Geometric Constructions

HISTORY OF MATHEMATICS Spring 2005 Geometric Constructions Notes, activities, assignment; #3 in a series. Note: I m not giving a specific due date for this somewhat vague assignment. The idea is that it should give you a better feeling for what we will be discussing when dealing with the three problems of classical Greek geometry. In these constructions, the only tools allowed are an unmarked ruler (straightedge) and a compass. Here is what is legal: 1. One usually starts with at least two given points because given a single point, all one can do is draw a circle around that point. That s about it, so if you are only given a single point, there isn t much you can do. You cannot use the straightedge in this case. 2. Given two points, one can draw the straight line joining the points. One can draw a circle centered at one point having the segment between the points as radius. 3. If it is legal to draw two lines, any point at which they intersect can be added to the points you have and becomes an additional given point for future constructions. Similarly, if it is legal to draw a circle and a line, or two circles, then the points of intersection of these figures can be added to the points one has constructed, and become given points for the next construction. 4. It is also legal, given points P, Q, a line l and a point R on l, to take the compass and open it to a width P Q (place the one end at P, open so the other end is at Q) and then, keeping the compass open to width P Q, draw a circle of radius P Q with center at R. This gives us two segments of length P Q at both sides of R on l. We ll call this transporting the segment P Q to start or end at R on l. 5. To give a segment is equivalent to giving two points, the endpoints of the segment. To give a circle is equivalent to giving two points, one is the center and the segment between the points is the radius. It might be convenient to have some notation, some of which has already been used. We ll use capital letters such as A, B,..., P, Q,... to denote points. If P, Q are points, we ll denote by P Q the segment of endpoints P and Q, which is a constructible object if P, Q are given. We ll denote by P Q the ray emanating from P and going through Q; it is the half of the straight line determined by P, Q which begins at P and contains Q. An angle is determined by two rays emanating from the same point, the vertex of the angle. We ll write ABC for 1

the angle determined by the rays BA and BC. There is a slight ambiguity here; the angle considered is always supposed to be less than 180 o. Given a segment AB we denote its length by AB. As already mentioned, starting with one point, you get nowhere (except filling the plane with circles). But starting with two points; i.e., a segment, you get an infinity of new points. Here are some constructions that start with (at least) two given points (interpreted as segment or circle). They vary from from easy to hard. 1. Given two lengths a, b, construct a segment of length a + b. By transporting, we can assume both segments are part of the same line; we can even assume that both start at the same point O and extend to the right of O, the one of length a ends at some point R, the one of length b at some point S: a = OR, b = OS. Put the sharp end of the compass at O, open so the other end is at R. Then, keeping the compass with that opening, center at S and draw the circle. This circle intersects the line l at two points A, B. If B is the one that is farther away from O, then a + b = OB. 2. Bisecting a segment. Assume the segment is OA. With center at O draw the circle of radius OA. Then, draw the circle of radius OA with center at A. These two circles intersect in two points RS draw the line RS. This line is perpendicular to the segment OA and intersects OA in its middle. 3. Drawing the bisector of a segment. If AB is a segment, the bisector is a line perpendicular to the line determined by A, B and dividing the segment AB into two equal parts. This construction is the same as in 2. 4. Given a line and a point on the line, drawing the line perpendicular to the given line, through the point. For a line to be given, there must be at least two points given on the line!! Let A be a point on the line l, we want to construct a perpendicular to l through A. Let B be any other point on the line (it has to be a given point!). Draw the circle of radius AB centered at A. This circle intersects l at two points C, D. Noticing that A is the midpoint of the segment CD, now construct the bisector of this segment. That gives you the line through A perpendicular to l. 5. Given a line and a point not on the line, drawing a line through the point parallel to the original line. We use the fact that a perpendicular to a perpendicular is a parallel. Say the line is l, and R is the point not on l Recall that we have at least two points given on l, let O be one of them. With center at R draw the circle of radius RO. This circle intersects l in two points; one is O; call the other one Q. Draw the bisector m of the segment OQ; this bisector goes through R. Now draw the perpendicular through R to the bisector m. That is the desired parallel. For the sequel we assume given two points O and P, hence also the line l determined by them. We take the length OP to be the unit length; i.e., OP = 1. The previous constructions could be done without need of a unit of length, but to construct products and divisions, the unit of length is needed. So these objects are given. Moreover, when we give a segment we will always assume that one endpoint is at O, the other endpoint to the right of O on ell; we can always transport it there. 2

6. Given segments of lengths a, b, construct a segment of length ab. Here we are given points R, S; OR = a, OS = b. Let Q be a point not on l. Such a point can always be constructed, for example when bisecting the segment OP one gets two points off the line l. On the ray OQ mark off distances 1 and b from O; that is, draw the circles of center O and radii OP, OS and let P, S be the respective intersections of these circles with OQ. Draw the line m determined by R and P. Draw the line through S parallel to m. This line intersects l at a point T. Then OT = ab. 7. Given segments of lengths a, b, construct a segment of length b/a. Again we are given points R, S; OR = a, OS = b. As for the product, draw a ray OQ from O which does not lie in l. On OQ mark off distances a and b from O; that is, draw the circles of center O and radii OR, OS and let R, S be the respective intersections of these circles with OQ. Draw the line m determined by P and R. Draw the line through S parallel to m. This line intersects l at a point T. Then OT = b/a. Exercise 1. Assume given the objects of the picture, with OP = 1. O P To the right of O construct points A, B, C, D, E such that a. OA = 3 b. OB = 1/2 c. OC = 4 d. OD = 12 c. OE = 4/3 DO NOT USE THIS SHEET, work on your own paper. BONUS PART. Can you explain why the rules for getting the product and the quotient work? 8. Given a segment of length a, construct one of length a. We assume given Q on l so that OQ = a. Let R be the point such that OR = a + 1. That is, open the compass to a width OP, then draw the circle of radius OP = 1 centered at Q. This circle intersects the line l in two points; let R be the one such that O is not on the ray QR. Now bisect the segment OR, let C be the midpoint of OR. Draw the circle c of center C, radius OC. Draw the perpendicular to l through Q. This perpendicular intersects the circle c in two points; select either on of them, call it S. The segment QS has length a. 3

Exercise 2. Either starting with the same objects as in Exercise 1, or using your constructions of that exercise, mark off a point F on OP such that OF = 3. Bonus part. Why does the construction for the square root work? 9. Bisecting angles. An angle is determined by three non-collinear points. If A, B, C are three such points, then the angle ABC is the angle determined by these points with vertex at B. To bisect, draw the circle of center B, radius BA. This circle intersects the ray BC at a point C. Bisect the segment AC and let D be the midpoint. The ray BD is the bisector of the angle. 10. Polygons. These are, of course, the most interesting of all the constructions. One is given a circle (which, as mentioned, is equivalent to giving two points) and one has to inscribe a regular polygon into the circle. Some are very easy; some are very hard. In our descriptions, O is the center of the circle, P is a another given point; we picture it to the right of O so that the line l determined by O, P is horizontal; we let OP be our unit of length; OP = 1. In all constructions, P is going to be a vertex of the polygon. The regular polygon of three sides, also known as the equilateral triangle. The line l intersects the circle at two points; the one on the right is P, let R be the one on the left. Draw the bisector of the segment RO. This bisector intersects the circle at two points P 1, P 2. The points P, P 1, P 2 are the vertices of the triangle. The regular polygon of four sides, also known as the square. Let R be the other point at which l intersects the circle (as for the triangle) and draw the bisector of RP ; i.e., a line perpendicular to l through O. This line intersects the circle in points we ll call P 1, P 3. The vertices of the square are, in order, P, P 1, R, P 3. Doubling the number of sides and other generalities. First of all, to construct a regular n-gon, all one needs to figure out is how to construct the first vertex to be encountered after P if one moves along the circle in a counterclockwise direction. If that vertex is P 1, then one simply sets the compass opening at width P P 1 and with that opening one draws a circle centered at P 1. This circle intersects the big circle at two points; one is P, the other one is P 2, the next vertex. In practice, one does not have to draw the full circle, just a small arc to intersect the big circle past P 1. Repeat at P 2 to get P 3 ; keep repeating until you are back at P n = P. With this out of the way, we observe that if one can construct a regular polygon of n sides, one can construct at once the one of 2n sides. Say the first vertex past P of the regular n-gon is the point P 1. Bisect the angle P OP 1 ; the bisecting ray will intersect the circle at a point Q 1.That is the first vertex of the regular 2n-gon. The regular polygon of six sides, also known as the regular hexagon. Since we know how to construct the triangle, all we need to do is apply the rule about doubling the number of sides. However, there is an even simpler way, because the side of the hexagon equals the radius of the circle. So with the compass opening set at OP draw an arc of circle with center at P so as to 4

intersect the circle of center O at a point P 1 above the line l That is the first vertex of the regular hexagon. The regular polygon of five sides, also known as the regular pentagon. This is the first tough one. It is closely related to the golden mean. Here is the rule; why does it work? Draw the line through O perpendicular to l, let S be the point above l where it intersects the circle. Bisect the segment OS and let D be the midpoint. Draw the ray bisecting the angle ODP ; this ray intersects the segment OP at a point N. Through N, draw the perpendicular to l and let P 1 be the point above l where it intersects the circle. That is the first vertex past P of the pentagon. What can be constructed? The construction of the pentagon was not easy and other constructions are even harder. In fact, it was one of the great surprises of 1796 when the then nineteenth year old Gauss figured out how to construct the regular 17-gon, also known as the decaheptagon or heptadecagon. People must have asked themselves questions such as: Is the fact that nobody had been able to figure out how to construct the regular enneagon (9 sides) due to it being impossible to do so, or due to nobody being yet clever enough to figure out a way? Incidentally, one can show that if one can construct a regular polygon of n sides and one of m sides and n and m are relatively prime (have no common divisor other than 1), then one can construct the polygon of nm sides. The result, due to Gauss, which classifies all constructible regular polygons, relies on a mistake by Fermat. Fermat declared that he thought that all numbers of the form 2 2n + 1, n = 0, 1, 2,... were prime. This is his only known result about which he expressed doubts saying he could not prove it! It turned out to be false. The first few numbers in question are 2 20 + 1 = 2 1 + 1 = 3, 2 21 + 1 = 2 2 + 1 = 5, 2 22 + 1 = 2 4 + 1 = 17, 2 23 + 1 = 2 8 + 1 = 257, 2 24 + 1 = 2 16 + 1 = 65537, 2 25 + 1 = 2 32 + 1 = 4294967297,...... While the numbers corresponding to n = 0, 1, 2, 3, 4 are, in fact, prime, the number corresponding to n = 5 is not. In Fermat s day this was not an easy number to handle and it was only about a century later that Euler seems to have given a talk in which he simply wrote 2 25 + 1 = 4294967297 = 641 6700417 5

and proceeded to verify the statement by multiplying out the numbers. In mathematics, if one can t prove something one makes a definition, and so one defines a Fermat prime to be a prime number of the form 2 2n + 1, for some integer n 0. So far, the only Fermat primes found, even using super-duper digital computers, are the five given by Fermat: 3, 5, 17, 257 and 65537. It is not known whether the number of Fermat primes is finite or infinite. We then have the following theorem, mostly due to Gauss. Theorem. The regular polygon of n sides can be constructed if and only if n decomposes in the form n = 2 k p 1 p r where k = 0, 1, 2,... and p 1, p 2,..., p r are distinct Fermat primes. Examples of values of n for which one can construct the regular n-gon: n = 3, 5, 17, 257, 65537; these are all Fermat primes. Also n = 12 = 2 2 3, n = 15 = 3 5, n = 30 = 2 3 5, n = 16 = 2 4 (the list of Fermat primes can be empty), etc. Examples of values of n for which one cannot construct the regular n-gon: n = 7; while 7 is prime it is not a Fermat prime. The same goes for n = 11, 13. n = 9 = 3 3; while 9 is a product of Fermat primes, it is not a product of distinct Fermat primes. Similarly for n = 25. The fact that we know that something is constructible does not mean that it is easy to actually construct it. The theorem gives an idea on how to proceed, however even for the pentagon the method suggested by the theorem would be quite nasty. The idea is to find short cuts, efficient constructions. There are quite a few efficient ways to construct the heptadecagon, the polygon of 257 sides was constructed by Richelot in 1832 (published in Crelle s journal) and an 1894 paper, existent only at Göttingen, by a Professor Hermes of Lingen details how the good professor dedicated ten years of his life to construct the regular polygon of 65537 sides. Exercise 3. Given a circle, construct the inscribed equilateral triangle, square, pentagon, hexagon and octagon. As an added challenge, construct the inscribed regular polygon of 15 sides. Use different circles for the different polygons. Bonus Exercise 4. Why does the construction of the pentagon work? Bonus Exercise 5. It can be shown that if r, s are relatively prime integers, one can find integers a, b such that ar + bs = 1. Explain why this could be relevant to proving the following theorem: If r, s are relatively prime and one can construct the regular r-gon and the regular s-gon, then one can construct the regular rs-gon. 10. The golden mean. A point G is said to divide a segment AB in extreme and mean ratio or in golden section when the longer of the two segments formed is the mean proportional between the shorter segment and the whole segment AB. For example if AG is the shorter and GB the longer segment, then we require AB GB = GB AG. 6

The ratio of the longer to the shorter segment (or of the shorter to the shortest segment) is called the golden mean and sometimes denoted by ϕ. That is From these equations we get ϕ = AB GB = GB AG. GB 2 = AG AB = ( AB GB ) AB = AB 2 GB AB ; dividing by GB 2 this becomes 1 = ϕ 2 ϕ; thus ϕ 2 ϕ 1 = 0. By the quadratic formula, ϕ = (1 ± 5)/2 and since it is clearly positive, ϕ = 1 + 5. 2 Very bonus exercise 6. If in the regular pentagon we draw all the diagonals; i.e. all the line segments joining non-consecutive vertices, the resulting figure is a star, known as the pentagram. Be careful when drawing it, it is supposed to have magical properties, not all of them benign. Now every diagonal intersects every other diagonal; show that it intersects it in a golden section. 7