Actually, you probably could devise some set of correspondence rules that would let you express algebraic results geometrically, and thus construct a “geometric proof” of the impossibility of trisecting the angle. The diagrams you’d draw in the course of that “geometric proof” probably wouldn’t remotely resemble any actual angle or an attempt at its trisection, though.
Now, see, that’s not quite right.
First off, there really is some arc with the properties I’m shooting for: it’s the arc you effectively compared the construction in my OP to. You noted that I was off by tenths of a degree for a 140-degree angle, off by hundredths of a degree for a 60-degree angle, off by thousandths of a degree for a 20-degree angle, and so on; you went well beyond merely saying that I was wrong, and quite accurately showed why the arc I’d suggested wasn’t the arc in question.
Second, there are a number of angles that can be exactly trisected with compass and straightedge: we can exactly trisect a 180-degree angle, we can exactly trisect a 120-degree angle, we can exactly trisect a 90-degree angle, and so on.
And, of course, we can construct an exactly trisected 135-degree angle from scratch: just straightedge a 180-degree angle, bisect it into 90-degree angles, bisect those into 45-degree angles, put the compass spike on the vertex, arc the pencil out at some arbitrary distance to produce five equidistant interception points, connect four adjacent ones with straightedged segments, and then connect the first and fourth with a straightedged segment we’ll call AD.
So exactly trisect the 180-degree angle that we’re calling segment AD. Place the compass spike on AD’s midpoint, arc the pencil from A to D, and note the exact points where the trisecting rays intercept it.
The vertex of the 135-degree angle is below that – and we can draw the vertex of a 90-degree angle below that, using AD as the hypotenuse of an isosceles right triangle. And then we can exactly trisect that 90-degree angle: we can place the compass spike on that 90-degree vertex, arc the pencil from A to D, and note the exact points where the trisecting rays intercept it.
We can do all of that as sure as we can exactly trisect segment AD.
There really is a specific arc of points marking where the trisection point hits: it’s the arc you compared to my original effort, an arc that stretches from the exact point we get for a trisected 180-degree angle through the exact point we get for a trisected 135-degree angle and through the exact point we get for a trisected 90-degree angle.
If that specific arc is centered on some point E, then we can find E: we can straightedge the exact segments that connect those exact points to each other, and we can straightedge exact perpendiculars out from their midpoints to note whether they all intersect at some exact point. If that specific arc isn’t centered on some such point, then they won’t so intersect.
I’d love those pointers, that help, said learning, and, um, the opposite of crackpottery. If you can help me compute the size of the error in this latest construction, then, yes, by all means, please walk me through whatever strikes you as not really being that bad.
Still doesn’t get it, does he?
Other Waldo, you might be interested in this book.
Even if there is a curve of such points, can you demonstrate that that curve is a circular arc?
Not that it matters, but…
No, you can’t. This would yield a 40 degree angle, which could then be bisected to yield a 20 degree angle, from which one could in turn yield the length ratio cos(20 degrees); as a 120-degree angle is constructible “from scratch”, this would amount to a method of producing cos(20 degrees) “from scratch”, which is, as has been noted, impossible.
The first thing I think you should do is really understand the problem. That is, first I think you should really understand that what you’re trying to do is provably impossible. I know I’m just repeating myself (and several other posters here), but just one more time: It Can’t Be Done. It doesn’t matter if you’re smarter than Gauss. The proof doesn’t rely at all on enumerating lots of different methods that one might use and showing that they fail (which would only work as a proof if you managed to prove that you’ve considered all possible methods); the proof en-compasses (sorry) everything that can possibly be done with a straightedge and compass.
The actual proof of this fact uses some results that are at a math-undergrad level. But the basic idea of the proof is pretty easy. There are really only a few things you can do in a construction:[ol][li]Mark a point at the intersection of two lines[]Mark a point at the intersection of a line and a circle[]Mark a point at the intersection of two circles[]Draw a line between two points[]Draw a circle with one point as center and passing through another point[/ol]You start with two given points. I’ll say that they have coordinates (0,0) and (1,0). Now consider the (x,y) coordinates of all of the points you can construct. With high-school algebra, you can check that in all of these cases, all you can do is take sums, differences, products, quotients, and square roots of these coordinates. Nothing you can do in these constructions can get you anything that looks like a cube root. (Constructing the cube root of 2 is called the Delian problem. Like trisecting the angle, It Can’t Be Done.) You can get fourth roots (the fourth root of 2 is constructible), but it turns out–this is the advanced part of the proof–that that’s not good enough.[/li]The second thing I think you should do is to understand how to prove (and disprove) alleged geometric constructions. The first part of this is to understand what a mathematician means by “proof.” In your posts here, you’ve presented some constructions, but you haven’t given any reason for your assertion that they are trisections, other than that they look kind of close. A typical proof of a geometric construction uses all of those techniques from high-school geometry: similar and congruent triangles, interior angles of circles, etc. If you sweep out a circle with radii VA and VD, then you’re allowed to say that VA=VD, and so on. Pick up a book on plane geometry if you need to refresh your memory on theorems and proof techniques.
A disproof, like the one I presented above, is often much easier, since all you have to do is a numerical calculation. All I had to do is some numerical coordinate geometry to compute the angles XVD and AVD. I made the image in PostScript, because I have some PostScript vector-math functions lying around, but you can use just about any package that lets you do math (and, ideally, make graphs to show you the results). Pick up a book on vector arithmetic or analytic geometry to remember the formulas for the intersections listed above (line-line, line-circle, and circle-circle). Now just encode your construction as a sequence of point definitions and you’re done. So for your construction, I did something like this:
Given: V, A, D
E = D + (D-A)
B = (2/3)*A+(1/3)*D
C = (1/3)*A+(2/3)*D
[note that these definitions of B, C, and E are all shortcuts from the way you’d actually make them in a geometric construction; for example, you’d actually make E as the second intersection of line AD with the circle centered at D and passing through A]
X = one of the intersections of the circle at E through C and the circle at V through A
<XVD = arcsin(VX x VD)
<AVD = arcsin(VA x VD)
To compute X, consider the triangle EVX. You can compute the lengths of all three sides (EV, because you know E and V, EX=EC, and VX=VA). so you can use the law of cosines to find the angle <XVE. Now with some simple trigonometry you can find the two intersections; pick the one lying in angle <AVD.
This should let you duplicate my table, and then you can do the same thing for any other construction you want to try. You can find the circle that passes through your three chosen points, so that it trisects three chosen angles; but the arc that trisects all of the angles won’t be this circle, so you still won’t have a trisection.
References: I don’t have any good references for elementary plane and analytic geometry handy. Anyone have any suggestions?
Schaum’s Outline of Geometry is worth checking out, and cheap.
If he doesn’t get it after that very cogent outline of the situation, he’s not gonna get it.
Sure we can: if the curve is a circular arc, then (a) any two points on that arc are the same distance from the center of the circle, and (b) the perpendicular of a segment connecting two such points will pass through that center point, such that (c) the perpendiculars of any two segments connecting any three such points will intersect at that center point. So long as we can use straightedge and compass to supply the exact location of three such trisection points – which is granted for a 180-degree angle and a 135-degree angle and a 90-degree angle – then we can draw two segments and their intercepting perpendiculars.
We can then draw a circle around that center point, with a radius extending out to any of those three points on that curve – and so long as we can use straightedge and compass to supply the exact location of various other points on the curve, we can demonstrate whether each point falls on that same circular arc; mere bisection can get us the exact trisection point for a 45-degree angle, and for a 67-and-a-half degree angle, and so on. Either they fall on that circumference or they don’t.
Indistinguishable points out that we can’t do this for a 120-degree angle, as I’d mistakenly claimed. Omphaloskeptic points out that “all you have to do is a numerical calculation. All I had to do is some numerical coordinate geometry to compute the angles XVD and AVD. I made the image in PostScript, because I have some PostScript vector-math functions lying around, but you can use just about any package that lets you do math (and, ideally, make graphs to show you the results).” I’ll go do that, and then “try to compute the size of the error in your latest construction” as Omphaloskeptic recommended.
Correction:
These should all be unit vectors; divide each by its magnitude before taking the cross product.
Yes, of course, you can construct a circle through three points; this is easy. What Chronos and the rest of us are asking is, Why do you think this arc is a circle? (It’s not.) Why do you think the circle you construct through these three points will pass through the other points you want? (It won’t.)
This goes back to the whole idea of a proof. If the arc is a circle, then you need to prove it to demonstrate that your construction is valid. A proof is more than just a numerical test, though an accurate numerical calculation is better than just doing the construction and eyeballing it.
OK, but don’t neglect to study a proof-based geometry book for long enough to understand what a proof is and how to prove the correctness of a construction.
<hijack>So I gather you can’t divide an angle by 5 either, since all you have is bisection.
I have long wished there was an easy straightedge-and-compass construction for a pentagon. It’s easy to do in the real world: just tie an overhand knot in a strip of paper, snug it down, and hold it up to the light. (You can even see the inscribed pentacle.)
Pity.</hijack>
The regular pentagon actually is constructible, though a general angle cannot be quintisected (that would require construction of roots of a quintic).
In fact, an n-gon is constructable iff n is a product of a power of 2 and any number of Fermat primes. So in addition to the triangle and pentagon, you can also construct a regular 17-gon, a 257-gon, and a 65537-gon (though I’d hate to have to draw the diagram for that last one).
Thanks to both of you.
Oddly enough, over in the Kobayashi Maru thread in Cafe Society, Le Ministre de l’au-delà just posted a link to a heptadecagon construction at http://mathworld.wolfram.com/, which contains a link to the pentagon. That page mentions the knot method, and cites a paper on a compass-only construction.
Thanks again.