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Piwith the Golden Ratio Approximating

Piwith the Golden Ratio Approximating

Approximating

with the Golden PiStacy L. Linn and David K. Neal round 225 BC, wrote the treatise , which contained the first derivation of a for- mula for the area of a circle and the first formal approximation of the Awe now call p. At that time, it was already known that the ratio of the areas of two circles equals the ratio of the of their diameters. This result had appeared in ’s Elements (Euclid 1956) as Proposition 12.2.

It follows that if circle O1 has diameter d1 and circle O2

has diameter d2, then

area()O area ()O 1 = 2 . 2 2 d1 d2

From this relationship, Archimedes understood that the ratio of the area of a circle to the of its diameter must be constant for every circle. He sought to find not only a formula for the area of a circle but also this constant ratio. To do so, Archimedes first noted that the area of a is ap/2, where a is the apothem (measured from the center of the polygon to the midpoint of a side) and p is the perime- ter. Using this result, Archimedes proved that the area of a circle is equal to the area of a right that has one side equal to the radius r of the circle and the other side equal to the circumference C. That is, A = rC/2. He then found his approximation of p by using the perimeter of regular polygons, beginning with a hexagon and then doubling the number of sides to 12,

472 TEACHER | Vol. 99, No. 7 • March 2006 Copyright © 2006 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NCTM. 24, 48, and 96. This Archimedean method has been well outlined in many sources (see Burton 2003, b Dunham 1990), and it was not improved upon until the development of calculus 1900 years later. a b Today, Archimedes’ method still provides a rich setting to show the interplay of algebra and . It can be used not only to study p but also to introduce a another geometric curiosity, called the golden ratio, or Φ. In this article, we will reemploy the Archimedean method, but we will begin with a regular a + b rather than a hexagon. Then, by adding a little trigonometry, we will show how to find an algebraic Fig. 1 with dimensions a × b and (a + b) × a approximation of p in terms of the golden ratio. using polygon perimeters survived for centuries ARCHIMEDES’ DISCOVERY OF THE thereafter. In 1579, Francois Viète (1540–1603) of “CIRCLE CONSTANT” France found p to nine places using polygons 16 Archimedes knew that the ratio of the area of a cir- with 6 • 2 = 393,216 sides. In 1610, Ludolph van cle to the square of its diameter d must be a con- Ceulen (1540–1610) of the Netherlands found p to 60 stant k for every circle. He eventually showed that 35 decimal places using polygons with 4 • 2 sides. this constant is about 11/14. Using his formula for The accuracy and efficiency of approximating p the area of a circle O, he obtained with Archimedes’ method has now been surpassed by methods that use calculus and infinite series. area(O )rC /24dC / C k ====, However, many lessons from geometry can still be d222d d 4d illustrated with Archimedes’ classical technique. which gives C/d = 4k. Rather than directly finding the constant k for the THE GOLDEN ratio of area to diameter squared, Archimedes in- The golden ratio can be motivated from a “propor- stead found an approximation of the constant 4k. We tionally pleasing” rectangle with sides of length a and now call this constant p, which is the ratio of circum- b such that the ratio a : b equals the ratio (a + b) : a ference to diameter in any circle. To find an estimate (see fig. 1). In this case, what is the value of a/b? of p, he presented the now-classical Archimedean If we have such a rectangle where a : b = (a + b) : Method of approximating the circumference of a a, then a2 = ab + b2, or a2 – ab – b2 = 0. Dividing by 2 unit circle with the perimeter pn of an inscribed n- b , we obtain sided regular polygon. In particular, he gave a a2/b2 – a/b – 1 = 0. method for finding the side length s2n of an inscribed 2n-sided regular polygon in terms of the side length If we let x = a/b, then we obtain the “golden quad- 2 sn of an inscribed n-sided regular polygon. ratic” equation, x – x – 1 = 0, which has as its only

Because the side s4 of an inscribed square and the positive solution side s6 of an inscribed regular hexagon are easily found, we can continue to double the number of sides to gen- x =+()/.152161 ≈ 88. erate the side lengths of the inscribed regular octagon, 16-gon, 32-gon, etc., and the side lengths of the in- Thus, a/b equals (1 + 15)/2. This value is called the scribed regular 12-gon, 24-gon, 48-gon, etc. That is, we golden ratio and is denoted by Φ. k can find the side lengths of regular 4 • 2 -sided poly- k gons and 6 • 2 -sided polygons that are inscribed inside THE GOLDEN TRIANGLE a unit circle that has a diameter d = 2. Then using such An with base that measure a polygon with n sides, we can approximate p by 72° also has sides that are in the golden proportion.

• Consider triangle ABC with m∠A = m∠C = 72°, AB = C pnsnn (1) π =≈ = . CB = a, and AC = b (fig. 2). We assert that a/b = Φ. d 2 2 If we extend the base of the triangle ABC by By starting with both inscribed and circumscribed length a, we obtain a new isosceles triangle DCB hexagons and doubling the number of sides four having equal legs of length a (fig. 3), and m∠DCB = times, Archimedes used 96-sided polygons to obtain 180° – 72° = 108°. Thus, its congruent base angles C 211, 875 each must measure ≈≈3.. 141635 d 67, 441 180°− 108 ° =°36 . 2 This Archimedean method of approximating p

Vol. 99, No. 7 • March 2006 | MATHEMATICS TEACHER 473 B B

36° 36° 36°

a a a a

72° 72° 72° 72° 108° 36° A b C A b CDa

Fig. 2 The golden triangle Fig. 3 Two similar golden

But taking the two triangles together defines a larger triangle BDA that has base angles measuring s 72° and vertex D measuring 36°. Thus, 5 ᭝BDA is similar to ᭝ABC. By this , we 2 have a : b = BC : AC = DA : BA = (a + b) : a, which 1 is the ratio that defines Φ. Therefore, as with the 36° , we obtain a/b = Φ. 72° APPLYING SOME TRIGONOMETRIC IDENTITIES

Applying the and a double-angle for- s5 mula to triangle ABC, we can find a trigonometric form of the golden ratio. Then we can write cos 36° and sin 36° in terms of Φ. To do so, we first recall that the law of sines gives us sin∠A sin∠∠B sin A a Fig. 4 Triangles inside the inscribed pentagon = or = . a b siin ∠B b by equation (2), we have Then applying the formula sin(2 q) = 2sin q cos q, = °= −Φ2 we obtain s5 2siin36 4 . ° Φ ==a sin72 FINDING THE NEXT INSCRIBED SIDE LENGTH ° b sin36 Next, we look at Archimedes’ method for determining sin(236• ° ) 23636sin°° cos = = =°236cos . the new polygon side length upon doubling the number sin36° sin36° of sides. We begin by assuming that a regular n-sided polygon is inscribed inside a circle that has a radius of Φ 15+ length 1 and that we know the length s of its sides (fig. Thus, cos36°= = . 5). Then we inscribe a regular 2n-sided polygon inside 2 4 the same circle. We aim to find the length t of the new Because sin2 36° + cos2 36° = 1, we also obtain sides in terms of the original side length s (see fig. 6).

2 To find t, we first consider one isosceles triangle of ⎛ ΦΦ⎞ − 2 2 4 (2) sin36°= 1 − cos 36 °= 1 −⎜ ⎟ = . the original n-sided polygon that has sides of length 1, ⎝ 2 ⎠ 2 1, and s, and apothem of length a. The extension of the apothem to the circle creates a radius of length 1 THE LENGTH OF THE INITIAL INSCRIBED that is a common side of two isosceles triangles within PENTAGON SIDE the 2n-sided polygon. The apothem also bisects the This result allows us to find the side length of a reg- side of length s and creates a with sides ular pentagon inscribed in a unit circle. An in- of length a, s/2, and 1, as shown in figure 7. scribed regular pentagon creates five isosceles trian- Applying the Pythagorean to this right gles that each have a vertex angle of 360°/5 = 72° triangle, we obtain opposite the side of length s5 (fig. 4). By bisecting ⎛ ⎞ 2 22s 221 one of these triangles and then applying right-trian- 1 =+a ⎜ ⎟ =+as. ⎝ 2⎠ 4 gle trigonometry, we obtain sin 36° = s5/2. Hence,

474 MATHEMATICS TEACHER | Vol. 99, No. 7 • March 2006 t 123 t 1 – a s s 2

1 a 1 1

Fig. 7 One section of the inscribed polygons

==−−2 Fig. 5 An inscribed n-sided polygon inside a circle of radius 1 (4) s2n 24sn .

APPROXIMATION OF p t Because we know the side length s of the initial in- s 5 scribed pentagon, we can apply equation (4) re- peatedly to find the side lengths of inscribed regular polygons with 10, 20, 40, 80, 160, etc., sides. In 1 doing so, we obtain a pattern of nested radicals in- volving the golden ratio:

=−−2 ss1024 5

2 =−−24( 4 −Φ2 )

=−2 Φ

=−− 2 Fig. 6 An inscribed 2n-sided polygon inside a circle of radius 1 s20 2 4 s10

2 Then solving for a, we obtain =−−24( 2 −Φ ) − 2 1 2 4 s =−+Φ (3) as=−1 = . 22 4 2 Similarly, we obtain 4 − s2 Therefore, 11−=−a . 2 =−++Φ s40 222 ,

Next, we use the right triangle with of length t and sides of length s/2 and 1 – a to obtain =−++++Φ s80 2222 , 2 2 ⎛ ⎞ ⎛ − 2 ⎞ 2 s 221 4 s t = ⎜ ⎟ +−()1 as = +⎜1 − ⎟ ⎝ 2⎠ 4 ⎝⎜ 2 ⎠⎟ and =−++++Φ s160 22222 . 1 ⎛ 4 − s2 ⎞ = ss22+−14 − +⎜ ⎟ ⎝ ⎠ 4 4 Then using equation (1), we obtain our approxima- =−24 −s2 . tions of p in terms of the golden ratio Φ. See table 1. The value of p to 35 decimal places is Hence, ts=−−242 . 3.14159265358979323846264338327950288.

More specifically, if sn denotes the side length of a regular n-sided polygon that is inscribed inside a To obtain an approximation this accurate, we unit circle, then would need to double the number of sides 58 times

Vol. 99, No. 7 • March 2006 | MATHEMATICS TEACHER 475 TABLE 1 When we double the number of sides, we then Decimal Approximations of p in Terms of Φ obtain × nsn n π = Decimal Approximation 24ns − ssnss2 4 − 2 2 (5) A = 2n 2nnn= 22. 2n 4 2 54− Φ2 5 2.93892626146236564584 But if we make the substitution 2 ss=−−24 2 , 1052− Φ 3.09016994374947424102 2nn then equation (5) becomes 2010 2−+ 2 Φ 3.12868930080461738020 ns A = n . 2n 2

4020 2−++ 2 2 Φ 3.13836382911379780132 The details are left as an exercise. That is, the area of a regular 2n-sided polygon inscribed in a unit circle is numerically equal to half the perime- 80402222−+++Φ 3.14078526072548872166 ter of a regular n-sided polygon inscribed in a unit circle. Because each value approximates p, it is 1608022222−++++Φ 3.14139079370052833368 one step more efficient to use the perimeter of the polygons, rather than the area, to find our ap- proximations of p. The reader is left with the fol- 58 and use an inscribed regular polygon with 5 • 2 = lowing exercises. 1,441,151,880,758,558,720 sides. Fortunately, these values now can be generated easily with a EXERCISES

short computer program that would have saved 1. Find the initial side length s4 of a square that is Van Ceulen many years of work. inscribed in a unit circle, and then use equation (4)

to find the side lengths s8, s16, s32, s64, and s128. Then USING THE AREA TO APPROXIMATE p use equation (1) to find approximations of p using k Archimedes discovered the area of a circle to be A = these polygons with 4 • 2 sides. rC/2, which is not our modern, commonly used for-

mula. But he also found that the ratio of circumfer- 2. Find the initial side length s6 of a regular hexa- ence to diameter is constant for any circle. This gon inscribed in a unit circle. Then find the side

constant was labeled p in 1706 by the English lengths s12, s24, s48, and s96, and the resulting approxi- Sir William Jones (1675–1749). mations of p obtained from using these polygons k This symbol became the standard notation after the with 6 • 2 sides. great (1707–1783) adopted it in his own work beginning in 1748 (Burton 2003). So 3. For a triangle with sides of length a, b, and c, now we say p = C/d as well as C = pd. Substituting Heron’s formula gives the area as into Archimedes’ formula for circular area, we ob- tain our modern formula: ss()()(),−−− a s b s c

1 1 1 2 ArCrdr==()π =(()ππ• 2rr= where s = (a + b + c)/2. A regular n-sided polygon 2 2 2 inscribed in the unit circle creates n congruent tri-

Using r = 1, we see that p also equals the area of angles that have sides of length 1, 1, and sn. Apply

a unit circle. Thus, we could use the area An instead Heron’s formula to show that the area of the poly- of our inscribed polygons to approximate p. From gon is given by Archimedes’ formula, we can find this polygonal area if we know the apothem and perimeter of the ns 4 − s 2 n n . polygon. But the perimeter of an n-sided regular 4 • polygon with side length sn is simply n sn, and

equation (3) gives us the apothem in terms of sn for 4. Using the fact that regular polygons inscribed in a unit circle. Hence, =−−2 ss2nn24 1 ns4 − s 2 π ≈=Aap =nn. nn2 4 for regular polygons inscribed in a unit circle, show that

476 MATHEMATICS TEACHER | Vol. 99, No. 7 • March 2006 Hint: Use right-triangle trigonometric Boston: McGraw-Hill, 2003. ns4 − s 2 ns identities and the closed-form values of Dunham, William. Journey through Genius: 22nn= n . 42 sin 36° and cos 36°. The Great of . New York: John Wiley & Sons, 1990.

5. Let Sn be the side length of a regular 8. Apply exercise 5 to find S10, and then Euclid. The Thirteen Books of the Elements. polygon that is circumscribed about a find the resulting approximation of p ob- Translated and with an introduction by unit circle. For doubling the number of tained by using the perimeter of a cir- Sir Thomas Heath. Vol. 3. New York: sides, show that cumscribed regular . Dover, 1956. ∞

2 −+ 424 +S 9. Find S for a regular hexagon circum- S = n . 6 STACY LINN, slinn@ 2n scribed about a unit circle. Then find S Sn 12 b-g.k12.ky.us, teaches and the resulting approximation of p ob- mathematics at Bowling

6. Find S4 for a square that is circum- tained by using the perimeter of a regular Green High School in scribed about a unit circle, then use the 12-sided polygon circumscribed about a Bowling Green, KY 42101. result from exercise 5 to find S8. Find unit circle. She enjoys teaching alge- the approximation of p that is obtained bra and geometry and by using the perimeter of a circum- 10. Show that the area of a regular n- studying the history of scribed regular octagon. Repeat the sided polygon circumscribed about a mathematics. DAVID NEAL, • process for a regular 16-sided polygon. unit circle equals (n Sn)/2. In other [email protected], is a professor of words, with circumscribed polygons mathematics at Western Kentucky 7. For a regular pentagon circumscribed there is no difference between using the University, Bowling Green, KY 42101. about a unit circle, show that area or the perimeter to approximate p. He enjoys teaching all levels of mathematics and doing research in 2 224−Φ REFERENCES probability theory. Photographs by Brian Linn S = . 5 Φ Burton, David M. The History of and James Barskdale; all rights reserved Mathematics: An Introduction. 5th ed.

300TH ANNIVERSARY OF p

In 1706, Sir William Jones labeled the familiar value of 3.141592 . . . with the Greek letter p. On March 14, join mathematics educators and students around the country in celebration of this 300th anniversary of p. Use this date to engage students in activities related to the history and concept of p and enrich and deepen the students’ understanding of the concept. More information about p and about ways to mark the day can be found on the fol- lowing Web sites:

archive.ncsa.uiuc.edu/edu/RSE/RSEorange/ buttons.html mathwithmrherte.com/pi_day.htm joyofpi.com/pilinks.html eveander.com/triviap members.aol.com/loosetooth/pi.html www-groups.dcs.st-and.ac.uk:80/~history/ HistTopics/Pi_through_the_ages.html mathforum.com/t2t/faq/faq.pi.html www.eduref.org/cgi-bin/printlessons.cgi/ Virtual/Lessons/Mathematics/Geometry/ GEO0001.html

Vol. 99, No. 7 • March 2006 | MATHEMATICS TEACHER 477

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