One of the most well known problems from ancient Greek mathematics was that of trisecting an angle by straightedge and compass, which was eventually proven impossible in 1837 by Pierre Wantzel, using methods from Galois theory.

Formally, one can set up the problem as follows. Define a *configuration* to be a finite collection of points, lines, and circles in the Euclidean plane. Define a *construction step* to be one of the following operations to enlarge the collection :

- (Straightedge) Given two distinct points in , form the line that connects and , and add it to .
- (Compass) Given two distinct points in , and given a third point in (which may or may not equal or ), form the circle with centre and radius equal to the length of the line segment joining and , and add it to .
- (Intersection) Given two distinct curves in (thus is either a line or a circle in , and similarly for ), select a point that is common to both and (there are at most two such points), and add it to .

We say that a point, line, or circle is *constructible by straightedge and compass* from a configuration if it can be obtained from after applying a finite number of construction steps.

Problem 1 (Angle trisection)Let be distinct points in the plane. Is it always possible to construct by straightedge and compass from a line through thattrisectsthe angle , in the sense that the angle between and is one third of the angle of ?

Thanks to Wantzel’s result, the answer to this problem is known to be “no” in general; a *generic* angle cannot be trisected by straightedge and compass. (On the other hand, some *special* angles can certainly be trisected by straightedge and compass, such as a right angle. Also, one can certainly trisect generic angles using other methods than straightedge and compass; see the Wikipedia page on angle trisection for some examples of this.)

The impossibility of angle trisection stands in sharp contrast to the easy construction of angle *bisection* via straightedge and compass, which we briefly review as follows:

- Start with three points .
- Form the circle with centre and radius , and intersect it with the line . Let be the point in this intersection that lies on the same side of as . ( may well be equal to ).
- Form the circle with centre and radius , and the circle with centre and radius . Let be the point of intersection of and that is not .
- The line will then bisect the angle .

The key difference between angle trisection and angle bisection ultimately boils down to the following trivial number-theoretic fact:

*Proof:* Obvious by modular arithmetic, by induction, or by the fundamental theorem of arithmetic.

In contrast, there are of course plenty of powers of that are evenly divisible by , and this is ultimately why angle bisection is easy while angle trisection is hard.

The standard way in which Lemma 2 is used to demonstrate the impossibility of angle trisection is via Galois theory. The implication is quite short if one knows this theory, but quite opaque otherwise. We briefly sketch the proof of this implication here, though we will not need it in the rest of the discussion. Firstly, Lemma 2 implies the following fact about field extensions.

Corollary 3Let be a field, and let be an extension of that can be constructed out of by a finite sequence of quadratic extensions. Then does not contain any cubic extensions of .

*Proof:* If contained a cubic extension of , then the dimension of over would be a multiple of three. On the other hand, if is obtained from by a tower of quadratic extensions, then the dimension of over is a power of two. The claim then follows from Lemma 2.

To conclude the proof, one then notes that any point, line, or circle that can be constructed from a configuration is definable in a field obtained from the coefficients of all the objects in after taking a finite number of quadratic extensions, whereas a trisection of an angle will generically only be definable in a cubic extension of the field generated by the coordinates of .

The Galois theory method also allows one to obtain many other impossibility results of this type, most famously the Abel-Ruffini theorem on the insolvability of the quintic equation by radicals. For this reason (and also because of the many applications of Galois theory to number theory and other branches of mathematics), the Galois theory argument is the “right” way to prove the impossibility of angle trisection within the broader framework of modern mathematics. However, this argument has the drawback that it requires one to first understand Galois theory (or at least field theory), which is usually not presented until an advanced undergraduate algebra or number theory course, whilst the angle trisection problem requires only high-school level mathematics to formulate. Even if one is allowed to “cheat” and sweep several technicalities under the rug, one still needs to possess a fair amount of solid intuition about advanced algebra in order to appreciate the proof. (This was undoubtedly one reason why, even after Wantzel’s impossibility result was published, a large amount of effort was still expended by amateur mathematicians to try to trisect a general angle.)

In this post I would therefore like to present a different proof (or perhaps more accurately, a disguised version of the standard proof) of the impossibility of angle trisection by straightedge and compass, that avoids explicit mention of Galois theory (though it is never far beneath the surface). With “cheats”, the proof is actually quite simple and geometric (except for Lemma 2, which is still used at a crucial juncture), based on the basic geometric concept of monodromy; unfortunately, some technical work is needed however to remove these cheats.

To describe the intuitive idea of the proof, let us return to the angle bisection construction, that takes a triple of points as input and returns a bisecting line as output. We iterate the construction to create a quadrisecting line , via the following sequence of steps that extend the original bisection construction:

- Start with three points .
- Form the circle with centre and radius , and intersect it with the line . Let be the point in this intersection that lies on the same side of as . ( may well be equal to ).
- Form the circle with centre and radius , and the circle with centre and radius . Let be the point of intersection of and that is not .
- Let be the point on the line which lies on , and is on the same side of as .
- Form the circle with centre and radius . Let be the point of intersection of and that is not .
- The line will then quadrisect the angle .

Let us fix the points and , but not , and view (as well as intermediate objects such as , , , , , , ) as a function of .

Let us now do the following: we begin rotating counterclockwise around , which drags around the other objects , , , , , , that were constructed by accordingly. For instance, here is an early stage of this rotation process, when the angle has become obtuse:

Now for the slightly tricky bit. We are going to keep rotating beyond a half-rotation of , so that now becomes a *reflex angle*. At this point, a singularity occurs; the point collides into , and so there is an instant in which the line is not well-defined. However, this turns out to be a *removable singularity* (and the easiest way to demonstrate this will be to tap the power of complex analysis, as complex numbers can easily route around such a singularity), and we can blast through it to the other side, giving a picture like this:

Note that we have now deviated from the original construction in that and are no longer on the same side of ; we are thus now working in a *continuation* of that construction rather than with the construction itself. Nevertheless, we can still work with this continuation (much as, say, one works with analytic continuations of infinite series such as beyond their original domain of definition).

We now keep rotating around . Here, is approaching a full rotation of :

When reaches a full rotation, a different singularity occurs: and coincide. Nevertheless, this is also a removable singularity, and we blast through to beyond a full rotation:

And now is back where it started, as are , , , and … but the point has moved, from one intersection point of to the other. As a consequence, , , and have also changed, with being at right angles to where it was before. (In the jargon of modern mathematics, the quadrisection construction has a non-trivial monodromy.)

But nothing stops us from rotating some more. If we continue this procedure, we see that after two full rotations of around , all points, lines, and circles constructed from have returned to their original positions. Because of this, we shall say that the quadrisection construction described above is *periodic with period *.

Similarly, if one performs an octisection of the angle by bisecting the quadrisection, one can verify that this octisection is periodic with period ; it takes four full rotations of around before the configuration returns to where it started. More generally, one can show

Proposition 4Any construction of straightedge and compass from the points is periodic with period equal to a power of .

The reason for this, ultimately, is because any two circles or lines will intersect each other in at most two points, and so at each step of a straightedge-and-compass construction there is an ambiguity of at most . Each rotation of around can potentially flip one of these points to the other, but then if one rotates again, the point returns to its original position, and then one can analyse the next point in the construction in the same fashion until one obtains the proposition.

But now consider a putative trisection operation, that starts with an arbitrary angle and somehow uses some sequence of straightedge and compass constructions to end up with a trisecting line :

What is the period of this construction? If we continuously rotate around , we observe that a full rotations of only causes the trisecting line to rotate by a third of a full rotation (i.e. by ):

Because of this, we see that the period of any construction that contains must be a multiple of . But this contradicts Proposition 4 and Lemma 2.

Below the fold, I will make the above proof rigorous. Unfortunately, in doing so, I had to again leave the world of high-school mathematics, as one needs a little bit of algebraic geometry and complex analysis to resolve the issues with singularities that we saw in the above sketch. Still, I feel that at an intuitive level at least, this argument is more geometric and accessible than the Galois-theoretic argument (though anyone familiar with Galois theory will note that there is really not that much difference between the proofs, ultimately, as one has simply replaced the Galois group with a closely related monodromy group instead).

** — 1. Details — **

We now make the argument more rigorous. We will assume for sake of contradiction that for every triple of distinct points, we can find a construction by straightedge and compass that trisects the angle , and eventually deduce a contradiction out of this.

We remark that we do not initially assume any uniformity in this construction; for instance, it could be possible that the trisection procedure for obtuse angles is completely different from that of acute angles, using a totally different set of constructions, while some exceptional angles (e.g. right angles or degenerate angles) might use yet another construction. We will address these issues later.

The first step is to get rid of some possible degeneracies in one’s construction. At present, nothing in our definition of a construction prevents us from adding a point, line, or circle to the construction that was already present in the existing collection of points, lines, and circles. However, it is clear that any such step in the construction is redundant, and can be omitted. Thus, we may assume without loss of generality that for each , the construction used to trisect the angle contains no such redundant steps. (This may make the construction even less uniform than it was previously, but we will address this issue later.)

Another form of degeneracy that we will need to eliminate for technical reasons is that of *tangency*. At present, we allow in our construction the ability to take two tangent circles, or a circle and a tangent line, and add the tangent point to the collection (if it was not already present in the construction). This would ordinarily be a harmless thing to do, but it complicates our strategy of perturbing the configuration, so we now act to eliminate it. Suppose first that one had two circles already constructed in the configuration and tangent to each other, and one wanted to add the tangent point to the configuration. But note that in order to have added and to , one must previously have added the centres and of these circles to also. One can then add to by intersecting the line with and picking the point that lies on ; this way, one does not need to intersect two tangent curves together.

Similarly, suppose that we already had a circle and a tangent line already constructed in the configuration, but with the tangent point absent. The centre of , and at least two points on , must previously have also been constructed in order to have and present; note that are not equal to by hypothesis. One can then obtain by dropping a perpendicular from to by the usual construction (i.e. drawing a circle centred at with radius to hit again at , then drawing circles from and with the same radius to meet at a point distinct from , then intersecting with to obtain ), thus avoiding tangencies again. (This construction may happen to use lines or circles that had already appeared in the construction, but in those cases one can simply skip those steps.)

As a consequence of these reductions, we may now assume that our construction is *nondegenerate* in the sense that

- Any point, line, or circle added at a step in the construction, does not previously appear in that construction.
- Whenever one intersects two circles in a construction together to add another point to the construction, the circles are non-tangent (and thus meet in exactly two points).
- Whenever one intersects a circle and a line in a construction together to add another point to the construction, the circle and line are non-tangent (and thus meet in exactly two points).

The reason why we restrict attention to nondegenerate constructions is that they are *stable with respect to perturbations*. Note for instance that if one has two circles that intersect in two different points, and one of them is labeled , then we may perturb and by a small amount, and still have an intersection point close to (with the other intersection point far away from ). Thus, is locally a continuous function of and . Similarly if one forms the intersection of a circle and a secant (a line which intersects non-tangentially). In a similar vein, given two points and that are distinct, the line between them varies continuously with and as long as one does not move and so far that they collide; and given two lines and that intersect at a point (and in particular are non-parallel), then also depends continuously on and . Thus, in a nondegenerate construction starting from the original three points , every point, line, or circle created by the construction can be viewed as a continuous function of , as long as one only works in a sufficiently small neighbourhood of the original configuration . In particular, the final line varies continuously in this fashion. Note however that the trisection property may be lost by this perturbation; just because happens to trisect when are in the original positions, this does not necessarily imply that after one perturbs , that the resulting perturbed line still trisects the angle. (For instance, there are a number of ways to trisect a right angle (e.g. by bisecting an angle of an equilateral triangle), but if one perturbs the angle to be slightly acute or slightly obtuse, the line created by this procedure would not be expected to continue to trisect that angle.)

The next step is to allow analytic geometry (and thence algebraic geometry) to enter the picture, by using Cartesian coordinates. We may identify the Euclidean plane with the analytic plane ; we may also normalise to be the points , by this identification. We will also restrict to lie on the unit circle , so that there is now just one degree of freedom in the configuration . One can describe a line in by an equation of the form

(with not both zero), and describe a circle in by an equation of the form

with non-zero. There is some non-uniqueness in these representations: for the line, one can multiply by the same constant without altering the line, and for the circle, one can replace by . However, this will not be a serious concern for us. Note that any two distinct points , determine a line

and given three points , , , one can form a circle

with centre and radius . Given two distinct non-parallel lines

and

their unique intersection point is given as

similarly, given two circles

and

their points of intersection (if they exist in ) are given as

and

and the points of intersection between and (if they exist in ) are given as

The precise expressions given above are not particularly important for our argument, save to note that these expressions are always algebraic functions of the input coordinates such as , defined over the reals , and that the only algebraic operations needed here besides the arithmetic operations of addition, subtraction, multiplication, and division is the square root operation. Thus, we see that any particular construction of, say, a line from a configuration will locally be an algebraic function of (recall that we have already fixed ), and this definition can be extended until one reaches a degeneracy (two points, lines, or circles collide, two curves become tangent, or two lines become parallel); however, this degeneracy only occurs in an proper real algebraic set of configurations, and in particular for in a dimension zero subset of the circle .

These degeneracies are annoying because they disconnect the circle , and can potentially block off large regions of that circle for which the construction is not even defined (because two circles stop intersecting, or a circle and line stop intersecting, in , due to the lack of a real square root for negative numbers). To fix this, we move now from the real plane to the complex plane . Note that the algebraic definitions of a line and a circle continue to make perfect sense in (with coefficients such as now allowed to be complex numbers instead of real numbers), and the algebraic intersection formulae given previously continue to make sense in the complex setting. The point now is allowed to range in the complex circle , which is a Riemann surface (conformal to the Riemann sphere after stereogrpahic projection). Furthermore, because all non-zero complex numbers have square roots, any given construction that was valid for at least one configuration is now valid (though possibly multi-valued) as an algebraic function on outside of a dimension zero set of singularities, i.e. outside of a finite number of exceptional values of . But note now that these singularities do not disconnect the complex circle , which has topological dimension two instead of one.

As mentioned earlier, a line given by such a construction may or may not trisect the original angle . But this trisection property can be expressed algebraically (e.g. using the triple angle formulae from trigonometry, or by building rotation matrices), and in particular makes sense over . Thus, for any given construction of a line , the set of in for which the construction is non-degenerate and trisects is a *constructible set* (a boolean combination of algebraic sets). But is an irreducible one-dimensional complex variety. As such, the aforementioned set of is either *generic* (the complement of a dimension one algebraic set), or has dimension at most one. (Here we are implicitly using the fundamental theorem of algebra, because the basic dimension theory of algebraic geometry only works properly over algebraically closed fields.)

On the other hand, there are at most countably many constructions, and by hypothesis, for each choice of in , at least one of these constructions has to trisect the angle. Applying the Baire category theorem (or countable additivity of Lebesgue measure, or using the algebraic geometry fact that an algebraic variety over an uncountable field cannot be covered by the union of countably many algebraic sets of smaller dimension), we conclude that there is a single construction which trisects the angle for a generic choice of , i.e. for all in outside of a finite set of points, there is a construction, which amongst its multiple possible values, is able to output at least one line that trisects .

Now one performs monodromy. Suppose we move around a closed loop in that avoids all points of degeneracy. Then all the other points, lines, and circles constructed from can be continuously extended from an initial configuration as discussed earlier, with each such object tracing out its own path in its own configuration space. Because of the presence of square roots in constructions such as the intersection (1) between two circles, or the intersection (2) between a circle and a line, these constructions may map a closed loop to an open loop; but because the square root function forms a double cover of , we see that any closed loop in , if doubled, will continue to be a closed loop upon taking a square root. (Alternatively, one can argue geometrically rather than algebraically, noting that in the intersection of (say) two non-degenerate circles , there are only two possible choices for the intersection point of these two circles, and so if one performs monodromy along a loop of possible pairs of circles, either these two choices return to where they initially started, or are swapped; so if one doubles the loop, one must necessarily leave the intersection points unchanged.) Iterating this, we see that any object constructed by straightedge and compass from must have period for some power of two , in the sense that if one iterates a loop of in avoiding degenerate points times, the object must return to where it started. (In more algebraic terminology: the monodromy group must be a -group.)

Now, one traverses along a slight perturbation of a single rotation of the real unit circle , taking a slight detour around the finite number of degeneracy points one encounters along the way. Since has to trisect the angle at each of these points, while varying continuously with , we see that when traverses a full rotation, has only traversed one third of a rotation (or two thirds, depending on which trisection one obtained), and so the period of must be a multiple of three; but this contradicts Lemma 2, and the claim follows.

## 64 comments

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10 August, 2011 at 2:52 pm

Qiaochu YuanNice argument! I’m just a little confused about how you’re stating Corollary 3. As far as I know, this is true in all characteristics and follows directly from the fact that dimension is multiplicative in towers, with no need to appeal to Galois theory.

10 August, 2011 at 3:22 pm

Terence TaoHmm, you’re right. (I thought one needed to ensure the big field was Galois in order for it to contain the full splitting field of the cubic extension, but now that I think about it, this is unnecessary for the argument.) I’ve reworded the text accordingly.

10 August, 2011 at 3:02 pm

Sujit NairYou might also be interested in the following book.

http://www.springer.com/mathematics/algebra/book/978-1-4020-2186-2

V. I. Arnold uses monodromy of curves on Riemann surfaces of functions representable by radicals to prove the Abel–Ruffini theorem. It is a very “beautiful” argument :)

10 August, 2011 at 3:27 pm

Terence TaoThanks for the reference, which I was not aware of! From a cursory inspection it does indeed seem that the argument here (based on replacing Galois groups by monodromy groups) is essentially the same as Arnold’s, though in a slightly different context.

10 August, 2011 at 3:30 pm

Aaron SheldonDoes the initial assumed contradiction need to be: there exists an angle that can be trisected?

This is the logical negation of: no angle can be trisected

10 August, 2011 at 3:57 pm

Franciscus RebroSeems to me the assumption up for contradiction is that ANY angle can be trisected, i.e. there exists some construction which will trisect angle CAB as C sweeps around A in a circle (except for possibly some of those points leading to removable singularities).

10 August, 2011 at 6:50 pm

The only post whose title I understand… « Definitely Maybe[…] time, but awkwardly, today is the ONLY blog post whose title I understand…Here is the blog on angle trisection by straightedge and compass, a primary-school geometry theorem for Chinese […]

10 August, 2011 at 8:29 pm

paramanandsVery wonderful proof indeed. Even if we leave the “details” part the proof is very convincing to a reader who is at high-school level. I wonder if many other advanced theorems of mathematics in general could be proved using ideas within the realm of high-school mathematics.

10 August, 2011 at 8:38 pm

Allen KnutsonOf course some angles can be trisected — 180 degree angles, for example. Or 0. The traditional one to prove impossible is 60 degrees.

Just to be an ungrateful jerk, Terry, I want to suggest that the right way to encompass (ha!) the first half of that exposition is with a _movie_ of the monodromy. Of course, there should be scary music as you pass through the removable singularity.

10 August, 2011 at 8:49 pm

Terence TaoYes, 60 degrees is indeed the traditional example of a non-trisectable angle, though I don’t see a way to adapt the monodromy argument to show that this specific angle cannot be trisected, so I decided to omit this example from the discussion.

A movie (or at least an animated image) would indeed be the best way to illustrate the process, but I don’t have the experience to make such a thing. Of course, if another reader would volunteer to make such an image (with or without scary music), that would be much appreciated :-)

[Edit: it also occurs to me that an applet would also do the job nicely. -T]11 August, 2011 at 11:05 pm

math_readerI’d like to recommend the excellent free program called geogebra http://www.geogebra.org/cms/en which allows to create easily so-called dynamic worksheets which are simply animated applets. It is used by many high school teachers worldwide and should be well-suited to your needs here. In particular the page of their wiki contains lots of examples http://www.geogebra.org/en/wiki/index.php/English Finally their forum is very active (e.g. to provide help with installation on linux).

12 August, 2011 at 3:55 am

math_readerAddendum: of course there’s also a nice youtube channel with lots of advanced tutorials http://www.youtube.com/geogebrachannel and I’ve just found this other nice video tutorial to get started and save as a java applet http://www.youtube.com/watch?v=SVSNHhCeukw

12 August, 2011 at 10:10 am

azjpsI took the liberty of creating such an animation in GeoGebra, minus the music: http://imgur.com/a/N3pvo (the red figures being those of period 2).

12 August, 2011 at 10:48 am

Terence TaoThanks! (it seems that it is the green figures that are of period 2, rather than the red ones, though. In any event, colouring objects by their period is a neat idea.)

16 December, 2012 at 3:35 pm

WAYNEcheck out: http://www.math.umbc.edu/~rouben/Geometry/trisect-baker.html

Rouben may help with your applet

11 August, 2011 at 5:44 am

Joshua Zelinsky(This was undoubtedly be one reason why, even after Wantzel’s impossibility result was published, a large amount of effort was still expended by amateur mathematicians to try to trisect a general angle.)I think there’s a typo here- “be” seems to be unnecessary. Also I suspect that this is not the major reason for why amateurs expended and still expend some work trying to trisect. This is not the only example where crankish or simply very amateur individuals try to do something that has been proven to be impossible. The most obvious such example is those who want to get a 1-1 correspondence between the integers and the reals. There are a lot of them. And yes, Cantor’s proof that you can’t is really quite straightforward.

[Corrected, thanks – T.]11 August, 2011 at 6:49 am

Ian TobascoRichard Courant also discusses the topic of constructable numbers in “What is Mathematics?”

http://www.amazon.com/Mathematics-Elementary-Approach-Ideas-Methods/dp/0195105192

There he demonstrates that the angle 60 deg cannot be trisected. The main idea is Corollary 3, but this is proved without appealing to Galois theory or a dimension argument.

15 August, 2011 at 2:03 pm

Martin CohenPoor Herbert Robbins – ignored again!

25 July, 2014 at 9:38 am

Ghulam SarwarTrisection is a matter of construction, not calculation.You can trisect a line but numbers fail to calculate it, and the same case is with trisection of angle.

11 August, 2011 at 8:02 am

AlexWhat software did you use to draw the diagrams?

[Inkscape – T.]12 August, 2011 at 8:34 am

Slipper.MysteryIt looks like same argument works to show not only impossibility of trisection but impossibility of splitting by anything other than a power of 2, since only powers of 2 divide powers of 2?

12 August, 2011 at 9:33 am

Terence TaoYes, that is correct. (This is one small advantage of this argument over the standard field-theoretic argument, which can reach the same conclusion but only after a bit of fiddling around with minimal polynomials.)

12 August, 2011 at 1:03 pm

ImNotSureHi Tao,

A question, do you have any references? a book o somewhere to read more about this proof

bye

12 August, 2011 at 4:56 pm

Linda JanourovaNeat — and I’m sure I’m missing something obvious, but the “intuitive” version of the proof doesn’t seem quite “complete” to me (if that even means anything!). I’m happy to accept everything up to the last step — but once we get to that step, is it really so clear that because the trisecting line {\ell} only makes a one-third (120 degree) rotation for every full rotation of {C}, the whole construction must have a period that is a multiple of 3?

Couldn’t the original construction have contained another line {\ell_1} that is coincides with {\ell} (rotated through 120 degrees) — so that a full rotation of {C} just brings {\ell} to {\ell_1}?

12 August, 2011 at 5:40 pm

Terence TaoOne should view the construction as a

labeledconstruction, with each point, line, or circle in the construction given a label such as or . If one arrives at a construction with the same set of lines and circles, but for which the labels have been swapped, this is not considered the same construction as the original construction for the purposes of determining periodicity. So even if the construction contains the 120-degree rotated lines to , it will only return to the original construction after a number of rotations equal to a multiple of 3.12 August, 2011 at 5:55 pm

A SERIOUS geek out on mathmatics (via What’s new) « Normanomicon[…] One of the most well known problems from ancient Greek mathematics was that of trisecting an angle by straightedge and compass, which was eventually proven impossible in 1837 by Pierre Wantzel, using methods from Galois theory. Formally, one can set up the problem as follows. Define a configuration to be a finite collection of points, lines, and circles in the Euclidean plane. Define a construction step to be … Read More […]

13 August, 2011 at 8:56 am

A geometric proof of the impossibility of angle trisection by straightedge and compass « What’s new « kronikoles[…] A geometric proof of the impossibility of angle trisection by straightedge and compass « What’s n…. Like this:LikeBe the first to like this post. […]

13 August, 2011 at 4:42 pm

Walking Randomly » 80th Carnival of Mathematics[…] Tao gives us a geometric proof of the impossibility of angle trisection by straightedge and compass while the Geometry and the imagination blog discusses Rotation numbers and the Jankins-Neumann […]

15 August, 2011 at 8:13 am

fedjaI was once challenged with presenting the proof of the impossibility of the angle trisection to undergraduates. It took me a while to figure out how to do it but I finally got away without Galois though I needed the notion of the quadratic field extension. Here is what I did:

1) Introduce complex numbers as points on the plane

2) Define constructible numbers for an initial configuration of points

3) Show that any constructible number belongs to some quadratic extension of a quadratic extension of … the field generated by the original points.

4) Now, suppose that is in the original field and the last extension was obtained by joining with . Then , . If , lies in the previous extension, so the last extension was vacuous. If , was in the previous extension already. If , then , so if was in the original field, we can take the square root of there and the last extension was vacuous again. Thus, if is in the first field in the chain, the whole chain is unnecessary and we should have in the original field.

5) Take the 120 degrees angle with (so you get that there at once). If for , then, since , we must have but the equation has no rational roots, so , which is absurd.

To be honest, this still takes a lot of time (three 50 minute lectures, to be exact), but it appeals only to the elementary algebra.

16 August, 2011 at 10:56 am

plmI have not read it but this article posted today may be relevant:

http://arxiv.org/abs/1108.2793

19 August, 2011 at 7:26 am

Menelaus theorem by way of Reidemeister move 3 | chorasimilarity[…] I was in fact motivated to draw the figures and explain all this after seeing this very nice post of Tao, where an elementary proof of a famous result is given, by using “elementary” graphical […]

20 August, 2011 at 4:53 pm

A geometric proof of the impossibility of angle trisection by straightedge and compass « Nico For Math[…] standard way in which Lemma 2 is used to demonstrate the impossibility of angle trisection is via Galois theory. The implication […]

27 August, 2011 at 7:16 pm

Absolute Value (mathematics) « Jeinrev[…] Geometric proof that angle trisection by straightedge and compass is impossible (terrytao.wordpress.com) […]

2 September, 2011 at 4:21 pm

QuoraHas a negative result ever published in a high-profile academic journal (i.e. Nature, or Science)?…In the field of mathematics, negative results are treated probably with the same regard as positive results. After all, if a proof has been determined to be impossible to solve either at all or in a certain way, it does help scope the problem in a non-…

14 October, 2011 at 5:13 pm

JoeTotally new to this site. Interested in, not obsessed by, the trisection problem.

Could I have your permission to email, in a ACDC attachment, a model of a proposed trisection construction for comment?

7 November, 2011 at 11:05 am

AnonymousRead my method of this problem at http://www.scribd.com/doc/62318863/Trisection-of-an-Angle-English

20 April, 2012 at 5:01 pm

mohammed nowairanannouncement : my father found out a method with prove to trisect any angle no matter what’s the degree !! i know it looks like a joke but for who interested i mean official centers or official Representatives of math , please contact me on my father’s email !! The research consists of 46 pages written by hand in the Arabic language only unfortunately , including 16 explanatory figures, but does not include a preface or an epilogue.

m.nowairan@gmail.com

24 May, 2013 at 7:00 am

sunwukongTrisection of an angle.

(20130523, pak)

(graphic did not paste)

Consider an arbitrary angle (say for instance 32 degrees)

If you draw two lines thru a point (a1) at this angle for 50 cm or so.

If you then draw a circle around the point (p) at some arbitrary distance (say r=10 cm)

If you then mark out points along both of the lines at r distance (10 cm)

Then swing arcs from the point thru each of these “r” distance points

The distances between the points where each “r” distance crosses the two lines (32 degrees) apart will be sin(32)*r.

Then where the base is 3*r the small base of the triangle will be three times the base of the smallest triangle. Simply mark r distance from each of the sides and you have trisected the angle between the two long sides of the triangle.

how is this not a trisection?

24 May, 2013 at 7:11 am

Terence TaoTrisecting a chord is not quite the same thing as trisecting an angle, though it gives a reasonably good approximation. Equating the two happens to be a rather commonly made error in first attempts to trisect an angle: see e.g. http://mathworld.wolfram.com/AngleTrisection.html

It is instructive to work through the trigonometry (using the sine rule) for an explicit example (e.g. the 32 degree angle you mention) to see that the angles constructed by trisecting the chord are not quite a perfect trisection (the angles will be close to, but not exactly, 32/3 = 10.666… degrees).

23 September, 2013 at 2:38 pm

AnonymousAngle trisection is possible by Euclidean methods in 3 dimensional space.

7 May, 2014 at 5:41 am

Manthos MonogiosA solution to the geometric trisection of any angle may have been found.

1 November, 2014 at 3:07 pm

AnonymousIt has. IOSR JOURNAL OF MATHEMATICS volume 10, issue 5, version 6, pages 48 – 51

8 April, 2015 at 11:47 pm

Danish RugjeeHello there,

If you’re proof is correct, so how did i trisect an angle???

Please have a look at the website i posted

26 May, 2015 at 11:04 am

Chenjia LinI was trying out this problem two days ago with only a straight edge and compass, and for some reason, I was able to trisect a 60 and 45 degree angle. Any attempt afterwards didn’t work for some reason.

28 July, 2015 at 5:39 am

William (Bill) Greig. Bsc(St And., PGCE,Sorry y’all, but I have done it. My method depends on an extension of the series of circle equalitiy theorems: In any given circlle, equal chords, Implies equal arcs, implies equal angles at the centre, implies equal half size angles at the circumference ( subtended bu the same of equal arcs) which extends to equal angles at any point on the plane: The proof hangs on triangle congruences: It is then used back track from : ning the angle into an isosoceles triangle. Drawing a semicircle with the third side as diameter. Cutting three 60degree arcs on this semicircle, and joining the ends to the original vertex. Detailed proof supplied to any who have the grace to ask

1 January, 2016 at 6:30 am

sunwukongI have a method that seems to work

I have a .jpg slide that shows the method how do I u/load a .jpg?

9 July, 2016 at 5:11 am

AjAybut I can prove that we can easily trisect an angle!

I have developed a New approach to trisect an angle!

:)

9 July, 2016 at 8:17 pm

Jonathan CrabtreeDear Dr. Tao,

Thank you for your wonderful website. How privileged we are that you care to share so much!

Have you seen the published proof of angle trisection via origami? (Not by compass and straightedge!)

Trisecting an angle, Karim Noura, Vinculum, Vol. 51, No. 4, pp. 8-10, 2014. http://search.informit.com.au.ezproxy.slv.vic.gov.au/documentSummary;dn=633791700982280;res=IELAPA

Slides for the above can be seen at

Angle Trisection (AAMT KN 2015)fromKarim NouraThanks again!

Best wishes,

Jonathan Crabtree

9 July, 2016 at 10:32 pm

AjAybut i have my own way to prove about trisection of an angle…

19 August, 2016 at 1:34 am

Eulogio GarciaAndri Lopez refutes the Abel-Rufffini theorem.

Shows that Galois equations are solved in based their coefficients; forthis, we chose the two coefficients of major value.

Let (e) and (d) such that:

Example: the quintic equation.

Solution: $\latex (x = 4)$

See: http://www.hrpub.org/journals/jour_info.php?id=24 Vol 4 (2) 2016

26 August, 2016 at 8:56 am

Nk CThanks a lot I like this posts

12 September, 2016 at 12:48 pm

DavidI could be wrong, but could someone please explain to me why this linked method doesn’t accomplish the goal?

I believe it follows the rules of compass and straight edge. It could only be an approximation, but if so, please explain why.

Thank you.

12 September, 2016 at 4:51 pm

DavidAnd to extrapolate, if this is a correct method, you can use it to divide an angle n-times. And if you can do that, you can construct an n-gon. Which i believe is also currently considered to be impossible. I think the whole impossibility factor is when cubes and cube roots are involved. But this method takes a completely different approach I think. But again, I could be wrong and probably am.

12 September, 2016 at 9:26 pm

AnonymousThe problem starts at 4:46 (the process of creating 3 equal circles which divide the arc DE into 3 equal arcs can’t be done in finitely many steps!)

12 September, 2016 at 9:32 pm

DavidThank you for the reply. But I’m not sure what you mean. The division of arc DE is marked by the compass radius. Can you please elaborate about the “finitely many steps”?

12 September, 2016 at 10:16 pm

AnonymousThe problem is how to determine (in finitely many steps) the exact(!) compass radius needed for the division of the arc DE into 3 equal arcs (which is equivalent to trisecting a corresponding angle).

13 September, 2016 at 6:45 am

DavidThe radius that I constructed at KJ is exactly 1/6 the length of arc DE, so that is able to divide the arc into 6 equal parts. If it can divide by 6, then it can be divided by 3, which is how the arc can be divided into 3 equal parts by my method. What am I missing?

13 September, 2016 at 8:24 am

DavidI realize the logic in my proof at the end of the video is incorrect. In the first, I calculate the length of a straight line, in the next I calculate the length of the arc, so naturally they will not match exactly. Here is another try:

We know that AK is equal to AJ, and the angle KAJ is ½ the total angle BAC. And we know the length of AK 1/3 of AD.

We can express: circumference with radius KJ = AD/3 x 2pi

If we plug in an arbitrary angle for BAC, say 30, the angle for KAJ is half, or 15 degrees. The arc is 1/24 the circumference of the entire circle with radius KJ.

So the length of KJ = AD x .0872

Does this match up to the length of LG?

If we say that our angle BAC is 30 degrees, we know that the arc DE = 1/12 the total circumference of the circle with radius AD. And we know that KJ is 1/6 of arc DE.

From our previous argument, we know KJ = AD x .0872

We can express:LG = 2pi x AD / 12 / 6 = AD x .0872

What about the angle DAL? It should equal 10 degrees if it’s 1/3 of the 30 degree angle BAC. Is it?

Since AD and AL are equal length, they form an isosceles triangle DAL. We know KJ is ½ the length of DL. We can also bisect line DL to divide DAL into 2 right triangles with hypotenuse of length AD.

sin(5) = KJ /AD = .0871557 so DAL has an angle of 10 degrees and is 1/3 the angle BAC!

13 September, 2016 at 10:49 pm

AnonymousAs a simple check of your last construction, your construction should give sin(5 degrees) in terms of rational numbers and square roots only.

This expression can be computed to sufficiently many significant digits in order to see its deviation from the exact value of sin(5 degrees) (which can’t be expressed using only rational numbers and square roots!)

14 September, 2016 at 10:14 am

anonymous2Your construction would work if: 2 sin (A/4) = 6 sin (A/12) where A is the angle in radians. In other words, the corresponding implied polygons’ sides’ lengths are not integer multiples of each other. It is remarkably close, but not close enough for mathematicians. Looks like the error is less than half a percent. Probably why you didn’t notice it.

14 September, 2016 at 5:02 pm

DavidThank you for the explanation. I will check it out.

15 September, 2016 at 11:06 pm

AnonymousLet . Given and a sufficiently good (initial) approximation of , it is possible by (a geometric version of) the Newton-Raphson iteration to generate (geometrically!) a quadratically(!) convergent sequence of approximations to .

15 September, 2016 at 9:01 pm

DavidWhy does the method of trisecting an angle (using a marked ruler) break the rules when you can construct the same thing using a compass and straightedge that is “unmarked”?

https://en.wikipedia.org/wiki/Angle_trisection

15 September, 2016 at 9:41 pm

DavidNevermind. Getting late and I made an incorrect assumption.