 One of the highest achievements of Islamic mathematics occurred in the work of Omar al-Qayyami, who lived in the 11th century. Qayyami is probably best known as the author of a collection of quatrains known as vrubayats. These are four line poems that gained some notoriety in the 19th century when they were translated into English. Qayyami is more important as a mathematician and possibly the architect of the North Dome Chamber of the Friday Mosque in Isfahan. Qayyami wrote of treaties explaining how to solve cubic equations. His basic approach was to identify curves whose intersections would yield the solution to a cubic equation. His basic tool is the theory of proportionals. So remember that one of our problems is that in order to make geometric sense of an algebraic expression we must interpret all terms as representing the same type of object. So for a cubic expression x cubed plus ax squared plus bx plus c, this means that all terms must be viewed as solid objects. The cube of the unknown, x cubed, is a cube with unknown side length. The quadratic term ax squared, we can take that as a solid with square base of unknown side length and height a. The linear term bx, well that has to be a solid with unknown side x, and it's convenient if we assume its base is square, so bx is really p squared x. And finally the constant term c is some rectangular prism, and it's convenient if we assume one side has the same dimensions as the known dimensions of the linear term. So c is p squared q. In order to understand Qayyami's approach it's helpful to understand quite a bit of geometry and in particular the theory of proportionals. So some of our useful results. Suppose a is to b as c is to d, then ad is equal to bc and conversely. This is easiest to see algebraically if we treat these ratios as fractions. So algebraically, a is to b as c is to d, well that's really saying that a over b is equal to c over d, and we can multiply across to get ad equal to bc. And so that's the forward direction. Conversely, if we have ad equal to bc, we can work our way backwards to the original proportionality. Another proposition, if a is to b as c is to d, then am is to bm as c is to d and conversely. Again we can view this algebraically. Our proportionality gives us a over b equals c over d, and if we, in modern terms, multiply numerator and denominator by the same thing, we get a fraction equivalence that we can also convert back into a ratio. And the converse will be left as an exercise for the viewer. And again, if a is to b as c is to d, then am is to bm as cm is to dn, and conversely, a somewhat more obscure result involves what is known as a mean proportional. Suppose a is to b as b is to c, then a squared is to b squared as a is to c. That's the squares on the first to the square on the second is the first to the third. And what's the point of introducing a whole bunch of propositions if we don't use them? Let's go ahead and do this entirely within the area of proportionality. So we know that a is to b as b is to c. One of our propositions says that we can multiply the first and the second terms in a proportionality by different amounts. So we'll multiply the first by a and the second by b. And another one of our propositions says that we can remove a common factor. And so here on the right side, we see that both have this factor b, which we can remove. And we can extend this idea to get what's called a continued proportional. Suppose a is to b as b is to c, and b is to c as c is to d. Then a squared is to b squared as a is to c. And we can also show that a squared is to b squared as b is to d. And all of this holds conversely. The significance of this is that if we can express a ratio of squares, we can unpack it and obtain a proportionality. And we can convert that back into a conic section or a circle. In order to do that, it helps to have some useful geometry about the conic sections and circles. So let a b be the diameter of a circle and c h be perpendicular. Then a c is to c h as c h is to c b. And conversely, if a c is to c h as c h is to c b, then c h is perpendicular to the diameter a b of a circle. We'll prove the converse algebraically. Let a b equal q, the diameter, a c equals x, and c h equals y. And note that this means c b is q minus x. And so we're assuming that a c is to c h as c h is to c b. x is to y as y is to q minus x. So we can rewrite this as a rational equation and do a little work and rewrite this as the equation of a circle. It also helps to know something about the parabola and hyperbola. So we know that for any parabola, there is a line p called the parameter. So that for any point x on the parabola, the square on yx, the line drawn ordinate-wise, is equal to the rectangle on p and o y. Or algebraically, if we let yx be x and o y be y, then p y equals x squared. And for the hyperbola, we have the following result. For any hyperbola, there is an area, q squared, so that for any point y on the hyperbola, the rectangle on o x and xy is equal to q squared. And again, algebraically, if we let o x equal x and xy equal y, then q squared equals xy. And we refer to this as a rectangular hyperbola.