You can use a picture to show that a particular method doesn’t work, but there’s no picture you can draw to show that no method works. For that, you need the algebraic proof. chrisk’s outline captures all of the relevant ideas, so refer to that if you get caught up in the details.
A straightedge and a compass. Here’s how to trisect a line:
Let’s say you have a horizontal line segment AB. From A, draw a diagonal towards the top right. It doesn’t matter what the angle is. Spread your compass to any arbitrary length, and use it to mark off three equal distances along this diagonal. Starting at A, call these points C, D, E.
What you need now is to draw a diagonal, going down and to the left, starting at B, and parallel to the other diagonal. Once you’ve done that, use your compass to copy the lengths along ACDE. Label these as F, G, H, where F is the closest to B.
Now simply draw lines CG and DF, which trisect the original AB. Hmmm… I now see that E and H are superfluous.
(How to make a parallel line? I don’t remember exactly, but I think this will work: Drop a perpendicular from B to ACDE. Then, at B, draw a perpendicular to the line you just made. This line will be parallel to ACDE.)
There will be a lot of confusing and overlapping lines, but it will work.
Wow, is that ever a lot easier to do than mine!!!
But I’m not sure why it works. It obviously has something to do with a 30-60-90 triangle, but I don’t have the brainpower for it right now. …Need… more… coffee!!!
That’s not the simplest known construction. There’s a nice website here that contains a lot of detail on the number of steps needed for various constructions, and they have a nice applet that lets you play around with them.
While the method shown on Wikipedia is relatively simple, it only works for trisection. The method described by Giles and Keeve (and Euclid, for that matter) can be used to divide a segment into n equal pieces, for any integer value of n.
Thanks Giles (and others).
It violates the rules of construction, but if you actually need to get some work done, trisecting (or n-secting) an angle using the compass as a divider is fairly easy. “Walk” the divider along an arc, and tweak the setting until you hit exactly (as near as you can tell) n steps.
It is the trial and error aspect of this that violates the rules of construction.
The Wikipedia method is as general as Keeve’s. All it’s doing is drawing an equilateral-triangular lattice and using the lattice lines to perform the subdivision, like this (bottom; the top picture is a quintisection).
If you have straightedge, you don’t need to compass as much at shows because you don’t drink.
And the Wikipedia method is better, because I suggested an arbitrary line, and the easily constructed line at an angle of 60 degrees is both more specific and gives an easy way of constructing the required parallel lines.
Hm, interesting, I’d never seen that method before.
Well, it’s a little from Column A and a little from Column B. I of course accept the conclusion of said mathematicians; that’s why I lack confidence in any procedure that runs afoul of it. And so if it ever looks to me like I’ve come up with such a solution, I simply assume there’s a flaw that other folks can find.
For example, I’m now twice as confused as when I started this. As has been handily stated, trisecting a 60-degree angle with compass and straightedge is impossible in a way that trisecting a 90-degree angle isn’t; a 90-degree angle is special like that, a 60-degree angle isn’t. So how much skepticism is appropriate if, as far as I can tell, a number of different methods seem to work just fine on the special properties of a 60-degree angle?
Here, let me move that from hypothetical to actual:
Given a 60-degree angle, place the compass spike on the vertex V and make a circle – sweeping the pencil from ray to ray at points A and D. Trisect segment AD into AB, BC, and CD. Straightedge a perpendicular from B out to point X on the circumference. Straightedge a perpendicular from C out to point Y on the circumference. Straightedge from V through X, and from V through Y.
-or-
Given a 60-degree angle, place the compass spike on the vertex V and make a circle – sweeping the pencil from ray to ray at points A and D. Trisect segment AD into AB, BC, and CD. Placing the compass spike at A and the pencil at B, sweep to intercept the circumference at point X. Placing the compass spike at D and the pencil at C, sweep to intercept the circumference at point Y inside the original angle. Straightedge from V through X, and from V through Y.
-or-
Given a 60-degree angle, place the compass spike on the vertex V and make a circle – sweeping the pencil from ray to ray at points A and D. Trisect segment VA into VG, GH, and HA. Place the compass spike at point A, with the pencil at point H, and sweep to intercept the circumference at point X inside the original angle. Trisect segment VD into VJ, JK, and KD. Place the compass spike at point D, with the pencil at point K, and sweep to intercept the circumference at point Y inside the original circle. Straightedge from V through X, and from V through Y.
As far as I can tell, any of those would successfully exploit the fact that the short side of a 20-80-80 triangle is one-third the length of either long side. Given the impossibility of using a compass and straightedge to trisect a 60-degree angle, I could be wrong about said fact, or about whether those techniques exploit said fact – or, yes, about the impossibility, but since I (a) can’t spot the error that would require of all those mathematicians, and likewise (b) can’t spot the error I’m making otherwise, it seems more sensible to just ask the folks here on the SDMB whether they can spot whatever the heck it is that I’m missing.
And with any luck, this’ll be the end of it. Well, for geometry, anyhow.
This isn’t true; the ratio is sin(20 degrees) : sin(80 degrees), which is approx. 1 : 2.88, rather than 1 : 3.
See, now, that’s exactly what I was looking for. Much obliged.
No, I think we can find it quite easily with compass and straightedge – if it exists. The only reason we wouldn’t be able find point E with compass and straightedge would be if such a circle can’t exist. Bear with me just a little longer:
We know that some few angles are special, in that they can be trisected with an ordinary compass and an unmarked straightedge: a 90-degree one is a classic, a 180-degree one likewise – and, if you merely trisect segment AD, we get what Omphaloskeptic’s diagram and your calculations verify as being just about perfect for just about zero.
Isn’t that enough?
First trisect the 180-degree angle that is segment AD: shoot two 60-degree angles out from its midpoint, such that they intercept the circle you’d draw if placing the compass spike at that midpoint and swinging the pencil from A to D; call those points X and Y, respectively. Then trisect the length of segment AD into AB, BC, and CD and straightedge from B to X and C to Y. Then use AD as the hypotenuse of a right triangle, with its 90-degree vertex on the side of AD opposite from X and Y; trisect that angle, put put the spike at its vertex, and then note where those rays intercept the arc formed by swinging the pencil from A to D – call 'em points I and J, respectively.
Then bisect segment CY, and shoot a perpendicular out from its midpoint; if you put the spike at any point on that line, you can arc the pencil from C to Y. Then bisect CJ and shoot a perpendicular out from its midpoint; from any point on that line, you can arc the pencil from C to J. Where those two perpendiculars intercept would be point E: it’s the one spot where an arc would go from C to J to Y.
And you can do likewise for some point F, with a circumference passing through known points B and I and X.
So it’s irrelevant that we can only trisect some three special angles with a compass and straightedge; we only need three points on a circle to act as endpoints of segments with perpendiculars that intercept at the center, because point E is equidistant from C and from J and from Y.
So it’s trivially easy to construct point E and its circle for those three angles. The question is whether such a circle also hits just right for other angles in general and whatever arbitrary one we built AD off to begin with. And we can test that: given some arbitrary angle, put the compass spike at its vertex V, sweep the pencil from points A to D on its rays, straightedge A to D for a segment before doing all that other stuff I just said: completely ignoring angle V while trisecting the 180-degree that is AD, and trisecting segment AD, and using segment AD as the hypotenuse of a trisected right triangle. And then shoot out perpendiculars to center a circle on point E, such that C and Y and J are points on the circumference of that circle –
– and then, finally, come back around to the original angle, and see whether the point where a ray trisecting angle AVD intercepts that first arc at the point where it likewise intercepts the circle centered on E (or F).
If it doesn’t, then our diagram can show that, no, we can’t draw such a circle with compass and straightedge; it’s not that kind of an arc. (If it does, then our diagram would show that we could draw such a circle with compass and straightedge.)
The point you seem to be missing is that we don’t have to test anything, because we know from first year algebra that trisecting an angle with a compass and straightedge is impossible. Do you agree that the algebraic proof is valid? If not, where do you see an issue?
First off, I’d thought that (a) there’s currently no geometric proof to accompany the algebraic one, though (b) it’d be useful to supply a geometric proof, and (c) this would provide that geometric proof. If so, then sketching out the arc in question would helpfully show that it can’t be centered on a fixed point: in the kind of arc we’d need, the perpendicular of a segment connecting any two points on the arc would intercept with the perpendicular of a segment connecting any other two points on that arc – and so we’d construct some of those few angles that can be trisected with compass and straightedge, and they’d hit points on the arc in question, and we’d connect those points with segments, and we’d straightedge out perpendiculars from those segments, and those perpendiculars would meet at various points instead of some one point E, such that geometry demonstrates what algebra indicates.
Second, though, I’m at an impasse: what should I do if I can’t find a flaw in the algebraic proof and yet also can’t find a flaw upon sketching out a circle centered on the one point where the various perpendiculars meet? Obviously one of my failures to find a flaw is incorrect; obviously I’m incompetent to realize which it is; obviously I’ve already seen Omphaloskeptic spot a flaw I couldn’t find; obviously I’ve already seen Indistinguishable spot a separate but related flaw in my reasoning; why not just put my suggested method for finding E up for discussion, so that someone can pick something apart whenever time and inclination line up just right for them?
I deny ©. It’s easy enough to demonstrate, “geometrically” if you like, that a particular method does not, in fact, actually trisect a 60 degree angle. There’s nothing more to it than working out what angle is actually produced by the method under investigation and observing that it is not 20 degrees. But that just shows that that particular method fails, a result which is usually not particularly interesting, there usually not having been any good reason to have initially suspected the method would instead succeed. To prove the interesting result that there isn’t any compass-and-straightedge method that can construct a 20 degree angle, I do not see any way to avoid bringing in some abstract algebra; at least, you haven’t yet outlined any way of doing so with a “purely geometric” proof.
Proof. You keep using that word; I do not think it means what you think it means.
All you would prove this way is that your particular method of construction doesn’t work. Just because the points you’re trying to construct don’t all lie on a circle doesn’t mean, in general, that their arc is not constructible. Tomorrow you’d have another idea: perhaps your point E needs to be at a different (but still constructible) distance along the line depending on the size of the angle; or maybe it needs to be on a different line; or after drawing the circle about E you need to add another epicycle.
The impossibility of trisection is (I want to stress this again) a much more general result than this. It encompasses every allowed straightedge-and-compass operation, even the ones you haven’t thought of yet. I have a hard time even imagining how a geometric proof could proceed.
Well, you could start by trying to understand what mathematicians consider a “flaw.” You can’t expect to prove or disprove the correctness your construction by actually performing the construction and eyeballing the result; mathematicians aren’t impressed by getting “close” to a trisection. You haven’t even attempted to prove that any of your constructions are exact trisections, so there’s not even anything to point at as a flaw; all a mathematician can say is, No, you’re wrong.
If you are interested in learning, rather than crackpottery, then you might try to compute the size of the error in your latest construction, or in any of the others. This requires a bit of trigonometry or analytic geometry, but it’s not really that bad. If you want help post and I’ll give you some pointers.
My dad wasted a bit of his free time on this. Same with squaring the circle. Of course, he didn’t really succeed, but it cut down on the time he spent yelling at Mom.