Notes — Chapter 14: Regular Polygons and the Circle
To the Student. The equilateral triangle is the simplest regular polygon; next comes the square, the pentagon, the hexagon,…. All are symmetric by rotations about their centers and ahve line symmetries by reflections in lines through their centers. The more sides a regular polygon has, the more symmetries it has and the closer it is a circle. We find the area of a circle, mr? as the limit of the areas of inscribed regular polygons. The limit of the perimeters of these polygons gives us the circle’s circumference, 27r.
ll“ Limits and Bounds. Archimedes showed that the area of a circle is equal to that of a right triangle whose altitude is the radius r of the circle and whose base is equal to the circumference c of the circle. Algebraically: Area = (1/2)rc. This is Proposition 1 of his Measurement of the Circle. The ratio 7 is Circumference/Diameter = c/d, and d = 2r, so Archimedes result gives our Area = mr’.
A good source on Archimedes’ methods is Sherman Stein’s Archimedes: What Did He Do Besides Cry Eureka!, Mathematical Association of America, Washngton DC, 1999. Continuing his work, Archimedes got the following bounds for the value of 7: 37° iy oe 33. Archimedes used bounds; he used inscribed and circumscribed regular polygons to find upper and lower bounds for 7. In this chapter, we use limits, defining the area and circumference of a circle as the limit of the areas of inscribed polygons and the circumference as the limit of the perimeters of inscribed polygons.
Calculators give us the easy informal path to limits used in this chapter. The method of finding upper and lower bounds following in Archimedes footsteps is still of interest. See the supplementary notes on this chapter for details. The idea is sketched below.
If you inscribe a regular n-gon in the unit circle, it will have a perimeter 2n sin (282°), as given by Theorem 75, on page 581 of Geometry. If you circwmscribe a regular n-gon about the unit circle so that it is tangent to the circle at the vertices of the inscribed n-gon, you can see that the perimeter of this circumscribed n-gon is 2n tan (48°). The circumference of the unit circle, 27, must lie between the perimeters of these inscribed and circumscribed n-gons. So
° ° (*) 2nr sin( ) <The 2nr tan ( ).
This form relates directly to the perimeters and circumference, but dividing by 2r upper and lower bounds for
° °
nsin( ) <i ntan(—— ).
With a calculator we can work these out as numbers and see that they bracket 7 and are close to 7 when n is
large. With n = 6000 the average of these bounds gives ™ to six decimal places on my calculator. There is something self-referential in this confirmation because calculators, in effect, use a stored value of 7 when they find trigonometric functions. Our understanding of these. calculations is based on the geometry of area and circumference developed in this chapter.
The circumference of a circle of radius r may be thought of as the single number lying between the upper and lower bounds established by (*) for every n. Alternatively, it may be thought of as C = limp soo 2nr sin(#8°). Either way, its formula is C = 2mr, and we evaluate it by calculator. For practical purposes, we need not concern ourselves with the fact that our machine only gives the answer to ten decimal places. In the physics and engineering, few things can be measured to even six decimal place accuracy.
Lesson 1. Regular Polygons. A polygon is regular if it is convex and if all of its edges are congruent to each other and all of its angles are congruent to each other. These are local conditions. If a convex polygon on a soccer field is convex, you can check its regularity even in a dense fog. Pace around the polygon. Check that the sides are the same length. Check that all the angles are equal.
If local measurements show the fog-bound convex polygon is regular, then Theorem 74 tells you that it has an incenter, and you can work out its shape and size from your measurements. The proof shows how the local geometry determines the global structure. Local regularity implies that the polygon has symmetries interchanging any two of its sides or angles.
This is the pattern for regular figures in general. We require that the various faces and edges of regular polyhedra be equivalent, a local condition, and we find that there are global symmetries that interchange these elements as well. The study of regular figures is tied up with the effort to find local characterizations that force the existence of global symmetries.
A regular polygon with n sides is symmetric by reflection in lines through its incenter and a vertex. It is also symmetric under rotations about its center by any multiple of the central angle subtended by one of its sides from its center.
Set I. These are exercises in seeing regular polygons and analyzing how they fit into other patterns. Set I covers the basic material in this lesson.
Set II. Mason’s Triangle is a beautiful construction. The angles are not quite right, being off by less than a degree. This might be a good test of technology: Can students see what is up using Geometer’s Sketchpad™?
In Folding an Octagon (33-43) the problem is to see that the octagon is regular. What follows is keyed to the figure on page 577 of Geometry. The figure is symmetric
by reflections in the perpendicular lines GE and HF, which run through O, the center of the square EFGH, as well as by reflections in the diagonals AC and DB which also run through O. These symmetries show that
OH = 0G = OF = OF:
because each of these is segments can be carried any other of them by one of these reflections. For the same reason,
Oh OJ SOhia.0 1
We see GCFO is a square with diagonals GF and OC which intersect at a point we label M. Diagonal OC divides square GCFO into isosceles right triangles, one of which is △GCO. The folding shows ∠CGK = ∠KGM, so both of these are 22.5° angles. It follows that ∠KGO = 45° + 22.5° = 67.5°. Because ∠GOK = 45° it follows from the Angle Sum Theorem that ∠GHO = 67.5° as well, and △GOK is isosceles. Hence, OG = OK, and all of the segments in the displays above must be equal. The reflective symmetries can send any of the central angles the octagon GK- FJEIHL makes at its center O into any other of these central angles, so they are all congruent, and as we have seen, all of these vertices are equidistant from O, so this is a regular polygon in the global sense.
A rotation by 45° about O sends octagon GKFJEIHL onto itself. So it is equiangular and equilateral, and it is clearly convex, so it meets our local definition of a regular polygon.
Hexagonal Fastener (44-50) looks at the geometry that allows wrenches to work. What is important is the relative difference between the radius of the incircle and that of the circumcircle of the head of the fastener. Square and nuts are common, fire hydrants are designed for pentagonal wrenches (to discourage frivolous uses), and we do not see septagonal or octagonal nuts. An interesting source on design and engineering of common things, including fasteners, is The Evolution of Useful Things by Henry Petrovski, Vintage reprint, 1994.
Inscribed Pentagon (51-61) develops key facts about the regular pentagon. Referring to the labels on page 578 of Geometry, this figure is symmetric by reflection in the line FE. The calculations here provide a good review of secant angles. An interesting property of the 36°, 72°, 72°, triangle △AFB is that there is a point X on side AB such that △AFB ∼ △FXB and AX = XF = FB. Let the ratio of similarity be r such that rBX = FB. Then BA = BX+XA = BX+BF = BX+rBX = r7BX, because scaling BX up by r gives us BF and scaling BF up by r gives us BA. Geometrically: BX is to XA as XA is to BA. The ratio r is then the Golden Number, ¢, the positive root of r² —r−1=0, Bee
Given segment AB the point X such that BX is to XB as XB is to BA, is said to divide AB in eztreme and mean ratio. The construction for dividing AB in extreme and mean ratio is given in Proposition 11, Book II of Euclid. Extreme and mean ratio reappears in at
the start of Book XIII leading to constructions of the regular pentagon, the dodecahedron, and icosahedron. The diagram in these exercises is found in Heath’s Euclid for Proposition 8, Book XIII. Once X has been constructed on AB, the regular pentagon ABCDE is easily constructed. The construction of this triangle is found in Euclid, Proposition 10, Book IV. Euclid, Proposition 11, Book IV gives a construction for inscribing a regular pentagon in a given circle.
The fact that the 36°, 72°, 72° triangle can be split into a similar triangle plus an isosceles triangle leads to the possibility of recursive constructions, including tilings. The aperiodic Penrose tilings are related to pentagons and 36°, 72°, 72° triangles, and the Golden Number is intimately linked to these aperiodic tilings. See, for example, Figure 10.3.14 of Tilings and Patterns by Branko Griinbaum and G. C. Shephard, W. H. Freeman and Company, New York, 1987.
Set III. A Close Construction is a mixture of construction and analysis, with calculators playing a role.
Lesson 2. The Perimeter of a Regular Polygon. Here the concrete example of a regular pentagon shows us the situation for a regular n-gon. If you want an upper bound on the circumference as well, circumscribe a regular pentagon about the circle with its sides tangent to the vertices of the inscribed regular pentagon. Its sides have length 2rtan 180
▶ so its perimeter is 5 x 2rtan 180 For the general n-gon, replace 5 by n and get
° fe)
2nr sin( ) < Circumference < 2nr tan( zo)
for the circumference of a circle of radius r. Here we use the principle Archimedes used: If one bounded convex region lies inside another, then the inner region has the smaller area and smaller perimeter.
Set I. Perimeter Equations (1-9) gives practice with the formula developed and shows how N tends to 7 as the number of sides increases so that these regular polygons approach a circle. The rest of Set I covers the basic ideas of the lesson. Hydrogen Fluoride (19-29) is a nice application. The Angstrém is 107!° meters; this is not a Systeme International (SI) unit but it lies between the SI nanometer (107° meter) and the picometer (10-’* meter). For a better understanding of the sizes of things in the universe, students should see Powers of Ten, either the movie by the Office of Charles and Ray Eames or the SCIENTIFIC AMERICAN Library book, New York, 1983, by Philip and Phylis Morrison and the Office of Charles and Ray Eames. These are highly recommended, and there is a good deal of related material on the Web.
Set II. Applications to cell phones (33-40). Ten Pentagons (41-48) combines geometric thinking and use of
a calculator. Euclid’s Discovery (49-58) is Proposition 10 of Book VIII. This is a discovery exercise.
Set III. Golden-Heather. Here we see the 36°, 72°, 72° triangle mentioned above in connection with Inscribed Pentagon. If we look at the ratio of AE to CE, which is the ratio of the distance between nonconsecutive petals to that between consecutive ones, this is the same ratio as BA/BF because △ABF (of page 578) ≅ △CAE (of page 578). As the discussion of Inscribed Pentagon in the notes for the previous lesson shows, this ratio is ¢, the Golden Number or Golden Ratio.
A broader pattern is
sin 20 = 2sin@cos@ = 2sin @sin (90° — 8),
from which the first trigonometric surprise flows. Recognizing 2 sin 54° as ¢, the positive root of r?—-r−1 = 0, we can see by algebra that ¢ = 1 + (1/¢). Successive replacements of ¢ by 1 + (1/¢) give us the continued fraction expansion of ¢:
g=1+
1+
Lesson 3. The Area of a Regular Polygon. The dissection by Jézef Kitirscék is very beautiful, and it is explored further in Kiirscék Triangles, Set II, 41-51.
The area formula has the extra cosine factor and is not as simple as the result for perimeters. Archimedes did his approximations for the case of perimeters when he estimated 7. This is because the situation for perimeters is simpler. But Archimedes had already shown for a circle of radius r: Area = πr² by a rigorous argument using exhaustion. The Greeks defined 7 as the ratio of the diameter of a circle to its circumference — a geometric definition, independent of estimating 7 by limits or bounds.
Set I covers all the basic ideas. In Area Equations (+ 7) the calculator tempts us to plug in a very large value for n to see if our result looks like πr². Formulas should be understood as well as used, and Inscribed Polygons (8-11) and The Area of a Square (12-19) build an understanding of the formulas of this lesson. Close But Not Quite (20-28) gives a lesson in the ways calculation goes beyond what can be seen.
Set II begins with Honeycomb Geometry (29-40), which includes reviewing the K, which uses 30°, 60°, 90° triangles. This builds the trigonometric muscles.
Kiurscak Triangles (41-51). The assumption that all of the yellow triangles are congruent is made in order to simplify these exercises. But the geometry is determined by insisting that ABCD be a square and the purple triangles be equilateral, with the layout as shown.
Here is why: △IBC ∼ △HBA by SSS. Then by congruence and properties of isosceles triangles, ∠IBC = 4ICB = ∠HAB = ∠HBA = 15°. Where we use the fact that ∠IBC + ∠IBH + ∠HBA = 90°. Similar reasoning on angle sums shows that ∠GCD = ∠EAD = 15°, so yellow triangles △EAD ∼ △GCD & △IBC ∼ △HBA by SAS. Now, by adding up the angles about points E and F we can conclude ∠DEH = ∠DGI = 150°, which is the size of the obtuse angles of all the yellow triangles we have so far proven congruent. The short sides of △DEH and △DGI are equal and equal to the short sides of all the yellow triangles so far proven congruent. The obtuse angles of all these triangles are 150°, so these two yellow triangles are congruent to the lot already analyzed. The last pair of yellow triangles is △FHD and △FID. These are congruent by SSS, and congruent to the rest of the yellow triangles.
The students are told to assume all the yellow triangles are congruent in order to avoid this long argument.
Rats! (52-58) involves conjecture and computation. Calculators will be used here.
Set III. Congruent Pentagons. Marjory Rice has done a good deal of interesting work on tilings, particularly with pentagons. You can find information on her work by searching the Web.
Lesson 4. From Polygons to Pi. The circumference of a circle is the limit of the perimeters of the polygons that can be inscribed in it. More generally this approach allows us to define the length of the boundary of any bounded, convex region on the plane. The limit is taken as the number of sides of the inscribed polygons is allowed to go to infinity. It is not necessary that the inscribed polygons be regular. For circles, regularity allows us to have a formula giving the perimeter of the inscribed polygon as a function of n. This makes it easy to apply the idea of a limit.
▶ Archimedean Alternative. Archimedes used the principle that when one convex region is properly included in another, the smaller region has the smaller area and the smaller perimeter. This principle lets us see that the perimeter of a bounded, convex region R is the least upper bound of the perimeters of all convex polygons contained in R.
Given a point X and a convex region R not including X, there is always a unique X’eR that is closest to X, and this X’ is necessarily on the boundary of the smaller region. To see why this is true, think of blowing up a circular balloon centered on X until it just touches R at Xx’.
Archimedes’ principle about the perimeters of bounded convex regions can be proven by thinking about what happens when points on the boundary of the larger region are mapped to the closest point in the smaller region. If X and Y are on the boundary of the larger region, let X’ and Y’ be the corresponding closest points
in the smaller region. You can show that XY > X’Y’, and that some pairs of boundary points of the larger region must be mapped to pairs on the boundary of the smaller region that are actually closer together than the original two points. This means that the length of the boundary is shrunk by this mapping, and the smaller region must have that smaller perimeter. The boundary shrinks as it is drawn inward.
Set I covers all basic ideas. Viéta’s Calculation would carry us through 16 doublings of the number of sides, starting from a hexagon. The 5 doublings shown in my appendix to this chapter’s notes, give 3 decimal places accuracy (4 if you average the upper and lower bounds), and each doubling reduces the error to 1/4th of what it was before, so the additional 11 doublings of Viéta would give something like 10 place accuracy.
Set IT has nice applications, including Videotape (52- 56) and distances along the earth’s orbit in Going in Circles. Both help students get a sense of sizes and distances and how things can be worked out using geometry.
Set III. πre Change is practical and within the capabilities of average students. Nothing tricky here.
Lesson 5. The Area of a Circle. One way to get circles or spheres is to pump up a uniformly stretchy boundary. For spheres, a soap bubble is the nicest example. A drop of oil spreads out on smooth water to form a circle.
Here we see why a city built against a relatively straight boundary and defended by as short a wall as can be managed will be defended by a (roughly) circular wall. The problem of getting the most land for the least wall (or stretched ox hide) is called Dido’s Problem and has been studied widely.
Here is a way of proving that for a given length L, the area bounded on one side by a curve C of length L and on the other by a line ℓ can be a maximum only if C is a semicircle whose diameter lies on @. I refreshed my memory on this by looking at The Enjoyment of Mathematics by Hans Rademacher and Otto Torplitz, Princeton University Press, 1957, Sections 21 and 22. This book is highly recommended and available from Dover Publications in 2003.
You have to draw the diagram to see this. Let the curve C start at point A on line ℓ, which is parallel to the bottom of your sheet of paper. C curves upward and returns to @ at point B, to the left of A. Suppose that C encloses as large an area as possible for the given length. Let X be any point on C other than A or B. Then we have △AXB, and the arcs of C above the chords AX and BX. We regard these arcs as rigidly fixed above their chords, but the chords are to be regarded as hinged at X. If ∠AXB # 90° we can increase the area of the arrangement while keeping the length of
the upper bounding curve the same by opening or closing this angle to make it 90° while sliding its ends along é. Thus, if C encloses the maximum area for the given arc length ∠AXB = 90° for every X on C. This means that C is the locus of all points above line ℓ such that ∠AXB = 90°; in short, C is a semicircle whose diameter lies on ℓ, as was to be proven.
This is an isoperimetric problem, that is, a problem of maximizing or minimizing one quantity while holding some other parameter constant. One good source on such problems is Stefan Hildebrandt and Anthony Tromba’s Mathematics and Optimal Form, SCIENTIFIC AMERICAN Books, New York, 1986, later revised as The Parsimonious Universe, Copernicus Books, New York, 1996.
For isoperimetric problems I also recommend Mathematical Snapshots by Hugo Steinhaus, available from Dover Publications and Oxford University Press in 2003. This book has so many wonderful things in it, most of them presented with figures, drawings, or pictures. (The First Edition, printed in Poland, had a pocket in the back containing 3-D glasses, a pop-up dodecahedron, and so on. This is a hands- and minds-on book. Now you have to make your own models, but that may be an advantage.)
When thinking of the area of a circle, you should allow yourself to think of a circle as a triangle whose apex is at its center and whose base is its circumference. This is something of a geometric distortion, but it gives you a correct understanding of the relationship between the circumference and the area. See Equal Areas, Exercise 17. It might be easier to think of the circle in this exercise as consisting of thin flexible concentric layers. If you sliced from C to A in this figure and let the layers uncurl on a line tangent to the lowest point on the circle, you would get a triangle with apex at A, symmetrical about the vertical line AC and having a base of length 2nr, which is the same as the length AB seen on page 601 of Geometry.
Set I gives practice seeing circular areas in various situations and using the formulas we have to work problems. The photograph in Ripples shows the complex structure to the expanding ring of capillary waves from a stone dropped into still water. Careful observation of these spreading rings of waves will give food for thought. There are a few nice photographs of capillary waves on the Web, but I found no simple, definitive source on this.
Central-Pivot Irrigation (18-24) is a guestion that has appeared in abstract form on the MCAS mathematics test and in NCTM materials. The flash of insight is that we are looking at the same ratio for both these cases, the portion of a square area filled by the square’s incircle — and this is a correct intuition. In these exercises, the student will see by a calculation that both
arrangements water the same fraction of the surrounding square. A question for teachers is: Should you insist on a calculation here or would you give a student full credit for the insight alone?
Circle on a Grid (25-30). Observation, calculation, and judgment are all needed in these exercises. This matter is given serious attention by David Hilbert and Stephan Cohen-Vossen in Geometry and the Imagination, in 2003 an American Mathematical Society Chelsea reprint. See the beginning of Chapter II, Regular Systems of Points. These questions lie in an area known as the geometry of numbers and extensively studied by Hermann Minkowski. Hilbert and Cohen-Vossen get the following neat result using these ideas: On a square lattice, a parallelogram with lattice points as vertices that contains no lattice points in its interior has area 1 square unit.
Set II. Area Problems (31-36) are those wonderful exercised where you see and calculate the answer. Cable Disaster (37 and 38) is a lesson in the importance of understanding geometry. You cannot expect 2 cables of 3 inch diameter to have the strength of one cable 1 inch in diameter. Even using 4 cables 5 inch in diameter in place of 1-inch cable might not be prudent. If the load were to come unequally on one of these 4, then if the over-stressed cable failed, then the other 3 would be at risk.
Set III. Lull’s Claim. It is rewarding for students to see that they can get something right that a sage so famous as Lull got wrong. Modern tools trump fuzzy thinking.
Lesson 6. Sectors and Arcs. It is reasonable to suppose that mathematical objects ezist and we merely investigate and discover some of their properties. In a general way, this is the Platonist view. But it is also reasonable to take the view that we need to develop theories of the length of curves and areas of regions in order for these notions to have a definite meaning. In an introductory course, we are Platonists. We have neither the time nor tools to construct notions like area from scratch — and the students would not appreciate the effort. The areas and perimeters of bounded convex regions and of regions that can be broken down into a finite number of such regions (or their complements, intersections, and so on) all have perfectly definite meanings, and these can be worked out by methods of the this chapter.
This is the lesson where students get to use whatever methods they can bring to bear to calculate areas and lengths. Proportionality is the guide in working these things out for sectors of circles:
Sector Angle _ Sector Area _ Sector Arc
Full Circle are - Zr
lt Fractal Troubles? Paradoxes will arise if we assume that the boundary of any region can be assigned a finite length. Fractals have so seeped into the general consciousness that we know that the Koch Snowflake has a boundary of infinite length, and we know that - trying to define the length of the coast of Britain is a tricky business. See The Fractal Geometry of Nature by Benoit Mandelbrot, W. H. Freeman and Company, New York, 1982, or other books. Plate 47 in Mandelbrot’s book is an interesting example. Six congruent areas surround a seventh, also congruent, to form a similar figure enlarged by a linear scale of /7. But if you look at the boundary of the scaled up region, you will see that it consists of 6 curves each of which is congruent to half of the boundary of the original region. If the boundaries of these regions had finite length, we would expect the enlarged region to have a boundary of length 6 x 4x (Boundary Length Of The Smaller Regions) on the one hand. On the other hand, we would expect the boundary length to scale up linearly with the diameter of the set, and hence, that it would be multiplied by a factor of 7. But 6 x 4 03 56 J/7. So our assumption that this region’s boundary has a finite length must be wrong.
Set I. Many applications here: slicing oranges, dolphin sonar, latitude, driveways, hearing and location of sounds, and Reuleaux triangles (these are used in Wankel engines). We meet The Radian (26-30). Some discussion with students of the various modes on calculators would be in order.
Set II. The range of examples here is also great. Four Crescents (49-56) is an extended see-and-compute series with a surprising finale.
Set III. Pizza Puzzle is a nice exploration. It would be ideal for a Geometer’s Sketchpad™ project. After you discover what the answer is, there is still the challenge of finding a reasonable way to prove you are correct. Jacobs has a nice discussion in his answer section and his reference gives further details.
Summary and Review. One theorem here is that all regular polygons are cyclic. This implies that they are symmetric by easily found reflections and rotations. While we have focussed on the details needed to carry us through this material, there is a larger picture, which involves ideas about symmetry. This can be mentioned as well in summing up the chapter.
Set I. Molecule (1-3). The conventions in representing organic molecules include not showing the hydrogen atoms, which is one reason the line formula looks simpler that the space-filling model. There may be some flexibility in the molecule, and structures may be able to rotate about a single bond, so one molecule may appear in several different looking forms, either as a space model or as a diagram.
Regular 15-gon (4-14). This is the figure for Proposition 16 of Book IV in Thomas L. Heath’s edition of Euclid. The construction is given there. This series reviews the formulas developed and involves calculation and calculators. A nice added question here is: How do you locate the other vertices of the regular 15-gon?
Penny Farthing Bicycle (15-20) uses calculators. Area Comparisons (23-26) has a nice Aha! discovery at the end. Gothic Arch involves both exact calculations and the use of a calculator. Mathematical constructions of ornamental features are found in every place and age. Moorish tilings come to mind immediately, but examples of arches and traceries like this one are also common. See, for example, John L. Heilbron’s Geometry Civilized, Oxford University Press, 1998, Chapter 5, From Polygons to Pi.
Time Zones (39-43) helps students build a geometric understanding of the earth and its geography. There is 3-dimensional thinking is needed here.
From Decagon to Square (44-49) is a dissection with a lesson: In dissections areas are preserved, although perimeters may not be. In Exercise 45, I note that
180°; 180° in 180° a 1
2sin =“S = SU, sane
12 ariel? 6 2
Students will not know this, but my calculator gives 0.5 for this answer to 10 decimal places.
Chord and Ring (50-54) is a classic. If you assume the question has a definite answer, then you can find that answer by considering a degenerate case — here when the inner circle has 0 as radius. The 3-dimensional version of this conundrum involves a ball with a cylindrical hole centered on the ball’s center. The length of the hole is L. What is the remaining volume of the ball?
Biting region (63 and 64). Beware of the dog!
▶ How Archimedes Measured the Circle. This is Archimedes with improvements by Andrew M. Gleason. Archimedes begins by observing that if a bounded convex region B is properly contained in the bounded convex region C, then the perimeter of B is smaller than that of C, and likewise for their areas.
Here is our setup: A regular n-gon with side length s = 2 is circumscribed about a circle of radius ay, here “an” stands for apothem (the word for the radius of the incircle of a regular polygon). We let b, stand for the radius of the circumcircle for our regular n-gon. Gleason observes:
Incircle C Regular n-gon C Circumcircle
giving, for perimeters 27a, < 2n < 27bny.
By algebra we get the following bounds on 7
nm S sk n (*) Fi = T ==. Be
When n = 6 we have a regular hexagon with side length s = 2, then the radius is bg = 2 and the apothem is ag = V3. This gives us the crude bounds
6
cpap Sec tema
V3 2
rounding up on the high side.
Archimedes was careful and made sure that at each stage where he made an approximation, he made it in the direction that preserved the inequality he was interested in. As we move to decimal numbers, I will treat the calculator answer as exact, although I should follow Archimedes in his careful rounding. We will not push this so far that rounding at the 10th decimal would have any noticeable effect. If we wished to go further, every rounding or truncation would have to be made with due care.
These estimates can be easily improved by a series of steps, each step doubling the value of n. To do this draw a diagram showing a vertical side of our inscribed regular n-gon, s, of length 2, with the apothem, an, going to the left horizontally from the midpoint of s to the center O of the incircle and the circumcircle. Then add radii, b,, from O to the endpoints of s. To double the number of sides to 2n, you extend a, through O by a distance b, to get our new center, O’, which is on the circumference of the circumcircle of our original n-gon. Then the angle subtended by s at O’ is half the angle subtended by s at O, and s will be a side of our new 2n-gon with center O’. From the figure you can see that
(Scheme) don = Gn + bp and bon = 1/a², +1.
This shows us how to compute agp, and be, from ap, and b,. Two additions, one squaring, and one extraction of a square root are all we need. These operations were routine even in ancient times. The Babylonians knew how to extract square roots. Guess-and-test is a good method, if you don’t have to do too many square roots. Another method is to make a guess at the square root of a positive number d; call your guess g. You know that Vd lies between g and d/g, if these are equal, your first guess was on the mark. If not let your second guess be $(g + as This method can be carried out using approximate decimal arithmetic or using exact fractional arithmetic. It converges rapidly and can be handled by machine or by paper and pencil.
Gleason’s scheme is a way of calculating the value of 7 to high precision using methods known before Euclid. It is easy to program a spreadsheet or programmable calculator to compute a, and b, and establish successive limits on 7 using this method. It is based on Archimedes’ approach of successive doublings of the number of sides, but it uses the trick of getting its upper bound for the number m by inscribing a circle in our n-gon, rather
than circumscribing an n-gon outside our fixed circle. It allows the circles to approximately double in linear size at each step, while the sides remain at length 2. This simplifies the algebra. And we use algebra, which makes our work easier than that of Archimedes. Finally, we let the computer to do the messy stuff for us.
What exactly are these ans and b,s² Drawing the diagram of one side s of our n-gon. The apothem ay goes from the center O of the n-gon to the midpoint M of our side s. Let P be one end point of the side s and P’ be its other endpoint. The distance OP is b,. Because
fe} if ° we have a regular n-gon, ∠MOP = Hee = 180° So we have the right triangle △OPM with a, = OM, MP = 1, and OP = b,. Trigonometry gives us
2 180°
bn sin( = ) =i] wand aun cos( ) = Os.
n n
When (*) is rewritten in trigonometric terms it becomes
nye igellie 180° (t) nsin( ) <n <ntan( )
n n
If you look at Theorem 75, page 581 of Geometry, you will see that the left-most quantity is Jacobs’s N.
To understand (t+) in direct geometric terms, we need another diagram. Let ray m bisect ∠POP’. This ray m follows the apothem of our regular n-gon to M, the midpoint of chord PP’, then cuts the circumference of the unit circle at a point we label N. The tangent t to the radius OP of our unit circle at P cuts the ray m at point we call T. Then ∠POT = a0” = 180” We have three convex regions:
△POP’ c Sector POP’ C Quadrilateral POP’T.
Their perimeters must be unequal in the same order as the set inclusions, by Archimedes’ assumptions. They all share OP and OP’ as sides, so we can ignore these and see that
Chord PP’ < Arc Length PP’ < PT + TP’.
But Chord PP’ = 2sin 18°, Arc Length PP’ = “=e, and PT+TP’ =2tan Bo So the above inequality is
ESOS) S827 180°
2sin <— < 2tan,
n n
Multiplying by n/2, we get (f).
If we shift from degrees to radians, then ae =
* radians, and (f) gives us the inequality
ek cite, T
SIN =<y=ixt tame orssin O<oe<atanG,
jal n
which holds where @ is in radians and lies in the range 0<6@< 71/2. To go from submultiples of z to arbitrary angles @ requires observing that the length of the arc
bounding sector with central angle θ, in radians, is just @r. This is intuitive and can be proven once you have the formal tools for defining the lengths of curves.
Archimedes started with a hexagon, our ag = V3 and bg = 2, which comes from 6 congruent. pieshaped pieces, each of which is an equilateral triangle of side length 2. Our table of Archimedes/Gleason Bounds gives a, and b, and bounds am for n = 6, 12, 24, 48, 96, and 192. Our best value for 7 based on these computations would be the average of the upper and lower bounds on 7m for n = 192. This estimate is 3.1416627612, to my HP 48G+’s 10 places. To 10 places 7 = 3.1415926536, so the average of our upper bound and lower bounds exceeds 7 by 0.0000701076 or roughly 1 part in 45,000.
Let’s get an upper bound on the gap between the upper bound, U,,, and the lower bound, L, that we get for 7. For a given n > 4, we have
be ae
L, = nsin.— <a < ntan—< U,,.
n n
Now we put an upper bound on U,, — L, that holds for n > 6. In this calculation we use the fact that nsin(a/n) < 4, which is true because the left side of this inequality is a lower bound for 7 and hence less than 4. We also use that fact that cos(1/n) > cos(m/4) = 1/vV2, which holds because 0 < 2/n,7/4. Finally, we use the fact that 1 — cos(/n) < $(m/n)*, which more familiar in calculus, but is part of geometry as well.
Un — Ln = (nsin
= | (— )
< (a) {Use nsin = < 4.}
T if
=< 47 V/2(1 2(1 — cos(m/n)) {Use U cos(*) ee 7 }
lee ave 322
Pe av3( 5/2) ) < 2v3(=) ae
Putting this on one line, we have
32/2 (t) Un —In <
nz
This shows that limn..Un—Ln = 0. We know Ln < Lon < 7 < Uan < Un, so if we proceed by doubling the number of sides our lower bounds increase from one step to the next and our upper bounds decrease. The common limit along such a sequence of doubling of n’s must be 7. Indeed, 7 could be defined analytically as this common limit. We have chosen to define 1 geometrically, so these calculations are simply a way of working out the value of m numerically — a reworking of the calculations Archimedes did so long ago.
This inequality (t) implies that |L, — | < 32V/2/n² and |U,−2| < 32/2/n², so the upper and lower bounds both converge to 7 as n — oo. In fact, the lower bounds increase as n increases while the upper bounds decrease as n increases. This is not as obvious as it is when we double the number of sides, but it is true.
Archimedes/Gleason Bounds for 7
Djaeeras. b, Bounds on 7
6 V3 2 3.465 Uppers 6 3 Lowerg 12° 9948/3 IA/ 2/3 3.2153903092 12 3.1058285412 24 ~=7.5957541127 7.6612975756 3.1589113607 24 3.1326286138 48 15.2570516883 15.2897882987 3.1460862151 48 3.1393502030 96 30.5468399870 30.5632039091 3.1427145996 96 3.1410319509 192 61.11004389661 61.1182253095 3.1418730500 192 3.1414524723