Chapter V of Algebra

(Kasir translation)

*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.

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

*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:

- 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 [3]) = (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."

Eves, Howard. "Omar Khayyam's Solution of Cubic Equations", Mathematics Teacher 51 (Apr., 1958): 285-86.