Well, “the impossibility of using an unmarked straightedge and an ordinary compass to trisect some arbitrary angle,” anyhow. (On the one hand, I wasn’t sure just how big a title we’re allowed; on the other, it looks a heck of a lot clunkier all spelled out like that.)
Be that as it may. Like any number of prior crackpots, I’d gotten curious about the problem and started playing around with diagrams; I’m now at the point where I can’t find a flaw in my approach, which means I can sure use the assistance of folks with more know-how. And since the SDMB is the only place I lurk, I couldn’t readily think of another place to look for help.
So.
First you bisect the angle in question, creating a ray-to-ray segment AD that’s perpendicular to the angle’s bisector. Then (1) trisect that segment into AB, BC, and CD before you (2) use point D as the center of a circle with its radius extending to A, (3) straightedge AD out to double its length at point E on that circle, (4) use point E as the center of a circle with its radius extending to C, and then (5) straightedge from the original angle’s vertex through the point where the arc connecting A to D intercepts the circle centered on E. Thanks to any and all with the time and inclination to take this apart – and all apologies for not working in the obligatory item regarding Opal.
So A is on one ray of the angle that you’re trisecting, and D is on the other ray of the angle you’re trisecting, and both are the same distance from the vertex.
Well, see, back when you’re bisecting the original angle to set up step 1, you do it by putting the compass spike at the vertex of that angle and sweeping the pencil through an arc from some point on one ray to some point on the other ray. (As it happens, those points are then used as points A and D for the segment.)
If you’ve got some compass-and-straightedge method for bisection that doesn’t involve making that arc, then I guess you can add it at the end: put the compass spike at the original angle’s vertex after following steps 1-5, and then create an arc by sweeping the pencil from A to D.
B and C are the points you’d get after you trisect segment AD. So if AD is six inches long, then B will be two inches in from point A, and C will be two inches past that, such that if you go another two inches you’d reach D.
E is the point you’d get if you then go another six inches past D – by using AD as the radius of a circle with its center at D, and then straightedging to give that circle a twelve-inch diameter of AE.
(Of course, you wouldn’t know this stuff about “two inches” and “six inches” and “twelve-inch diameter” if you’re just using an ordinary compass and an unmarked straightedge – but I hope that conveys what I’ve got in mind, anyhow.)
Okay. I see your diagram. Now let’s call the vertex of the original angle V. And we’ll use the name CC for the “point where the arc connecting A to D intercepts the circle centered on E.”
Angles AVB and CVD are equal, but angle BVC is slightly larger than them. The problem is that CCVD is even smaller than CVD, and the inequality is even worse than before.
Besides which, why would you think that the three sections are equal? Is it merely that they LOOK equal, or is there some underlying mathematical formula to what you’re doing?
(A) Proof that BVC is larger than CVD: Let AVD be an angle 179 degrees wide. AVB and CVD will be razor thin, and BVC will be practically the whole thing.
(B) Proof that CCVD is smaller than CVD: Drop a perpendicular to AD at point C. All points on the circle (the big one centered on E) will be on the D side of the perpendicular, not on the B side (except for point C itself, obviously). Therefore, CC is closer to D than to B, and the angle CCVD is smaller than CVD.
Instead of trying to come up with a process that should – for some reason – produce a trisection, I cheated with a protractor to make a bunch of trisections. At that point, it’s trivially easy to trisect a 150-degree angle into little 50-degree angles, and to trisect a 9-degree angle into little 3-degree angles, and to trisect a 90-degree angle into little 30-degree angles, and so on.
And, well, that’s pretty much it, really; I trisected a 180-degree angle into little 60-degree angles, even, and just kept noting where each point would hit for a trisection. It formed what looked like an arc, and so I figured I’d see what sort of arc it was: would you get it from a circle with a small radius, or a big one? I still have no idea why it should be that arc instead of some other one; why not put point E three times as far out, or five times as far out? Beats the heck out of me.
If you can trisect a line then you only need to trisect a line between two points equidistant from the vertex along each edge.
I would first draw a circle from the vertex of the angle with the compass (alternatively you could use the full length of the straight edge depending on its length/the size of the compass/the angle). The circle will intersect the edges of the angle at two points. Draw a straight line between those two points with the straight edge and trisect the line using the compass. Then it’s simply a matter of drawing a line with the straight edge from those two points on the line to the vertex.
Forget a 179-degree AVD for a moment, and just consider a 180-degree AVD. In other words, it’d pretty much just be a segment – which, for the sake of explanation, let’s say it’s six inches long, with point V right at the three-inch mark. So it’s A, then B at the two-inch mark, then V at the three-inch mark, then C at the four-inch mark, then D at the six-inch mark. (And, eventually, E at the twelve-inch mark, once you’ve straightedged out to it.)
Put the compass spike at V. Put the pencil at A and sweep it over to D. You should now have a semicircle stretching from A to D. Put the compass spike at point E, put the pencil at point C, sweep it up – and where that arc hits the semicircle, you should now have a point where you’d straightedge from V to create a 60-degree angle.
CC is supposed to be closer to D than to B – except at point C itself, which is where it needs to be to trisect the closest thing to a zero-degree angle. (Which is the only time you could trisect the angle by just marking points B and C.) As the angle gets bigger and bigger with AD still serving as the same-length ray-to-ray segment, point CC simultaneously goes up and out.
I can’t imagine what your diagrams look like. In your diagrams, is CVD smaller than BVC? And is CCVD even smaller still? If not, why not? And if so, then how do you see it as being a third of AVD?
CVD is, generally, irrelevant. And so, too, is BVC.
If AVD is a 180-degree angle, then CCVD should be a 60-degree angle. If AVD is a 90-degree angle, then CCVD should be a 30-degree angle. And so on.
So, yes, CVD is smaller than BVC. But I’m not really making any claims about angle CVD, just about CCVD.
Picture AD as a perfectly horizontal segment.
CCVD isn’t smaller than CVD because – well, yes, admittedly, point CC goes out from C, but it also goes up. The question is merely whether it goes up high enough.
But have you verified your results beyond looking at the angles and saying “they look equal”? Even if you don’t want to do a geometric proof, you should at least do the trigonometry. For example, let’s say that you start out with AVG being an angle of 150 degrees, and AV and VG are each one meter long. From that, and careful use of Sine and Cosine tables and that sort of thing, you should be able to calculate the exact position of CC, and confirm that the angle BVCC is 25 degrees (half of 50 degrees).
It’s not a solution, as even the OP acknowledges. It’s simple enough to prove that no compass-and-straightedge method can produce an angle of 20 degrees (and thus, cannot trisect arbitrary angles).
Applying the OP’s method to a 60 degree angle will produce something other than a 20 degree angle. I guarantee it. The only question is why one might otherwise think it would work, and what the flaw in such reasoning is. In this case, there doesn’t seem to be any particular such reasoning, other than “It seemed to work in some cases.”
Will it produce something bigger than or smaller than a 20-degree angle?
As far as I can tell, applying the method to a 60-degree angle produces a 20-degree angle; if it didn’t produce such an angle, I wouldn’t have posted. Obviously you could be right, but you’ve got an option I don’t: apply the method to a 20-degree angle and tell me whether it comes out too big or too small.
Well, yeah – with the proviso that it seemed to work in the case of trisecting a 60-degree angle.
