Back to the papersThere were three problems that the ancient Greeks (600BC to 400AD) tried unsuccssfully to solve by Euclidean methods, all of which were proven unsolvable by these means as much as two thousand years later, as a result of progress in algebra, and the idea of analytic geometry in the sense of Descartes. The Greeks wanted to solve these problems using only a Euclidean constructions, or as they themselves called them, "plane" methods. Though they were never able to do so ( as they cannot be done this way, they did find a series of remarkably clever constructions using more powerful techniques, involving so-called "solid" and "mechanical" methods, as well as a technique called "verging". Then, in the 19th century, the impossibility of finding purely Euclidean constructions for these problems was finally proved.
The three classical Greek problems were problems of geometry: doubling the cube, angle trisection, and squaring a circle. Duplication of the cube is the problem of determining the length of the sides of a cube whose volume is double that of a given c ube. A cube by definition is a three dimensional shape comprised of a height, width, and depth all of the same magnitude s. To find its volume, one multiplies the length (s) by the width (s) and then by the depth (s): the volume is s(s(s or s3.
Diagram not converted, here and below
In Cube A, red lines are of length s. They comprise the height of the cube. Green lines are also of length s and comprise the depth of the cube. Blue lines are also of length s and comprise the width of the cube. Twice the volume of a cube with volume s3 would be 2s3. Let this new cube of twice the volume of the original have sides of length y. Then, y3 equals 2s3. The problem of the duplication of the cube is then to find is the value of y. So, solving for y one gets
y3 = 2s3
y = [2(s3)](1/3)
y = 2(1/3)s 2(1/3) is an irrational number. Irrational numbers cannot be expressed as the quotient of two integers. The Greek mathematicians thus needed to find a way to construct this quantity geometrically. Angle trisection means dividing a given angle into three equal angles. The problem may have come up after geometers constructed a pentagon. The geometers wanted to be able to construct more polygons but realized that the only way to make ones that had n ine sides or an amount of sides that was a multiple of nine involved having to trisect an angle. Thirdly, squaring a circle (quadrature of a circle) is the construction of a square whose area is equal to the area of a given circle. "Since a circle with radius r has the area pr2, the problem of constructing a square with area equal to that of a given circle whose radius is the unit length 1 amounts to the construction of a segment of length p(1/2) as the edge of the required square. This segment will be constructible of and only if the number p is constructible" [7, p140]. Since p is an irrational number it cannot be expressed as the quotient of two integers, which is how the difficulty arises. The Greeks were unable to solve these problems under specific restrictions - with "plane" proofs. We will now discuss their classification of problems into three levels of complexity, as explained by the late commentator Pappos. The Greeks classified prob lems by the types of curves entering into the constructions used to solve them. The three classes of problems were plane, solid, and linear, corresponding to three types of curves. Plane problems could be solved only by the use of "plane curves": straig ht lines and circles. A straight line has the formula y=mx+b where m is the slope, b is the y-intercept, and (x,y) is a point through which the line passes. Circles are of the form x2+y2=1 where (x,y) is a point where the circle passes through. These ar e specific curves whose equations have degree 1 or 2. "Solid problems" can be solved by means of one or more "solid curves", meaning conic sections. Conic sections are "the nondegenerate curves generated by the intersections of a plane with one or two nappes [a nape is one of the single cones that makes up a double cone] of a cone. For a plane perpendicular to the axis of the cone, a circle is produced. For a plane which is not perpendicular to the axis and which intersects only a single nappe, the curve produced is either an ellipse or a parabola (Hilbert and Cohn-Vossen 1999, p. 8). The curve produced by a plane intersecting both nappes is a hyperbola (Hilbert and Cohn-Vossen 1999, pp. 8-9)." The equation of a parabola is of the form y=ax2+bx+c where a, b, and c are constants and (x,y) is a point of inte rsection. Ellipses have the equation (x2/a2)+(y2/b2)=1 where a and b are constants and (x,y) is a point of intersection. A hyperbola has the equation (x2/a2)-(y2/b2)=1 where again a and b are constants and (x,y) is a point of intersection. Diagram from http://mathworld.wolfram.com/ConicSection.html and edited by me. In diagrams A and B is a conic section made up of two cones, or nappes, with their pointed bases touching. In diagram A, the top cone has a plane transverse it at an angle. T his plane cuts through from the base of the cone to a side. Where the plane meets the edges of the cone forms a parabola. In the bottom cone, a plane cuts through it perpendicular to the vertical axis of symmetry. Where this plane meets the edges of t he cone forms a circle. Also in the bottom cone, where a plane cuts through its sides, at an that is not perpendicular to the vertical axis and does not cut through a base, an ellipse is formed where the plane and sides of the cone meet. In diagram B, a plane cuts through both cones forming a hyperbola where the sides and plane meet. So, proofs of solids contained the use of more than just straight lines and circles. Linear problems needed to be solved by the methods already mentioned and by more complicated curves than those of solids. These curves included such involved shapes a s spirals, quadratrices, conchoids, and cissoids. There are different types of spirals; they have different formulas but all confirm to the definition: "The locus of a point P which winds around a fixed point O (called the pole) in such a way that OP is monotonically decreasing." (*) For example, the Spiral of Archimedes has the polar formula r = a. Diagram from http://www-groups.dcs.st-and.ac.uk/~history/Curves/Spiral.html The equation of the Conchoids have the equation x2y2=(y+1)2(4-y2). Below is an example of a conchoid.
Diagram from http://www-groups.dcs.st-and.ac.uk/~history/Curves/Conchoid.html and edited by me. Cissoids have the equation y2 = x3/(2a - x). Diagram from http://www-groups.dcs.st-and.ac.uk/~history/Curves/Cissoid.html From these three shapes and their equations one can see how much more elaborate the curves used in linear are compared to solid and plane. Therefore, the different type of equations and thus shapes the ancient Greeks had were classified into three distin ct groups of proofs of various complexity.
The mathematicians wanted to solve the Classical Greek problems with proofs using plane; they just wanted to use circles and straight lines. However, "it is evident that the Greek geometers came very early to the conclusion that the three problems in q u estion were not plane, but required for their solution either higher curves than circles or constructions more mechanical in character than the mere use of the ruler and compasses" (4). Many geometers of different schools tried to solve the problems with the limits of just straight lines and circles but none succeeded: "Geometers took up the question and made no progress for a long time" (5). Instead, they had to resort to proofs that would classify the problems as solid. Hippocrates of Chios made find ing solutions to the duplication of the cube possible. He found that the problem could be simplified to that of finding two mean proportionals that were in continued proportion between two given straight lines: "Hippocrate showed that the problem could b e reduced to that of finding two mean proportionals: if for a given line segment of length a it is necessary to find x such that x3 = 2a3, line segments of lengths x and y respectively may be sought such that a:x = x:y= y:2a; for then a3/x3 = (a/x)3 = (a/ x)(x/y)(y/2a) = a/2a = 1/2" (6). So, first one starts with a given length a. When a is cubed (a)(a)(a) and this is multiplied by 2, the product is equal to x cubed (x)(x)(x). One also needs to find the line segments x and y that fulfill the ratios that a divided by x is equal to x divided by y which is then also equal to y divided by 2a. If one finds x,y, and a such that to fulfill these ratios then a33/x3 could then be set equal to the ratios we were trying to fulfill since (a/x)3 means (a/x)(a/x)(a/ x). And the ratios were that (a/x) = (x/y) = (y/2a). So by substitution (a/x)3 = (a/x)(x/y)(y/2a). Simplifying the right hand side we are left with 1/2. Therefore a3/x3 = 1/2. This calculation to find the solution today is now taught to be divide eac h side by 2 to get 1/2x3 = a3. Then divide both sides by x3 to get 1/2 = a3/x3.
Menaechmos was a Greek mathematician from the period of approxiametly 380- 320 B.C. He is one of the geometers who came up with a proof for the duplication of the cube. He uses reduction to the two mean proportionals in his proof. Also, "Menaechmus' solution is probably the first one using the conic sections and their properties" (7) His problem can be classified as plane because he uses such curves as parabolas and hyperbolas in his proof. Following is a translation of his proof to demonstrate t he use of the reduction to the two mean proportionals and the use of linear methods. Let AO, OB, be placed at right angles, represent the two given straight line segments, AO being the greater. Suppose the problem is solved, and let the two mean proportionals be OM measured along BO and ON measured along AO. Complete the rectangle OMPN. Then, since AO:OM = OM:ON = ON:OB, (i.e. a:x = (x:y) = y:b is equialent to a:y = (y:x) = x:b inversed), we have (1) OB.OM = ON2 = PM2, (multiply out the last equality) so that P lies on a parabola with vertex O, with axis OM, and with latus rectum OB. Again (2) AO.OB = OM.ON = PN.PM, (i.e. ab = xy), so that P lies on a hyperbola with O as a centre and OM, ON as assymptotes, and such that the rectangle contained by the straight lines PM, PN drawn from any point P of the curve parallel to one assym ptote and meeting the other respectively is equal to the given rectangle AO.OB. If, then, we draw the two curves in accordance with the data, we determine the point P by their intersection, and AO:PN = PN:PM = PM:OB (a:x = x:y = y:b) Proof from http://www.cs.mcgill.ca/~cs507/projects/1998/zafiroff/ and edited by me.
The ancient Greek mathematicians also could not find a proof for angle trisection that would classify the problem as a plane: "We are told that the ancients attempted, and failed, to solve the problem by 'plane' methods, i.e. by means of straight line and circle; they failed because the problem is not 'plane' but 'sold'. Moreover they were not yet familiar with conic sections, and so were at a loss; afterwards, however, they succeeded in trisecting an angle by means of conic sections, a method to which t hey were led by the reduction of the problem to another, of the kind known as vergings" (8). The reduction to verging method is used to trisect acute angles, the proof of trisecting right angles only involves drawing an equilateral triangle. Verging in volves drawing a line of a given length that is twice that of another given length, between to perpendicular lines so that the line being drawn verges, slopes or inclines, towards a certain point. Below is a diagram of line ED that that is twice the leng th of AB and is between AE and AC, which are perpendicular, in such a way that ED verges toward point B. Diagram from Sir Thomas Heath's A History of Greek Mathematics The proof of why the verging produces a trisection of an angle is as follows with reference given to the diagram above.
Let ABC be the given angle, and let AC be drawn perpendicular to BC. Complete the parallelogram ACBF, and produce the side FA to E. Suppose E to be such a point that, if BE be joined meeting AC in D, the intercept DE between AC and AE is equal to 2AB. Bisect DE at G, and join AG. Then DG = GE = AG = AB. Therefore angle ABG = angle AGB = 2(angle AEG)
= 2 angle DBC, since FE, BC are parallel. Hence angle DBC = 1/3 angle ABC Proof from Sir Thomas Heath's A History of Greek Mathematics
The squaring of a circle also could not be solved using only a straight lines and circles in the proof. The Greek mathematicians realized this: "Hippocrates [470 - 410 BC] was aware that 'plane' methods would not solve it" (9). However, the mathematicia ns produced other proofs to solve the problem. In order to prove the solution, p had to be approximated. The approximation originated with Antiphon (480 - 411 B.C.) who "originated the idea of exhausting an area by means of inscribed regular polygons wi th an ever increasing number of sides" (10). This can be restated as drawing inside the circle a regular polygon with as many sides as possible, as the number of sides increase, they get smaller and thus fit closer to the shape of the circle. Below one can compare the amounts of area of the circle (shaded) that is not included in the polygon depending on the number of sides of the polygon (assume the sides that make up the polygon are equal). With Antiphon's idea, Archimedes (287 - 212 BC) was able to approximate pi. Archimedes produced a polygon with ninety-six sides inside a circle. So, he was then able to approximate p as between 3 10/71 and 3 1/7, which is a range of 1/497. Once was s o closely approximated, mathematicians were able to produce a square whose area was almost equal to that of a circle's. Archimedes did produce a proof of squaring a circle that contains the use of pi. A description of Archimedes' proof is as follows: It [Archimedes' proof] states that the area of the circle is equal to the area of a right triangle whose perpendicular height is equal to the circle's radius and the base length is equal to the circle's circumference. This can easily be demonstrated algeb raically to hold true, but note that this is not a proof of equality. The base length can be found by rolling the circle down a line for exactly one full revolution Description and diagram from http://www.perseus.tufts.edu/GreekScience/Students/Tim/SquaringCircle.html Notice how p is used in the equation.
In 1837 Pierre Wantzel proved that the classical Greek problem of a squaring a cube could not be solved with the restriction of using only straight lines and circles. He published the proof in Liouville's Journal of: the means of ascertaining whether a geometric problem can be solved with ruler and compasses. Wantzel proved that a cube could not be duplicated under the restrictions because the proof requires a ratio not constructible with just straight lines and circles. In order to construct the proo f, ones needs to find the ratio of sides of the original cube. This ratio is the "Delian constant" 21/3. This number is not an Euclidean number meaning that it is defined as not being able to be constructed using classical geometric constructions. Wantzel also proved that an angle could not be trisected with the restrictions in 1837. Below is a reconstruction of his proof. The term "Lemma" is used in the reconstruction; it is a theorem that states that for a cubic polynomial equation with integer coefficients (ax3 + bx2 + cx + d = 0) either one of its roots is rational or none of its roots are constructible. "Following Wantzel, I'll show that a 60 degree angle is not trisectable... cos(3a) = cos(a)cos(2a) - sin(a)sin(2a) = cos(a)(cos2(a) - sin2(a)) - 2sin2(a)cos(a) = cos(a)(2cos2(a) - 1) - 2(1 - cos2(a))cos(a) = 4cos3(a) - 3cos(a) For a = 20o, cos(3a) = cos(60o) = 1/2 and the equation becomes 8cos3(a) - 6cos(a) - 1 = 0. Replacing cos(a) with x we finally get (2) 8x3 - 6x - 1 = 0 The substitution v = 2x transforms (2) into (3) v3 - 3v - 1 = 0 By Lemma, we only have to show that this equation has no rational roots. Assume on the contrary that v = p/q is a rational solution to (3) in lowest terms. Then, p3 - 3pq2 - q3 = 0. First, rewrite this as p(p2 - 3q2) = q3. It follows that p|q3. If p were not a unity (i.e. 11) then any prime factor of p would divide q. Since the fraction p/q is irreducible, p ought to be a unity. On the other hand, we may also write q(3pq + q2) = p3, which, by the same argument, shows that q|p, i.e. q = 11. It appears that the only rational solution equation (3) might have is either +1 or -1. By direct inspection, neither satisfies (3). By Lemma, (3) has no constructible solutions and neither does (2). In addition, in 1882 Lindemann proved that the restrictions prohibited the construction of squaring a circle. He devised a proof that p is transcendental, meaning that it satisfies no algebraic equation with integral coefficients. Since quadrature of a circle means the Euclidean construction of vp, and p satisfies no algebraic equation, it follows that neither p nor its square root can be constructed by "plane" methods. In conclusion, the classical Greek problems could not be solved because the mathematicians put limits on how they could prove them. When they classified the problems as plane, meaning they could only use straight lines and circles to solve them, they res tricted themselves. However, when the Greek geometers removed the restrictions they were able to constructs proofs. The reason why they could not construct proofs with the restrictions was because it is impossible as proven in the nineteenth century whe n mathematicians were able to construct proofs of the impossibly of solving the classical Greek problems.
Back to the papers