by PerCLupi
Thu Jul 3rd, 2008 at 05:58:40 PM EST
The Measurement of the Circle is a little book by Archimedes. This has only three propositions. It seems to be a summary for teaching purposes and therefore the real book was lost. In addition, it has come to us with many problems.
In the Proposition 1, in a very intuitive way, Archimedes establishes the equivalence between the area of a circle and the area of a rectangle triangle. After, he demonstrates the proposition by the method of reduction ad absurdum. He also used the famous method of exhaustion.
In my annotated translation, I have analysed the figures appearing in this Proposition 1. The study of the use of letters of the alphabet, which appear poorly in the figures, allows reasonably repair a defect of textual transmission and confirm how Archimedes worked.
Before offering in ET my complete edition of this little book, I propose a game to reason. I ask you that you make the analysis of these letters in the figures, which I have done to amend a bit the text. If your reasoning reaches the same conclusion that I, the test is very valuable.
[editor's note, by Migeru] Fold inserted here
I explain:
The order of the letters of the ancient Greek alphabet is as follows:
Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ Σ Τ Υ Φ Χ Ψ Ω
Looking at the double image that follows (which always appears in the manuscripts and editions, without studying),
we see clearly that there is, at least, one gap in the use of the letters:
a) They are used Α Β Γ Δ to indicate the four corners of the square inscribed in the circle.
b) The Ε is used for the rectangle triangle.
c) Ζ is used to indicate the midpoint of the arch AB
But from here, it's going to use Μ Ν Ξ Ο Π Ρ, so that the letters Η Θ Ι Κ Λ are clearly missing. There is an obvious gap. That letters after the P are missing it is debatable: Archimedes would have worked on the corner O of the square circumscribing the circle, and it would have been assumed by him that this would also apply to the other three sides of the aquare, although there might be a second gap from P until the last letter needed (Σ Τ Υ Φ Χ Ψ Ω).
The question that is raised is as follows: How can we understand that it has been done by Archimedes, for maintaining the letters that appear in manuscripts (that is, from A to Z, and then followed by M) in their places and for filling the gap mentioned (H to Λ) from Z to M?
To reason about that, we must have the actual text of Archimedes. Here goes, as it appears in editions. After, I'll explain the text with my textual proposals, obtained precisely from the reasoning on the figures and on the problem of the gap in the series of letters of the alphabet.
This is the Proposition I of The Measurement of the Circle:
ΚΥΚΛΟΥ ΜΕΤΡΗΣΙΣ
αʹ.
Πᾶς κύκλος ἴσος ἐστὶ τριγώνῳ ὀρθογωνίῳ, οὗ ἡ μὲν ἐκ τοῦ κέντρου ἴση μιᾷ τῶν περὶ τὴν ὀρθήν, ἡ δὲ περίμετρος τῇ βάσει.
Ἐχέτω ὁ ΑΒΓΔ κύκλος τριγώνῳ τῷ Ε, ὡς ὑπόκειται· λέγω ὅτι ἴσος ἐστίν.
  THE MEASUREMENT OF THE CIRCLE
1.
Each circle is equivalent to a rectangle triangle, where both the radius is equal to one side comprising the right angle, as the circumference is equal to the base.
Be the ΑΒΓΔ circle to the triangle E, as it is assumed: I say that is equal.

  
Εἰ γὰρ δυνατόν, ἔστω μείζων ὁ κύκλος, καὶ ἐγγεγράφθω τὸ ΑΓ τετράγωνον, καὶ τετμήσθωσαν αἱ περιφέρειαι δίχα, καὶ ἔστω τὰ τμήματα ἤδη ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κύκλος τοῦ τριγώνου· τὸ εὐθύγραμμον ἄρα ἔτι τοῦ τριγώνου ἐστὶ μεῖζον. Εἰλήφθω κέντρον τὸ Ν καὶ κάθετος ἡ ΝΞ· ἐλάσσων ἄρα ἡ ΝΞ τῆς τοῦ τριγώνου πλευρᾶς. Ἔστιν δὲ καὶ ἡ περίμετρος τοῦ εὐθυγράμμου τῆς λοιπῆς ἐλάττων, ἐπεὶ καὶ τῆς τοῦ κύκλου περιμέτρου· ἔλαττον ἄρα τὸ εὐθύγραμμον τοῦ Ε τριγώνου· ὅπερ ἄτοπον.   Indeed, if possible, be largest the circle, and be inscribed the ΑΓ square, and be bisected the arcs of circumference, and the segments of circles be already smaller than the excess by which the circle beyond the triangle: well then, the rectilinear figure is still major than the triangle. Be centre the N point, and be drawn the ΝΞ perpendicular: well then, the ΝΞ is less than the side of the triangle. But the perimeter of the rectilinear figure too is lower than the other side, from the moment that too it is smaller than the circumference of the circle: therefore, the rectilinear figure is less than the E triangle, which is just absurd. 
Ἔστω δὲ ὁ κύκλος, εἰ δυνατόν, ἐλάσσων τοῦ Ε τριγώνου, καὶ περιγεγράφθω τὸ τετράγωνον, καὶ τετμήσθωσαν αἱ περιφέρειαι δίχα, καὶ ἤχθωσαν ἐφαπτόμεναι διὰ τῶν σημείων· ὀρθὴ ἄρα ἡ ὑπὸ ΟΑΡ. Ἡ ΟΡ ἄρα τῆς ΜΡ ἐστὶν μείζων· ἡ γὰρ ΡΜ τῇ ΡΑ ἴση ἐστί· καὶ τὸ ΡΟΠ τρίγωνον ἄρα τοῡ ΟΖΑΜ σχήματος μεῖζόν ἐστιν ἢ τὸ ἥμισυ. Λελείφθωσαν οἱ τῷ ΠΖΑ τομεῖ ὅμοιοι ἐλάσσους τῆς ὑπεροχῆς, ᾗ ὑπερέχει τὸ Ε τοῦ ΑΒΓΔ κύκλου· ἔτι ἄρα τὸ περιγεγραμμένον εὐθύγραμμον τοῦ Ε ἐστὶν ἔλασσον· ὅπερ ἄτοπον· ἔστιν γὰρ μεῖζον, ὅτι ἡ μὲν ΝΑ ἴση ἐστὶ τῇ καθέτῳ τοῦ τριγώνου, ἡ δὲ περίμετρος μείζων ἐστὶ τῆς βάσεως τοῦ τριγώνου. Ἴσος ἄρα ὁ κύκλος τῷ Ε τριγώνῳ.   Be the circle, however, if possible, lower than the E triangle, and be circumscribed the square, and be bisected the arcs of circumference, and be drawn tangents by the points: then, it is right the ΟΑΡ angle. Well then, the ΟΡ is greater than the ΜΡ (since the ΡΜ is equal to the ΡΑ): therefore, also the ΡΟΠ triangle is greater than the half of the figure ΟΖΑΜ. Be the similars to the ΠΖΑ sector lower than the excess by which the E beyond the ΑΒΓΔ circle: Well then, in addition, the circumscribed rectilinear figure is less than the E, which is just absurd (indeed, it is major, because not only the NA is equal to the perpendicular side of the triangle, but also the perimeter is greater than the basis of triangle). Therefore, the circle is equal to E triangle. 
(This is a very literal translation: this is the text of editions and manuscripts. The figure is necesary. I don't know how it will apear, but it will apear soon. This is an advance for impatients. With my poor English, it is a boldness to do what I do here. Everything is for Europe and friends!
Notes
One problem is that the Greeks could not "measure" curves: the proceedings are finding equivalencies with "straight" forms. The intuition (intuitive rectification of the circumference) of Archimedes, in the Proposition 1, can be seen in the figure below:
If r (radius of the circle) is equal a a (height of the rectangle triangle) and c (circumference, ie "perimeter" of the circle the early designation of π comes from περί μετρος ) equals b (base of the rectangle triangle), then C (area of the circle) = T (area of the rectangle triangle). Indeed, those relationships established by hypothesis, then T = (a*b)/2 (something known) and C = (r*c)/2; then C = T. Archimedes the only thing he has to do is to demonstrate such a relationship
Proposition 1 is organized by classical manner:
1. Enunciated (πρότασις) [Each circle is equivalent to a rectangle triangle, where both the radius is equal to one side comprising the right angle, as the circumference is equal to the base.]
2. Explicit expression (ἔκθεσις) and 3. Determination of limits or conditions (διορισμός) [Be the ΑΒΓΔ circle to the triangle E, as it is assumed: I say that is equal.]
The demonstration, which it is divided into two symmetrical parts, continues. Each of the two parts consists of classic items [4. Preparation (κατασκευή)
5. Demonstration (ἀπόδειξις)
6. Partial conclusions (συμπέρασμα)]
And the proposition is closed with the classic conclusion: quod erat demonstrndum. [Therefore, the circle is equal to E triangle.]
Archimedes begins the demonstration using the "reduction ad absurdum" method, assuming the hypothesis that, the relations being maintained, is C>T. He prepares (he reasons), according to this hypothesis and using the "method of exhaustion" until that, according to the reasoning, it is found a polygon [he says "inscribed rectilinear figure": here, as a hypothesis, reaches the octagon], which is less than the C, but this is already bigger than the T (C>P>T); and, then, he demonstrates more geometrico that there is a contradiction, because C>P and T>P: then, he concludes that the hypothesis C>T is absurd.
The reasoning is:
Be C>T, relations being maintained.
It obtained a polygon (P), such that if C=T+x, then C=P+y, being x>y.
Then, C=T+x = P+y, and it is P>T.
But the geometric demonstration states that T>P, because the perimeter of the polygon is smaller than the circumference (which, by assumption, is equal to the base of the rectangle triangle) and the apothem is less than the radio (which, by hypothesis , is equal to the height of the rectangle triangle). Therefore, this [T>P] contradicts the earlier result that it is P>T. Then, it is absurd and it can not be C>T.
In the second part of the ad absurdum demonstration, Archimedes works as at first, but assuming the hypothesis that it is T>C. From circumscribed square, Archimedes follows a similar reasoning that in the first part. He reaches a similar conclusion that it is absurd that it is T>C.
The reasoning is: Be T>C, relations being maintained.
It obtained a polygon (P), such that if C+x=T, then C=P+y, being x>y.
Then, C=Tx = P+y, and it is T>P.
But the geometric demonstration states that P>T, because the perimeter of the polygon is longer than the circumference (which, by assumption, is equal to the base of the rectangle triangle) and the apothem is equal than the radio (which, by hypothesis , is equal to the height of the rectangle triangle). Therefore, this [T>P] contradicts the earlier result that it is P>T. Then, it is absurd and it can not be T>C.
And the final conclusion is: Therefore, if it is not C>T or T>C, it can only be C=T, as it was stated in the definition.
The text is corrupted. What is the "straight figure"? and other issues. Looking at the first part of the demonstration, we can see that, after speaking of inscribed square and the bisection of the arches, only establishes the point Z. And then, going to geometric demonstration was begun by the N center with the M point identified as another midpoint of the arc AΓ, without it having been said.
My question was: How Archimedes was the original drawing, to move to M, N from Z?
A scientist as Archimedes would not use the letters of arbitrary way. Is there any scientist would do this today? I do not think so.
And the study of the figure led me to establish "a just manner" for acting as a philologist. That is the game that I ask you: How do you reason about this?
The remedy of this problem we can fix some corrupt aspects of the text. The authors talk about problems with text, but they have not studied the issue from the analysis of the poor figures. And if any of you thinks the same way as me and he reaches the same conclusion, this would be a definitive test for me.
Archimedes drew in a sand box. This bronze statue of Archimedes is at the Archenhold Observatory in Berlin. It was sculpted by Gerhard Thieme and unveiled in 1972. [From Wikipaedia. Archimedes]
So our text would incorporate improvements. :D
Update [2008723 2:38:50 by PerCLupi]: Solving the problem (?):
If Archimedes had continued designating with letters from the Z, according to his explanatory process, the inscribed rectilinear figure should have to be the AZ, ZB, B<*Θ>, <*Θ>Γ, Γ<*Η>, <*H>Δ, Δ<*Ι>, and, then, the <*I> should be where the M is:
Fig. 1
Then he did not proceed in this manner. Or at least, this reasoning does not solve the problem of the gap of letters, which actually exists. If there is no mistake (and it should not, because we have consistency between text and figure, and if error had existed, we would have to assume that a copyist corrected figure, or even text and figure, for a such consistency being obtained; but this assumption should be discarded, because if someone corrects, he corrects properly, without an incongruous gap), we must look for another procedure. And there is only one plausible procedure, in my opinion. Let's see: in his sandbox, Archimedes has a drawing like this:
Fig. 2
He had drawn a circle of any radius, had inscribed a square, and had circumscribed another one (both required for his method of exhaustion), and he had drawn a rectangular triangle (obviously, from a methodological point of view, there is no reason for having the exact catheti in relation to the circle), as it is shown in Fig. 2.
The circle can now be designated with points Α, Β, Γ, Δ as usual and the triangle, with the letter Ε. Then he said that a square is inscribed, whose procedure is already done, because drawing a circle and inscribing a square in a circle are made through the same procedure: therefore, Archimedes does not have to do more than drawing straight lines joining the cutoff points of two perpendicular diameters with the circumference, and letters corresponding to the corners are the same as he used to designate the circle.
Now, the interesting point begins: immediately after, the arches of circumference, whose cords are sides of the inscribed square, must be bisected. When, at first, the square was circunscribed, the need was already done, and now the corresponding points are designated with letters: the midpoint of intersection with the circumference between A and B is designated with the letter Z, which is the letter immediately following to last used (E). When he goes to designate the midpoint of intersection between Β and Γ, he has before the lower left corner of the circumscribed square, and this corner is designated with the letter <Η>: the letter <Θ>, therefore, is used for that midpoint. When he goes to designate the midpoint of intersection between Γ and Δ, he has before the lower right corner of the circumscribed square, and this corner is designated with the letter <Ι>: the letter <Κ>, therefore, is used for that midpoint. When he goes to designate the midpoint of intersection between Δ and Α, he has before the upper right corner of the circumscribed square, and this corner is designated with the letter <Λ>: then, this midpoint must be designated with the next letter, which is precisely the M (and the M already appears in the text).
Archimedes is in the first half of the demonstration, and now he does not need the circumscribed square, so he does not designate the upper left corner of the circumscribed square. The more geometrico demonstration of the first part is what follows now. The apothem of inscribed octagon should be drawn. The center of the circle is designated precisely with the letter N, which is the next to last used (M) and the NΞ line is the apothem.
The second part of the demonstration starts with the circumscription of the square and the letters Ο, Π, and Ρ are necessary. And the problem has been solved: the letters remain in their existing locations and the gap has been reconstructed in a very credible way. The original figure is therefore this:
Fig. 3
This having been established with methodological verisimilitude, we can try certain textual proposals, based on the figure, rebuilt so solid. Without this, it is impossible to propose any conjecture sufficiently substantiated. And this is what we bring to our edition.