Khayyam and His Solutions of the Cubic

Solution Commentary:

Solution of Main Problems: 1. With parallel lines providing the AAA criteria, triangles ABD and ACE are

similar. Thus, by proportionality of sides of similar triangles, AB/AC = AD/AE. By substitution, AB/(AB+BC) = AD/(AD+DE) or (AD+DE)/AD = (AB+BC)/AB. However, this implies that 1 + DE/AD = 1 + BC/AB or AB/BC = AD/DE.

2. To solve the linear equation mx = n, view m and n as lengths of segments. Then, by this construction and Problem #1:

x1

n

m

E

A B

D

C

We have 1/x = m/n, or its equivalent mx = n, with the solution being length x. 3. First, m/n = n/z implies that z = n2/m. Second, m/z = n/w implies that w =

(zn)/m or w = [(n2/m)n]/m. Thus, by simplification, w = n3/m2. 4. The underlying justification are as follows:

(EK)(KJ) = (EM)(MH) Points H and J are on hyperbola (BG)(ED)=(BE)(AB) From proportion used to construct point G (EK)(KJ) = (BE)(AB) Transitive property as EM=BG and MH=ED (BL)(LJ)=(EK)(BE)+(EK)(KJ) Distributing with BL=EK and LJ=BE+KJ (BL)(LJ)=(EK)(BE)+(AB)(BE) Substituting using (EK)(KJ)=(BE)(AB) (BL)(LJ)=(BE)(AL) Substituting EK+AB = AL (BL)2(LJ)2=(BE)2(AL)2 Squaring both sides (LJ)2=(AL)(LC) LJ mean proportional between AL and LC (BE)2(AL)2=(BL)2(AL)(LC) Substitution (LJ)2=(AL)(LC) (BE)2(AL)=(BL)2(LC) Cancellation of common AL length (BE)2(BL+AB)=(BL)2(BC-BL) Substituting AL=BL+AB and LC=BC-BL m2(BL+ n3/m2)=(BL)2(p-BL) Substituting BE=m, AB = n3/m2, BC = p (BL)3 + m2(BL) + n3 = p(BL)2 Simplifying to remove denominator term x = BL is the desired root Surprise, it works!

Extension 1: Two rectangular parallelepipeds of volumes m2n and p2q are as follows:

m

q

p

p

n

m

The proportion m/p = p/k implies that k = p2/m, while the proportion m/k = q/n implies that q = (mn)/k. Substituting for k, q = (mn)/[p2/m] = (m2n)/p2. Thus, m2n = p2q.

Extension 2: Segment CD is the mean proportional between segments OD and DA, which implies that CD2 = (OD)(DA) = (OD)(OA OD) = (OD)(OA) (OD)2. But by substituting length OA, we have 22

2 )()( ODm

nODCD

= or 2222 )()( ODmODnCDm = .

And, point C on parabola implies that CD = (1/m)(OD)2 or by substituting, 22

222 )()( ODmODn

m

ODm =

or 222

42 )()( ODmODn

m

ODm =

or

( ) 224 )()( ODmODnOD = . Cancelling a common term OD gives the desired cubic relationship (OD)3 + m2(OD) = n.

Open-Ended Exploration: Students need to play with the construction dynamically. The pdf http://www.math.ntnu.no/~hanche/blog/khayyam.pdf provides a slightly different take on the problem and the roots being revealed.

Teacher Commentary:

Eves (1983) served as the model for this investigation of the cubic x3+ m2x + n3 = px2. The key portion as originally published in Eves (1958).

When investigating the cubic equation cubic x3+ m2x + n3 = px2, emphasize Khayyams view that each of the four elements are solids (e.g. m2x is the volume of a rectangular parallelepiped with side lengths m, m, and x). Thus, using four different rectangular parallelepipeds, Khayyam is combining three solids to equal a fourth solid. This view probably influenced Khayyams view of either the existence or the usefulness of negative and imaginary roots.

When exploring the Lemma in Extension Problem #1, make sure to connect it back to the first two steps in the original construction. Also, as a good follow-up question, ask students to determine if the Lemma is true for general parallelepipeds, where a parallelepiped is defined to be a solid with three pairs of parallel faces. Also, Berggren (1986, pp. 120-121) is the source of the Lemma.

Having students consider each step of Khayyams proof may seem tedious, but in my experience, it is worth it. That is, students gain an appreciation for both Khayyams interesting construction and his overall commitment to solve all types of cubic equations geometrically. Because he used conic sections, his strategy was to show a solution was obtainable in theory but not using Euclidean tools.

Finally, considering the given problem and its proof, students should reflect on the fact that Omar used no symbols. Rather, he wrote everything out in words. For example, he occasionally would uses shay for unknown, jidr for root, and dirham

for number. However, when searching for an unknown, his often used the special term mal, which literally means treasure.

As a good source of writing projects, students can explore any of the following ideas relative to Khayyams geometric solution of cubic equations:

Referring to a cubic equation, Khayyam once said: This cannot be solved by plane geometry, since it has a cube in it. For the solution we need conic sections. Reacting to this statement on his web site (http://www.mathpages.com/home/kmath448.htm), Kevin Brown suggests: Here we might credit Khayyam with anticipating the eventual proof of the unsolvability of the Delian problem (duplicating the cube) by straight-edge and compass, but it seems to me this comment may also shed some light on his statement that the cubic cannot be "solved algebraically". Do you agree with this broad suggestion and its implications?

Mardia (1999) suggests that Omar might have foreshadowed the analytical geometry of DescratesDescartes methods are similar to those of Omar but Descartes realized that certain intersection points represented imaginary rootsand thus gave a complete solution for negative, positive and imaginary roots. First, do you agree with this Descartes-built-on-Omar idea? Second, did Omar ignore imaginary roots while Descartes did not?

Investigate Khayyams geometrical solutions of other types of cubics. Other than an involvement of conic sections, is there a general pattern or is each unique in its approach? Berggren (2007) provides the solution to the cubics x3+bx=ax2+c and x3+bx=ax2+c, Berggren (1986) provides the solution to the cubic x3+mx=n, and Lumpkin (1978) provides the solution to the cubic x3=bx+a.

Additional References:

Berggren, J. (1986). Episodes in the Mathematics of Medieval Islam. Springer. Pp. 118-124.

Berggren, J. (2007). Mathematics in Medieval Islam in V. Katzs (ed.) The Mathematics of Egypt, Mesopotamia, China, India, and Islam. Princeton University Press, pp. 556-560.

Eves, H. (1958). Omar Khayyam's Solution of Cubic Equations, Mathematics Teacher. Vol. 51, pp. 285 - 286.

Eves, H. (1983). Omar Khayyams geometrical solution of cubic equations (ca. 1090) in Great Moments in Mathematics: Before 1650. pp. 148-159.

Lumpkin, B. (1978). A mathematics club project from Omar Khayyam. Mathematics Teacher. December, pp. 740-744.

Mardia, K. (1999). Omar Khayyam, Ren Descartes and solutions to algebraic equations. Paper presented to Omar Khayyam Club (London) and is available at http://www.amsta.leeds.ac.uk/~sta6kvm/omar.pdf.