What I’ve done is – well, okay, start with that trisected segment; it’s two inches from A to B, two inches from B to C, two inches from C to D. Then straightedge a segment out from the center point of AD (which is V, at the three-inch mark) to what you’re calling CC. Then straightedge a segment out from the center point of AD to what I guess you’d call BB. Then straightedge segments connecting D to CC, and CC to BB, and BB to A, such that you should be looking at, uh, the top half of a hexagon, if you see where I’m going with this.
A to BB? That’s three inches. BB to CC? Also three inches. CC to D? Also three inches.
That’s if AVD is a 180-degree angle. If it’s anything less than 180, then A to BB (and BB to CC, and CC to D) should be less than three inches but more than two inches. The angle at ADCC should, of course, get smaller and smaller likewise: it’s like you’re watching the half-a-hexagon get flatter and flatter while the sides get shorter and shorter.
Well, I started working it out numerically for applying your method to 90 degrees instead, and what I get is that you end up with approximately 30.3013 degrees, instead of exactly 30. So it looks like it comes close, which is why it seemed to work, but doesn’t hit right on.
I suppose I could figure out what happens for 60 degrees as well, but I’m, at the moment, lazy. I’ll grant you that it works perfectly for 180 degrees, though.
Well, for the sake of argument I’d grant that (a) it’s off a bit at 90 degrees, and that (b) it works perfectly at 180. But if you do get past any momentary laziness, I wouldn’t mind hearing three or so other scraps of information from you:
1) Does it also work just about perfectly at just about zero?
As the angle gets ludicrously small, as the vertex gets farther and farther away from segment AD, would you agree that it’ll eventually be such that A to BB to CC to D will be virtually indistinguishable from A to B to C to D? If so, then
2) By your calculations, to what extent is it off for, say, a 120-degree angle?
-and-
3) By your calculations, to what extent is it off for, say, a 60-degree angle?
If, for the sake of argument, you’re right – well, then I’m obviously not competent to work those out; when I try, I get results that look correct. But if you’re capable of running it for a couple of other values, then it’d all come down to whether you find the arc gets proportionately more off as it approaches some key point, and then gets proportionately less off after that. If so, then it’s just a matter of finding a slightly better point to locate E; we just need a somewhat bigger circle, and we can find it. But if not, then, yes, the whole method is fatally flawed – which means you can disprove it, or correct it, in short order.
First you should understand that your whole method is certainly fatally flawed, in the sense that you can’t do any finite compass-and-straightedge construction to exactly trisect an angle. This isn’t just a matter of not being clever enough; it can’t be done. Results that “look correct” are meaningless to a true trisection.
If all you want is to get close, then you can add more steps and get as close as you like, but this is no more a “trisection” in the sense the geometers are talking about than is just eyeballing it. If you want you can subdivide the segment VA into a million equal pieces, transfer these lengths onto a normal, use that to estimate the sine of the angle, and look up the result in a trig table. It’s still not good enough for a mathematician.
As for numerical results, here’s an image showing the way I understand your construction to work, along with a table of its error for angles from 0° to 180°. (Note that your method, as I understand it, actually gives a nonzero error at 180°.)
That’s what the link would seem to indicate – but then why are Indistinguishable and I both coming up with no error for 180? (And what is that a link to, exactly? Did you just whip that up, or is that something that already existed? And if it’s the latter, do you likewise have a handy link to an image of the correct arc?)
I just made the picture and table. I don’t know why my results at 180° don’t agree with yours. Let’s see if I understand. You draw a horizontal line 12 inches long. A is at 0in, B at 2in, V at 3in, C at 4in, D at 6in, and E at 12in along this line. Now you draw a circle of radius 3in centered at V and a circle of radius 8in centered at E, intersecting at X (or CC, in up-thread notation), right?
This gives you a triangle with sides EV=9, VX=3, and EX=8. The law of cosines gives cos <XVE=13/27, which is not quite 1/2, so the angle is a little bigger than 60°.
It occurs to me, staring at my algebra, that I slightly messed up the calculation for applying your method to a 90 degree angle. Alas. Also, I was wrong when I said things worked out exactly for 180 degrees; I misconstructed E when verifying that. Not a great track record, eh?
But, all the same, to reiterate what Omphaloskeptic is saying, if exact trisection is what you are after, then, yes, your method is fatally flawed. Sure, there will be some point you can put E where the circle will be big enough to ensure the intersection will be at the trisection, but that is nearly empty tautology; finding such a point will remain difficult, by which I mean, not possible with a compass and a straightedge. As I thought you were yourself aware of from the OP, general compass-and-straightedge trisection is impossible [not just that no one has figured out how to do it, but that it’s been proven that it cannot be done].
I’m curious as to what point that is, though. If there is a circle that would correctly sweep through all the intersection points, then, yes, I suppose it would be impossible to locate such a point with compass and straightedge alone; the reason why it’s impossible to trisect an angle (with compass and straightedge) would follow from why it’s impossible to locate that point (with compass and straightedge).
In roundabout fashion, I suppose it’s what I was just saying above: we can explain why it’s impossible, but a diagram marking the proper location for E would show why it’s impossible. But so long as I was figuring that the proper location for “E” was a matter of doubling the length of AD, the diagram was showing a point that of course can be located with a straightedge and a compass, which is incorrect.
That’s right. The problem is, I didn’t start off with that; I guess I started off by putting the compass spike on D, the pencil on A, and sweeping out; putting the compass spike on A, the pencil on D, and sweeping out; putting the spike on V, putting the pencil on A, and sweeping out; and then putting the spike on V, putting the pencil on D, and sweeping out. I’d figured that straightedging from V to each of those intersection points would generate 60-degree angles, and then figured that the circle from E would sweep from C to CC or B to BB.
The impossibility of constructing such an E is not going to be any more directly apparent from describing it in some other manner than the impossibility of carrying out the trisection already is. If there is a fundamental stumbling block, it’s the impossibility of constructing a length ratio of cos(20 degrees) with compass and straightedge. But none of these immediately leap out pictorially with “Aha! Now I see why it can’t be done”; so far as I can see, you’re not going to be able to apprehend the impossibility of any construction with directly geometric intuition (what kind of geometric intuition would suggest that some point could not be reached with compass and straightedge?). Instead, for the classical impossibility results, you’ll need to delve into abstract algebra.
Well, that’s just it: the algebraic proof is pretty danged abstract, which is why I figured a geometric demonstration might be more apparent; it might not, but, hey, who knows until you draw it up and take a look, right? I mean, say you sketched out the correct location for point E and measured it to be a little over 3.14159 units away from a three-unit-long segment AD; wouldn’t that start to suggest an obvious explanation of why it can’t be done with a compass and straightedge?
In any event, thanks to one and all for the quick and sensible replies.
Ah. Of course, it’s not immediately obvious why π can’t be constructed either [sure, π is transcendental, but that’s not obvious, is it?]. Of course, once given that knowledge, you’ve established the impossibility of such things as squaring the circle, and it’s not an unreasonable hope that the same transcendentality might explain this impossibility as well; unfortunately, as it happens, it won’t work out.
In this case, the non-constructible ratio of interest is, like I said, that cos(20 degrees); using the cosine addition formula, we have that 1/2 = cos(60 degrees) = cos(20 + 20 + 20) = 4cos(20)^3 - 3cos(20). The impossibility of constructing this number isn’t because it’s transcendental; it clearly isn’t transcendental, being as it is a root of the polynomial 4x^3 - 3x - 1/2. No transcendental πs are going to appear anywhere within this construction and mediate its impossibility, I’m afraid. Rather, it’s the fact that 4x^3 - 3x - 1/2 is a 3rd degree irreducible polynomial that presents the problem here; any constructible number will arise from a minimal polynomial whose degree is a power of 2 (because the equations defining circles are quadratic). But I don’t think anyone could look at a number like cos(20) = 0.939692621… and directly sense that problem.
Which does, however, make it (relatively) closely related to the third of the three great unsolvable classic problems, that of doubling the cube (which boils down to constructing the cube root of 2).
Exactly; in both cases (trisecting the angle and doubling the cube), one is trying to construct a constant of “minimal degree” 3 over the rationals, which can’t be done, as 3 isn’t a power of 2.
And I don’t mean to imply that the impossibility of squaring the circle is entirely unrelated; that again happens because this would allow the construction of a ratio (π) which is not of minimal degree some finite power of 2; only in this case, it’s actually, furthermore, not even of any finite degree. So, as concerns the circle-squaring problem, one can take a shortcut and hit the same impossibility result with less work; showing that constructible numbers have to be roots of polynomials is trivially easy, while showing that furthermore there must be a minimal such polynomial with degree a power of 2 requires a bit more legwork, avoidable when discussing π but unavoidable when discussing the other two results. So, that distinction is all I was trying to draw in my last post, and why I don’t think you’ll find, from expressing the distance to E in any other way, something that leaps out at you and explains the impossibility in a more intuitive, non-abstract-algebra manner.
What the heck do you mean by “as far as I can tell” and “it seemed to work”? Do you have confidence in your procedure or not? If you believe that your procedure works, that’s great, but show more confidence in it! But if you have any reason at all to think that your procedure doesn’t work, then I suggest you might be better off accepting the conclusion of hundreds (thousands?) or years of mathematicians who say it can’t be done.
PS: Using a straightedge and compass, it is trivially simple to construct angles of 30, 60, 90, and 120 degrees. Angles of 0 or 180 degrees can be made with just the straightedge, without resorting to the compass. It seems to me that in all your testing, these are the angles you’re chosen to trisect. With all sincere respect, perhaps you’ve simply been lucky. Try an angle of a different size and see what happens.
By the way, this was bugging me enough that I actually looked for the 20-degree proof you mentioned, because it seemed to defy common sense for me. “Of course you can produce an angle of 20 degrees with a compass and straightedge - you can bisect arbitrary angles. So if you start with a 40 degree angle, you can construct a 20 degree angle. It’s all a question of what the input is.”
The actual proof, as I now understand, runs more like this:
You can produce a 60 degree angle FROM SCRATCH. This can easily be demonstrated.
It’s mathematically provable (I didn’t follow just how,) that you can’t produce a 20 degree angle from scratch.
If you had a method of universal trisection, you could produce a 60 degree angle from scratch, and then trisect it. This would be a way of producing a 20 degree angle from scratch and violate point 2.
Thus, no method of universal trisection through compass and straightedge is possible.
I’m repeating this here to confirm that I’ve got it right, and for the benefit of anybody else who wasn’t understanding the ‘20 degree argument.’
Yes, that’s all exactly correct. Sure, if you start with a 40 degree angle, you can construct a 20 degree angle; and if you start with a 20 degree angle, you’ll have an even easier time constructing a 20 degree angle! But you can’t produce either of those from scratch.
You can divide any line segment AB into any number of equal parts with straight edge and compass. Basically, you construct another line segment AC originating at one end of the line segment divided into equal parts (using the compass), then construct lines through the points dividing AC parallel with the line BC.
