Euclid and pi

Pi, also know as Archimede's constant, is not mentioned in Euclid's Elements. The closest that Euclid comes is Proposition II in Book XII which states that two circles are to each other as the squares of their diameters.

Postulate 1: Law of Trichotomy

For any two values x,y, there are only three possible states:
(a) x = y
(b) x is less than y
(c) x is greater than y

This is one of the postulates of real numbers. See here for details on constructing real numbers.

Lemma 1: Similar polygons inscribed in circles are to one another as the squares on their diameters





























Proof:

(1) By assumption, polygon ABCDE is similar to polygon FGHKL with BM and GN being the diameters of circles.

(2) From the property of similar polygon (see definition above), we have:

∠ BAE ≅ ∠ GFL

BA/AE = GF/GL

(3) triangle ABE is equiangular to triangle FGL [See Lemma 1 here for details]

(4) Since they are equiangular, we know that:

∠ AEB ≅ ∠ FLG

(5) Since both angles open on the same length of the circumference (see here):

∠ AEB ≅ ∠ AMB
∠ FLG ≅ ∠ FNG

(6) From (4) and (5), we can conclude that:

∠ AMB ≅ ∠ FNG

(7) Since BM and GN are both diameters of the circle (see here),
we can conclude that both ∠ BAM and ∠ GFN are right angles.

(8) Since the angles of a triangle add up to 180 degrees (see Lemma 4 here), we can conclude from step #6 and step #7 that triangle ABM is equiangular to triangle FGN.

(9) From the properties of equiangular triangles (see Lemma 3 here), we know that:
BM/GN = BA/GF

(10) From (9), we can conclude that:
(BM/GN)² = (BA/GF)² = BM²/GN² = BA²/GF²

(11) From similar polygons (See Theorem here), we know that:
(Area of ABCDE)/(Area of FGHKL) = BA²/GF²

(12) Putting this all together, gives us:
BM²/GN² = (Area of ABCDE)/(Area of FGHKL)

QED

Lemma 2: if A/B = C/D with A greater than C, then D is less than B.

Proof:

(1) Let A/B = C/D with A greater than C.

(2) So that AD = BC

(3) Now, D ≠ B since if D = B, then AD is greater than BC which contradicts step #2.

(4) Now, D cannot be greater than B since then AD is greater than BC which contradicts step #2.

(5) So, by the Law of Trichotomy (see Postulate above), we can conclude that D is less than B.

QED


Theorem: Two circles are to each other as the squares of their diameters.













































Proof:

(1) Let C₁ be the circle formed with diameter BD and area A₁.

(2) Let C₂ be the circle formed with diameter FH and area A₂.

(3) Assume that A₁/A₂ ≠ (BD)²/(FH)²

(4) There exists an area S such that: (BD)²/(FH)² = A₁/S

(5) Assume that S is less than A₂

(6) Then, there exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion, see Lemma 2.]

(7) We can inscribe a similar polygon into circle C₁ [See here for details on this construction]

(8) From Lemma 1 above, we can conclude:

BD²/FH² = (Area polygon AOBPCQDR)/(Area polygon EKFLGMHN)

(9) But then, from step #4:

BD²/FH²= A₁/S

(10) So we can conclude that:

(Area polygon AOBPCQDR)/(Area polygon EKFLGMNH) = A₁/S

(11) Since A₁ is greater than Area polygon AOBPCQDR, we can conclude from step #15 that S is greater than Area polygon EKFLGMN from Lemma 3 above.

(12) But this is impossible since in step #6 we showed that S is less than this same regular polygon so we have a contradiction and we reject our assumption in step #5.

(13) Now, let's assume that S is greater than A₂

(14) So this means that (FH)²/(BD)² = S/A₁

(15) Let T be the area such that S/A₁ = A₂/T

(16) We can see that T is less than A₁ from Lemma 2 above.

(17) There exists a regular polygon that is greater in area than T but smaller than the area of the circle by the Method of Exhaustion (see Lemma 2)

(18) We can inscribe a similar polygon in A₂.

(19) From Lemma 1 above, we can conclude:

FH²/BD² = (Area polygon EKFLGMHN)/(Area polygon AOBPCQDR)

(20) Likewise from step #14, we have:
(FH)²/(BD)² = S/A₁

(21) And from step #15, this means that:

(Area polygon EKFLGMNH)/(Area polygon AOBPCQDR) = A₂/T

(22) Now A₂ is greater in area than polygon EKFLGMNH so that T must be greater than the area of polygon AOBPCQDR

(23) But this is impossible from step #17 so we have a contradiction and we reject step #13.

(29) We now apply the Law of Trichotomy (see Postulate above) and we are done.

QED

References

• Phil Schultz, Eudoxus's Proof that the area of a circle is a constant times diameter squared
• David Joyce, Euclid's Elements