Chapter 5 Athens Let no one ignorant of geometry enter my door-plato s Academy

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1 43 Chapter 5 Athens Let no one ignorant of geometry enter my door-plato s Academy It was perhaps the devastating discovery of the irrationality of the square root of that sent such quivers down the Greek mathematical spine that they are, for several centuries, going to insist on geometric constructions as the only source of true knowledge a view that will be held by some until Newton's time. Since they could do much with their tools: the straightedge and compass, their reliability on them wasn't an overwhelming limitation. And even perhaps, just a few centuries after Pythagoras, some Greek minds realized the foolishness of the insistence on straight edge and compass alone, without even one single marking on the ruler. From the historical point of view, however, there were some constructions they could not accomplish, and these became a famous and important part of their legacy, as well as a meaningful component of their mathematical activity of the period. In particular, three problems and a general construction have been left to posterity. These impossible constructions were not successfully disposed of until the nineteenth century: The Trisection of the Angle. When any angle is given, one was required to trisect it. The key word in the problem is the word any. In other words, they desired a technique applicable to any angle. Some angles, such as 90, could be trisected easily. The Quadrature of the Circle. Given a circle, to build a square with the same area. As we will see below they could square any polygonal area, and some curvilinear forms, but not the circle. Archimedes would later even square the parabola. The Duplication of the Cube. Given a cube, to construct a cube with twice the volume. As we will see, they could double the square, but not the cube. Of course, as mentioned above, all of these constructions were to be done with straightedge and compass alone. Their insistence on that narrow as it was forbade even one arbitrary marking on the straightedge.

2 44 Along with the three problems mentioned above, there is a fourth one: Which regular polygons were constructible? The heptagon, the 7-sided polygon, was particularly fascinating to them since it was the one with the fewest number of sides that could not be done. Around 1800, the young Gauss would explain completely which polygons are constructible and which are not. But in order to appreciate these problems, as well to study the Greeks attempts, we need to review what is constructible by their tools: straightedge and compass. 1. To bisect a segment, or equivalently, to draw the perpendicular bisector. Draw the two circles centered at each of the end points and going through the other, then join the intersection of the two circles, and this line will actually will be the perpendicular bisector of our original segment.. To drop the perpendicular to a line from a point not on the line. First draw any circle centered at the point and crossing the line. The two intersections mark a segment than when bisected by the previous construction will give the desired perpendicular. 3. To drop the perpendicular to a line at a point on the line. Draw any circle centered at the point and bisect the segment that is formed by the two intersections of this circle with the line. 4. To build the equilateral triangle on a segment, or to build the square on a given segment. Both of these were easy extensions of previous constructions and we will let the pictures speak for themselves. 5. To bisect any angle. Starting with any angle. Draw an arbitrary circle centered at the vertex of the angle and draw the corresponding chord. Bisect the chord. The angle has also been bisected. Surprisingly, this does not work for trisections.

3 45 6. Given a line and a point not on it, to draw the parallel to the line through the point. This was a common construction, and there were several ways to accomplish it. One of them was as follows. Start with a line and a point off it. Choose an arbitrary point on the line, and centered there, draw a circle that goes through the point not on the line. From the intersection of the circle and the line, draw a circle with the same radius, and also draw a circle with the same radius centered at the point off the line. Consider the point where these two circles intersect. Then the line going through this point of intersection and the point off the line is parallel to the original line. An alternative way is to draw two perpendiculars. 7. To divide a segment into n equal parts: Given a segment, divide it into n equal parts. We will exemplify the general procedure with n = 5. Take an arbitrary segment. Draw an arbitrary line through one of the end points of our segment to be divided. And mark off 5 equal segments of arbitrary length in that line. Draw the line connecting the end point of our fifth segment to the other end point of our original segment. Draw the parallels to that line from each of the other points. Similarity of triangles then gives the proportionality of the segments. 8. Doubling the square: Given a square to build the square with twice the area. This was explained by Plato in one of his dialogues. Take your original square and build a square with twice the side. This new square has four times the original area. If we then join the midpoints of the sides, we have exactly half of the big square, or equivalently, twice the little one. 9. Given two segments, to find their arithmetic mean, in other words given a and b, a+ b find. This is just the bisection of the segment a+ b. 10. Given two segments, to find their geometric mean: Given two segments a and b, find ab. This construction was very important to them since it involved one of their dearest means. Starting with the two segments a and b, build a semicircle with diameter a+ b, and erect the perpendicular where the two segments meet until it touches the circle. This segment is the geometric mean.

4 46 hence x = ab, and this is what we wanted. To prove that this line is indeed of the appropriate length, construct the triangle, which by Thales' Theorem we know it is a right triangle, so by Pythagoras' Theorem, a + x = y, x + b = z, and y + z = ( a+ b), so adding the first two equations and substituting, we get a + x + x + b = ( a+ b) = a + ab+ b, 11. Squaring a rectangle: Given a rectangle, to find the square with the same area. This is just the geometric mean of the two sides of the rectangle, for if b is the base and h is the height, then the square with side bh has the same area as the rectangle with base b and height h. 1. Squaring a triangle: Given a triangle, to find a square with the same area. Start with a triangle and draw its altitude. Then take the geometric mean of its altitude and its base, and build the square on that geometric mean. If b denotes the base of the triangle and h its altitude, then the geometric mean is bh, so the square has area bh which is twice what we want, hence if we take half of the square we will have a square of the same area as our original triangle. 13. Squaring any polygonal area, or equivalently, transforming any polygonal area to a triangle (without changing the area): Given any polygonal area, to find the square with the same area. The idea was to cut off one vertex at a time until we are down to a triangle and then square the appropriate triangle. The key was to exchange two triangles with the same base and height and thus of equal area. We illustrate by transforming a pentagonal polygon into a quadrilateral. Take any three consecutive vertices. Draw the chord joining the two outer ones. Draw the parallel through the middle vertex to that chord. Any triangle with the same base (the chord) and the same height will have the same area. Extend one of the sides in order to form a new triangle. Make the swap, and, presto, a corner has disappeared, and we have a quadrilateral of the same area.

5 Inscribing the hexagon, or the triangle, or the square. Given a circle, inscribe these regular polygons in it. To inscribe a polygon what one needs is to build the appropriate angle at the center of the circle, thus the chord that subtends that angle will be the side of the polygon. The hexagon, triangle and square were the easiest of all regular polygons to inscribe. First the hexagon. We need to break 360 into 6 equal parts so we need 60. That is, we need an equilateral triangle built at the center (two radii and the side of the hexagon) hence that side is the radius, it will fit exactly 6 times around the circle. For the triangle, once we have the hexagon, just join every other side. Finally for the square (we need 90 ), so just draw two perpendicular radii, and the hypotenuse will be the desired side of the square. We come now to the polygon that was dearest to the Greeks: 15. Inscribing the pentagon: Given a circle, inscribe the regular pentagon in it. This construction is basically equivalent to that of the construction of the golden mean (see construction 16). Start with a circle. Construct two perpendicular diameters. Then bisect any of the radii to obtain the point X. Construct a circle with center X and going through one of the points of the perpendicular diameter in the circle, for example the point A. Find the point of intersection B of this circle with the diameter that X is on. Then the segment AB is the side of our pentagon. A X B A X B A 16. To break an arbitrary segment into the Golden Ratio. A segment was said to be broken into the Golden Ratio if the short piece was to the long piece as the long piece was to the whole. Starting with the segment AB.

6 48 We find the midpoint M and the perpendicular at A. Then we find the point P on the perpendicular whose distance to A is the same as the distance from A to M. Then centered at P, we find the point Q whose distance to P is the same as the distance from to P to B. Then we find the point X in the segment AB is the same as the distance from A to Q. We claim X is the desired point. That is, we need to prove that A B BX AX =. AX AB P From the construction we know that A B A B A B M M X AX = AQ, AP = AM = BM = 1 AB, PQ = BP, PQ = AP + AQ, Q Q and from the Pythagorean theorem: ( BP) ( AB) = + ( AP). Substituting we get: hence ( AM)( AQ) ( AQ) ( AB). Since ( AP + AQ) = ( AB) + ( AP), ( AP) + ( AP)( AQ) + ( AQ) = ( AB) + ( AP) or + = AQ = AX and ( AM) = AB, ( AX) = ( AB) ( AB)( AX), so ( AX) ( AB)( BX), which is what we wanted. = As mentioned above, the Greeks were particularly fascinated with the pentagon. One of the reasons for this interest was the fact that the Golden Ratio naturally occurred inside the pentagon. Namely, if we look at any two diagonal of the pentagon then they cut each other by the Golden Ratio in other words, in the picture, we should have that α β α A X Y B AX: XB = XB: AB The proof of this is not hard. Consider the picture on the left. We know that AOB=108, we know too, OAB= OBA=α. But then, α = , and so α=36, and so β=36 also. Thus, OBX is isosceles, and we have XB=OB. Also, the two triangles OXA and AOB are similar, hence AX:AO=AO:AB, but BO=AO=XB, and so AX:XB=XB:AB as desired., A X B 17. Inscribing other regular polygons: Given a circle inscribe other regular polygons in it, in particular the decagon (10 sides) and the pentadecagon (15 sides). Both of these constructions are easy derivatives of previous constructions. The decagon is directly obtainable from the pentagon, and the general observation that once we could get an n -gon, then one can easily get the polygon with twice the number of sides by just bisecting a side and going on a radius on that bisection. The example from the pentagon to the decagon is typical. Start with a pentagon.

7 49 Bisect any side. Extend the radius through the middle point. Join the new point of the circle to one of the two vertices of the pentagon. That is the side of the decagon. The pentadecagon was more interesting and consisted on imposing a 3-gon (a triangle) and a 5-gon (a pentagon) in the right fashion: and thus obtaining the side of our 15-gon: In closing, then, what was remarkable to the Greeks was that: they could bisect any angle, yet they could not trisect every angle; they could square any polygonal area, yet they could not square the circle; they could double the square, yet they could not double (duplicate) the cube; and while they could build the pentadecagon, they could not construct the heptagon (7 sides). Naturally we must not lose sight of the arbitrariness of the restrictions, and later in their history they express reservations about the limitations. For example, Archimedes will, in the near future (next chapter), give a trisection by putting a single arbitrary mark on the straightedge.