aa r X i v : . [ m a t h . HO ] S e p MODERNIZING ARCHIMEDES’ CONSTRUCTION OF π DAVID WEISBART
Department of MathematicsUniversity of California, Riverside
In memory of my mentor and dear friend, Professor V.S.Varadarajan.
Abstract.
In his famous work, “Measurement of a Circle,” Archimedes described a pro-cedure for measuring both the circumference of a circle and the area it bounds. Implicitin his work is the idea that his procedure defines these quantities. Modern approaches fordefining π eschew his method and instead use arguments that are easier to justify, but theyinvolve ideas that are not elementary. This paper makes Archimedes’ measurement proce-dure rigorous from a modern perspective. In so doing, it brings a rigorous and geometrictreatment of the differential properties of the trigonometric functions into the purview of anintroductory calculus course. Contents
1. Introduction 12. Approximation by Regular m n -gons 33. Edge Length Comparison Theorems 74. Circumference and Area are Intrinsic 145. Application to the Circular Trigonometric Functions 23References 251. Introduction
Archimedes’ estimate of the value of π in “Measurement of a Circle” is one of his greatestachievements [2]. His approach anticipated foundational ideas of modern analysis as formu-lated in the 19th and early 20th centuries, making this work one of the first major analyticalachievements. Archimedes implicitly defined the circumference of a unit circle axiomati-cally as a number always greater than the perimeters of approximating inscribed polygonsand always less than the perimeters of approximating circumscribed polygons, where theapproximating polygons are regular refinements of a regular hexagon. By calculating the E-mail address : [email protected] . π perimeters of a regular inscribed and circumscribed polygon of 96 sides, he arrived at thefamous estimate < π < . He also found upper and lower bounds for the area of a disk respectively given by the areasof regular refinements of circumscribed and inscribed squares. He showed that the areaof the disk should be the area of a right triangle with one leg having length equal to thecircumference of the unit circle and the other having the length of the radius.The fact that the approximation procedure for the circumference appears to depend bothon the method of refinement and on the type of the polygons used in the first stage ofthe approximation presents a difficulty from a modern point of view. The circumferenceshould be intrinsic to the circle and independent of any specified approximation procedure.It is important to modernize Archimedes’ construction in a way that is maximally accessi-ble to a contemporary of Archimedes for both aesthetic and practical reasons. A practicalconsequence is that such a modernization cleanly and efficiently resolves the principle dif-ficulty that arises in developing the infinitesimal theory of the trigonometric functions, thecalculation lim x → sin( x ) x = 1 . (1.1)Unger notes in [5] some difficulties in the approach common in differential calculus text-books. Richman also points out in [3] some of these difficulties and directs the reader tothe approach of Apostol [1], that avoids common circular arguments by using twice the areaof a sector rather than arclength as the argument of the trigonometric functions. This isa natural point of view, especially when considering the direct analogy with the hyperbolictrigonometric functions. Although Apostol’s approach is compelling, it remains desirableto connect this nonstandard way of defining the trigonometric functions with the standardapproach. Moreover, Apostol defines the area of a sector rather than deriving it from alimiting procedure using inscribed and circumscribed polygons. So there is still some workto be done to connect the viewpoints.There are several approaches to computing (1.1) in the literature. Authors commonly usean analytical approach to solve the problem by defining the trigonometric functions as powerseries or as solutions to a system of differential equations. Some authors begin by defining theinverse of the tangent function as an integral. Unfortunately, these approaches lack geometricmotivation. In [6], Vietoris used the sum of angles formula for the trigonometric functionsto prove (1.1) directly. The notion of an arbitrary fraction of a circle requires a groupstructure on the circle and defining such a structure is certainly natural and unavoidable.The argument of the sine and cosine functions will be a multiple of the fraction of a circlethat a particular arc represents. However, taking the multiple to be the length of an arcrequires a definition of the length of an arc. Defining and computing the length of the arc isthe primary difficulty. ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π If one follows Vietoris, one must define π in some way independent of it being the area ofthe unit circle, or half its circumference. If one uses the approach of Zeisel in [7] that followsVietoris’—at least philosophically—then one can dispense with the alternate definition of π .Taking π as the limit of the area given by regular circumscribed n -gons and taking themultiple of the fraction in the argument to be π will imply the limit (1.1). However, withoutsome further work, even this elegant approach of Zeisel does not establish the meaning ofthe area and circumference of the unit circle as well as their relationship. In fact, one muststill show that the limit of the areas he discusses actually exists.Using only basic euclidean geometry, we prove that for any arc A that is less than half ofa circle, if n is greater than m and if ℓ n and ℓ m are the lengths of chords that respectivelysubtend arcs that are an n th and an m th of A , then mℓ n > nℓ m . We prove an analogous but reverse inequality for lengths corresponding to edges of cir-cumscribed polygons. These inequalities imply that the circle is a rectifiable curve. Theyfurthermore give the classical relationship between the circumference of a circle and the areathat it bounds in a way that is both rigorous and accessible to freshman calculus students.Using these inequalities, we show that π is the limiting value of any sequence of perimetersof approximating polygons. This approach is a simple modernization of the approach ofArchimedes and makes rigorous the usual geometric arguments in calculus textbooks thatprove (1.1). It is also elementary enough to be accessible, at least in principle, to a freshmancalculus student. Acknowledgements.
I thank Professor V.S.Varadarajan for the insights and suggestionshe gave me during our many discussions involving this paper. From him I learned to love,among many things, the history of our subject, and so I dedicate this paper to his memory.I thank Dr. Alexander Henderson for editing this paper and for his helpful comments.2.
Approximation by Regular m n -gons Denote by C the unit circle centered at O . Unless otherwise specified, n will be in N ≥ ,the set of natural numbers greater than or equal to three, and m will be in N , the naturalnumbers with zero. All sequences henceforth indexed by m will be indexed over the set N .2.1. Regular Inscribed and Circumscribed Polygons.Definition 1. An inscribed polygon of C is a simple polygon all of whose vertices lie on C and circumscribed polygon of C is a simple polygon all of whose edges intersect C tangentially.An inscribed polygon of C is regular if all of its edges are of equal length. A circumscribedpolygon of C is regular if each of its edges intersects C tangentially at its midpoint.While circumscribed polygons are often assumed by definition to be convex, this assump-tion is redundant. Furthermore, our definition of regularity implies that all edges of a regularcircumscribed polygon are congruent. Denote respectively by g ( m, n ) and G ( m, n ) the reg-ular inscribed and circumscribed polygons with m n edges. Up to congruency, there is only MODERNIZING ARCHIMEDES’ CONSTRUCTION OF π one such inscribed and one such circumscribed polygon for each choice of m and n . Denoterespectively by ℓ n ( m ) and L n ( m ) the edge length of g ( m, n ) and G ( m, n ) . The respectiveperimeters of g ( m, n ) and G ( m, n ) are p n ( m ) and P n ( m ) , where p n ( m ) = 2 m nℓ n ( m ) and P n ( m ) = 2 m nL n ( m ) . Let P and Q be adjacent vertices of g ( m, n ) and let A and B be adjacent vertices of G ( m, n ) .Denote respectively by α n ( m ) and α n ( m ) the areas of the triangles △ P OQ and △ AOB . Therespective areas of g ( m, n ) and G ( m, n ) are a n ( m ) and A n ( m ) , where a n ( m ) = 2 m nα n ( m ) and A n ( m ) = 2 m nα n ( m ) . The triangle inequality and the additivity of area together imply the following proposition.
Proposition 2.1.
For fixed n , the sequences ( p n ( m )) and ( a n ( m )) are both increasing andthe sequences ( P n ( m )) and ( A n ( m )) are both decreasing. The parallel postulate implies the following useful lemma. Refer to Figure 1 to clarify thestatement of the lemma.
AD EBO CMF
Figure 1
Lemma 2.1.
Suppose that C and B are points on C so that the counterclockwise orientedarc from C to B is less than half of C . If L B and L C are lines tangent to C respectively at B and C , then L B and L C intersect at a point A and the right triangles △ OAB and △ OCA are congruent. If F is the point of intersection of OA with BC , then ∠ AF B is a right angle.Let M be the point at which OA intersects C and L M be the line tangent to C at M . Denoteby D the point at which the line L M intersects BA and denote by E the point at which L M intersects CA . The angle ∠ AM D is congruent to ∠ EM A and both are right angles.
Area and Perimeter Bounds.
While Proposition 2.1 guarantees for fixed n the strictmonotonicity of ( p n ( m )) , ( a n ( m )) , ( P n ( m )) and ( A n ( m )) , it does not guarantee that thesequences are bounded. The following proposition provides the desired bounds. Proposition 2.2.
For each fixed n , p n ( m ) < P n ( m ) and a n ( m ) < A n ( m ) . Since ( p n ( m )) and ( a n ( m )) are increasing and bounded above by P n (0) and A n (0) , andsince ( P n ( m )) and ( A n ( m )) are decreasing and bounded below by p n (0) and a n (0) , all foursequences are convergent, implying Proposition 2.3. ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Proposition 2.3.
For each fixed n , there are real numbers p n , P n , a n and A n such that p n ( m ) → p n , P n ( m ) → P n , a n ( m ) → a n , and A n ( m ) → A n . Convergence of the Approximations.
Proposition 2.3 gives for each fixed n respec-tive limiting values for the areas and perimeters of regular refinements of regular inscribedand circumscribed polygons with n edges. It does not prove the equality of the respectivelimits. Theorem 1 (Heron’s Theorem) . If T is a triangle with side lengths a, b, c and A ( T ) is thearea of T , then A ( T ) = p s ( s − a )( s − b )( s − c ) where s = a + b + c . For any points A and B in the plane, denote by ℓ ( AB ) the length of the line segment AB .Denote by h n ( m ) the distance from a vertex of G ( m, n ) to C . Proposition 2.4.
Given A n and a n above, ( i ) a n = 12 p n , ( ii ) A n = 12 P n , ( iii ) P n = p n , and ( iv ) A n = a n . (2.1) Proof.
Since the sequences ( p n ( m )) and ( P n ( m )) are both convergent and are respectivelyequal to (2 m nℓ n ( m )) and (2 m nL n ( m )) , both ℓ n ( m ) and L n ( m ) tend to zero as m tends toinfinity. Heron’s theorem implies that α n ( m ) = q(cid:0) ℓ n ( m ) (cid:1) (cid:0) − ℓ n ( m ) (cid:1) (cid:0) ℓ n ( m ) (cid:1) = ℓ n ( m ) q − ℓ n ( m ) , and so a n ( m ) = 2 m n ℓ n ( m ) q − ℓ n ( m ) = p n ( m ) q − ℓ n ( m ) → p n . (2.2)Since C is a unit circle, α n ( m ) is equal to L n ( m ) and so A n ( m ) = 2 m nα n ( m ) = 2 m − nL n ( m ) = P n ( m ) → P n . (2.3)Suppose that G is a vertex of G ( m, n ) , that A and H are points on C where the edges of G ( m, n ) with endpoint G are tangent to C , and that △ GAH is counterclockwise oriented(Figure 2). Line segments AG and GH are half edges of G ( m, n ) , so ℓ ( AG ) is equal to L n ( m ) . Take N to be a point of C so that AN and N H are edges of g ( m + 1 , n ) . Theintersection, C , of lines tangent to C at A and N is a vertex of G ( m + 1 , n ) . Take P to bethe intersection of AH with OP and M to be the intersection of AN with OC . Let I be thepoint of C that intersects OC . Lemma 2.1 implies that ∠ GP A and ∠ CM A are right angles.Denote by B intersection of line tangent to C at A with line tangent to C at I . Denote by J the intersection of the line tangent to C at I with line tangent to C at N . The points B and J are neighboring vertices of G ( m + 2 , n ) . Angle ∠ GN A is obtuse because ∠ GN C isright, hence L n ( m ) > ℓ n ( m + 1) . MODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Extend BJ to meet OG at a point L . Lemma 2.1 implies that ∠ OM N and ∠ OIJ areboth right angles and so BJ and AN are parallel. The point L therefore lies between N and G . The line segment N G has length h n ( m ) and ℓ ( IC ) is equal to h n ( m + 1) . O HA G NIC JB L PMKDE F
Figure 2Line segments BI , IJ , and J N are congruent as half edges of G ( m +2 , n ) . The hypotenuseof the right triangle △ J N L is J L , so ℓ ( J L ) is greater than ℓ ( BI ) . Let K be a point on J L such that
J K and BI are congruent. Furthermore, let E and F be points lying on AG such that KE , LF , and IC are parallel. The similarity of △ BIC , △ BKE , and △ BLF and the equality of ℓ ( BK ) and ℓ ( BI ) together imply that ℓ ( KE ) is equal to ℓ ( IC ) . Since ∠ OAC is right, ∠ OCG is obtuse and so ∠ LF G is as well, implying that ℓ ( LG ) is greaterthan ℓ ( LF ) . Since ℓ ( BL ) is greater than ℓ ( BK ) , h n ( m ) = ℓ ( N G ) > ℓ ( LG ) > ℓ ( LF ) > ℓ ( KE ) = 3 ℓ ( IC ) = 3 h n ( m + 1) . Since m was arbitrary, h n ( m ) < (cid:0) (cid:1) m h n (0) . The triangle inequality implies that h n ( m ) + ℓ n ( m + 1) is greater than L n ( m ) , hence < L n ( m ) − ℓ n ( m + 1) < h n ( m ) < (cid:0) (cid:1) m h n (0) , and so < m nL n ( m ) − m +1 nℓ n ( m + 1) = P n ( m ) − p n ( m + 1) < n (cid:0) (cid:1) m h n (0) . (2.4)The estimate (2.4) implies that < P n ( m ) − p n ( m ) = ( P n ( m ) − p n ( m + 1)) + ( p n ( m + 1) − p n ( m )) < n (cid:0) (cid:1) m h n (0) + ( p n ( m + 1) − p n ( m )) → . ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π The difference P n ( m ) − p n ( m ) tends to zero as n tends to infinity, hence P n is equal to p n .Of course, (2.4.iv) follows immediately from (2.3) and (2.2). (cid:3) Edge Length Comparison Theorems
Comparing the edge lengths of inscribed and circumscribed segments corresponding todifferent regular subdivisions of an arc is the key to proving that the definition of π isindependent of any approximation scheme. This section will present such a comparison. Definition 2.
A ( counter ) clockwise oriented partition P of an arc A is a finite sequenceof points of A that is (counter)clockwise ordered, and the first and last points of P are theendpoints of A . Such a finite sequence is regular if all adjacent points of P are equidistant.Temporarily ignore the previous restrictions on the natural numbers m and n . O P n +1 P n P m +1 P m P P A B
Figure 3Suppose that m and n are natural numbers with n greater than m . Let A be an arc of C that is less than half of C and let ( P , . . . , P n +1 ) be a regular clockwise oriented partitionof A . Let B be the point of intersection of the lines tangent to C at P and P n +1 and let A be the point of intersection of the lines tangent to C at P and P M +1 . Denote by ℓ m , ℓ n , L m , and L n the respective lengths of P P m +1 , P P n +1 , P B and P A . We will show that nℓ m > mℓ n and nL m < mL n . General Symmetry Considerations For Polygonal Segments.
Assume that A , B , C , and D belong to a collection of points that form a regular partition of A (Figure 4),that A and C are adjacent, that B and D are adjacent, and that C and D lie on the same sideof the line AB and opposite the side where O lies. Suppose furthermore that the quadruple ( A, B, D, C ) forms a counterclockwise oriented partition of A . MODERNIZING ARCHIMEDES’ CONSTRUCTION OF π O BDCA EF
Figure 4
O BA EC
Figure 5Standard arguments using the SSS theorem imply Proposition 3.1, whose proof is left asan exercise.
Proposition 3.1.
The quadrilateral (cid:3)
ABDC ( Figure 4 ) is an isosceles trapezoid and AB and CD are parallel. Furthermore, in the degenerate case ( Figure 5 ) when C is equal to D ,angles ∠ BAC and ∠ ABC are congruent and CO bisects AB . P n,n − ... P n, P n − , P n, P ,n − P ,n − P ,n − P ,n − ... P , O P n P n − . . . P P P P ,n P n, Figure 6Suppose that A is an arc of C that is less than half of C and suppose that ( P , . . . , P n ) isa regular clockwise orientated partition with n greater than 3 (Figure 6). For each naturalnumber i in [1 , n ] , let L i be the line tangent to A and intersecting P i . Since A is less thanhalf of a circle, Lemma 2.1 implies that the lines L i and L j intersect for each i not equal to j ; call this point of intersection P i,j . Notice that P i,j is equal to P j,i . The key theorem of thenext section is to prove that j > i implies that ℓ ( P ,i P ,i +1 ) < ℓ ( P ,j P ,j +1 ) . ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Refer to Figure 7. Suppose that ( A, B, C, D ) is a clockwise ordered partition of an arc A of C and that AB has the same length as CD . Let L A , L B , L C and L D be lines tangent to A that intersect A , B , C , and D respectively. Since A is less than half of a circle, Lemma2.1 implies that L A and L D intersect at a point P AD , L A and L C intersect at a point P AC , L B and L D intersect at a point P BD , and L B and L C intersect at a point P BC . The SAS andSSS theorems along with Proposition 3.1 and Lemma 2.1 imply the following proposition,whose proof is left to the reader as a standard exercise. P CD L C P BD L B P BC P AC P AB O D L D CBA E F L L A P AD Figure 7
Proposition 3.2.
The line segment P AD P BC can be extended to a ray, L , that meets O and bisects and is perpendicular to P AC P BD , AD , and BC . Furthermore, L bisects ∠ P AC P AD P BD . Finally, triangles △ P AD P AC P BC and △ P AD P BD P BC are congruent, as are △ P AD P BC P AB and △ P AD P BC P CD . Denote by m ( ∠ ABC ) the degree measure of ∠ ABC . Refer to Figure 6 for Proposition 3.3.
Proposition 3.3.
Suppose that n is in N ≥ and ( P , . . . , P n ) is a regular clockwise orientedpartition of an arc A that is less than half of C . If m ( ∠ P i OP i +1 ) is equal to θ ◦ , then (1) m ( ∠ P ,n − P ,n P n, ) = (180 − nθ ) ◦ ;(2) m ( ∠ P ,n − P ,n − P ,n ) = m ( ∠ P n, P n, P ,n − ) = ( n − θ ◦ ;(3) m ( ∠ P n, P ,n − P ,n − ) = 180 ◦ − ( n − θ ◦ . Proof.
Both ∠ P ,n P n O and ∠ OP P ,n are right angles, implying that m ( ∠ P n OP ) + m ( ∠ P P ,n P n ) = 180 ◦ . Since m ( ∠ P n OP ) is equal to nθ ◦ , the fact that the sum of the interior angles of the quadri-lateral formed by P , P n , O and P ,n is ◦ implies (1). Since P , P ,n − and P ,n lie on the π same line, ∠ P P ,n − P ,n − and ∠ P ,n − P ,n − P ,n are supplementary. By (1), m ( ∠ P , P ,n − P ,n − ) = (180 − ( n − θ ) ◦ and so m ( ∠ P ,n P ,n − P ,n − ) = ( n − θ ◦ . Of course, an analogous argument proves the same result for ∠ P ,n − P n, P n , implying (2),although another argument is not necessary in light of Proposition 3.2. Finally, since thesum of angles of a quadrilateral is ◦ , (3) follows from (1) and (2). (cid:3) Length Estimates For Inscribed Polygonal Segments.
Let A again be an arcof a unit circle that is less than half of the circle. Suppose that ( P , . . . , P n +1 ) is a regularclockwise oriented partition of A (Figures 8 and 9). For each P i where i is a natural numberin [2 , n ] , there is a line perpendicular to P P n +1 that intersects P i at a point q i on P P n +1 .Figure 8 shows an example where n is even and Figure 9 shows the qualitative differencein an odd example. Set q to be equal to P and let q n +1 be equal to P n +1 . For each i between and n + 1 , let q i be the point on P P n +1 so that P i q i is perpendicular to P P n +1 . OX X X X X P P P P P P P P P q q q q q q q Figure 8 OX X X X X X P P P P P P P P P P q q q q q q q q Figure 9
Proposition 3.4. If i and j are natural numbers in (cid:2) , n +12 (cid:3) with i less than j , then ℓ ( q i q i +1 ) < ℓ ( q j q j +1 ) and ℓ ( q i q i +1 ) = ℓ ( q n +1 − i q n +2 − i ) . Proof.
Proposition 3.1 implies that each of the line segments P i P n +2 − i is parallel to P P n +1 .Since P i q i and P i P n +2 − i are perpendicular for i in (cid:2) , n (cid:3) , ℓ ( P i P n +2 − i ) = ℓ ( q i q n +2 − i ) . Let X be equal to q and, for each i in (cid:2) , n (cid:3) , let X i be the point of intersection of P i P n +2 − i and P i +1 q i +1 . Let X n − be equal to q n and, for each i in (cid:2) n + 1 , n (cid:3) , let X i be the point ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π of intersection of P i +2 P n − i and P i +1 q i +1 . For each i in (cid:2) , n (cid:3) , Proposition 3.1 implies that ℓ ( P i X i ) is equal to ℓ ( X n − i P n +2 − i ) , which is equal to ℓ ( q i q i +1 ) and ℓ ( q n +1 − i q n +2 − i ) .Let i and j be in (cid:2) , n (cid:3) with i less than j . To show that ℓ ( P i X i ) is less than ℓ ( P j X j ) ,it suffices to show that if i and i + 1 are in (cid:2) , n (cid:3) , then ℓ ( P i X i ) is less than ℓ ( P i +1 X i +1 ) .Extend the line segment P i P i +1 to a ray, R , originating at P i . Extend the line segment q i +2 P i +2 to a ray, R , that meets R at a point A . The ray R passes through the point X i +1 and l ( X i +1 A ) is greater than l ( X i +1 P i +2 ) , while angles ∠ X i +1 P i +1 A and ∠ X i P i P i +1 are congruent, implying that ∠ X i P i P i +1 is greater than ∠ X i +1 P i +1 P i +2 . Since △ X i P i P i +1 and △ X i +1 P i +1 P i +2 are both right triangles with hypotenuses that have the same length, ∠ P i P i +1 X i is greater than ∠ P i +1 P i +2 X i +1 , implying that ℓ ( P i X i ) is less than ℓ ( P i +1 X i +1 ) .Since i and j are integers, the statement of the proposition is different when n is odd andwhen n is even.Note that if n is odd, then the above arguments show that q n +12 q n +12 +1 is the longest of allof the segments q i q +1 . (cid:3) Since the lengths of the line segments q i q i +1 are strictly increasing up to the midpoint of A for even n and the middle segment for odd n , the following corollary is immediate. Corollary.
Let k be the largest integer less than or equal to n . If n is greater than and s is a natural number in (1 , k ] , then ℓ ( q q s +1 ) < ℓ ( P P n +1 ) n s. Remark.
The above corollary implies that the average length of a segment of q q k +1 is lessthan the average length of a segment of q q n if the arc between P and P k +1 is less than halfof A .From this corollary follows the following key theorem. Theorem 2.
Suppose that A is an arc that is less than half of a circle and that ( P , . . . , P n +1 ) is a regular clockwise oriented partition of A . Denote by ℓ n the length of the line segment P P n +1 , and by ℓ m the length of P P m +1 . If m is less than n , then nℓ m > mℓ n . Proof.
Assume first that m is greater than n and refer to Figure 10. Let Q be the pointof intersection of the circle of radius ℓ m centered at P with the line segment P P n +1 . Thelength of P Q is equal to ℓ m . If s the length of the line segment QP n +1 , then s = ℓ n − ℓ m . There are n edges with vertices on the arc between P and P n +1 . There are m + 1 vertices ofthe regular subdivision between P and P m +1 on A . Since m is greater than n , P m +1 liesclockwise from the intersection, M , of a line bisecting P P n +1 and the circle. For the samereason, the point P n − m lies counterclockwise from M . π Proposition 3.4 implies that the average length of the segments q q , q q , . . . , q n − m q n − m isthe same as the average lengths of the segments q n +1 q n , q n q n − , . . . , q m q m . However, forany edge P i P i +1 between the vertices P n +1 − m and P m +1 , the projection of the edge, q i q j on theline P P n +1 , is longer than the longest of the segments q j q j +1 with j in [1 , n − m ) ∪ ( m + 1 , n ] .Denote by d the length of the line segment q m +1 q n +1 to obtain the inequality d < n − mn ℓ n . Since ℓ m is the length of the hypotenuse of a triangle with a leg of length ℓ n − d and ℓ m isequal to ℓ n − s , the length s is less than d and so s < n − mn ℓ n . Since, ℓ m + s is equal to ℓ n , ℓ m + n − mn ℓ n > ℓ n , and so nℓ m > mℓ m .OP m +1 P n − P n P P P P m d sl n P = q P n +1 l m q m +1 Q Figure 10Suppose that m is equal to n . Since the sum of the lengths of two sides of a non-degeneratetriangle are greater than the length of the third, ℓ m is greater than ℓ m , and so nℓ m = 2 mℓ m > mℓ m = mℓ n . (3.1)In the case when m is less than n , there is a natural number k with k m < n ≤ k +1 m which together with (3.1) implies that nℓ k m > k mℓ n . ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π The points P and P k − m +1 have k − m segments between them as do the points P k − m +1 and P k m +1 . Thus, ℓ ( P P k − m +1 ) = ℓ ( P k − m +1 P k m +1 ) = ℓ k − m . Therefore, ℓ k − m is equal to the sum of the lengths of two sides of △ P P k − m +1 P k m +1 andso is greater than ℓ k m . Therefore, nℓ k − m > nℓ k m > k mℓ n and so nℓ k − m > k − mℓ n > mℓ n . Applying the above argument k − more times yields the inequality nℓ m > mℓ n . (cid:3) Length Estimates For Circumscribed Polygonal Segments.
Use the notation ofProposition 3.3 for the statement and proof of the proposition below.
Proposition 3.5. If k and l are natural numbers, then ≤ k < l ≤ n − implies that ℓ ( P ,k P ,k +1 ) < ℓ ( P ,l P ,l +1 ) .AB P i, P i − , P i, P ,i − P ,i − P ,i − P ,i − OP ,i P ,i Figure 11
O P , P P P , P , B P , P , MP , P Figure 12
Proof.
Suppose that i is a natural number in [3 , n ] . Proposition 3.2 implies that the line P ,i − P ,i − is a perpendicular bisector of P ,i − P ,i − . Denote by A this intersection. Toprove the proposition, it suffices to show that ∠ P ,i P ,i − P ,i − is greater than a right angle.In this case, there is a line that intersects P ,i − and is perpendicular to P ,i − P ,i − thatintersects P ,i − P ,i at a point B . The triangles △ P ,i − AP ,i − and △ P ,i − P ,i − B aresimilar, but since P ,i − P ,i − is twice the length of P ,i − A , P ,i − B is twice the length of P ,i − P ,i − , and so ℓ ( P ,i − P ,i ) > ℓ ( P ,i − B ) = ℓ ( P ,i − P ,i − ) . π Proposition 3.2 implies the congruency of the angles ∠ P ,i P ,i − P ,i − and ∠ P i, P ,i − P ,i .Proposition 3.3 implies that m ( ∠ P ,i P ,i − P ,i − ) = 90 ◦ − i − θ ◦ . Proposition 3.2 implies that P ,i − P ,i − is perpendicular to P ,i − P ,i − and furthermore that P ,i − P ,i − bisects ∠ P ,i − P ,i − P ,i − . Therefore, ∠ P ,i − P ,i − P ,i − and ∠ P ,i − P ,i − P ,i − are congruent. Proposition 3.3 implies that m ( ∠ P ,i − P ,i − P ,i − ) + 180 ◦ − ( i − θ ◦ = 180 ◦ . Therefore, m ( ∠ P ,i − P ,i − P ,i − ) is equal to i − θ ◦ and so m ( ∠ P ,i P ,i − P ,i − ) = m ( ∠ P ,i P ,i − P ,i − ) + m ( ∠ P ,i − P ,i − P ,i − )= 90 ◦ − i − θ ◦ + (cid:0) ( i − θ ◦ − i − θ ◦ (cid:1) = 90 ◦ + θ ◦ > ◦ . Consider the case where i is (Figure 12). As in the argument above, the key is to showthat m ( ∠ P , P P ) > ◦ . Proposition 3.2 implies that P , P , is tangent to A , hence perpendicular to P , P . Itsuffices to show that m ( ∠ P , P P ) does not equal 0. Since m ( ∠ P OP ) = θ ◦ , ∠ P P , P has measure (180 − θ ) ◦ . Since the angles ∠ P , P P and ∠ P , P P are congruent,both have measure θ ◦ , which is greater than . (cid:3) Theorem 3.
Suppose that n is a natural number greater than two and ( P , . . . , P n +1 ) is aregular clockwise orientated partition of an arc A . If L n and L m respectively denote thelength of the line segment P P ,n +1 and the length of the line segment P P m +1 , then m < n implies that nL m < mL n . Proof.
The lengths of the segments P ,i P ,i +1 are increasing in i , so the average length of asegment of L n is larger than the average length of a segment of L m . Therefore, L n n > L m m , implying the desired result. (cid:3) Circumference and Area are Intrinsic
Definition 3. A circuit is a finite sequence of points ( P , . . . , P n +1 ) on C that is counter-clockwise ordered and with the property that P is equal to P n +1 .Without loss in generality and in order to simplify the exposition, assume that the arcconnecting adjacent points of a circuit is less than half of a circle. ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Definition 4. A refinement of a circuit S is a circuit S such that, as functions on finitesets, the range of S is a subset of the range of S .Denote by S ( C ) the set of all circuits on C . Suppose that ( P , . . . , P n +1 ) is in S ( C ) . Theinscribed polygon g ( P , . . . , P n +1 ) is the polygon whose vertices are the points in the circuitand the circumscribed polygon G ( P , . . . , P n +1 ) is the polygon whose vertices are the pointsgiven by the intersections of the lines tangent to C at adjacent points of the circuit. Definition 5.
For any polygon P , denote by π ( P ) the perimeter of P . Let min( P , . . . , P n +1 ) denote the minimum edge length of g ( P , . . . , P n +1 ) and denote by max( P , . . . , P n +1 ) themaximal edge length of g ( P , . . . , P n +1 ) . OP = P n +1 P n P n − P n − P P P V n V n − V n − V V V Figure 134.1.
Approximation By Rational Circuits.
Let the length ℓ be less than 2. Constructrecursively a sequence of points and a corresponding piecewise linear path in the followingway. Take P to be a point on C . Let P n be the n th term in a sequence of points on C so that if P i and P i +1 are adjacent points in the sequence, then ℓ ( P i P i +1 ) is equal to ℓ andthe triangle △ OP i P i +1 is counterclockwise oriented. The circle C ℓ,n of radius ℓ with center P n intersects C in exactly two places. Let P n +1 be the first of these points counterclockwisefrom P n . The line segment P n P n +1 is of length ℓ . If the sequence has the property that forsome k , P k is equal to P , then ℓ is a rational length and there is a first such k that is the numerator of ℓ , denoted by N ( ℓ ) . The closed piecewise linear path Γ( ℓ ) is the finite sequenceof edges Γ( ℓ ) = (cid:0) P P , . . . , P N ( ℓ ) − P N ( ℓ ) (cid:1) . Let OP be the line from the origin to the point P . Denote by D ( ℓ ) , the denominator of ℓ ,the number of times the lines P i P i +1 intersect the line segment OP , where i in (1 , N ( ℓ )] . The π denominator counts how many times Γ( ℓ ) wraps around the origin. A length ℓ is integral if D ( ℓ ) is equal to 1, implying that ℓ is the edge length of a regular inscribed polygon. A circuitis said to be integral if all of its edge lengths are integral. A regular circuit is an integralcircuit in which the distance between any two adjacent points in the circuit is the same. Thecounterclockwise ordered set of intersections of any circumscribed regular polygon with C is a regular circuit. Furthermore, any regular circuit is the set of intersections of a regularcircumscribed polygon with C . Proposition 4.1.
Let ℓ and m be rational lengths. Let L and M denote the edge lengths ofcircumscribed edges corresponding to the inscribed edges ℓ and m respectively. If m is lessthan ℓ , then N ( ℓ ) D ( ℓ ) ℓ < N ( m ) D ( m ) m and N ( ℓ ) D ( ℓ ) L > N ( m ) D ( m ) M. Remark. If ℓ and m are integral lengths, then D ( ℓ ) and D ( m ) are both equal to 1 and theproposition gives a comparison of the perimeters of the regular polygons of side lengths ℓ and m . In the rational but non-integral case, it is not reasonable to compare the lengths ofthe curves Γ( ℓ ) and Γ( m ) because of the dissimilar wrapping of the curves Γ( ℓ ) and Γ( m ) around the circle. Dividing the lengths of the above curves by their respective denominatorsgives a way to compare the average length of the curves on a single wrapping around thecircle. It is this idea that will prove important in the next proposition. Proof.
Since length is invariant under rotation, assume that both paths Γ( ℓ ) and Γ( m ) startwith the same initial point. All the points on the paths Γ( ℓ ) and Γ( m ) are points on theregular polygon with N ( ℓ ) N ( m ) sides containing the point P . Since ℓ may be realizedas the edge length of a straight line segment traversing D ( ℓ ) segments of a regular N ( ℓ ) -gon and m may be realized as the edge length of a straight line segment traversing D ( m ) segments of a regular N ( m ) -gon, ℓ may also be realized as the edge length of a straightline segment traversing D ( ℓ ) N ( m ) segments of a regular N ( ℓ ) N ( m ) -gon and m may alsobe realized as the edge length of a straight line segment traversing D ( m ) N ( ℓ ) segments of aregular N ( m ) -gon. Since ℓ is larger than m , D ( ℓ ) N ( m ) > D ( m ) N ( ℓ ) . By assumption, both lengths are less than a diameter. Theorem 2 therefore implies that D ( ℓ ) N ( m ) m < D ( m ) N ( ℓ ) ℓ, and this proves the proposition.The case of circumscribed polygons is similar except that instead of comparing the lengths ℓ and m , compare the lengths L and M and appeal to Theorem 3 to obtain the reverseinequality. (cid:3) Proposition 4.2. If ( P , . . . , P k +1 ) and ( Q , . . . , Q n +1 ) are two rational circuits, then min( P , . . . , P k +1 ) > max( Q , . . . , Q l +1 ) ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π implies that (1) π ( g ( Q , . . . , Q l +1 )) > π ( g ( P , . . . , P k +1 )) and (2) π ( G ( Q , . . . , Q l +1 )) < π ( G ( P , . . . , P k +1 )) . Proof.
Denote by ℓ i and m i the lengths ℓ i = ℓ ( P i P i +1 ) and m j = ℓ ( Q j Q j +1 ) . Abuse notation and denote again by ℓ i and m i the respective segments of length ℓ i and m i . There are k segments ℓ , . . . , ℓ k for g ( P , . . . , P k +1 ) and n segments m , . . . , m n for g ( Q , . . . , Q n +1 ) . Let ℓ be the shortest of the segments of g ( P , . . . , P k +1 ) and m n be thelongest of the segments of g ( Q , . . . , Q n +1 ) . Take Λ to be the product Λ = N ( ℓ ) · · · N ( ℓ k ) N ( m ) · · · N ( m n ) and take Λ copies of both polygons. Compute the respective perimeters of these polygonsto obtain Λ π ( g ( P , . . . , P k +1 )) = Λ( ℓ + ℓ + · · · + ℓ k )= Λ N ( ℓ ) N ( ℓ ) ℓ + · · · + Λ N ( ℓ k ) N ( ℓ k ) ℓ k = D ( ℓ ) Λ N ( ℓ ) N ( ℓ ) D ( ℓ ) ℓ + · · · + D ( ℓ k ) Λ N ( ℓ k ) N ( ℓ k ) D ( ℓ k ) ℓ k < D ( ℓ ) Λ N ( ℓ ) N ( ℓ ) D ( ℓ ) ℓ + · · · + D ( ℓ k ) Λ N ( ℓ k ) N ( ℓ ) D ( ℓ ) ℓ = (cid:18) D ( ℓ ) Λ N ( ℓ ) + · · · + D ( ℓ k ) Λ N ( ℓ k ) (cid:19) N ( ℓ ) D ( ℓ ) ℓ = Λ N ( ℓ ) D ( ℓ ) ℓ . The inequality follows from Theorem 2 since ℓ is the minimum length of the segments of g ( P , . . . , P k +1 ) . The ultimate equality follows from the fact that k X i =1 D ( ℓ i ) Λ N ( ℓ i ) = Λ . To justify the last assertion, note that given Λ copies of the polygon g ( P , . . . , P k +1 ) , thesegments in all of these copies form a path that wraps around C exactly Λ times. However,this collection of polynomial contains Λ copies of each segment ℓ i , hence Λ N ( ℓ i ) copies of eachpath Γ( ℓ i ) which wraps around C exactly D ( ℓ i ) times. Summing each of these Λ N ( ℓ i ) D ( ℓ i ) wrappings gives the total number of wrappings, Λ , implying the above formula for the sum. π Similarly, use the reverse inequality and make use of Proposition 4.1 to obtain the inequal-ity Λ π ( g ( Q , . . . , Q n +1 )) > Λ N ( m n ) D ( m n ) m n > Λ N ( ℓ ) D ( ℓ ) ℓ > Λ π ( g ( P , . . . , P k +1 )) . Dividing both sides of the inequality by Λ finishes the proof for inscribed polygons.The proof of the result for circumscribed polygons is done in the same way but appeals tothe appropriate inequality in Proposition 4.1, reversing the inequalities analogous to thosegiven above. (cid:3) Definition 6.
Following Archimedes, define π to be equal to p (as defined in Proposi-tion 2.3).Denote by Q ( C ) the set of all rational circuits on C and suppose that P is in Q ( C ) .Denote the maximal edge length of P , the mesh of P , by µ ( P ) . Theorem 4.
For any positive real number ε there is a positive real number δ such that µ ( P ) < δ implies that < π − π ( g ( P )) < ε and < π ( G ( P )) − π < ε. Proof.
Let ε be a positive real number. Both π ( g ( m, and π ( G ( m, tend to π as m tends to infinity, and so as long as m ′ is a large enough natural number, < π − π ( g ( m ′ , < ε and 0 < π ( G ( m ′ , − π < ε. (4.1)Let P be a rational circuit. Take δ to be a positive real number less than the side lengthof g ( m ′ , , so that the maximum possible side length of G ( P ) is less than the side lengthof G ( m ′ , and the maximum possible side length of g ( P ) is less than the edge length of g ( m ′ , . Since µ ( P ) is less than δ , Proposition 4.2 implies that π ( g ( m ′ , < π ( g ( P )) and π ( G ( m ′ , > π ( G ( P )) . (4.2)Take m ′′ to be a large enough natural number so that the edge length of π ( g ( m ′′ , is smallerthan the smallest edge length of g ( P ) and the edge length of π ( G ( m ′′ , is smaller than thesmallest edge length of G ( P ) . Such a choice of m ′′ is possible because the side lengths ofboth π ( g ( m, and π ( G ( m, tend to zero as m tends to infinity. Proposition 4.2 impliesthat π ( g ( P )) < π ( g ( m ′′ , < π and π ( G ( P )) > π ( G ( m ′′ , > π. (4.3)Inequalities (4.1), (4.2), and (4.3) together imply the theorem. (cid:3) Denote by π ( n ) the perimeter of a regular n -gon inscribed in C and denote by Π( n ) theperimeter of a regular circumscribed n -gon. Corollary (Corollary to Proposition 4.2) . If n is greater than m and m is greater than 2,then (1) π ( n ) > π ( m ) and (2) Π( n ) < Π( m ) . ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Corollary (Corollary to Theorem 4) . For any natural numbers n and m larger than 2, p m = p n , P m = P n , a m = a n , and A m = A n , where the notation follows Proposition 2.3.Remark. The next section presents results somewhat more general than those of the abovetheorem. Until this point we have only used the convergence of bounded monotone sequences,the axioms of Euclidean geometry, and the properties of the natural numbers. In this nextsection, we use a cardinality argument to show that not all edge lengths are rational lengthsand so this section relies on a consideration that is highly unlikely to have been consideredby a contemporary of Archimedes, but is necessary from a modern perspective.4.2.
Approximation by General Circuits.
Each rational length corresponds to a pair oftwo integers, a numerator and a denominator, making the set of rational lengths a count-ably infinite set. However, there are uncountably many possible edge lengths and so thisnecessitates a study of circuits that are not necessarily rational.
Proposition 4.3.
Given any circuit ( R , . . . , R n +1 ) and a positive real number ε , there is arational circuit ( P , . . . , P n +1 ) such that | π ( g ( R , . . . , R n +1 )) − π ( g ( P , . . . , P n +1 )) | < ε. Proof.
Suppose that a circuit R has n distinct points and thus corresponds to a vertex setof an inscribed polygon with n sides. Let ε be a positive real number. There is a naturalnumber m such that the edge length ℓ of g ( m, is smaller than ε n . Take R to be a pointof g ( m, . To each point R i of R , take P i to be R i if R i is a vertex of g ( m, . Otherwise,take P i to be the first vertex of g ( m, that is counterclockwise from R i . With such a choiceof P i , the length ℓ ( P i R i ) is less than ℓ . The circuit P that is equal to ( P , . . . , P n +1 ) is arational circuit. Furthermore, (cid:12)(cid:12)(cid:12) ℓ ( P i P i +1 ) − ℓ ( R i R i +1 ) (cid:12)(cid:12)(cid:12) < ℓ < εn for each natural number i in [1 , k + 1] . Therefore, | π ( g ( P )) − π ( g ( R )) | < ε. (cid:3) Corollary.
For any circuits ( P , . . . , P k +1 ) and ( Q , . . . , Q l +1 ) on C , min( P , . . . , P k +1 ) > max( Q , . . . , Q l +1 ) = ⇒ π ( Q , . . . , Q l +1 ) ≥ π ( P , . . . , P k +1 ) . Proof.
Suppose that P is the circuit ( P , . . . , P k +1 ) and that Q is the circuit ( Q , . . . , Q l +1 ) .Suppose further that min( P , . . . , P k +1 ) is greater than max( Q , . . . , Q l +1 ) . For any positivereal number ε , there are rational circuits ˆ P and ˆ Q such that | π ( g ( ˆ P )) − π ( g ( P )) | < ε | π ( g ( ˆ Q )) − π ( g ( Q )) | < ε π and furthermore such that min( ˆ P , . . . , ˆ P k +1 ) > max( ˆ Q , . . . , ˆ Q l +1 ) . We leave the straightforward details of the proof to the reader. Proposition 4.3 implies that π ( ˆ Q ) is greater than π ( ˆ P ) , and so π ( Q ) + ε > π ( P ) . Since the above inequality holds for any positive ε , π ( Q ) is greater than or equal to π ( P ) . (cid:3) Given any two points on C , there is an m large enough so that the arc between the twopoints contains two vertices of g ( m, . The following lemma follows from this fact and thefact that rotations preserve the ordering of points on C . Lemma 4.1.
Given points P , Q , and R on C in counterclockwise order, there is a point S between Q and R such that ℓ ( P S ) is a rational length. Establish the following notation for the statements and proofs of Lemma 4.2 and Lemma 4.3.Suppose that A , B and C are points on C in counterclockwise order and the arc from A to C is less than half of C . For any points X and Y on C , denote by P X,Y the intersection ofthe lines tangent to C at X and Y (Figure 14 ). P A,Q P A,B PP B,C
O AQBCP
A,C
Figure 14
Lemma 4.2.
There is an order preserving bijection from the points on the line segment P A,B P A,C and the points on the arc between B and C .Proof. Suppose P is a point on P A,C P B,C . Let C P be the circle of radius ℓ ( CP ) centered at P . The circle C p intersects C at precisely two points, at the point C and at a point Q thatlies on the arc from A to B . Define the circles C P B,C and C P A,C in the same way as C P . Thecircles C P B,C , C P , and C P A,C all intersect at C and, since they have different radii, they can ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π meet at no more than two points. Since OC is tangent to all three circles, the three circlesmeet only at C . The radius of C P A,C is greater than the radius of C P , which is greater thanthe radius of C P B,C , therefore ℓ ( BC ) < ℓ ( QC ) < ℓ ( AC ) . Since all three points lie counterclockwise from A on an arc less that half of a circle, Q isclockwise from B and counterclockwise from A .Suppose that Q is a point between A and B on the arc from A to B , that B is clockwisefrom C , and that the arc from A to C is less than half of a circle. Let L be the line tangentto C at Q . Since ∠ QOC is less than a straight line, L intersects the line tangent to C at C at a point P and the intersection occurs on the same side of C as P A,C and P B,C . The point Q is between A and B , and so ℓ ( BC ) < ℓ ( QC ) < ℓ ( AC ) , hence m ( ∠ BOC ) < m ( ∠ QOC ) < m ( ∠ AOC ) . The line segments BO , QO , and AO are all radii, implying the equality of ℓ ( BO ) , ℓ ( QO ) ,and ℓ ( CO ) and, therefore, the inequalities ℓ ( CP B,C ) < ℓ ( CP ) < ℓ ( CP A,C ) . The point P therefore lies on the line segment P A,C P B,C . Note that Lemma 2.1 implies that
P Q and CP are congruent and so the map we initially constructed will map P to the point Q .Let φ be the map taking points on P A,C P B,C to points on the arc from A to B and let ψ be the map taking points on the arc from A and B to point on P A,C P B,C . These functionsare inverses of each other, and so both are invertible, hence bijective. (cid:3)
Lemma 4.3.
Let ε be a positive real number. There is a Q on the arc from A to B suchthat Q is not equal to B and (cid:12)(cid:12) ℓ ( AP A,B ) + ℓ ( P A,B B ) + ℓ ( BP B,C ) + ℓ ( P B,C C ) − (cid:0) ℓ ( AP A,Q ) + ℓ ( P A,Q Q ) + ℓ ( QP C,Q ) + ℓ ( P C,Q C ) (cid:1)(cid:12)(cid:12) < ε. (4.4) Furthermore, any point Q ′ on the arc from Q to B will also satisfy the above estimate where Q is replaced by Q ′ .Proof. Pick a point z on AP A,B such that ℓ ( zP A,B ) is less than ε . Lemma 4.2 guarantees theexistence of a point Z on the arc between A and B such that z is the point of intersectionof the line tangent to C at Z and the line segment AP A,B . Pick a point q on zP A,B suchthat ℓ ( qP A,B ) is less than ε . Lemma 4.2 guarantees the existence of a point Q on the arcfrom Z to B such that q is the point of intersection of the line tangent to C at Q and theline segment AP A,B . Using the established notation, q is the point P A,Q and, furthermore, Q will satisfy the estimate (4.4). Since the bijection given in Lemma 4.2 is order preserving,if Q ′ is any point not equal to B in the arc from Q to B , then the point Q ′ will also satisfythe estimate (4.4). (cid:3) π Proposition 4.4.
Suppose that ( P , . . . , P k +1 ) is a circuit with the property that the arcbetween any adjacent points is less than a fourth of a circle. Suppose ε is a positive realnumber. There is a rational circuit ( Q , . . . , Q l +1 ) such that | π ( G ( P , . . . , P k +1 )) − π ( G ( Q , . . . , Q l +1 )) | < ε. Proof.
Set Q equal to P . Suppose that ( Q , Q , . . . , Q i , P i +1 , . . . , P k +1 ) is a sequence ofpoints where each Q i lies on the arc between P i and P i +1 , each Q i lies on the same regularpolygon, each edge is less than a quarter of the circle, and | π ( G ( P , P , . . . , P i , P i +1 , . . . , P k +1 )) − π ( G ( Q , Q , . . . , Q i , P i +1 , . . . , P k +1 )) | < ε. Lemma 4.3 implies that there is a Q i +1 not equal to P i +1 but on the arc between Q i +1 and P i +2 such that for any Q on the arc between Q i +1 and P i +2 , | π ( G ( P , P , . . . , P i , P i +1 , . . . , P k +1 )) − π ( G ( Q , Q , . . . , Q i , Q, P i +2 , . . . , P k +1 )) | < ε. Take Q to be a vertex of a regular polygon on which Q through Q i lie by choosing the regularpolygon to be a sufficiently fine refinement of the regular polygon on which Q through Q i are vertices. Recursively construct in this way a polygon G ( Q , . . . , Q k +1 ) , with Q and Q k +1 both equal to P , which approximates G ( P , . . . , P k +1 ) in the sense that | π ( G ( P , . . . , P k +1 )) − π ( G ( Q , . . . , Q k +1 )) | < ε. (cid:3) Corollary. If ( P , . . . , P k +1 ) and ( Q , . . . , Q l +1 ) are two arbitrary sequences of points on C then min( P , . . . , P k +1 ) > max( Q , . . . , Q l +1 ) implies that π ( G ( Q , . . . , Q l +1 )) ≤ π ( G ( P , . . . , P k +1 )) . Proof.
The above corollary is proved exactly as the first corollary to Proposition 4.3 is provedabove except that the inequalities are reversed because they are reversed in Proposition 4.2. (cid:3)
Theorem 5.
Suppose that P is a general partition of C . For any positive real number ε there is a positive real number δ such that µ ( P ) < δ implies that < π − π ( g ( P )) < ε and < π ( G ( P )) − π < ε. Proof.
The theorem is proved in the same way as Theorem 4 but by appealing to Proposi-tion 4.2 and the corollary to Proposition 4.4 rather than Proposition 4.2 and the corollaryto Proposition 4.3. (cid:3)
ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π Let P be a general partition of C . Denote by α ( g ( P )) the area of the inscribed polygon g ( P ) and α ( G ( P )) the area of the circumscribed polygon G ( P ) . Theorem 6.
For any positive real number ε there is a positive real number δ such that µ ( P ) < δ implies that < π − α ( g ( P )) < ε and < α ( G ( P )) − π < ε. Proof.
Let ε be a positive real number. Theorem 5 implies that there is a positive realnumber δ so that if P is a circuit ( P , . . . , P n +1 ) with mesh less than δ , then < π − π ( g ( P )) < ε and < π ( G ( P )) − π < ε. (4.5)Restrict δ so that π δ is less than ε . The equality α ( G ( P )) = 12 π ( G ( P )) together with (4.5) implies that < α ( G ( P )) − π = 12 π ( G ( P )) − π < π + ε − π = ε . Denote by ℓ i the length of the segment P i P i +1 . Heron’s Theorem implies that α ( g ( P )) = n X i =1 ℓ i r − ℓ i . For each i , ℓ i (cid:18) − µ ( P ) (cid:19) ≤ ℓ i (cid:18) − ℓ i (cid:19) ≤ ℓ i r − ℓ i ≤ ℓ i . Since the sum of all of the ℓ i ’s is the perimeter of an inscribed polygon and therefore lessthan π , the bound on δ implies that π ( g ( P ))2 − ε < π ( g ( P ))2 − µ ( P ) n X i =1 ℓ i ≤ α ( g ( P )) ≤ π ( g ( P ))2 . This proves the theorem. (cid:3) Application to the Circular Trigonometric Functions
Arclength of an Arc and Area of a Sector.
Suppose that A and B are points on C and that A is the counterclockwise oriented arc from A to B . If the length ℓ ( AB ) is arational length, then for some natural number m , there is a regular inscribed n -gon g (0 , n ) such that A has on it exactly k + 1 vertices of g (0 , n ) and both A and B are vertices of g (0 , n ) . Denote by ℓ n the edge length of g (0 , n ) to obtain the equality kℓ n = kn π ( g (0 , n )) . π The inscribed polygon g ( m, n ) will have m n + 1 vertices on A . Denote by π m,n ( A ) theapproximation of the arclength of A given by π m,n ( A ) = 2 m kℓ n = 2 m k m n π ( g ( m, n )) = kn π ( g ( m, n )) → πkn as m → ∞ . This approximation of the length of an arc of C depends only on the fraction of C that A represents and so defines the length of an arc for any arc whose endpoints form a rationallength. Define similarly the area of a sector of C , the area bounded by the lines OA and OB and the counterclockwise oriented arc A , so that α m,n ( A ) = kn α ( g ( m, n )) → πkn as m → ∞ . If the length ℓ ( AB ) is not a rational length, then there is a sequence of rational lengthsthat approximate it. In particular, take A to be a point on g (0 , n ) for some n where n is alarge enough natural number so that at least one vertex of g (0 , n ) lies on A between A and B . For each m , choose the vertex B m of g ( m, n ) that is closest to B . The counterclockwiseoriented arc A m from A to B m will have rational length ℓ ( AB m ) and will be a fraction φ ( AB m ) of a circle, where φ ( AB m ) is an increasing bounded above sequence with lim m →∞ φ ( AB m ) = φ ( AB ) , where φ ( AB ) is an irrational number. Obtain as m tends to infinity the limits π m,n ( A m ) = 2 πφ ( AB m ) → πφ ( AB ) and α m,n ( A m ) = πφ ( AB m ) → πφ ( AB m ) , where the limits are independent of the sequence ( B m ) that approximates B .Using the notation above and given any counterclockwise arc A from a point X on C toa point Y on C , define the angle measure θ ( A ) by θ ( A ) = 2 πφ ( XY ) . The length θ is the length of the arc A . The area, α ( A ) of the sector bounded by OX , OY ,and the counterclockwise oriented arc A is given by α ( A ) = θ ( A )2 . Limit of the Sine Function.
The argument frequently given by authors of calculustexts for calculating the limit (1.1) is perfectly valid, although it does require some expla-nation since we have not yet introduced a coordinate system nor defined the trigonometricfunctions. View the unit circle as the subset C of the plane given by C = { ( x, y ) : x + y = 1 } . Define the trigonometric functions, as is customary, as functions of the argument θ in [0 , π ) ,so that the point (cos( θ ) , sin( θ )) is the point on the unit circle so that the counterclockwiseoriented arc A from (1 , to (cos( θ ) , sin( θ )) has length θ . The area of the sector defined by ODERNIZING ARCHIMEDES’ CONSTRUCTION OF π A is θ . Take θ to be less that π . Bound the area of the sector above and below respectivelyby the areas of the triangles △ (0 , , (cid:16) , sin( θ )cos( θ ) (cid:17) and △ (0 , θ ) , θ ) , sin( θ )) to obtain the inequality
12 cos( θ ) sin( θ ) ≤ θ ≤
12 sin( θ )cos( θ ) implying that ≤ θ sin( θ ) ≤ θ ) . Take limits and use the sandwich theorem to obtain the limit (1.1) as a one sided limit. If ( x, y ) is a point in the first quadrant so that the signed length of the arc from (1 , to ( x, y ) is θ , then define the signed length of the arc from (1 , to ( x, − y ) to be − θ . Given thissigned argument, the sine function is an odd function of θ and the cosine function is even.Extending the definitions of the trigonometric functions in this way to negative signed arclengths permits the limit (1.1) to be viewed as a two sided limit and the oddness of the sinefunction implies this two sided limit. While this is a standard approach to calculating thelimit in many calculus texts, [4] for example, the approach is non-rigorous because thesetexts omit a discussion of the equivalence of π defined as half the circumference of a unitcircle and as the area of a unit circle and so as well the relationship between the length ofan arc and the area of the sector that the arc defines. Our current discussion closely followsArchimedes and remedies this shortcoming. References [1] Apostol, T. M.:
Calculus . Blaisdell, Waltham, Mass., (1967). (Referred to on page 2.)[2] Heath, T. L.:
The Works of Archimedes , Cambridge University Press (1897). (Referred to on page 1.)[3] Richman, F.:
A Circular Argument . The College Mathematics Journal, Vol. 24, No. 2. (Mar., 1993), pp.160-162. (Referred to on page 2.)[4] Stewart, J.:
Calculus . Brooks Cole; 6 edition (2007). (Referred to on page 25.)[5] Unger, P.: Reviews, American Mathematical Monthly 93 (1986) 221–230. (Referred to on page 2.)[6] Vietoris, L.:
Vom Grenzwert lim sin x . Elem. Math. 12 (1957), pp. 8–10. (Referred to on page 2.)[7] Zeisel, H.: lim x → sin xx and the definition of ππ