Measurement of a circle
Proposition 3
From Thomas L. Heath, The Works of Archimedes 

I. Let AB be the diameter of any circle, 0 its centre, AC the tangent at A; and let the angle AOC be one-third of a right angle.

 Then OA : AC = B 3 : 1 > 265 : 153..... (1),

 and OC : CA = 2 : 1 = 306 : 153.............. (2).

 First, draw OD bisecting the angle AOC and meeting AC in D.

 Now CO: OA = CD: DA, [Eucl. VI. 3]

 so that CO + OA : OA = CA: DA, or

 CO + OA : CA = OA :AD.

 Therefore [by (1) and (2)]

 OA : AD > 571 :153 .................................. (3).

 Hence OD ² : AD ² = (OA ² + AD ²) : AD ²

 > (571 ² + 153 ²) : 1 53 ²

 > 349450 :23409,

 so that OD: DA > 591 1/8 : 153........... (4).

 Secondly, let OE bisect the angle AOD, meeting AD in E.

 Then DO : OA = DE : EA,

 so that DO + OA : DA = OA : AE, 

Therefore OA : AE > (591 1/8 + 571) : 153, by (3) and (4)

           > 11 62 1/8 : 153................................ (5).

 It follows that

 OE ² : EA ² > {(1 162 1/8 ) ² + 153 ²} : 153 ²

                  > (1350534 33/64 + 23409) : 23409

                  > 1373943 33/64 : 23409.

 Thus OE : EA > 11 72 1/8 : 153......... ....(6).

 Thirdly, let OF bisect the angle AOE and meet AE in F.

 We thus obtain the result, corresponding to (3) and (5) above, that

 OA :AF > (1 162 1/8 + 11 72 1/8) : 153

             > 2334 1/4 : 153....................... (7).

 Therefore OF ² : FA ² > (2334 1/4) ² + 153 ² : 153 ²

                                 > 5472132 1/16 : 23409.

 Thus OF : FA > 2339 1/4 : 153............ (8).

 Fourthly, let OG bisect the angle AOF, meeting AF in G.

 We have then

 OA : AG > (2334 1/4 + 2339 1/4) : 153, by means of (7) and (8)

              > 4673 1/2 : 153.

 Now the angle AOC, which is one-third of a right angle, has been bisected four times, and it follows that

 angle AOG = 1/48 of a right angle.

 Make the angle AOH on the other side of OA equal to the angle AOG, and let GA produced meet OH in H.

 Then angle GOH = 1/24 of a right angle.

 Thus GH is one side of a regular polygon of 96 sides circumscribed to the given circle.

 And, since OA : AG > 4673 1/2 : 153,

 while AB = 20A and GH = 2AG,

 it follows that AB : (perimeter of polygon of 96 sides)        > 4673 1/2 : 153 X 96 

                 > 4673 1/2 : 14688.

 But (14688)/(4673 1/2) = 3 + (667 1/2)/(4673 1/2) < 3 1/7.

 Therefore the circumference of the circle (being less than the perimeter of the polygon) is a fortiori less than 3 1/7 times the diameter AB.

Explanation 

 265:153 < B 3:1 < 1351:780. 

It is probable that these were calculated by anthyphaeresis, which was known to the Greeks and which we call the method of continuous fractions.

 If a = a₀ is a magnitude, let a₀ = [ a₀] and a₁= 1/( a ₀ - a₀) , so a₁ > 1 . Let a₁ = [ a₁ ] and a₂= 1/( a₁ - a₁) . Continue and you get a sequence {a_n} of integers such that a = a₀ + (1/(a₁ + 1/(a₂ + 1/.... 

Calculating such a sequence could be done either numerically or geometrically by the Greeks, depending on how a is presented. The process is called anthyphaeresis.

 The convergents of the sequence are the successive truncations: c₀ = a₀, c₁ = a₀ + 1/a₁, c₂ = a₀ + 1/(a₁ + 1/a₂)),.... 

They have the following properties:

1.  The convergents are rationals which are alternately < and > a .
2. Those which are < a (the odds) increase and differ from alpha by less than 1/ denominator. Those which are > a (the evens) decrease and differ from alpha by less than 1/ denominator.
3. If a is rational, the sequence terminates. If a is a root of a quadratic, the sequence is periodic.

1 < 5/3 < 19/11 < 71/41 < 265/153 < B 3 < 1351/780 < 362/209 < 97/56 < 26/15 < 7/4 < 2 

so this is probably how Archimedes got his approximations. 

(An alternative derivation is given in Katz, p.122)

 Now the idea is to compute the ratio of the radius to the perimeter of an inscribed regular polygon of 96 sides.

 After this the method is to use the lower approximation to estimate the perimeter of a regular 96-gon circumscribing a circle by starting with a hexagon one side tangent to a given radius, so central angle 30 degrees, side B 3 times the radius, using the upper bound mentioned above. He then halves the central angle 4 times, getting 12-, 24-, 48- and 96-gons. At each step, the side is calculated using Euclid's theorem that when the vertex angle of a triangle is bisected, the ratio in which the base is divided is the ratio of the corresponding sides; together with Pythagoras' Theorem. Since the perimeter of the circumscribing 96-gon : diameter < 3 1/7 , so also is circumference of circle : diameter , 3 1/7 .

 Next A. constructs an inscribed regular 96-gon by starting with a hexagon and halving the central angle 4 times, at each step computing a lower bound for the ratio perimeter : diameter by the same two theorems and using the lower bound for root 3 , eventually proving that the perimeter of the inscribed 96-gon : diameter > 3 10/71 , so also is circumference of circle to diameter > 3 10/71 .

 For details of the proof, click here.

Archimedes' Quadrature of the parabola
From Thomas L. Heath, The Works of Archimedes 

 If Qq be the base of a segment of a parabola, and V the middle point.of Qq, and if the diameter through V meet the curve in P, then P is the vertex of the segment.

 For Qq is parallel to the tangent at P [Prop. 1]. Therefore, of all the perpendiculars which can be drawn from points on the segment to the base Qq, that from P is the greatest. Hence, by the definition, P is the vertex of the segment.

 If Qq be a chord of a parabola bisected in V by the diameter PV, and if RM be a diameter bisecting QV in M, and RW be the ordinate from R to PV, then

 PV= 4RM.

 For, by the property of the parabola,

 PV : PW= QV²: RW²

            = 4RW² : RW²,

 so that    PV=4PW,

 whence    PV= 4/3 RM. 

 If Qq be the base, and P the vertex, of a parabolic segment, then the triangle PQq is greater than half the segment PQq.

 For the chord Qq is parallel to the tangent at P, and the triangle PQq is half the parallelogram formed by Qq, the tangent at P, and the diameters through Q, q'.

 Therefore the triangle PQq is greater than half the segment.

 Corollary. It follows that it is possible to inscribe in the segment a polygon such that the segments left over are together less than any assigned area.

 If Qq be the base, and P the vertex, of any parabolic segment, and if R be the vertex of the segment cut off by PQ, then

         triangle PQq=8 triangle PRQ.

 The diameter through R will bisect the.chord PQ, and therefore also QV, where PV is the diameter bisecting Qq. Let the diameter through R bisect PQ in Y and QV in M. Join PM.

 By Prop. 19, PV = 4/3 RM.

     Also PV=2YM.

    Therefore YM=2RY,

     and triangle PQM=2 triangle PRQ.

     Hence triangle PQV = 4 triangle PRQ,

     and triangle PQq = 8 triangle PRQ.

 Also, if RW, the ordinate from R to PV, be produced to meet the curve again in r, RW = rW,

 and the same proof shows that,

     triangle PQq = 8 triangle Prq. 

 If there be a series of areas A, B, C, D,... each of which is four times the next in order, and if the largest, A, be equal to the triangle PQq inscribed in a parabolic segment PQq and having the same base with it and equal height, then

     (A+B+C+D+...) < (area of segment PQq).

 For, since triangle PQq = 8 triangle PRQ = 8 triangle Pqr, where R, r are the vertices of the segments cut off by PQ and Pq as in the last proposition,

     triangle PQq = 4(triangle PQR + triangle Pqr).

 Therefore since     triangle PQq = A,

     triangle PQR + triangle Pqr = B.

 In like manner we prove that the triangles similarly inscribed in the remaining segments are together equal to the area C, and so on.

 Therefore A + B + C + D + ... is equal to the area of a certain inscribed polygon, and is therefore less than the area of the segment. 

 Given a series of areas.A, B, C, D,..., Z, of which A is the greatest and each is equal, to four times the next in order, then

         A + B + C + ...+ Z + 1/3 Z = 4/3 Z.

 Take areas b, c, d, ... such that

         b = 1/3 B,

         c = 1/3 C,

         d = 1/3 D, and so on.

 Then, since b = 1/3 B,

 and     B = 1/4 A,

         B + b = 1/3 A.

 Similarly,     C + c = 1/3 B.

         and so on.

 Therefore 
     B + C + D + ... + Z + b + c + d + ... +z = 
     1/3(A + B + C + ... + Y).

 But b + c + d + ... + y = l/3(B + C + D + ... + Y).

 Therefore by subtraction, 

        B + C + D+ ...+ Z + z = 1/3 A.

    or A + B + C + ... + Z + 1/3 Z = 4/3 A. 

 Every segment bounded by a parabola and a chord Qq is equal to four-thirds of the triangle which has the same base as the segment and equal height.

 Suppose     K= 4/3 triangle PQq,

 where P is the vertex of the segment; and we have then to prove.that the area of the segment is equal to K.

 For, if the segment be not equal to K, it must either be greater or less.

 1. Suppose the area of the segment greater than K.

 If then we inscribe in the segments cut off by PQ, Pq triangles which have the same base and equal height, i.e. triangles with the same vertices R, r as those.of the segments, and if in the remaining.segments we inscribe. triangles in the same manner, and so on, we shall finally have segments remaining whose sum is less than the area by which the segment PQq exceeds K.

 Therefore the polygon so formed must be greater than the area K; which is impossible, since [Prop. 23]

         A + B + C +...+ Z < 4/3 A,

 where         A = triangle PQq.

 Thus the area of the segment cannot be greater than K.

 2. Suppose, if possible, that the area ofthe segment is less than K.

 If then triangle PQq = A, B = 1/4 A, C = 1/4 B, and so on, until we arrive at an area X such that X is less than the difference between K and the segment, we have

         A + B + C + ...+ X + I/3 X = 4/3 A [Prop. 23]

             = K.

 Now, since K exceeds A + B + C +...+ X by an area less than X, and the area of the segment by an area greater than X, it follows that

         A + B + C+ ... +X > (the segment);

 which is impossible, by Prop. 22 above.

 Hence the segment is not less than K.

 Thus, since the segment is neither greater.nor less than. K,

     (area of segment PQq) = K = 4/3 triangle PQq.

Explanation 

1.  Vertex B of largest inscribed triangle lies on line parallel to vertex through midpoint of chord. 
2. The tangent at B is parallel to chord. His proof is geometric, but we can see this by analytic geometry.
3. If you cut off the maximum triangle from the remaining segment continually, the area of new triangle is 1/4 area of preceding triangles. 
4. If you add to a magnitude A continually 1/4 A, at each step you get 4/3 A - remainder where remainder is 1/3 of the last added magnitude. We would prove this by induction, starting with 5/4 = 4/3 - 1/12.

Proofs

1.  Suppose parabola is y = x² and chord is y = cx + d . Then height of inscribed triangle is h(x) = cx + d - x². So h'(x) = c + 2x = 0 if and only if x = c/2 . On the other hand the intersections of chord and parabola are c/2 plus or minus discriminant/2, so maximum height occurs at midpoint of chord.
2. "Symptoms of the parabola" are the Greek versions of analytic geometry. They state that if parallel chords A Gamma and T Phi meet the chord bisectors parallel to the vertex at Delta and Pi and meet the parabola at B, then B Delta B Pi = (A Delta)² : (T Pi)².

3. Construct an inscribed polygon as follows:

 Construct the largest triangle, and on each of the other two sides as chord, construct the largest inscribed triangle and so on. On the n-th step there are 2ⁿ chords; on each construct the largest inscribed triangle, giving 2ⁿ triangles and 2ⁿ⁺¹ new chords.

 4. At each stage, the area of the new triangles constructed is 1/4 times the area of the previously constructed traingles [Check]. Hence if the area of the original triangle is A, then the area constructed at stage n is: A(n) = A(1+1/4+(1/4) ² + ... (1/4)ⁿ) = A(4/3 - 1/3(1/4)ⁿ). [Check]

 [4/3 - 1/3(1/4)ⁿ = {1-(1/4)ⁿ⁺¹}/ 3/4 ].

 5. By the usual Method of Exhaustion, the area of the parabolic segment is not > 4/3 A and not < 4/3 A , so it must = 4/3 A .

Archimedes also proved this result as a demonstration of how mechanical methods, i.e. centres of gravity can be used to prove geometrical theorems. For this proof, click here.

 For more on Archimedes, click here.

Exercise

Prove that if P is a point on a parabola, T is the tangent at P and C is any chord parallel to T, then the line through P parallel to the axis of the parabola meets C at its mid--point.