Modernizing Archimedes’ Construction of π

: In his famous work, “Measurement of a Circle,” Archimedes described a procedure for measuring both the circumference of a circle and the area it bounds. Implicit in his work is the idea that his procedure deﬁnes these quantities. Modern approaches for deﬁning π eschew his method and instead use arguments that are easier to justify, but they involve ideas that are not elementary. This paper makes Archimedes’ measurement procedure rigorous from a modern perspective. In so doing, it brings a rigorous and geometric treatment of the differential properties of the trigonometric functions into the purview of an introductory calculus course.


Introduction
Archimedes' estimate of the value of π in "Measurement of a Circle" is one of his greatest achievements [1]. His approach anticipated foundational ideas of modern analysis as formulated in the 19th and early 20th centuries, making this work one of the first major analytical achievements. Archimedes implicitly defined the circumference of a unit circle axiomatically as a number always greater than the perimeters of approximating inscribed polygons and always less than the perimeters of approximating circumscribed polygons, where the approximating polygons are regular refinements of a regular hexagon. By calculating the perimeters of a regular inscribed and circumscribed polygon of 96 sides, he arrived at the famous estimate: He also found upper and lower bounds for the area of a disk, respectively given by the areas of regular refinements of circumscribed and inscribed squares. He showed that the area of the disk should be the area of a right triangle with one leg having a length equal to the circumference 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 both on the method of refinement and on the type of polygon used in the first stage of the approximation presents a difficulty from a modern point of view. The circumference should 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 accessible to a contemporary of Archimedes for both aesthetic and practical reasons. A practical consequence is that such a modernization cleanly and efficiently resolves the principle difficulty that arises in developing the infinitesimal theory of the trigonometric functions, the calculation lim x→0 sin(x) Gearhart and Schultz discuss in [2], the sinc function, its connection to a variety of topics, and some of its many applications that give important motivation for the independent study of (1). Unger notes in [3] some difficulties in the approaches to treating this limit that are common in differential calculus textbooks. Richman also points out in [4] some of these difficulties and directs the reader to the approach of Apostol [5], which avoids common logically circular arguments by using twice the area of a sector rather than arc length as the argument of the trigonometric functions. This is a natural point of view, especially when considering the direct analogy with the hyperbolic trigonometric functions. Although Apostol's approach is compelling, it remains desirable to connect this nonstandard way of defining the trigonometric functions with the standard approach. Moreover, Apostol defines the area of a sector rather than deriving it from a limiting procedure using inscribed and circumscribed polygons, so there is still some work to be done to connect the viewpoints.
There are several approaches to computing (1) in the literature. Authors commonly use an analytical approach to solve the problem by defining the trigonometric functions as power series or as solutions to a system of differential equations. Some authors begin by defining the inverse of the tangent function as an integral. Unfortunately, these approaches lack geometric motivation. Cipra studied in [6] the derivatives of the sine and cosine functions by considering the properties of integrals of these functions. He then determined the circumference of the unit circle by using the formula for the length of an arc of a parameterized planar curve. However, the argument that he uses for the sine and cosine function is the arc length, so the length of an arc of a circle still requires a geometric definition independent of his approach.
In [7], Vietoris used the sum of angles formula for the trigonometric functions to prove (1) directly. The notion of an arbitrary fraction of a circle requires a group structure 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 circle that a particular arc represents. However, taking the multiple to be the length of an arc requires a definition of the length of an arc. Defining and computing the length of the arc is the primary difficulty. If one follows Vietoris, one must define π in some way independent of it being the area of the unit circle, or half its circumference. If one uses the approach of Zeisel in [8] that follows Vietoris'-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 the multiple of the fraction in the argument to be 2π will imply the limit (1). However, without some further work, even this elegant approach of Zeisel does not establish the meaning of the area and circumference of the unit circle or their relationship. In fact, one must still show that the limit of the areas he discusses actually exists.
The study of the properties of inscribed and circumscribed polygons and their generalizations are still of current interest, and Apostol and Mnatsakanian present in [9] some interesting results about circumgons, where the notion of a circumgon generalizes the notion of a circumscribed polygon. The novelty of the current study is our modernization of Archimedes' approach to approximating the circle by inscribed and circumscribed polygons. Using only basic euclidean geometry, we prove our main technical result: for any arc A that is less than half of a circle, if n is greater than m and if n and m are the lengths of chords that respectively subtend arcs that are an nth and an mth of A , then m n > n m .
We prove an analogous but reverse inequality for lengths corresponding to edges of circumscribed polygons. These inequalities imply that the circle is a rectifiable curve. They, furthermore, give the classical relationship between the circumference of a circle and the area that it bounds in a way that is both rigorous and accessible to freshman calculus students. Using these inequalities, we show that 2π is the limiting value of any sequence of perimeters of approximating inscribed or circumscribed polygons. This approach is a simple modernization of the approach of Archimedes and makes rigorous the usual geometric arguments in calculus textbooks that prove (1). It is also elementary enough to be accessible, at least in principle, to a freshman calculus student.

Approximation by Regular 2 m n-Gons
Denote by C the unit circle centered at O. Unless otherwise specified, n will be in N ≥3 , the set of natural numbers greater than or equal to three, and m will be in N 0 , the natural numbers with zero. All sequences henceforth indexed by m will be indexed over the set N 0 .

Regular Inscribed and Circumscribed Polygons
Definition 1. An inscribed polygon of C is a simple convex polygon all of whose vertices lie on C . An inscribed polygon of C is regular if all of its edges are of equal length.

Definition 2.
A circumscribed polygon of C is a simple convex polygon all of whose edges intersect C tangentially. A circumscribed polygon of C is regular if each of its edges intersects C tangentially at its midpoint and all of its edges are of equal length.

Remark 1.
The assumption of convexity in the above definitions is redundant, as is the assumption in the case of circumscribed polygons that all edges have the same length. We leave the verification of these assertions to the reader as an exercise.
Denote respectively by g(m, n) and G(m, n) the regular inscribed and circumscribed polygons with 2 m n edges. Up to congruency, there is only one such inscribed and one such circumscribed polygon for each choice of m and n. Denote respectively by n (m) and L n (m) the edge length of g(m, n) and G(m, n). The respective perimeters of g(m, n) and G(m, n) are p n (m) and P n (m), where p n (m) = 2 m n 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 POQ and AOB. The respective 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 1. For fixed n, the sequences (p n (m)) and (a n (m)) are both increasing and the 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 the statement of the lemma.

Lemma 1.
Suppose that C and B are points on C so that the counterclockwise oriented arc 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 ∠AFB 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. Denote by 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 ∠AMD is congruent to ∠EMA and both are right angles.

Area and Perimeter Bounds
While Proposition 1 guarantees for fixed n the strict monotonicity of (p n (m)), (a n (m)), (P n (m)) and (A n (m)), it does not guarantee that the sequences are bounded. The following proposition provides the desired bounds.

Proposition 2.
For each n and each m, 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), and since (P n (m)) and (A n (m)) are decreasing and bounded below by p n (0) and a n (0), all four sequences are convergent, implying Proposition 3.

Proposition 3.
For each 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 3 gives for each fixed n respective limiting values for the areas and perimeters of regular refinements of regular inscribed and circumscribed polygons with n edges. It does not prove the equality of the respective limits.
Theorem 1 (Heron's Theorem). If T is a triangle with side lengths a, b, c and A(T) is the area of T, then For any points A and B in the plane, denote by (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 4. For each n,
(i) a n = 1 2 p n , (ii) A n = 1 2 P n , (iii) P n = p n , and (iv) A n = a n .
Proof. Since the sequences (p n (m)) and (P n (m)) are both convergent and are respectively equal to (2 m n n (m)) and (2 m nL n (m)), both n (m) and L n (m) tend to zero as m tends to infinity. Heron's theorem implies that and so a n (m) = 2 m n 1 2 n (m) 1 − 1 4 n (m) 2 = 1 2 p n (m) 1 − 1 4 n (m) 2 → 1 2 p n .
Since C is a unit circle, α n (m) is equal to 1 2 L n (m) and so A n (m) = 2 m nα n (m) = 2 m−1 nL n (m) = 1 2 P n (m) → 1 2 P n .
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 (AG) is equal to 1 2 L n (m). Take N to be a point of C so that AN and NH are edges of g(m + 1, n). The intersection, C, of lines tangent to C at A and N is a vertex of G(m + 1, n). Take P to be the intersection of AH with OP and M to be the intersection of AN with OC. Let I be the point of C that intersects OC. Lemma 1 implies that ∠GPA and ∠CMA 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 adjacent vertices of G(m + 2, n). Angle ∠GN A is obtuse because ∠GNC is right; hence Line segments BI, I J, and JN are congruent as half edges of G(m + 2, n). The hypotenuse of the right triangle JNL is JL, so (JL) is greater than (BI). Let K be a point on JL such that JK 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 (BK) and 3 (BI) together imply that (KE) is equal to 3 (IC). Since ∠OAC is right, ∠OCG is obtuse and so ∠LFG is as well, implying that (LG) is greater than (LF). Since (BL) is greater than (BK), The triangle inequality implies that h n (m) + n (m + 1) is greater than 1 2 L n (m); hence and so 0 < 2 m nL n (m) − 2 m+1 n n (m + 1) = P n (m) − p n (m + 1) < 2n 2 3 m h n (0).
The estimate (5) implies that 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, Proposition 4 part (iv) follows immediately from (3) and (4).

Edge Length Comparison Theorems
Comparing the edge lengths of inscribed and circumscribed segments corresponding to different regular subdivisions of an arc is the key to proving that the definition of π is independent of any approximation scheme. This section will present such a comparison.

Definition 3.
A (counter) clockwise oriented partition P of an arc A is a finite sequence of points of A that is (counter)clockwise ordered, and the first and last points of P are the endpoints 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 and refer to Figure 3 for the discussion that follows.
O P n+1 P n P m+1 P m P 2 P 1 A B Figure 3. Comparing edge lengths associated to different numbers of congruent segments.
Suppose 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 1 , . . . , P n+1 ) be a regular clockwise oriented partition of A . Let B be the point of intersection of the lines tangent to C at P 1 and P n+1 and let A be the point of intersection of the lines tangent to C at P 1 and P M+1 . Denote by m , n , L m , and L n the respective lengths of P 1 P m+1 , P 1 P n+1 , P 1 B and P 1 A. We will show that n m > m n and nL m < mL n .

General Symmetry Considerations for Polygonal Segments
Suppose that A, B, C, and D belong to a collection of points that form a regular partition of A ( Figure 4), that A and B are adjacent, that C and D are adjacent, and that B and C lie on the same side of the line AD and opposite the side where O lies. Suppose furthermore that the quadruple (A, B, C, D) forms a clockwise oriented partition of A . Standard arguments using the SSS theorem imply Proposition 5, whose proof is left as an exercise. Refer to Figure 6. 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, Lemma 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 and SSS theorems along with Proposition 5 and Lemma 1 imply Proposition 6, whose proof is left to the reader as a standard exercise. Figure 6. An isosceles trapezoid that defines a segment of a circumscribed polygon. Proposition 6. 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 .
Suppose that A is an arc of C that is less than half of C and suppose that (P 1 , . . . , P n ) is a regular clockwise orientated partition with n greater than 3 ( Figure 7). For each natural number i in [1, n], let L i be the line tangent to A and intersecting P i . Since A is less than half of a circle, Lemma 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 the next section is to prove that Proposition 7. Suppose that n is in N ≥3 and (P 1 , . . . , P n ) is a regular clockwise oriented partition 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 1,n−1 P 1,n P n,2 ) = (180 − nθ) • ; (2) m(∠P 2,n−1 P 1,n−1 P 1,n ) = m(∠P n,1 P n,2 P 2,n−1 ) = (n − 1)θ • ; (3) m(∠P n,2 P 2,n−1 P 1,n−1 ) = 180 • − (n − 2)θ • .
Of course, an analogous argument proves the same result for ∠P 2,n−1 P n,2 P n , implying (2), although another argument is not necessary in light of Proposition 6. Finally, since the sum of angles of a quadrilateral is 360 • , (3) follows from (1) and (2).

Length Estimates for Inscribed Polygonal Segments
Let A again be an arc of a unit circle that is less than half of the circle. Suppose that (P 1 , . . . , P n+1 ) is a regular clockwise oriented partition of A (Figures 8 and 9). For each P i where i is a natural number in [2, n], there is a line perpendicular to P 1 P n+1 that intersects P i at a point q i on P 1 P n+1 .  Figure 8 shows an example where n is even and Figure 9 shows the qualitative difference in an odd example. Set q 1 to be equal to P 1 and let q n+1 be equal to P n+1 . For each i between 1 and n + 1, let q i be the point on P 1 P n+1 so that P i q i is perpendicular to P 1 P n+1 .
Proof. Proposition 5 implies that each of the line segments P i P n+2−i is parallel to P 1 P n+1 . Since P i q i and P i P n+2−i are perpendicular for i in 1, n 2 , Let X 1 be equal to q 2 , and for each i in 2, n 2 , 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−1 be equal to q n , and for each i in n 2 + 1, n , let X i be the point of intersection of P i+2 P n−i and P i+1 q i+1 . For each i in 1, n 2 , Proposition 5 implies that (P i X i ) is equal to (X n−i P n+2−i ), which is equal to (q i q i+1 ) and (q n+1−i q n+2−i ).
Let i and j be in 1, n 2 with i less than j. To show that (P i X i ) is less than (P j X j ), it suffices to show that if i and i + 1 are in 1, n 2 , then (P i X i ) is less than (P i+1 X i+1 ). Extend the line segment P i P i+1 to a ray, R 1 , originating at P i . Extend the line segment q i+2 P i+2 to a ray, R 2 , that meets R 1 at a point A. The ray R 2 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 (P i X i ) is less than (P i+1 X i+1 ). Since i and j are integers, the statement of the proposition is different when n is odd and when n is even.
Note that if n is odd, then the above arguments show that q n+1 2 q n+1 2 +1 is the longest of all of the segments q i q +1 .
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 1. Let k be the largest integer less than or equal to n 2 . If n is greater than 2 and s is a natural number in (1, k], then (q 1 q s+1 ) < (P 1 P n+1 ) n s.

Remark 2.
The above corollary implies that the average length of a segment of q 1 q k+1 is less than the average length of a segment of q 1 q n 1 if the arc between P 1 and P k+1 is less than half of 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 1 , . . . , P n+1 ) is a regular clockwise oriented partition of A . Denote by n the length of the line segment P 1 P n+1 , and by m the length of P 1 P m+1 . If m is less than n, then n m > m n .
Proof. Assume first that 2m is greater than n and refer to Figure 10. Let Q be the point of intersection of the circle of radius m centered at P 1 with the line segment P 1 P n+1 . The length of P 1 Q is equal to m . If s the length of the line segment QP n+1 , then There are n edges with vertices on the arc between P 1 and P n+1 . There are m + 1 vertices of the regular subdivision between P 1 and P m+1 on A . Since 2m is greater than n, P m+1 lies clockwise from the intersection, M, of a line bisecting P 1 P n+1 and the circle. For the same reason, the point P n−m lies counterclockwise from M.
Proposition 8 implies that the average lengths of the segments q 1 q 2 , q 2 q 3 , . . . , q n−m q n−m is the same as the average lengths of the segments q n+1 q n , q n q n−1 , . . . , q 2+m q 1+m . However, for any 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 the line P 1 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 Since m is the length of the hypotenuse of a triangle with a leg of length n − d and m is equal to n − s, the length s is less than d and so Since, m + s is equal to n , m + n − m n n > n , and so n m > m m .
Suppose that 2m is equal to n. Since the sum of the lengths of two sides of a non-degenerate triangle are greater than the length of the third, 2 m is greater than 2m , and so In the case when 2m is less than n, there is a natural number k with 2 k m < n ≤ 2 k+1 m which together with (6) implies that n 2 k m > 2 k m n .
The inequality n m > m n .
follows from applying the above argument k − 1 more times.

Length Estimates for Circumscribed Polygonal Segments
Use the notation of Proposition 7 and refer to Figure 11 for the statement and proof of the proposition below. Figure 11. The apex of the intersection of circumscribed edges and their refinements.
Proof. Suppose that i is a natural number in [3, n]. Proposition 6 implies that the line P 1,i−1 P 2,i−2 is a perpendicular bisector of P 1,i−2 P 2,i−1 . Denote by A this intersection. To prove the proposition, it suffices to show that ∠P 1,i P 2,i−1 P 1,i−2 is greater than a right angle. In this case, there is a line that intersects P 2,i−1 and is perpendicular to P 1,i−2 P 2,i−1 that intersects P 1,i−2 P 1,i at a point B. The triangles P 1,i−2 AP 1,i−1 and P 1,i−2 P 2,i−1 B are similar, but since P 1,i−2 P 2,i−1 is twice the length of P 1,i−2 A, P 1,i−2 B is twice the length of P 1,i−2 P 1,i−1 , and so (P 1,i−1 P 1,i ) > (P 1,i−1 B) = (P 1,i−2 P 1,i−1 ). Proposition 6 implies the congruency of the angles ∠P 1,i P 2,i−1 P 1,i−1 and ∠P i,2 P 2,i−1 P 1,i . Proposition 7 implies that Proposition 6 implies that P 2,i−2 P 1,i−1 is perpendicular to P 1,i−2 P 2,i−1 and furthermore that P 2,i−2 P 1,i−1 bisects ∠P 2,i−1 P 2,i−2 P 1,i−2 . Therefore, ∠P 1,i−2 P 2,i−1 P 2,i−2 and ∠P 2,i−2 P 1,i−2 P 2,i−1 are congruent. Proposition 7 implies that Consider the case where i is 2 ( Figure 12). As in the argument above, the key is to show that Figure 12. The intersection of circumscribed edges associated to an arc with two segments.

Theorem 3.
Suppose that n is a natural number greater than two and (P 1 , . . . , P n+1 ) is a regular clockwise orientated partition of an arc A . If L n and L m respectively denote the length of the line segment P 1 P 1,n+1 and the length of the line segment P 1 P m+1 , then m < n implies that nL m < mL n .
Proof. The lengths of the segments P 1,i P 1,i+1 are increasing in i, so the average length of a segment of L n is larger than the average length of a segment of L m . Therefore, implying the desired result.

Circumference and Area Are Intrinsic
Definition 4. A circuit is a finite sequence of points (P 1 , . . . , P n+1 ) on C that is counterclockwise ordered and with the property that P 1 is equal to P n+1 (see Figure 13).
Without loss in generality and in order to simplify the exposition, assume that the arc connecting adjacent points of a circuit is less than half of a circle.

Definition 5.
A refinement of a circuit S 1 is a circuit S 2 such that, as functions on finite sets, the range of S 1 is a subset of the range of S 2 .
Denote by S(C ) the set of all circuits on C . Suppose that (P 1 , . . . , P n+1 ) is in S(C ). The inscribed polygon g(P 1 , . . . , P n+1 ) is the polygon whose vertices are the points in the circuit and the circumscribed polygon G(P 1 , . . . , P n+1 ) is the polygon whose vertices are the points given by the intersections of the lines tangent to C at adjacent points of the circuit. Definition 6. For any polygon P, denote by π(P) the perimeter of P. Let min(P 1 , . . . , P n+1 ) denote the minimum edge length of g(P 1 , . . . , P n+1 ) and denote by max(P 1 , . . . , P n+1 ) the maximal edge length of g(P 1 , . . . , P n+1 ). Figure 13. Inscribed and circumscribed polygons associated to a circuit.

Approximation by Rational Circuits
Let the length be less than 2. Construct recursively a sequence of points and a corresponding piecewise linear path in the following way. Take P 1 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 (P i P i+1 ) is equal to and the triangle OP i P i+1 is counterclockwise oriented. The circle C ,n of radius with center P n intersects C in exactly two places. Let P n+1 be the first of these points counterclockwise from P n . The line segment P n P n+1 is of length . If the sequence has the property that for some k, P k is equal to P 1 , then is a rational length and there is a first such k that is the numerator of , denoted by N ( ). The closed piecewise linear path Γ( ) is the finite sequence of edges Γ( ) = P 1 P 2 , . . . , P N ( )−1 P N ( ) .
Let OP 1 be the line from the origin to the point P 1 . Denote by D( ), the denominator of , the number of times the lines P i P i+1 intersect the line segment OP 1 , where i in (1, N ( )]. The denominator counts how many times Γ( ) wraps around the origin. A length is integral if D( ) is equal to 1, implying that is the edge length of a regular inscribed polygon. A circuit is said to be integral if all of its edge lengths are integral. A regular circuit is an integral circuit in which the distance between any two adjacent points in the circuit is the same. The counterclockwise 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 regular circumscribed polygon with C . By assumption, both lengths are less than a diameter. Theorem 2 therefore implies that and this proves the proposition. The case of circumscribed polygons is similar except that instead of comparing the lengths and m, compare the lengths L and M and appeal to Theorem 3 to obtain the reverse inequality.
Proof. Denote by i and m i the lengths i = (P i P i+1 ) and m j = (Q j Q j+1 ).
Abuse notation and denote again by i and m i the respective segments of length i and m i . There are k segments 1 , . . . , k for g(P 1 , . . . , P k+1 ) and n segments m 1 , . . . , m n for g(Q 1 , . . . , Q n+1 ). Let 1 be the shortest of the segments of g(P 1 , . . . , P k+1 ) and m n be the longest of the segments of g(Q 1 , . . . , Q n+1 ). Take Λ to be the product and take Λ copies of both polygons. Compute the respective perimeters of these polygons to obtain Λπ(g(P 1 , . . . , P k+1 )) = Λ( 1 + 2 + · · · + k ) The inequality follows from Theorem 2 since 1 is the minimum length of the segments of g(P 1 , . . . , P k+1 ). The ultimate equality follows from the fact that To justify the last assertion, note that given Λ copies of the polygon g(P 1 , . . . , P k+1 ), the segments 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 i ; hence, Λ N ( i ) copies each path Γ( i ) which wraps around C exactly D( i ) times. Summing each of these Λ N ( i ) D( i ) wrappings gives the total number of wrappings, Λ, implying the above formula for the sum.
Similarly, use the reverse inequality and make use of Proposition 10 to obtain the inequality 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 to the appropriate inequality in Proposition 10, reversing the inequalities analogous to those given above.

Definition 7.
Following Archimedes, define 2π to be equal to p 3 (as defined in Proposition 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 ).
Let P be a rational circuit. Take δ to be a positive real number less than the side length of g(m , 3), so that the maximum possible side length of G(P ) is less than the side length of G(m , 3) and the maximum possible side length of g(P ) is less than the edge length of g(m , 3). Since µ(P ) is less than δ, Proposition 11 implies that π(g(m , 3)) < π(g(P )) and π (G(m , 3)) > π(G(P )).
Take m to be a large enough natural number so that the edge length of π(g(m , 3)) is smaller than the smallest edge length of g(P ) and the edge length of π (G(m , 3)) is smaller than the smallest edge length of G(P ). Such a choice of m is possible because the side lengths of both π (g(m, 3)) and π (G(m, 3)) tend to zero as m tends to infinity. Proposition 11 implies that π(g(P )) < π(g(m , 3)) < 2π and π(G(P )) > π (G(m , 3)) > 2π.
Denote by π(n) the perimeter of a regular n-gon inscribed in C and denote by Π(n) the perimeter of a regular circumscribed n-gon.
Corollary 2 (Corollary to Proposition 11). If n is greater than m and m is greater than 2, then (1) π(n) > π(m) and (2) Π(n) < Π(m). 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 3.

Corollary 3 (Corollary to
Remark 4. The next section presents results somewhat more general than those of the above theorem. 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 next section, we use a cardinality argument to show that not all edge lengths are rational lengths, and so this section relies on a consideration that is highly unlikely to have been considered by a contemporary of Archimedes, but is necessary from a modern perspective.

Approximation by General Circuits
Each rational length corresponds to a pair of two integers, a numerator and a denominator, making the set of rational lengths a countably infinite set. However, there are uncountably many possible edge lengths and so this necessitates a study of circuits that are not necessarily rational.
Proof. Suppose that a circuit R has n distinct points and thus corresponds to a vertex set of an inscribed polygon with n sides. Let ε be a positive real number. There is a natural number m such that the edge length of g(m, 3) is smaller than ε 2n . Take R 1 to be a point of g(m, 3). To each point R i of R, take P i to be R i if R i is a vertex of g(m, 3). Otherwise, take P i to be the first vertex of g(m, 3) that is counterclockwise from R i . With such a choice of P i , the length (P i R i ) is less than . The circuit P that is equal to (P 1 , . . . , P n+1 ) is a rational circuit. Furthermore, for each natural number i in [1, k + 1], and so |π(g(P )) − π(g(R))| < ε.
We leave the straightforward details of the proof to the reader. Proposition 12 implies that π(Q) is greater than π(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 ).
Given any two points on C , there is an m large enough so that the arc between the two points contains two vertices of g(m, 3). The following lemma follows from this fact and the fact that rotations preserve the ordering of points on C . Lemma 2. Given points P, Q, and R on C in counterclockwise order, there is a point S between Q and R such that (PS) is a rational length.
Establish the following notation for the statements and proofs of Lemma 3 and Lemma 4. 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 of the lines tangent to C at X and Y (Figure 14).

Lemma 3.
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 (CP) centered at P. The circle C p intersects C at precisely two points, at the point C and at a point Q that lies 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 . The circles C P B,C , C P , and C P A,C all intersect at C, and since they have different radii, they can meet at no more than two points. Since OC is tangent to all three circles, the three circles meet only at C. The radius of C P A,C is greater than the radius of C P , which is greater than the radius of C P B,C , therefore (BC) < (QC) < (AC).
Since all three points lie counterclockwise from A on an arc less that half of a circle, Q is clockwise 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 clockwise from C, and that the arc from A to C is less than half of a circle. Let L be the line tangent to 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 (BC) < (QC) < (AC), hence m(∠BOC) < m(∠QOC) < m(∠AOC).
The point P therefore lies on the line segment P A,C P B,C . Note that Lemma 1 implies that PQ 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 functions are inverses of each other, and so both are invertible, and hence bijective.
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 (zP A,B ) is less than ε 4 . Lemma 3 guarantees the existence of a point Z on the arc between A and B such that z is the point of intersection of the line tangent to C at Z and the line segment AP A,B . Pick a point q on zP A,B such that (qP A,B ) is less than ε 4 . Lemma 3 guarantees the existence of a point Q on the arc from Z to B such that q is the point of intersection of the line tangent to C at Q and the line segment AP A,B . Using the established notation, q is the point P A,Q , and furthermore, Q will satisfy the estimate (10). Since the bijection given in Lemma 3 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 satisfy the estimate (10).
Proof. The above corollary is proved exactly as the first corollary to Proposition 12 is proved above except that the inequalities are reversed because they are reversed in Proposition 11. 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 0 < 2π − π(g(P )) < ε and 0 < π(G(P )) − 2π < ε.
Proof. The theorem is proved in the same way as Theorem 4 but by appealing to Proposition 11 and the corollary to Proposition 13 rather than Proposition 11 and the corollary to Proposition 12.
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 0 < π − α(g(P )) < ε and 0 < α(G(P )) − π < ε.
Denote by i the length of the segment P i P i+1 . Heron's Theorem implies that α(g(P )) = n ∑ i=1 i 2 1 − 2 i 4 .
Using the notation above and given any counterclockwise arc A from a point X on C to a point Y on C , define the angle measure θ(A ) by θ(A ) = 2πφ(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

Limit of the Sine Function
The argument frequently given by authors of calculus texts for calculating the limit (1) is perfectly valid, although it does require some explanation since we have not yet introduced a coordinate system nor defined the trigonometric functions. View the unit circle as the subset C of the plane given by Define the trigonometric functions, as is customary, as functions of the argument θ in [0, 2π), so that the point (cos(θ), sin(θ)) is the point on the unit circle so that the counterclockwise oriented arc A from (1, 0) to (cos(θ), sin(θ)) has length θ. The area of the sector defined by A is θ 2 . Take θ to be less that π 2 . Bound the area of the sector above and below respectively by the areas of the triangles (0, 0)(1, 0) 1, sin(θ) cos(θ) and (0, 0)(cos(θ), 0)(cos(θ), sin(θ)) to obtain the inequality 1 2 cos(θ) sin(θ) ≤ 1 2 θ ≤ 1 2 sin(θ) cos(θ) implying that 1 ≤ θ sin(θ) ≤ 1 cos(θ) .
Take limits and use the sandwich theorem to obtain the limit (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, 0) to (x, y) is θ, then define the signed length of the arc from (1, 0) to (x, −y) to be −θ. Given this signed 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 arc lengths permits the limit (1) to be viewed as a two sided limit and the oddness of the sine function implies this two sided limit. While this is a standard approach to calculating the limit in many calculus texts- [10], for example-the approach is non-rigorous because these texts omit a discussion of the equivalence of π defined as half the circumference of a unit circle and as the area of a unit circle, and so as well the relationship between the length of an arc and the area of the sector that the arc defines. Our current discussion closely follows Archimedes and remedies this shortcoming.
Funding: This research received no external funding.