# Talk:Angle trisection

WikiProject Mathematics (Rated C-class, Mid-importance)
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Mathematics rating:
 C Class
 Mid Importance
Field: Geometry (historical)

## Initial draft

I, user:dino, wrote this from a re-direct. Yes, the diagrams are not so hot. But I wanted something "out there," perfect or not. Anyone with better software than the Dia (software) I used is welcome to replace them.

dino 18:51, 4 August 2007 (UTC)

## Animated GIF

Would someone tell me how to create the nice animated gif like the bisection one. —Preceding unsigned comment added by 97.65.82.66 (talk) 20:43, 5 May 2011 (UTC)

## 2π/5

The article states that "For example, 2π / 5 radians (72°) may be constructed, and may be trisected. Also there are some angles, that are not-constructable, but (if somehow given) trisectable, e.g. 2π / 5." Surely 2π/5 cannot be both constructable and non-constructable? -Elmer Clark (talk) 10:05, 6 January 2008 (UTC)

Sure, correct example is 3π/7. It is not constructable, but (if somehow given) it is trisectable. —Preceding unsigned comment added by 213.171.48.226 (talk) 21:33, 23 January 2008 (UTC)
I think any angle 3π/N is trisectable for any integer N, right? Jokem (talk) 21:17, 15 August 2011 (UTC)
No. If this were the case, every regular 3n-gon would be constructible, which is known not to be the case. However, 3π/N is trisectable in cases where N isn't divisible by 3: copy it enough times to produce an angle of π/N or 2π/N with one of the arms of the original angle.
So 3π/3, 3π/6, 3π/12 are not trisecable? Jokem (talk) 21:20, 16 August 2011 (UTC)
Are you seriously confusing PQ with QP, or acting up to your name? — Smjg (talk) 09:22, 19 September 2011 (UTC)

## mistake in the statement of the "general theorem"

The correct statement of the theorem is: The angle $\theta$ may be trisected if and only if $q(t) = 4t^{3}-3t-cos(\theta)$ is reducible over the field extension $Q(cos(\theta))$. I think the confusion arises because in the theorem we want to construct $\theta/3$ from $\theta$, whereas in the previous section (where the important trigonometric identity appears) we construct $\alpha$ from $3\alpha$. Right? —Preceding unsigned comment added by 87.165.223.61 (talk) 16:30, 22 January 2009 (UTC)

## No criticism?

Haha, only on Wiki. —Preceding unsigned comment added by 80.213.165.232 (talk) 18:45, 29 January 2010 (UTC)

## Solution of a polynomial

What is a solution of a polynomial? I suspect you mean root of an irreducible polynomial of second order, or something related? Could you please also explain why? Quiet photon (talk) 20:30, 23 March 2010 (UTC)

"The proof would take us afield"? this is not encyclopedic.. proof is required :P --187.40.247.75 (talk) 04:23, 14 April 2010 (UTC)

## New material OK?

I'm planning on adding to this article significantly; I hope that's OK with everyone? I intend to cover not only Wantzel's proof that angle trisection cannot be done with compass and straightedge, but also the mathematical methods by which it can be done. I'm going to have to be slow, so please be patient with me! :) Help would be appreciated – thanks! Willow (talk) 02:07, 4 May 2010 (UTC)

## 90 Degree

Any right angle, divided or multiplied by 2 is trisecable, right? Jokem (talk) 21:33, 13 May 2010 (UTC)

Indeed, with SE&C you can construct on a given line any multiple of 30°, or construct 30° and bisect it as many times as you like. -- Smjg (talk) 10:53, 14 May 2010 (UTC)

## Trisecting analytically

Near the beginning the article says "Also, it is possible to trisect any angle analytically." Depending on how one interprets it, this sentence is either misleading or wrong. Trisecting analytically requires solving a cubic equation which has three real roots. But this cannot in general be done analytically unless you express the solution in terms of the cube root of a complex number; this is the casus irreducibilis. You can take the cube root of the complex number trigonometrically, but then you end up using an expression of the form cos[(1/3)cos^{-1}z]. So the analytic trisection ends up being expressed in terms of the analytic trisection, which is circular.

Does anyone object to my removing the indicated sentence?Duoduoduo (talk) 22:15, 18 May 2010 (UTC)

## What the blazes?

Cosine of a cosine? Yes, it should probably go, unless the is good support somewhere.

dino (talk) 00:13, 19 May 2010 (UTC)

## Some angles may be trisected

I believe that the statement "Further, an angle of k degrees (k an integer) can be trisected if and only if 3 divides k" is incorrect, the correct statement should read "... 9 divides k. An angle of 60 degrees is a counter-example.

In general, an angle x can be trisected if x/360° is a rational number whose denominator (in lowest terms) is not a multiple of 3. --Jwillekens (talk) 09:09, 23 May 2010 (UTC)

You're right, it's mistaken. The original source says it's constructible (not trisectible) if 3|k. I've deleted it. I'll leave it to you to put in something about 9|k if you want to, along with a reference. I'm not sure whether that is if and only if, or just if. Duoduoduo (talk) 15:44, 23 May 2010 (UTC)
The original source should do[might possibly do]... Since an angle is trisectible if and only if its trisection is constructible, the statement in Duoduoduo's source implies that the precise condition is 9|k [is a sufficient condition], indeed. JoergenB (talk) 21:41, 22 March 2012 (UTC)
When I thought over the details, in order to satisfy myself of the validity of the converse statement, I didn't quite succeed:-( in fact, I came up with a counterexample. Let α = 6π/11 = 98$\frac2{11}$°. α is non-constructible; but if for some reason we are given an exact α angle, then we also may construct 4α = 2π + β, where indeed 3β = α. Thus, α is trisectible, although β isn't constructible.
Thus, the problem of determining the trisectible angles is more intricate than I thought. We may employ just the original source for the claim that 9|k is sufficient, but not for the necessity. JoergenB (talk) 14:57, 26 March 2012 (UTC)

## marked straightedge

I suppose it would be cheating to just hold the compass on the straightedge in such a way as to indicate the distance CD? Technically you are still just using a strightedge and compass, though. Jokem (talk) 16:05, 10 June 2010 (UTC)

That is called a neusis construction and is mentioned in the article. 67.119.12.216 (talk) 18:52, 14 September 2010 (UTC)
OK, it is not quite the same as a 'marked' straightedge, but close enough I guess. Jokem (talk) 16:46, 24 September 2010 (UTC)

Is the straightedge assumed to be infinitely long? If not, would it be cheating to use the length of the straightedge itself, or would the ends be considered "marks?" Wayne (talk) 23:05, 20 April 2014 (UTC)

It is considered infinitely long. Dmcq (talk) 23:22, 20 April 2014 (UTC)

## Approximation, do you know of any better?

http://www.youtube.com/watch?v=fkgGDw8Kt-M Thank you 88.89.195.122 (talk) 13:54, 19 January 2012 (UTC)

Take a protractor and measure the angle. Divide by three. Much easier and just as accurate, maybe better. What's there is not particularly simple nor can I see why you'd want to waste time on it. Dmcq (talk) 16:03, 19 January 2012 (UTC)
Or, to put it another way, trisecting an angle exactly using compass and straightedge is interesting for historical reasons; both "trisecting exactly using other tools" and "trisecting approximately using compass and straightedge" don't have this motivation, and so really what's the point? It's also worth noting that one can approximately trisect angles to arbitrary accuracy using repeated bisection, since 1/4 + 1/16 + 1/64 + ... converges to 1/3. --Joel B. Lewis (talk) 16:20, 19 January 2012 (UTC)
The youtube video doesn't give any information about accuracy as a function of the measure of the original angle, and it's unvetted original research. But I don't think one should simply dismiss the topic of approximating trisection using compass and straightedge, or the topic of exact trisection by other means. Underwood Dudley's book The Trisectors devotes an entire chapter--Chapter 1--to exact non-Euclidean constructions. And the last two thirds of the book is devoted to giving Euclidean constructions which, while not exact as typically claimed by their originators, can be viewed as approximations. Dudley peppers these discussions with analyses of their accuracy over the range of possible starting angles: e.g. "unusually accurate, with a maximum error of only 3' for any angle from 0° to 230°" (p.90 of 1994 revised edition); "It is not a bad idea, the error being less than 20" for angles up to 81°, but it grows quickly thereafter, reaching 2' at 90° and a whole degree at 135°" (p. 132); etc., etc. Duoduoduo (talk) 16:59, 19 January 2012 (UTC)
Already the successive bisection method yields arbitrarily good approximations, with very simple means. The price is the number of operations you have to perform in order to get the error small. (For the repeated bisection method, each new bisection halves the remaining error.)
Thus, the proper question should be: Given a certain measure of the "complexity" of the construction (measured e.g. in its number of "elementary steps" needed - whatever that would be), how do we minimise the error for any given bound for the complexity? Such a question indeed should have some intrinsic interest. However, I'm not convinced that this article is the right place to discuss the problem - especially if it involves original research. JoergenB (talk) 21:54, 22 March 2012 (UTC)

## About trisection of an angle (0-180)

Little help here. I'm in +1. I found how to trisect angle with the Greek tools. Don't know what to do next. Need guidance. E-mail: [redacted]. Rahul2312160 (talk) 16:33, 5 July 2011 (UTC)

As the lead of Angle trisection says about the tools allowed by the Greeks: "With such tools, the task of angle trisection is generally impossible". This is a proven result and not just speculation. Either you are using tools they didn't allow (or using them in ways they didn't allow), or your method only works for some specific angles. If you post the method to Wikipedia:Reference desk/Mathematics then somebody will probably tell you what is wrong. This talk page is only intended for discussion about how to improve the article. PrimeHunter (talk) 17:19, 5 July 2011 (UTC)

## Trisection Of An Angle

Sir: My papers: 1. Formulae to find out radius of a given curve or curve section 2. Some Geometrical Theorems 3. Trisection Of An Angle (Constructions) Published in My Journal at: http://jayeshjaiswal.narod.ru/ Also on Scribd at: http://www.scribd.com/doc/63377826/Some-Geometrical-Theorems http://www.scribd.com/doc/63377785/Trisection-of-an-Angle-Constructions http://www.scribd.com/doc/63377744/Formulae-to-Find-Out-Radius-of-a-Given-Curve-or-Curve-Section On I-Proclaim Bookstore at: http://i-proclaimbookstore.com/trofanandott.html

Regards, Jayesh Jaiswal — Preceding unsigned comment added by 117.229.96.188 (talk) 11:28, 1 September 2011 (UTC)

It needs to be peer reviewed before it can be put in Wikipedia. see WP:Original research. Dmcq (talk) 11:34, 1 September 2011 (UTC)

## Formula style

Ozob, did you take a look what you're reverting to? The text looked like this:

Note that $\cos(\pi/3) = \cos(60^\circ) = 1/2$. Then by the triple-angle formula, $\cos(60^\circ) = 1/2 = 4y^{3} - 3y$ and so $4y^{3} - 3y - 1/2 = 0$. Thus $8y^{3} - 6y - 1 = 0$, or equivalently $(2y)^{3} - 3(2y) - 1 = 0$. Now substitute $x = 2y$, so that $x^{3} - 3x - 1 = 0$. Let $p(x) = x^{3} - 3x - 1$.

With [itex] style, presence of ^\circ, (which is unavoidable as far as I can tell) starts to render the formula in LaTeX (image) style, which stands out of the surrounding text, and sharply contradicts the size of the next formula, $4y^{3} - 3y - 1/2 = 0$. Since the formulas are within the text (rather than standalone) and simple enough, it is much better to use small text throughout. I wasn't sure what to do with $\mathbb{Q}$ so I left it in: the alternatives are '''Q''' (Q) and &#x211a; (ℚ) No such user (talk) 07:58, 22 September 2011 (UTC)

See discussion at Wikipedia_talk:WikiProject_Mathematics#.7B.7Bmath.7D.7D_versus_.3Cmath.3E -- you've changed the formulas from being ugly in one way to being ugly in a different way. Joel B. Lewis (talk) 17:08, 22 September 2011 (UTC)
Replied there. No such user (talk) 06:50, 23 September 2011 (UTC)

Ozob, what's your substantial reason for reverting? I did not make the change arbitrarily, but due to visible quality of rendering. Have you read the discussion? No such user (talk) 14:36, 23 September 2011 (UTC)

Yes, I have read the discussion. Please don't talk to me that way.
You are, I hope, faimilar with WP:ENGVAR and WP:RETAIN. I am applying the same principle here. WP:MOSMATH uses exactly the same principle in a very similar context, see WP:MOSMATH#Very simple formulae. This is a common practice on Wikipedia whenever there are two or more alternatives and there is no community consensus for one over the other.
When the discussion at WT:WPMATH finishes, and if there is consensus to use {{math}}, then you are welcome to revert the article to a version which uses it. If there is consensus for plain HTML without {{math}}, then you would be welcome to write a version which uses that instead. It seems unlikely to me that there will be consensus for the article as it was before your edits, but I think it is best to wait until the discussion concludes. Ozob (talk) 21:20, 23 September 2011 (UTC)
Sorry for my tone, but you certainly didn't seem to me that you have read it, because you twice cited "arbitrariness".
I don't like bureaucracy (yes, I'm familiar with WP:RETAIN and the underlying principles). When an edit clearly improves the encyclopedia, in this case visual layout, I find it utterly pointless to revert it, twice, just because it violates some guideline: WP:IAR is a policy. The spirit of both WP:ENGVAR and WP:RETAIN is to prevent edit warring just because of editors' personal styles and preferences. And I explained, several times, that I don't have a personal preference, and that I changed it because the previous layout was just awful. Yes, there is no deadline, and we should reach an agreement, and I explained that I'll gladly revert myself back to >math> if we agree so, but I really can't understand why using an inferior version in the meantime. No such user (talk) 08:04, 24 September 2011 (UTC)
Well, not everyone agrees that it's inferior -- I personally am much happier with LaTeX-based math no matter how the images display. Joel B. Lewis (talk) 16:02, 24 September 2011 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────I understand where you come from, but we must think of readers first. The mixed-style above is unacceptable, it looks just awful. Besides, the WP:MOSMATH is fairly clear:

and later, emphasis mine:

Can we agree to change only the problematic stuff -- formulas containing (^\circ) and \mathbb{Q}\? No such user (talk) 09:27, 26 September 2011 (UTC)

## If a trisectrix can be drawn with compasses and straightedge

If a trisectrix can be drawn with compasses and straightedge,does that mean we can trisect an arbitrary angle? — Preceding unsigned comment added by 114.241.39.212 (talk) 08:20, 31 December 2011 (UTC)

No. Dmcq (talk) 14:43, 31 December 2011 (UTC)

## Why was pie-sector folding solution labelled "ineffective" in the edits history?

I am trying to learn how to contribute. My first edit was deleted within 3 minutes. I went to the edits history, and the reasons cited were for being in the wrong place (OK, I see this now), and being ineffective (??). To paraphrase:

"An arbitrary angle may be trisected with scissors, straightedge and compass by this method: Draw the angle, then use the compass pinned to the vertex to draw an arc, enclosing a pie-sector shape. Cut out the pie-sec with scissors. Now fold the paper (don't make a crease yet!), the foldline going from the vertex to a point on the arc. Select the foldline such that the folded-over edge perfectly bisects the remaining arc. Use the compass as a caliper to verify equal arc lengths for the 2-ply and 1-ply arcs. This is an iterative algorithm. When the two arcs are equal, crease the foldline. The arc has been divvied up into equal thirds. "

My guess is it belongs on the "origami" page for trisection, as it involves paper folding.

Does the fact that it is an iterative construction also affect the classification?

Finally, I have tested the method above, and it is accurate to within the conventional accuracies attendant to other Greek constructions. It works with any angle <=360 except for the slimmest. Can someone explain what was meant by "ineffective"? — Preceding unsigned comment added by Pbierre (talkcontribs) 19:23, 16 January 2012 (UTC)

Firstly material put into Wikipedia needs to be WP:Verifiable which means you need to be able to provide a citation showing where somebody has described it. STuff done by an editor themselves is counted as WP:Original research and may not be put in. Dmcq (talk) 20:28, 16 January 2012 (UTC)
Secondly as far as effective is concerned that means can the method theoretically produce an exact result using the allowed methods? And no this method can't. It depends on "Select the foldline such that the folded-over edge perfectly bisects the remaining arc". How is one supposed to do that? Doing that is equivalent to trisecting the angle and you're basically saying trisect the angle by trisecting it. The article mathematics of paper folding shows how an angle may be trisected using a neusis construction that is allowed in origami. That has the essentials of setting out the allowed actions and showing how an exact result can be achieved using the allowed actions. Dmcq (talk) 20:28, 16 January 2012 (UTC)
The Huzita–Hatori axioms article describe the commonly accepted axiomms that underlie the mathematic of origami constructions. Dmcq (talk) 20:35, 16 January 2012 (UTC)

## Proof of impossibility

The proof if wrong. One has to proof that a root of this equation is not contained in a field extension whose degree is a power of 2. Proving that it is not in a quadratic extension is not sufficient. D.Lazard (talk) 10:47, 7 April 2012 (UTC)

Quite right, I'll see if I can do something to it. Dmcq (talk) 10:55, 7 April 2012 (UTC)

## Chris De Corte found a very close approximation to the angle trisection problem

This can be found on his paper "Approximating the trisection of an angle". — Preceding unsigned comment added by Chrisdecorte (talkcontribs) 04:34, 18 July 2013 (UTC)

Self-published slide show, and there is already a close approximation; bisect n times, trisect the chord, double n times. — Arthur Rubin (talk) 04:40, 18 July 2013 (UTC)