meno 87a

[Constantinus Philo]

εἰ μέν ἐστιν τοῦτο τὸ χωρίον τοιοῦτον οἷον παρὰ τὴν δοθεῖσαν αὐτοῦ γραμμὴν παρατείναντα ἐλλείπειν τοιούτῳ χωρίῳ οἷον ἂν αὐτὸ τὸ παρατεταμένον ᾖ, Is παρατείναντα n pl acc part ao active?

[mwh]

I don’t see how it can be, but I don’t see how it could be masculine either. I just don’t understand it. I think I’d just about understand παρατεῖναν (intransitive), but this, no, or not with my current headache. But I’ve always had problems squaring a circle.

Does your commentary help?

[Constantinus Philo]

the comment says: si est ejusmodi ut qui illud applivaverit deficiat. Does it help? I start guessing that παρατείναντα should agree with τις. Maybe

[mwh]

But τις is not there, and παρατείναντα would’t agree with it if it did. I presume applivaverit should be applicaverit, but I can’t reconcile this with the Greek, nor do I understand it. It appears to take παρατείναντα as masculine (and transitive, as the following τὸ παρατεταμένον suggests it will be), but in that case, as I said, I don’t get the syntax. I expect I’m being thickheaded but right now I’m baffled.

[Constantinus Philo]

the whole comment is too long and so much convoluted that i can hardly understand anything. I am retyping it to the best of my ability: Next , to the case of parateinanta: if the acc is retained, elleipein must be taken with a personal subject, ' you have a deficiency'. By a similar personalization we say 'I am short of money'. this view is taken by Buttman, who translates: si est ejusmodi ut qui illud applicaverit deficiat. Bennecke follows him observing that if Plato had written οιον... παρατεταμενον ελλειπειν the construction would have been smooth enough, but he wanted παρατεταμενον in the next line and desired to avoid repeating the word. Plato therefore changed the construction to the active. I believe that Plato chose the acc rather than the dat or the gen because it was distinctly masculine and could by no possibility be taken to agree with any other word in the sentence. We have here the principle that may be called 'dissimilation of cases' which prof. Jebb well illustrates by his note on Soph. Ant., 546: μηδ' ἃ μὴ 'θιγες ποιοῦ σεαυτῆς. He explains the unusual acc with θιγγάνω by saying that ὧν woul have been intolerable on account of the second genitive after ποιοῦ . ...Another possibility is that a word has fallen out, as παρατείναντα - εὑρεῖν - ἐλλείπειν. Does this help in any way?

[Barry Hofstetter]

LSJ takes it as transitive citing this usage, if that helps:

4. apply a figure to a straight line, Pl.Men.87a; abs., Id.R.527a.

Liddell, H. G., Scott, R., Jones, H. S., & McKenzie, R. (1996). A Greek-English lexicon (p. 1327). Oxford: Clarendon Press.

[Constantinus Philo]

ok I will risk my translation: if there exists this such figure, such that the one who extends its given line, leaves out by such a space as is the space extended itself, there is, it seems to me, one consequence, and again, another one, if it is impossible for these things to happen.

[mwh]

I still can’t make sense of this. But ἐλλείπειν, like παρατείναντα, is surely intransitive and won’t be referring to a person (pace Buttmann?); the figure (or whatever it is) falls short of reaching (the line?); or something of the kind. But παρατείναντα is grammatically baffling, and I can’t help suspecting the text.

Thanks Barry for the LSJ reference. I only wish it helped. Being ἀγεωμέτρητος I wouldn’t qualify for entrance to the Academy.

PS. A fearsomely difficult papyrus commentary (P.Oxy.1808) on the fearsomely difficult Rep.546 refers to this Meno passage and draws a diagram, apparently of (a corner of) a square constructed on the diagonal of a smaller square, the figure that Socrates is evidently generating here.

[Constantinus Philo]

I have checked Jacob Klein's commentary on Meno, his translation of the passage is almost identical with mine,

[jeidsath]

λέγω δὲ τὸ ἐξ ὑποθέσεως ὧδε, ὥσπερ οἱ γεωμέτραι πολλάκις σκοποῦνται, ἐπειδάν τις ἔρηται αὐτούς, οἷον περὶ χωρίου, εἰ οἷόν τε ἐς τόνδε τὸν κύκλον τόδε τὸ χωρίον τρίγωνον ἐνταθῆναι, εἴποι ἄν τις ὅτι “Οὔπω οἶδα εἰ ἔστιν τοῦτο τοιοῦτον, ἀλλ’ ὥσπερ μέν τινα ὑπόθεσιν προὔργου οἶμαι ἔχειν πρὸς τὸ πρᾶγμα τοιάνδε· εἰ μέν ἐστιν τοῦτο τὸ χωρίον τοιοῦτον οἷον παρὰ τὴν δοθεῖσαν αὐτοῦ γραμμὴν παρατείναντα ἐλλείπειν τοιούτῳ χωρίῳ οἷον ἂν αὐτὸ τὸ παρατεταμένον ᾖ, ἄλλο τι συμβαίνειν μοι δοκεῖ, καὶ ἄλλο αὖ, εἰ ἀδύνατόν ἐστιν ταῦτα παθεῖν.

εἰ οἷόν τε ἐς τόνδε τὸν κύκλον τόδε τὸ χωρίον τρίγωνον ἐνταθῆναι
That seems simple enough. "If it's possible to inscribe the following area into the following circle as a triangle."

παρὰ τὴν δοθεῖσαν αὐτοῦ γραμμὴν
This must refer to the diameter of the circle. There's no other line given.

εἰ μέν ἐστιν τοῦτο τὸ χωρίον τοιοῦτον οἷον παρὰ τὴν δοθεῖσαν αὐτοῦ γραμμὴν παρατείναντα ἐλλείπειν τοιούτῳ χωρίῳ οἷον ἂν αὐτὸ τὸ παρατεταμένον ᾖ, ἄλλο τι συμβαίνειν μοι δοκεῖ, καὶ ἄλλο αὖ, εἰ ἀδύνατόν ἐστιν ταῦτα παθεῖν.
If one the one hand this area [the given area to be inscribed] is such that along the given line [diameter of the circle] constructions [the triangle(s), the rectangle(s)] lack such an area as would be the same as the construction [the rectangle], it falls out one way, and another if that is impossible.

There would be a number of ways to construct triangles and rectangles here to pursue the problem. It's possible that παρατείναντα means a single triangle and a single rectangle, but it could also mean multiple triangles (the most obvious would be the two right triangles along the diameter) or multiple rectangles (several ways to lay this out, including a rectangle split by the diameter).

I can take a look at Euclid to see if there's a similar problem, but I don't think he's being super exact and "παρατείναντα" as neuter plural makes sense to me as the above, especially if it refers to two right triangles.

[Constantinus Philo]

Euclid uses παραβάλλω instead of παρατείνω . Ι have found Stork's comment in the internet archive where he says τινα must be added here, all other commentators agree that it is masculine.

[jeidsath]

Also, the largest triangle that can be inscribed in a circle is a equilateral triangle, which is made up of three isosceles triangles, which split into two 30/60/90 triangles. Each of those have sides:

r/2 : sqrt(3) * r / 2 : r

and area (1/2) * b * h:

r^2 * sqrt(3) / 4

Total area for the equilateral triangle is:

(3/2) * sqrt(3) * r^2 (about 83% of the circle)

Which is the answer for us algebraic types.

Geometrically, the question is going to be answered by the proportion of the rectangle's width * height to (3/2) * sqrt(3) * r^2. If it's larger, then the triangle can't be inscribed, if it's less, then it can. If you were doing this like an Athenian mathematician, you might construct the equilateral triangle, find the 30/60/90 component and construct the rectangle on the 30/60/90 triangle and do your comparisons that way, but there's probably a much simpler way to go about it.

[Constantinus Philo]

most comments say that here Socrates refers to a problem of building a triangle in the circle whose surface (the triangle's) is equal to a given rectangle built on the diameter. I dont see however how they get it all.

[jeidsath]

a given rectangle built on the diameter

Hmm. If this is a "construct the triangle" problem that could be reasonable. But if it's an existence problem (as stated), any rectangle that intersects the circle on its far side is trivial. The interesting case is when the height gets just beyond 1.5 times the radius of the circle.

[Constantinus Philo]

well im at my wits end here

[jeidsath]

I was thinking about this during the sermon today and realized that I had forgot to divide by two. The maximum rectangle along the diameter will actually intersect the arc. The equilateral triangle will have a height of 3/2 * r, and a base of sqrt(3) * r. The equivalent rectangle by area with a length of 2*r would have a height of (3/8) * sqrt(3) * r.

Any rectangle with a length of 2 * r and height less than or equal to (3/8) * sqrt(3) * r can be inscribed on a circle with radius r. Anything larger can't. (You can verify that an equilateral triangle is the largest possible isosceles triangle in a circle by trigonometry and a derivative. But it seems obvious enough, and I imagine that there is a geometric proof.)

If I have some time this evening, I'll try to see if I can come up with a construction. It feels like it should be possible.

[jeidsath]

The most general construction that I came up with begins with a diameter of a circle A B, and with a rectangle drawn on that circle with corners A B C D. First construct an arc of the circle that is 8/3rds the height of the rectangle (AC = BD). There are many ways to do this. Then construct a rectangle on that arc whose height is 3/2ths the radius of the circle. Again, there are many ways to do this. Connect the 8/3rds arc with one of the places that the new rectangle intersects the circle, and you have your inscribed triangle with the correct area.

In the most extreme case, the new rectangle will only intersect the circle once, and give you an equilateral triangle. Any larger than that and it means that your initial rectangle is too big to inscribe. The equilateral limit case is the reason to use 8/3rds. However, in my opinion, this "magic number" based approach is contrary to the spirit of this sort of proof. Of course, there would be various ways to hide what you are doing with fancy triangles to get the trisection and make it look good, but I really can't imagine an Athenian geometrician coming up with this approach.

Trying to create a construction based on an isosceles triangle fails. The height winds up being a 4th power polynomial on the given rectangle's height. x^4 + 2*r*x^3 - 2*r^3*x + 4*r^2*h^2 - r^4 = 0. Where h is the height of the rectangle, r is the radius of the circle, and r + x is the height of the isosceles triangle. Hellenistic mathematicians might have been able to deal with that, but not Athenian.

Unless you use the magical 8/3rds height in my initial construction (or get it as a limit, as in the isosceles triangle approach) it's easy to generate methods that will work for some range of rectangle sizes but not for others. Some of these less general methods may require only a few steps to work, but fail when the rectangle grows beyond a certain size. This, I think, is the sort of method that Plato refers to in the quote.

I don't have any new insight on παρατείναντα, except to say there there will likely be multiple steps/constructions required for the full inscription of the triangle, and neuter plural seems like a reasonable way to refer to them.

[Barry Hofstetter]

I used to think that the boat building passage in the Odyssey was the most annoying passage in ancient literature (Od. 5.234-53), followed closely by Caesar's description of building a bridge (DBG 4.17), and for students of the New Testament, the shipwreck described in Acts 27. This passage, however, is an order of magnitude higher on the annoyance scale.

[seanjonesbw]

Then construct a rectangle on that arc whose height is 3/2ths the radius of the circle.

Joel, you should win some kind of special achievement award for this.

I've spent a very enjoyable five minutes trying to work out how I should say "3/2ths".

[jeidsath]

I used to think that the boat building passage in the Odyssey was the most annoying passage in ancient literature (Od. 5.234-53)

I've been teaching myself to sail on a small handmade wooden boat the last few weeks, and have found myself wishing several times that the poet had spent more time describing the rigging.

This passage, however, is an order of magnitude higher on the annoyance scale.

What is important here, I think, is whether the annoyance is recollected or newly discovered.