Return to my Mathematics pages
Go to my home page

Trisection of an Angle

© Copyright 1997 and 2003, Jim Loy

Under construction (just kidding, sort of).

This page is divided into seven parts:

Part I - Possible vs. Impossible

In Plane Geometry, constructions are done with compasses (for drawing circles and arcs, and duplicating lengths, sometime called "a compass") and straightedge (without marks on it, for drawing straight line segments through two points). See Geometric Constructions. With these tools (see the diagram), an amazing number of things can be done. But, it is fairly well known that it is impossible to trisect (divide into three equal parts) a general angle, using these tools. Another way to say this is that a general arc cannot be trisected. The public and the newspapers seem to think that this means that mathematicians don't know how to trisect an angle; well they don't, not with these tools. But they can estimate a trisection to any accuracy that you want.

Certain angles (90° for example) can be trisected. A general angle cannot. In fact, a 60° angle cannot. It is usually much more difficult to prove that something is impossible, than it is to prove that something is possible. In this case, mathematicians had to show just what kinds of lengths could be constructed. And then they could show that other lengths could not be constructed, because they were not the right kinds of lengths.

What can be done with these tools? Given a length a, we can multiply this length by any integer, and divided it by any integer. Together, these allow us to multiply this length by any rational number. Given two lengths a and b (and sometimes a unit length), we can add them together, subtract them, multiply them, divide one by the other. Given a length a and a unit length, we can find the length that is the square root of a. These are the only things we can do, with these tools. This has been proven, but I won't do that here. We can do these operations in many many combinations of ways [like sqr(a+sqr(b+sqr(c))), for example, where sqr() is the square root function]. But, if a construction involves a length that cannot be expressed by some combination of ratios and sums and square roots, then it cannot be constructed.

Towards a proof: If we begin with two points, the distance between them can be called 1. With those two points, and compasses and straightedge, we can produce any integer (as we can do addition and multiplication with these tools). Since we can perform division with compasses and straightedge, we can produce any fraction (rational number). We can easily construct the square root of any number that we have already constructed, including earlier square roots, using the Pythagorean theorem. We can combine rationals and square roots with these tools. What more can we do? The answer is, "Nothing." We can draw lines through points and we can draw circles. These involve addition, subtraction, multiplication, division, and taking the square root (Pythagorean theorem). OK, I can create any number of a certain kind (rationals and square roots and their combinations) with these tools. I am now considering some points and I am going to create a new point. I can do this by using previous lengths of our certain kind and drawing arcs which intersect, or by drawing lines which intersect with lines or arcs. We find that if the positions of our points are represented by numbers of our certain kind (coordinates of some kind), and we find our new point using arcs with radii of our certain kind, then we always produce new lengths of the same certain kind. If we create a new point by the intersection of two lines or an arc and a line, which were based upon numbers of our certain kind, then we always produce new lengths of the same certain kind. We add no new numbers of any kind (like cube roots), by connecting points or drawing circles. I still have not proved that, but we are much closer. Our certain kind of number has other names, including "ruler and compass number."

One of the classic problems that cannot be done, is to construct a square the same area as a given circle. This is called squaring the circle. This problem amounts to taking a unit length and constructing a length equal to pi. There is no way, in a finite number of steps, to represent pi in the form of ratios and sums and square roots, as was shown in 1882, by Ferdinand Lindemann. Another classic problem is to construct a cube that is twice the volume of a given cube. This is called doubling the cube. This problem amounts to taking the cube root of a length. This too cannot be done in a finite number of steps, using the above operations, as was shown by Wantzel in 1836.

The trisection of an angle involves the solving of a cubic equation, something that cannot be done, in general, using the above operations. This too was shown by Wantzel in 1836. After Wantzel published his discovery, Gauss claimed that he had proved it in about 1800, but had never published. A surprising result of this was that a 60 degree angle cannot be trisected (in other words, a 20 degree angle cannot be constructed), and so a regular nonagon (a nine-sided polygon) cannot be constructed, as they involve the construction of a cube root. Karl Friedrich Gauss (Gauß) then showed just which regular polygons could be constructed, and he explained how to construct a regular 17-gon.

arc lengthsIn the diagram on the right, the length of the trisecting chord x is the root of the equation (in which x^3 means x cubed) AB=3x-x^3 (discovered, or maybe rediscovered from Arab sources, by Pitiscus about 1600). In general, a cubic equation cannot be solved by construction, with compasses and straightedge. This equation can apparently be deduced from the triple angle equation: sin(3a)=3sin(a)-4sin^3(a), a cubic equation which can be derived from the sine of the sum of two angles formula [sin(a+b)=sin(a)cos(b)-cos(a)sin(b)].

I said above that the trisection (using compasses and unmarked straightedge in a finite number of steps) of a general angle was shown to be impossible, back in the 19th century. While it is difficult to reproduce that proof (I may attempt it some day), that should be good enough to disprove any future attempted trisection. We shouldn't have to disprove each attempted trisection as they come up. They are already disproved. Of course these attempted trisections may be very close to real trisections (within small fractions of degrees). Some are very close for small angles. Others are very close for other angles. Maybe they are close enough for practical purposes. But none of them can be exact.

I received this email. I guess I am stifling innovation when I point out that some things are actually impossible:

I read your article on angle trisection. You seem intelligent enough, but possibly too pompous to be a mathematician. I solved this problem when I was 13. You certainly don't have to disprove every new idea that comes along, you probably don't even need to see most of them. But to rest on an incomplete 200 year old theory with holes in it and thereby attempt to stifle innovative thought is exemplary of the kind of arrogance which allows fools to call themselves educated. Good day.

Can't pompous people be mathematicians? Darn. I may indeed be pompous in other ways, but I was just reporting what geometry of today says. We have a proof that angle trisection (using certain simple tools) cannot be done. You cannot disprove a proof just by calling it outdated. In fact, you cannot expect people to believe you when you say you have done something considered impossible, without any proof whatsoever. Thank you for informing me that you can do the impossible; forgive me for not believing you. I told him, "Of course you did not solve this problem when you were 13." He seemed to take this as a cruel blow, and vowed to never write to me again. Too bad.

Stifling innovation? I may have a different temperament from the person who sent me that email. Personally, I am intrigued as all get out by a problem that I know to be impossible. How do we know it is impossible? That must be hard to prove. Why doesn't this idea work or this other idea? I've learned some geometry by working on these impossible problems.

Other people have sent their trisections to me. Fortunately, most of the methods have been fairly easy to prove to be incorrect. Some have been difficult to understand at first. But, once I could use the method to draw the proposed trisection of an arbitrary angle, the three angles have never been the same. Usually, it was easy to see that the trisection method was not working, just by trying to trisect a large angle (maybe well over 90 degrees).

The difficult ones are the ones that require cheating (like Archimedes method below). One person said he could trisect an angle using an ellipse. Well, you may be able to draw some points of an ellipse, using compasses and straightedge, but you cannot draw the ellipse with these tools. Anyway, I have never heard that an ellipse could help trisect an angle, so I am somewhat skeptical.

All these trisection attempts are experimental geometry. You try a method that might be close to a trisection, and try to measure the angles to see if they are the same. They seem to be the same angle. Ah, I've found a trisection. Wrong, that is not how geometry works. Mathematicians prove everything they do, because you cannot ever be sure that something is true, unless you can prove it. And trisection is a good example of this.

Four books which prove that trisection is impossible with compasses and straightedge are The Trisection Problem by Robert C. Yates, Geometric Constructions by George E. Martin, Euclidean Geometry and Transformations by Clayton W. Dodge, and Ruler and the Round: Classic Problems in Geometric Constructions by Nicholas D. Kazarinoff. The algebraic theory concerning this is called Galois theory (a part of group theory), and there are a lot of books about that. Also see:

Of course the proof is difficult. See An Analogy, at the bottom of this page, to get a simplified flavor of what the proof is actually about.

Part II - The Rules

The Rules: In using the classic tools (straightedge (and pencil) and compasses), it is legal to use them only in the simple and obvious ways. We deduce the rules of this game mostly from the constructions of Euclid. These rules are seldom spelled out, although we do usually spell out that we can't use marks on the straightedge. One of my emails suggested holding the compasses against the straightedge while moving the straightedge. This amounts to making a not so simple tool of the two pieces of equipment, which is obviously illegal. As I said at the top, we use compasses for drawing circles and arcs, and duplicating lengths, and we use a straightedge (without marks on it) for drawing straight line segments through two points. We can do nothing else. We cannot even use the compasses to verify if one distance is equal to another distance. And we cannot eyeball a point and a line to tell if the point is on the line (sometimes we have to go to great trouble to prove that a point is on a line). And, the mistake which is repeated many times on this page, you cannot use compasses to measure or duplicate the length of an arc on circles of different radii (see Part IV, trisection #6).

You cannot use these tools to design other tools like the tomahawk shown below; in other words, you cannot pick up parts of your drawing and move them around as tools in their own right. While you can draw some points of a complicated curve with these classic tools, you cannot actually draw these curves. We cannot fold up the plane like origami.

Below, you will see bogus trisections which come within a hundredth of a degree of being right. Isn't that close enough? You cannot measure 1/100 degree on your protractor. And you can't even be that accurate with real compasses and straightedge. Well, our crude diagrams are a poor imitation of the perfect constructions that we are describing. When I get out my compasses and straightedge, and bisect an angle, the result is inaccurate. But when I describe the bisection of an angle, by drawing arcs, what I describe is perfect, with perfect lines and perfect arcs, using fictional perfect compasses and straightedge. And I end up with a perfect bisection, regardless of the crudity of my drawings. This mathematical perfection is also part of the rules. We pretend that we are using perfect instruments which draw perfect lines and circles, so we can then prove things about what we have constructed. So it makes no sense to measure an angle with a protractor and say, "Yup, that's 20 degrees." We have to prove it.

Our straightedge is often called a "ruler" (even in many books on geometry), but (if we want to follow the rules) we must not use the markings on the ruler.

Part III - Close, But No Banana

Here we see some attempts at trisection which are close, but not exact.

popular trisection attempt1. Trisect the chord: This diagram shows the most common attempt at trisection. Points are marked on the sides of the angle, an equal distance from the vertex. A line segment is drawn between these two points. This segment is trisected (easy to do, see next paragraph), resulting in the marking of two points. Lines are drawn from these two points to the vertex. These two lines almost trisect the angle. It gets fairly obvious that this doesn't work, if the angle is fairly large. For a proof that this is not a real trisection, see below. Many other attempts at trisection are just this one in a more complicated guise (also see below).

trisecting a segmentBy the way, this diagram shows how to trisect a line segment (AB). Through A, draw an arbitrary line. On this line mark off an arbitrary segment, starting at A. Duplicate this segment twice, producing a trisected segment AC (see the diagram). Draw the line BC. Through the other two points on the trisected segment, draw lines parallel to BC. These two lines trisect the segment AB. You can use the same method to divide a segment into any number of congruent (equal lengthed) smaller segments. Also see Geometric Constructions. There are other ways to trisect a line segment.

proofHere is a proof that the attempted trisection is not a real trisection. We are trying to trisect angle A by trisecting the line segment EF. For this proof, extend AC to B so that BC=AC, then connect BE. Triangle ACD is an isosceles triangle with congruent sides AC and AD [Triangles ACE and ADF are congruent by SAS (see Congruence Of Triangles, Part I), and that makes angle ACD=angle ADC]. Triangle BCE is congruent to triangle ACD [SAS]. Angle ACD is an acute angle. Then angle ACE is an obtuse angle.

Now, let's assume that this actually is a valid trisection (for the purpose of finding a contradiction, and proving that it is not a valid trisection). Then angle CAE is congruent to angle CBE. Then triangle EAB is isosceles, and AE=BE [base angles are =]. Then triangle ACE is isosceles [AC=AE]. Triangle ACE is congruent to triangle ACD [SAS or other]. Then angle ACE=angle ACD, which is acute. But above we showed that ACE is obtuse. So we have our contradiction. And our attempted trisection is not a valid trisection.

From this proof, we can guess that the attempted trisection is really close when AE is about the same length as AC. This happens when the angle A is small. But we knew that already, because to trisect angle A, we need to trisect the arc EF, not the line segment EF. Well, when angle A is small, arc EF is very close (in more than one way) to segment EF. To see just how accurate this method is for small angles, see method #10, below.

Also, it is apparently easy to prove (in the same diagram) that if the angle is actually trisected (instead of the line segment EF), then CD is shorter than DF.

Here is another proof that trisecting the line segment does not trisect the angle. With that method, try to trisect an angle of 60 degrees. You should find that the three angles are measurably different. The middle angle is about 2 or 3 degrees larger. So we have a counterexample. Which proves that that method is not a real trisection. Angles larger than 60 degrees would make this proof even more obvious. Using this method to trisect 60 degrees, the smaller angle is 19.1066 degrees, which is 0.8934 degrees off. I calculated the angle with eight digits of accuracy, and then rounded off to six digits. Using this method on a 120 degree angle, we get 30 degrees (which may be exact?), which is 1/4 of the angle, not 1/3.

Calling the angle that we are trying to trisect a, and the smaller angle that is our estimated trisection b, here is an equation that I get, relating a and b: b = (sin a)/sqrt(9-8sin^2(a/2)), where sqrt() is the square root function, and sin^2(a/2) is the sine squared of a/2.

attempted trisection1a. The above method disguised: Here is an attempted trisection which was sent to me by email, a couple of years ago. I have redrawn it (and added some lines to help with my comments). It was originally done with compasses only. Two equal segments are marked off on each of the two rays of the angle out to B and C. For the purposes of my comments below, I have extended the rays out to I and J. Then the midpoint between B and C is determined with compasses. Then we mark off the two points between I and J, again using the compasses. These two points were then claimed to trisect the original angle A. I drew red lines to them (from A). It is easy to show (using the many small congruent triangles that I have drawn, some of which are mirror images of others) that this construction trisects line segment IJ. And that makes it equivalent to the attempted trisection #1 above, and is not a real trisection of angle A.

another attempted trisection1b. Disguised further: Here is another attempted trisection that I received by email recently. Lines AB, AC and BC are bisected. And then the various other lines are drawn. It was hypothesized that points G and H trisect angle A. There are a couple of parallelograms (BDFE and CDEF). And the diagonals of a parallelogram bisect each other. So, G and H are at the midpoints of DE and DF respectively. We can compare this diagram with the previous diagram. Many of the same congruent triangles appear in both. It becomes obvious that these two methods define the same lines (the ones that I have drawn in red). It is not immediately obvious that this method trisects line BC. But comparing this diagram with the previous one, it becomes obvious that this is the same trisection. Again, this is not a real trisection of angle A.

a simple method1c. And further: Here, with angle AOB and AO=BO, we bisect segment OB at C, then bisect segment AC at D, which nearly trisects the angle. This produces the same exact angle as method #1 above, and is close to a true trisection for small angles. That should not be difficult to prove. So if we try it on a 60 degree angle, we will get 19.11 degrees, which is 0.89 degrees off, better than some.

still further1d. Still further: Here, with angle AOB and AO=BO, we bisect segment OB at C, then trisect segment AO at D (AD = AO/3). The intersection of AC and BD (E) nearly trisects the angle. This produces the same exact angle as method #1 above, and is close for small angles. Again, trying it on a 60 degree angle, we will get 19.11 degrees, which is 0.89 degrees off.

a not so simple method1e. Even further: I got this by email, the sender's initials are A. D. . We are trisecting angle A. We make AB = AC, bisect AB at D, and draw the perpendicular bisector EF of CD at E. We draw the arc with center at C and radius CD, intersecting EF at F. We bisect CF at G, and draw DG. DG intersects EF at H. The bisector of angle DHE intersects BC at K, which supposedly trisects angle A. Instead, K trisects segment BC (and CD and DE). You might want to show that, before reading the next paragraph.

OK, if you draw segment DF, then triangle CDF is equilateral. That makes triangle DHE a 30-60-90 right triangle, or half of a small equilateral triangle; I have drawn the rest of the equilateral triangle faintly. DE and HK are then medians of the equilateral triangle, and medians trisect each other (Euclid). So K trisects DE, and CD, and BC. And this trisection method is a well-disguised form of method 1 above.

2. Geometric series: Here is another interesting attempt at trisection of an angle. Divide the angle into fourths, easily done by bisecting twice. To this fourth, add 1/16 of the original angle (1/4 of the fourth). Then add 1/64, and 1/256, etc. forever. We have an infinite sum that adds up to 1/3 of the original angle.

The equation is: 1/3 = 1/4+1/16+1/64+1/256... This is just a Geometric Series. It is easy to show that the sum is 1/3.

This construction is not something that you can actually do, because it takes an infinite number of steps. Such an approach is not normally stated as being against the rules. But, it is taken for granted that it is against the rules. Notice that my article said the various impossible problems could not be done, "in a finite number of steps."

You can stop well before you have done infinitely many steps, and your approximate trisection can be very accurate, as accurate as you want. Using the first three terms of the series to trisect 60 degrees, I get 19.6875 degrees. The first four terms give 19.9219, which is 0.0781 degrees off. We can, of course, use more terms to get more accuracy.

A similar trisection is to construct 1/4+1/12 = 1/3. Of course you would have to trisect a 1/4 to get 1/12. But it is a smaller angle, and the approximation is probably more accurate, depending on the approximate method of trisection that you choose (see method #10, below). 1/4+1/16+1/48 is also similar, where you have to trisect a 1/16 to get 1/48. And you can do the same with any number of terms of the above infinite series. It always involves trisecting some very small angle, and so is never exact. But it can be as accurate as you want.

There are other infinite series which converge to 1/3, including other geometric series: 1/3 = 1/2-1/4+1/8-1/16+...

close3. Mechanix Illustrated: Here is a compasses and straightedge method (which I have simplified) that was apparently given in Mechanix Illustrated, Feb. 1966. It is really really close for angles less than 90 degrees, within a few hundredths of a degree. We are trying to trisect angle AOB (having drawn a circle about O with A and B on the circle). OC bisects the angle. Then we draw two lines parallel to OC, through A and B. EC=CD, and D is the center of the large circle. The attempted trisection is defined as shown. Probably the easiest way to show that it is not a real trisection is to use it on some angle, and then deduce what the three smaller angles really are, showing that the center one is not equal to the two outer ones. Attempting to trisect a 60 degree angle, I find that the middle angle (FOG) is 20.0226 degrees, which is 0.0226 degrees off.

close but no banana4. Arcs with the same chord: Some attempts at trisection involve something like this diagram (I've divided the diagram into two pieces, upper and lower, to make it clearer). We are trying to trisect the larger red angle (LAM). We have not yet determinded where the points L and M are.

We estimate the trisection (somehow, maybe by eyeballing or by some complicated construction) and get the small red angle (CBD) in the lower half of the diagram, with its arc CD. We then triple this angle, producing angle CBF (which should be approximately the same size as LAM). We then duplicate the chord CF inside the angle LAM, producing the chord LM and fixing the points L and M. The upper left part of the diagram shows how this chord can be duplicated. Then we draw the arc LM, with center A. Then we duplicate the small circle centered at D, and place this duplicate centered at M. This gives us the point O, on this circle and on the arc LM, which very nearly trisects the angle LAM.

blow upChords CF and LM are of equal length. But arcs CF and LM have different centers and different radii. Nevertheless, people seem to think that the two arcs are of equal length, because their chords are of equal length. On the right we see the area around point O, blown up about 20 times. I have moved angle FBC so that chords FC and LM exactly coincide (the green line here). As you can see, the two arcs do not coincide. They are very very close (and D nearly coincides with O), but they do not coincide. After all, it would be quite a coincidence if my estimated trisection fit the original angle exactly. I think that it is fairly easy to show that if the chords are the same length, but the radii are different, that the arc lengths are different, in situations similar to the above situation. But you can see why mere measurement of the angles would imply that a trisection had been accomplished.

trisecting 60 degrees by trisecting 45 degreesA site on the WWW used this method to trisect a 60 degree angle by trisecting a 45 degree angle outside of it. Here we have my drawing of this. The red lines are the 60 degree angle and its attempted trisection. The blue lines are the 45 degree angle and its actual trisection. The attempted trisection gives an angle of 19.87 degrees, which is 0.13 degrees off. 45 degrees is not very close to 60 degrees, a better angle will produce better results.

a curve, not a line5. A straight line?: I have been corresponding with a person (initials H. W. S. ) who came up with this attempt at trisecting a 60 degree angle. In this diagram (a 30-60-90 degree triangle), point H is positioned so that dividing the right angle C into pieces of some proportion will divide the angle B into the same proportion. If we position point H just right, we can bisect both angles. And if we position H at some other place, we can trisect both angles. Since it is easy to trisect the 90 degree angle, then we can use H to trisect the 60 degree angle. We could also use other positions of H to divide the angles into any other equal number of smaller angles. This person had come up with a theorem which implied that all possible positions of point H lie on a straight line (the line defined by A and the incenter of the triangle; see The Centers of a Triangle). Using this theorem, it is then easy to trisect the 60 degree angle.

Unfortunately, all possible positions of H do not lie on that line (in other words, this person's theorem is false). In the first diagram above, I have drawn a green line representing the path of H as angle BCH changes. And it appears to be a curve, as one might guess. I think it can be proved that this curve is convex in one direction, and if so no three points on that curve can lie on the same straight line. But this curve is very close to being straight for much of its length, and so a person might be fooled into thinking that it is straight. Pretending that the curve is a straight line, in the second diagram above right, we find that angle EBC is 38.7940 degrees, which is 1.2060 degrees off.

This curve is a rather complicated trigonometric curve. A trigonometric curve depends upon angles and lengths and trig functions, not just lengths as do the more familiar curves (Conic Sections).

This attempted trisection can be used to attempt the trisection of other angles, by using different triangles. Such attempts produce other curves which are also nearly straight lines.

trial and error6. My method: I suspect that this method has been used by someone else, in ancient times. But, so far, I have seen no evidence of that. So, for the time being, this one is my invention, and it is extremely accurate.

We are trying to trisect angle BAC, and we choose a point D (by some method, either eyeballing it, or through any of the other trisection attempts) which might trisect the angle. We then duplicate this angle (BAD) twice, producing points E and F in the diagram. Well, D was just a guess, and the real trisection point lies somewhere between D and F (F can be on either side of D). In fact it is closer to D than it is to F. An excellent guess is 1/3 of the way from D to F. A new point G, 1/3 of the way from D to F is really close to a trisection, but it is not perfect.

What is wrong with G as a trisection point? Doesn't it seem like it should work? The entire angle (BAC) is 3x (with x being the angle of our first estimate) plus a small angle m: A = 3x+m. A true trisection is 1/3 that, exactly: A/3 = x+m/3. So if we add our first estimate to one third of our small angle m, we should get an exact trisection. But we chose point G by trisecting the chord DF, not the angle DAF. So we were just using method #1 above, which we already proved was inaccurate. The length of segment DG is close to the length of the arc DG, but it is never exact. We are using method #1 on a very small angle, and so the result should be very accurate. We can use G as an estimate, and repeat the above approximation, to get an even better approximation.

Using this method (with method #1 above as a first estimate) on a 60 degree angle, we get an estimate of 19.9999276, which is only 0.0000724 degrees off, making this the second best trisection construction that I have tested (see #25 below). We can't possibly draw our lines with that kind of accuracy, so isn't that close enough to be called exact? No, it's still not a real trisection, because it is not exact. We are assuming (pretending) that we are using perfect tools here. This method is an extremely good estimate, but it is still an estimate.

From New Theory of Trisection7. "New theory": Here we see an attempted trisection from New Theory of Trisection by F. C. (I have decided to use initials instead of names, for people who think that they have found the elusive trisection), a geometry teacher, and he claims to be a mathematician. On the back cover, he admits that the construction is impossible, and then says that he has done it. We are trying to trisect angle XOY. We draw circle O, with radius 1, which determines point S on line XO (on the other side of O from X). Draw SO2 perpendicular to line OY. Bisect SO2 at M1. Draw circle M1 with radius equal to segment O2N, defining point M on line OY as shown. Draw O1 as shown, so O1M = OM. Then angle OO1S is the trisection of angle XOY. This attempt is very accurate for angles less than 110 degrees. But there is a flaw in F. C.'s proof. In particular, his theorem #11 is sloppily done, and is false.

my calculationsHis method is exact for 90 degrees (and 0 degrees). Using simple trigonometry, a calculator with trig functions, and the Pythagorean theorem, testing his method on a 60 degree angle produces an angle of 20.05512386 degrees, which is 0.0551 degrees off, close but no banana. He has patented a tool based upon this method, and I suppose it is pretty darned accurate. Of course there are simpler tools that are more accurate (see Part V below).

You might be interested in seeing how I calculated that 20.05512386 angle. It's not very difficult. On the right we see my diagram for a 60 degree angle. These lengths can be deduced (mostly using the Pythagorean theorem):

Every value is exact, except the two decimal values of the tangents, which are accurate to ten decimal places. Taking the arctan of 0.3650601618, I get 20.05512386 degrees, which is not 20 degrees. I read the above tan(20) off a calculator. I have gotten an even more accurate value by estimating the solution of the cubic equation for the triple angle tangent function, and the above value is accurate to the last displayed digit.

This trisection is based loosely upon Archimedes' trisection (Part IV, method #1, below). F. C. doesn't believe my estimate of his error (0.0551 degrees) because, "mathematical reason has proved that an algebraic method is not appropriate method for discussing trisection." Hm? He doesn't allow trigonometry, either. So, his method cannot be disproved, even by counterexample.

Mr. F. C. makes some extraordinary claims, among these:

He also feels that people who believe as he does have been condemned by the mathematical community.

8. Trisect a circle:

I received a description of this by email. It is a variation of method #6, above. We divide our angle into four equal arcs (bisect it twice), then construct a square with the chord CD (in the diagram) as a side. We circumscribe the square with a circle. We pretend that the circumference of this circle is equal to the arc length BC of our angle. We then trisect the circle, by inscribing an equilateral triangle. Then we transfer the side of the equilateral triangle to our arc BC (B'C' on the right part of the above diagram), again pretending that the circumference of the circle is equal to the arc length BC. Trisecting a 60 degree angle like this, I get an 18.40 degree angle, which is 1.60 degrees off. This method was inspired by rolling an angle into a cone, and then trisecting the circular base. See paper folding, below.

not very close9. Not very close: Here is a trisection that I got in my email. This may be what that person meant. We want to trisect angle AOB. We trisect the side AO, and make AR = AO/3. We draw the circle with center O and radius R. Then we draw the circle through A (I assume) tangent to line OB at B (this circle is not very difficult to construct). These two circles intersect at a point X (within the angle). And angle AOX "appears to be the trisection of the angle." This is actually very close for small angles. Using this method on a 60 degree angle, I get a "trisection" of 24.73 degrees, which is 4.73 degrees off, not very close. Of course, I may have misunderstood, as this person did not say that the circle tangent to OB should go through A. But without that condition, the angle AOX could be almost anything.

my improvement10. My improvement: Here is my improvement of the above method. Instead of using arc RX above, I draw perpendicular CD (AC = AO/3), to intersect the same arc. Testing this out on a 60 degree angle, I get 21.61 degrees, which is 1.61 degrees off, a significant improvement. And it too gets very close for small angles.

from the LA Times11. LA Times: Here is a construction (which I have simplified slightly) from the LA Times (advertisement by D.W.A., Mar. 6, 1966). We are trying to trisect angle BAC. We draw the circle with center A and radius AC (and BC). We draw line segment BC and bisect it at O. We draw line AO. We draw the circle with center O and radius OC. This crosses segment AO at L. We draw lines BL and CL. We trisect segment BC at W and J. We draw the circle with center C and radius CO. This intersects circle O at Z as shown. Draw line LZ, which intersects circle L at M as shown. Draw the perpendicular bisectors of MZ and MJ, which meet at S, as shown. Draw a circle with center S and radius SZ. This crosses circle A at point T (as shown) which may trisect angle BAC. Using this method on a 60 degree angle, I get an angle of 19.97 degrees, which is 0.03 degrees off, not bad.

Above, I said that I simplified the construction slightly. The author of this method trisected the right angle BLC to get point M. Later he drew MZ. Well, it turns out that LM and MZ are the same line. I merely noticed that point Z trisects the right angle BLC. And the facts that angle BLC is a right angle, and line LZ trisects it did not need to be mentioned during the construction. This construction also appears in Geometry by Harold Jacobs.

bisect then trisect12. Bisect and trisect: This one was in Martin Gardner's Scientific American column in 1966, and he got it from Mathematical Snapshots by Hugo Steinhaus. We want to trisect angle AOB. We bisect it getting C. We trisect segment CA, getting D, which "trisects" the angle. Using this method to trisect a 60 degree angle, I get 20.11 degrees, which is 0.11 degrees off. This is just method #1 above, trisecting a chord. But the chord is shorter, and so the trisection is significantly more accurate.

bisect twice13. Bisect twice: A natural extension of the previous method is to bisect the angle twice and then subtract off an estimated trisection. Here we trisect angle AOB. First we bisect it, giving point C, then we bisect again giving point D. Then we trisect the segment DC, giving point E, which is our trisection point. Using this method on a 60 degree angle, I got 19.9872 degrees, which is 0.0128 degrees off. You can see that trisecting the chord of a smaller angle produces much improved results. Of course the chord (CD) is here much closer in length to that of the arc (CD).

an interesting method14. An interesting method: This comes from Survey of Geometry, by Howard Eves. We are trying to trisect the angle AOB, with AO = BO. We draw a semicircle on segment AB, facing the point O. We trisect the semicircle (easy to do), giving us points D and E. We bisect segment DE twice, giving us point G (EG = ED/4), which is our trisecting point. This method works fairly well for small angles, but blows up for big angles. Trying it out on 60 degrees, I get 13.9 degrees, which is 6.1 degrees off, making it the worst method (for that particular angle) of all the methods I have tested.

d'Ocagne's method15. d'Ocagne's method: Unlike most of these methods, this one (by M. d'Ocagne in 1934) was designed only as an approximation, as the inventor knew that an exact trisection was impossible. We have angle AOB in circle O. We bisect the arc AB at point C. We extend BO to D, also on circle O. We bisect OD at E. And angle CEB is the approximately 1/3 of our original angle. Using this method on a 60 degree angle, I get 20.10 degrees, which is 0.10 degrees off. This method is found in a couple of places.

Durer's method16. Durer's method: Here is Durer's method (1525), which is very accurate. We draw arc AB with center O, and chord AB. We trisect the chord AB, at C. Draw CD perpendular to AB, intersecting the arc at D. With center A and radius AD, we draw the arc DE, intersecting the segment AB at E. We draw point F on segment EC, so EF = EC/3. With center A, and radius AF, we draw the arc FG, intersecting the arc AB at G. line OG approximately trisects the angle. Trying this method on 60 degrees, I get an error of less than 0.01 degrees. I will try to get a better estimate of the error. This method is found in several places. Durer was a famous painter and mathematician.

Karajordanoff's method17. Karajordanoff's method: This is an intentional approximation. We draw two circles about O, one with A and B on it, and one with twice the radius. We bisect the arc AB at C. We draw line AC. We draw BD perpendicular to OB, which intersects line AC at D. Se draw DE parallel to OA, which intersects the larger circle at E, which approximately trisects the angle. Using this method on a 60 degree angle gives us 19.97 degrees, which is 0.03 degrees off. This method is found in The Trisection Problem, by Robert C. Yates.

Kopf-Perron method18. Kopf/Perron method: This is another intentional approximation. We bisect OC at D, and draw the perpendicular DE (to DA). DF is 1/3 DE. We extend OD out to H, so CH = OC. We draw a circle with center F and radius AF. Draw line CB which intersects this arc at G. Then angle AHG is approximately 1/3 of the oringinal angle. Testing this method on 60 degrees, it is approximately 0.01 degrees off. This method is found in The Trisection Problem, by Robert C. Yates.

JJG method19. JJG method: This was given in a book called The Mathematical Atom, by J. J. G. (a mathematician) in 1932, as a serious trisection. We bisect the angle with OC, then bisect AOC with OD (as shown). Then draw DE perpendicular to OD, intersecting OC at E. We draw the circle with center E and radius EO. The line OD intersects this circle at F. Draw EG parallel to OF, intersecting the large circle at G, as shown. Draw lines EF and DG, which intersect at H. OH nearly trisects the original angle. Trying this method on 60 degrees, I get 19.99 degrees, which is 0.01 degrees off. This method is found in The Trisection Problem, by Robert C. Yates, who apparently simplified J. J. G's. original diagram.

between one-half and one-fourth20. Between one-half and one-fourth: I received email from L. D. K. showing this trisection. With centers at the vertex of our angle, we draw arcs with radii of 1, 2, 3, 4, 5, etc., each intersecting the sides of our angle. We can bisect the angle which gives us a point on the arc with a radius of 2. We can bisect each of those smaller angles, giving us four points on the arc with a radius of 4. These give us the green points shown. With that info, we can draw the lines in our latice, and fill in the trisecting points on the arc with a radius of 3, and even divide the arc with a radius of 5 into 5 equal arcs. We can continue out farther, and divide our angle into any number of equal arcs. The originator of this method says that it doesn't work for angles greater than 45 degrees, which should be a clue that something is wrong. In fact, none of the lines (except the rays of the angle itself) is a straight line. If they go through the proper points, they are all curved. The curvature is just not very noticeable for small angles.

between one-half and one-fourthErasing the unnecessary lines and points, this method makes a nice approximation of a trisection. Here I trisect a 30 degree angle, and get 10.04 degrees, which is not bad. The inventor of this method was right, that it is worse for larger angles. Trisecting a 60 degree angle, I get 20.33 degrees, which is 0.33 degrees off, still not bad.

"early folly" "simplified"

21. "Early folly": Above left is a method which can be found at the site Early Folly. This is a method from back in 1951, when three men discovered a trisection method, had it notarized, and sent it around to universities. The instructions were somewhat vague, and it took me a while to come up with the diagram. The second diagram is my simplification, which is magnified. Here are the steps in my own words:

We want to trisect angle AOB, with segments AO = OB. Draw the arc AB with center O (this is done later in the original, but it makes more sense here, to make AO = OB). Trisect AB, producing point C (much of the original method was involved in trisecting this line). Draw the semicircle with diameter AB, on the side of AB away from O. Trisect semicircle, getting point D as shown. Draw CD, and bisect it at E. Draw EF perpendicular to CD. EF intersects the extended AB at some point F. Draw the arc CD with center F. This intersects the arc AB at some point G, which we are told is a trisection point. Draw OG.

Trying this on a 60 degree angle, I get a trisection of 20.05 degrees, which is 0.05 degrees off. The method is better for smaller angles, and not quite as good for obtuse angles.

I think this method is one of those where someone started with a near trisection (trisecting the chord), and then chose a point way off to the side (the farther the better, actually). Then an arc with this point as a center is very nearly a trisection for quite a ways along its length, so we just choose the best such point that we can come up with. No proof was ever imagined. Method #15 above gives the same impression.

four circles22. Four circles: I received this by email. We are trisecting angle ABC. We draw the first circle with center B, and label the intersections with the rays of the angle A and C. We draw the chord AC. We bisect the chord at D, and draw BD. We draw a second circle with center B and radius 2AB, and we draw a third circle with center D and radius 2AB (the two larger circles in the diagram). Line BD intersects this last circle at E. With center E and radius AB, we draw a fourth circle, which intersects the second circle at two points, F and G, the two trisectors. The trisection not very good. Using 60 degrees, I get a trisection of 18.01 degrees, which is 1.99 degrees off. This method never gets particularly close to a trisection. Using 3 degrees, I get a trisection of 0.89 degrees.

same as #1?23. Same as #1?: I got this one in my email. We are trisecting angle AOB, with AO = BO. We bisect AB at C. Then we draw circle C, with radius AC, and circle A with the same radius. These circles intersect at point D, on the side of AB away from point O. And this may be a trisection point. Trying it on a 60 degree angle, I get 19.11 degrees, which is 0.89 degrees off, the same as method #1, above. So is this equivalent to method #1? No, only when the angle is 60 degrees. This method works well for angles close to 90 degrees (it is perfect for 90 degrees and 180 degrees), but gets worse and worse for small angles. As the angles get close to zero, this method approaches 1/4, not 1/3.

same as above24. Same as above: I got this one in the mail from A. R. S. from India. In the diagram, we draw the arcs with centers C and D and radius CD. These meet at E, outside the arc CD. We bisect CE at X, which is our trisection point. A little bit of simple geometry shows that this point X is the same as point D in the previous method, and trying it on a 60 degree angle gives a trisection of 19.11 degrees, which is 0.89 degrees off.

Mark Stark's estimation25. Mark Stark's amazing estimation: This one was never meant to be a true trisection, but an estimate, and it is very very accurate. Find the interactive Java applet at Mark Stark's Angle "Trisection". We are trying to trisect angle AOB, with AO = BO. We draw the circle with center O and radius OA, and we draw the segment AB. We then choose an estimate of a trisection point D on line AB, somewhere between 1/2 and 1/4 of the way from A to B. Then we draw the circle with center A and radius AD, giving us point E where this circle intersects the arc AB. We extend line AO out to F as shown, so OF = 3AO. We draw the circle with radius OF and center O. Line DE intersects this circle at a point G, as shown. We then draw line GO, which intersects arc AB at E', which is a much better estimate of our trisection point than E was. Mr. Stark says that repeating this estimate one more time (using E' to get a new point E'') will have an error of less than 0.0000000001 degrees.

Trying this method on a 60 degree angle, and choosing D as AD = AB/3 (method #1 above), gives us an angle AOE' of 19.9999937 degrees, which is only 0.0000063 degrees off, making this the best method that I have tested.

One might assume that the length OG can maybe be almost any length. Above, OG = 3AO, but 4AO or 2AO would actually produce good estimates of our trisection point. However, it turns out that G is almost ideally positioned. If you take several different estimates D (between 1/2 and 1/4 on AB as described above) and draw their lines (DE), they will intersect at several points very near to the position of G in the diagram.

26. Missing theorem: There is a book called The Missing Theorem by L. O. R. (a later edition is called Angular Unity), about triangles which have one angle twice another angle (see Archimedes' method, below). The author claims to prove that any trisection is possible, using the classical tools in the classical way. But he cannot show us how that might be done. His proof is a little bizarre.

27. A natural attempt: I recently received this in my email (from J. L., who is not me), but I am sure that I have seen it before. We want to trisect angle AOB. We draw an arc AB with center at O. We trisect segment AB at C. We draw CD perpendicular to AB, and intersecting the arc as shown. And D is our trisection point.

When a person finds that C (in this diagram) is a rather poor trisection point, it is natural to seek a better one. D is the next most natural candidate, and is a distinct improvement. Using this method on a 60 degree angle, we get a trisection (angle AOD) of 20.40 degrees, which is 0.40 degrees off, still not very accurate. As with most of these methods, the accuracy improves for smaller angles.

Angle AOC is too small. AOD is too large, but closer. The next guess might be some point on segment CD, closer to D than to C. Any such point would be a better guess than C or D. Some of the above methods (methods 11, 16, and 21, for example) are very probably based on this reasoning.

28. A parallel angle: I received this in my email. We are trying to trisect angle AOB. We draw an arc AB. We make a guess of a trisecting chord CD, centered about the angle bisector as shown. We duplicate CD giving CE and DG as shown. We draw EF parallel to AO, intersecting the angle bisector at F. We draw FG. Angle EFG = angle AOB, and we would seem to have trisected it with the chord CD (and CE and DG). But these chords and their arc are not chords and arc belonging to angle EFG; the center (O) is the wrong center, which should be F. And so, the trisection is surely not exact.

The accuracy of this method depends on how good a guess we made when choosing chord CD. If the guess was exact, then the trisection is exact. Using this method on a 60 degree angle, and choosing a CD that is 90% of a perfect guess, I get a trisection of 19.95 degrees, or 0.05 degrees off. So this is a fairly accurate method.

Part IV - "Cheating" (violating the rules)

See Part II - The Rules.

Archimedes trisection method1. Archimedes' method: This method is exact, and is attributed to Archimedes. But has a small flaw. We are trying to trisect angle BAC. Draw a circle with its center at the vertex. It intersects the sides of the angle at some points B and C (as in the diagram). Draw line BED, as in the diagram, so E is on the circle, and D is on line AC, and DE is the same length as the radius of the circle (It may help you to mark the distance DE on the straight edge, in order to line these up). Angle BDC is a perfect trisection of angle BAC. Proof: Call angle BDC x. Triangle DEA is an isosceles triangle, so angle DEA=180-2x. That makes angle AEB=2x. Triangle EAB is an isosceles triangle, so angle EAB=180-4x. Angle EAD=x, so angle DAB=180-3x. That makes angle BAC=3x. So, angle BDC is 1/3 angle BAC. The only problem is that it is impossible to draw line BED without cheating (by making marks on the straight edge).

I received email saying that the above method doesn't work. Apparently that person measured segments EA and ED, and they weren't an exact fit. He said that assuming AE = ED was "a big mistake." He apparently did not read my description of the construction. Well the method does work, regardless of the accuracy of my drawing.

It doesn't matter how far apart the marks are on our

I see on the WWW, and from my email, that some people consider this method valid, since the rules for use of compasses and straightedge are seldom spelled out. See Part II - The Rules.

an ancient method2. An ancient method: Here is an ancient method shown in Heath's A History of Greek Mathematics. It is not shown as a construction of a trisection, but as a help in the analysis of the problem. ABC is the trisected angle, and DBC is the trisection. We draw a rectangle ACBE and then the point F is the intersection of BD and EA. We find that DF = 2AB. So if we can construct line BF so that DF = 2AB, then we can trisect the angle. And we can do this with a marked straightedge, but not with the classic tools. This method is equivalent to Archimedes' method, above.

trisecting 90 degreestrisecting a general angle3. Paper folding:We are told that an angle can be trisected using paper folding. The book Amazing Origami by Kunihiko Kasahara gives the method for trisecting a 90 degree angle, producing angles of 30 degrees and 60 degrees, but seems to give no general method. On the left, we see this trisection. We start with a square paper (we can fold any irregular paper into a square), which we fold in half (vertically in the diagram). We then fold one of the corners over onto this fold. It is easy to show that this produces the second side of an equilateral triangle, and a 60 degree angle, and thus a 30 degree angle.

Above right is a questionable (probably) method by which trisection can be done, using paper folding. We roll the angle into a cone, and slowly flatten it while making sure that we approach two folds and three equal portions of the circumference. The angle does not have to be cut from a circle, but that simplifies the process. With this method, we can divide an angle into any number of equal angles. A variation of this method is to roll the angle into a cone, trisect the circular base (with an equilateral triangle) which trisects the arc of our angle, and then unroll the angle.

trisecting a general angleHere we see a good paper folding method found in the book Geometric Constructions by George E. Martin and attributed to Hisashi Abe in 1980. Here we are trying to trisect angle ABC. We find D the midpoint of AB. We fold line a perpendicular at B to BC, then make two lines (lines EA and c in the diagram) perpendicular to line a, through A and D, as in the diagram. Then (the part which we cannot do with compasses and straightedge) we find the fold b so that E lies on line AB producing E' the image of E, and so that B lies on line c, producing B' the image of B. Line BB' trisects the angle ABC. See Origami Trisection of an Angle for the proof. Here we have found a reflecting line for two pairs of points. We can do that with compasses and straightedge only if we know where all four points are beforehand. Here each of the points could have been anywhere on a given line.

The above method suggests that we can also use the edge of the paper as a marked ruler. So, you can use any number of methods, probably including Archimedes' method above, to fold a trisection. Most such methods would involve folding the paper more than once simultaneously, to get your ruler to anywhere on the paper. The method shown at the link above moves the ruler to a specific place on the paper, because it involved only one fold to move the ruler.

trisecting a general angleHere is another paper folding method shown in Geometric Constructions by George E. Martin, and the method and proof are attributed to Dayoub and Lott, for a geometry tool called a Mira, a mirror which allows you to draw reflections. Paper folding produces the same reflections. Let me describe the trisection using the diagram. We start with an arbitrary angle ABC. We find D, the midpoint of AB. We fold line a, through D, and perpendicular to BC. We fold line b, through D, and perpendicular to line a. Now (the part we cannot do with compasses and straightedge) we find line c by folding so that A folds onto line a and B folds onto line b, producing A' the image of A, and B' the image of B. The line BB' trisects angle ABC. Apparently there are three lines which meet the criteria that I described (folding A onto A' and B onto B'), but only one of them intersects segment AD. See the above book for the proof.

Euclid Challenge4. "Euclid Challenge": At Euclid Challenge, we come to a diagram similar to this (I have supplied the question mark next to point T). The two arcs have the same center, point B, which is out of my picture, to the south. The lower arc has 3/4 the radius of the upper arc. The lines DD', CC', and PP' meet at a point B. In the previous pages, the author constructed these points and lines so that angle DBP is 3.75 degrees (1/16 of 60 degrees) and angle DBC is 15 degrees (1/4 of 60 degrees). We now are going to find point T, so that angle DBT is 5 degrees, an unconstructible angle according to mathematicians. Let me quote the author on this (his steps 14 through 19, rephrased slightly), the italics are mine:

Grasp with the left hand a compass leg above the legpoint (identify as leg R). Place legpoint of leg R on point D'. Grasp with the right hand the other compass leg above the legpoint (identify as leg S). Place legpoint of leg S on D. Move both legpoints at a uniform rate; legpoint of leg R along arc D'C' toards point C', and legpoint of leg S along arc DC towards point C. When the legpoint of R reaches point P', stop both legpoints. At the location of the legpoint of leg S on arc DC mark point T -- trisection point.

The author further explains his "uniform rate":

Confirmation of the "Uniform Rate": The maximum travel on arc D'C' = 5.625 degrees (1/8 of 45 degrees) and on arc DC = 7.5 degrees (1/6 of 45 degrees or 1/4 of 30 degrees). These two angles, 5.625 degrees and 7.5 degrees can be constructed with a compass, and used for test points by continuing the movement of the two legpoints beyond the 3.75 degrees on ard K'L' [arc D'C'], and 5 degrees on arc KL [DC], to see that legpoint R reaches test point 5.625 degrees at the same time that legpoint S reaches test point 7.5 degrees.

This clever sliding of compass points at a uniform rate is, of course, a great violation of the rules, even if it were possible. The idea is that arc D'C' (and any of the smaller arcs on it) is 4/3 as long as arc DC, which is true. And this author seems to think that you can measure fixed distances along an arc using compasses, which you cannot. If you could, then you could make a line segment the same length as pi (which I think he does on another page). In my diagram above, the length of the arc D'P' is supposedly the same as the length of the arc DT. If we could do this with compasses, then angle DBT would indeed be 5 degrees. But, we can't. See Part II - The Rules and the next "trisection" below.

If, instead of measuring arc lengths with our compasses, we make the chords D'P' and DT of equal length, then we get an angle DBT very close to 5 degrees, less than 1/100 of a degree off. But, as we have seen several times above, chord length and arc length are not related in such a simple way.

To summarize my objections:

  1. This method violates the normal rules.
  2. If we accept this method, there is a much simpler way to use it. See "5. Duplicating an arc length," below.
  3. It can't be done. There is no precise way to move two points at the same speed, on two different arcs, simultaneously. In fact, there is no precise, finite way to measure arc lengths without measuring angles.

duplicating an arc length5. Duplicating an arc length: Using the same idea (duplicating an arc length) as the previous "construction," we can devise a simpler trisection. In this diagram, we want to trisect angle AOB. We make segment OC three times segment OA. Then the length of arc CE is three times the length of arc AB. We duplicate arc length AB and get arc CD, which trisects angle AOB.

I have never seen this "construction" given as a valid trisection. But this duplicating of an arc length is the main flaw in many of the trisection attempts shown above. Of course this trisection is perfect; it just can't be done with the classic tools. If we use chords instead of arcs, this "trisection" gives the same lengths as the one in Part III, "trisection" #1.

Just how are chord length and arc length related? With an angle A and a radius r, the chord length = r sin A/cos (A/2), and the arc length = 2 pi rA/360. You may remember from trigonometry or calculus that sines and cosines are not particularly simple functions of angles. They can usually only be estimated, and to evaluate them we use infinite series.

6. Tripling an angle: One common mistake is to choose an angle (or construct it), then triple it, getting a larger angle, and then claim that you have discovered a way to trisect the larger angle. Of course, you can easily triple any angle. But to go the other way, and trisect an arbitrary angle is what we are trying to do here. If we triple some angle, then the larger angle is not arbitrary in any way (at least not as mathematicians use the word). The Trisection Problem, by Robert C. Yates tells of a university president (J. J. C) who went public to the newspapers with such a trisection. Here is that trisection:

JJC method Draw lines BC and DF parallel to each other. Choosing arbitrary points C and D on the two lines, we draw the arc CF with center at D as shown. We draw lines DC and BF. With center at F, we draw the arc DB. We draw angle DCE equal to angle DCB. We draw AD parallel to CE and DE parallel to BF; these two lines intersect at a point E. Then lines DF and DC trisect the angle ADE.

How true. But angle ADE is not arbitrary. All the above accomplished was to construct an arbitrary angle DCB and then triple it, giving us angle ADE. Angle DCB is arbitrary (depending on where we put arbitrary points D and C), but it would be difficult to draw a really small angle, as points D and C would have to be very far apart. There are easier (somewhat) ways to triple an angle. You might want to verify that all of the small angles in the diagram are equal.

7. With a ruler: The tangent of 20 degrees is roughly 0.363970234. Using a ruler with markings that fine (or less accurate, if you're not so picky), draw a right triangle with one leg exactly 1, and the other leg 0.363970234, and you've got a 20 degree angle. Of course, the tangent of 20 degrees is not exactly 0.363970234, but that is more accurate than any ruler you will ever find. So isn't that good enough? Nope, it's still an approximation, no matter how accurate it is.

Part V - "Cheating" (using other tools)

With compasses and straightedge, only circles (and circular arcs) and lines can be drawn. Abandoning compasses and straightedge, it is simplicity itself to trisect any angle. Use a protractor (see the drawing). I assume CAD programs can trisect angles at will. We can measure the length of an arc with a curved string or piece of metal (like spring steel). A person could design any number of mechanical devices which can trisect any angle. And curves have been drawn, which make trisection easy (Part VI below). Or hard.

a real trisection1. A trisection tool: Here is a little tool (apparently invented by C. A. Laisant in 1875) which trisects angles. It is exact, assuming the line segments are exact. The dots are hinges. The two hinges at the far right slide along their trisection lines. We have two rhombuses and two of their diagonals.

We cannot use compasses and a straightedge to trisect a general angle. That has been proven. But we can make other tools (such as the device drawn here) to trisect such an angle. In fact, we can use a pair of compasses and a straightedge to make this device. All that this device does is triple the smallest angle.

tomahawk2. A tomahawk: Here we have a clever drafting tool called a tomahawk (or shoemaker's knife (there is another shoemaker's knife in geometry), invented before 1835), which can be used to trisect angles, as shown in the diagram. For small angles, the handle may have to be extended. The top end is marked off in three equal pieces, one of the marks being the center of the semicircle. We can prove that the trisection is exact by drawing three congruent right triangles, as shown here.

Why can't we just draw a tomahawk using compasses and straightedge, and then move that part of our diagram over to our angle, adjust its orientation, and then trisect the angle. Well that amounts to using a tool; we are cutting out a part of the plane and moving it, which is a complicated process and a violation of the rules. But also, there is no way to position our drawn tomahawk with perfect precision; in other words, we cannot draw our tomahawk in place, on top of the angle, and trisect the angle. That may be hard for you to believe.

carpenter's square3. A carpenter's square: A similar trisection can be done with a carpenter's square shown here. First we draw the horizontal line shown, which is parallel to the horizontal side of the angle, and the width of the square from that side of the angle. Then we position the square as shown, with equal lengths marked off on the righthand edge of square itself. And two of those points trisect the angle. We can verify that with the same three congruent triangles that we used with the tomahawk.

Part VI - Cheating (using curves other than circles)

There are several curves that can be used to trisect angles. Such a curve is called a trisectrix. These trisections are exact. But of course, you cannot draw these curves with compasses and straightedge. I drew most of these curves by drawing an angle and its triple, and then generating a curve (locus of points) by plotting the trisection point while varying the angle. Most of these curves are of the fourth degree or higher.

a trisectionlimacon1. Limacon: On the left, we see a special kind of limacon (limaçon), the purple curve that you see. Its polar equation is r = 1/2+cos(theta), where theta is the central angle. The origin is where the curve intersects itself. A general limacon has the equation r = b+a cos(theta). A limacon is also called a limacon of Pascal (named after the father of the famous Blaise Pascal). I have discovered five ways to use this limacon to trisect angles:

(i) On the right above is a simple method, just using the little loop of the limacon. We want to trisect angle ABC, we make AB the right length to fit in the loop, and make BC = AB. We then draw line AC, and where it crosses the limacon is a trisection point. If trisecting AC was a true trisection of the angle (part III, method #1, above), then this curve would be a circle. This seems to be the standard, well-known method. But limacon #4, below has also been published.

(ii) Above left, I am trying to trisect angle ABC, where AB is the right length to fit in the limacon as shown.. Make BC = AB, and draw line AC. Trisect line AC at D. Draw DE perpendicular to AC. The point E where this intersects the limacon is a trisection point..

another limacon a fourth method a fifth method

(iii) Above left is yet another way which I discovered for using a limacon to trisect an angle. Again we want to trisect angle ABC. Here we adjust the size of AB so that segment BD (half the length BA, as shown in the diagram) fits inside the small loop. We then draw AC and bisect it twice, giving AE = AC/4. Where line DE intersects the large part of the limacon is a trisecting point.

(iv) Above middle is yet another way to use a limacon to trisect an angle ( found in The Trisection Problem, by Robert C. Yates). It is closely related to method #3. We trisect the angle AOB merely by drawing BC. Then angle OBC is 1/3 angle AOB (angle BCA is 2/3 the same angle).

(v) Above right is yet another way to use a limacon to trisect an angle. Here OA is the same length as from O to the bottom of the limacon. Our angle cuts the limacon (as shown) at C. We draw a line OD a distance of OC from point A (again, as shown), and OD is our trisection.

Maclaurin's trisectrixanother way2. Maclaurin's trisectrix: On the left is Maclaurin's trisectrix. We want to trisect the angle ABC. We choose length AB so that AB = AO/3. I then label the point where our angle crosses the loop of the curve C. Then angle AOC is the trisection of angle ABC. The triangle BOC can be seen in Archimedes' method (part IV, method #1, above). But point A is not part of that method.

(ii) Above right is another way to use the Maclaurin trisectrix. We are trisecting angle AOB. Here B is chosen so angle ABO is a right angle. CD the perpendicular bisector of AB (at C) intersects the loop of the curve at a trisection point. Notice that it doesn't matter where A is (somewhere on the horizontal line), as constructing the perpendicular bisector of AB will always produce the same line CD.

conchoid of Nicomedes4. Conchoid of Nicomedes?: One WWW site calls this the conchoid of Nicomedes, but (as far as I know) that looks quite different. We are trying to trisect the angle AOB. We make the length AO the same as the radius of the circle O. The side OB of the angle intersects with the part of the curve that curls off to the left at a point B. We duplicate length OB as AC perpendicular to OA, and C is a trisection point for our angle. The polar equation of this curve is r = sec(theta/3) = 1/cos(theta/3).

quadratrix of Hippias5. Quadratrix of Hippias: This curve is the quadratrix of Hippias, and it allows us to divide an angle into any number of equal parts. Here we want to trisect angle AOB. The left ray of the angle crosses the curve at point B, and we draw the horizontal line BC, with C on the y-axis. Then we divide the y value by 3, and we get the y value of point D, also on the y-axis. Then we draw the horizontal line DE, which crosses the curve at E, which is the trisection point.

An equation for this curve is x = y cot(pi y/2) with x being a function of y.

Above we divided the y value of point B by 3, to get the trisection point. If we instead divided by 5, we would have found the point which divides angle AOB into five equal parts. Divide by 7 and get seven equal parts, etc.

This diagram, and the next one, were drawn with Geometer's Sketchpad and Paint Shop Pro.

Catalan's trisectrix6. Tschirnhausen cubic: Here is Tschirnhausen cubic, also called the Catalan's trisectrix, and L'hospital's cubic. Its polar equation is a = r cos^3(theta/3). I don't know how this can be used to trisect an angle. This curve is related to a parabola with focus at the origin, opening toward the left in the diagram, and the two curves intersect at (1,0) in this diagram. Apparently, both the parabola and the cubic curve are used to trisect an angle. I will try to figure this one out.

a cubic parabola7. A cubic parabola: This is a cubic parabola with the equation y = (x^3)/2. We are trying to trisect angle AOB. Along with the curve, we have two circles of radius 1 and 2 (the two larger circles in the diagram). BO intersects the smaller circle at C. We draw CD perpendicular to AO. OE = OD, as in the diagram. We draw EF with a slope of 3/2. It intersects the curve at G. We draw GH perpendicular to AO. GH intersects the larger circle at H, which is the trisection point.

There are other cubic equations that can be used in this way, with different slopes of FG. But this one has the simplest equation. In my diagram, it may look like OC is parallel to FG; it is not.

I found this trisection in The Trisection Problem, by Robert C. Yates. I found other cubic equations by botching the curve drawing.

my curve8. The "cycloid" of Ceva: Here is a curve discovered by Ceva in 1699. We want to trisect the angle ABC, which we fit into the blue circle (the radius of the circle is determined by the smallest loop of the curve). Then we draw CD parallel to BA until it meets the curve to the right for the last time, giving us point D, which trisects the angle. The intersecting point can be deduced to be x = sin(theta) and y = cos(theta/3). A polar equation is r = 1+2cos(2 theta).

I assumed that I was not the original discoverer of this curve. But most books don't seem to mention it. I finally found it in The Trisection Problem, by Robert C. Yates. In that book, the position of our angle is somewhat different.

a hyperbola9. A hyperbola: This one is apparently a hyperbola. It was apparently discovered by Pappus of Alexandria, who lived in about the year 300. O is the center of the hyperbola. We are trisecting angle ABC. AB = AO = OX. We draw AC perpendicular to AB, and CE perpendicular to AC. We draw AD = 2BC and intersecting the curve at D. We draw DE perpendicular to CE, and intersecting CE at E. Draw line BE, which trisects angle ABC.

We didn't need the last couple steps, as we already had an angle (OAD) which trisects angle ABC. I guess that if a hyperbola can be used to trisect an angle, then maybe an ellipse can too.

a rose curve10. A rose curve: A three-leafed rose curve has the right equation for the job: r = a sin(3 theta). For angles less than 180 degrees, we just use the upper right petal (leaf). We want to trisect angle AOB. We draw a circle half the radius of the rose curve, with center O (with B on the circle). We draw BC perpendicular to AO, at C. We draw a larger circle with radius OA+BC. This circle intersects the curve at two points, and we choose one of them (D) depending on which quadrant B is in.

11. A parabola: This is the parabola y=x^2. We have drawn a circle about the origin with radius = 2 (the purple circle). We want to trisect the angle AOB, with A on the circle. We draw AC perpendicular to the y-axis. We draw ED perpendicular to the y-axis, with E two units above the origin and ED = AC/2, as shown. We draw a circle with center D and radius OD. This intersects our parabola at F. We draw FB perpendicular to OB. This perpendicular intersects our original circle at G, which is the trisection point.

This method is found in The Trisection Problem, by Robert C. Yates.

The following are curves which I discovered while experimenting with trisections. I will begin the numbering again:

another curvesame curve1. A more complicated curve: On the left are four pieces of an intersting curve that I discovered, which can be used to trisect an angle. We want to trisect angle ABC. We fit segment AB into circle B. CD is perpendicular to BC. Where CD intersects with the curve is a trisection point. Above right is the same curve, as seen from a greater distance. If angle BCD is some other fixed angle, besides 90 degrees, then we get different positionings of the same curve.

Each branch of our graph may be a hyperbola. I will have to deduce an equation, in order to tell.

yet another curve2. Another complicated curve: Here is another curve that I discovered, that can be used to trisect an angle. Just the outside loop is used. We are trisecting angle ABC. We make segment AB the right length to fit into the curve, then make BC = AB. We make rhombus ABCD. We draw AC and trisect it, giving us point E. Line DE crosses the curve at a trisection point, F. Like the limacon above, this curve is very nearly a circle for small angles.

another complicated curve3. Yet another complicated curve: Here is another curve that I discovered. We want to trisect angle ABC. We choose length AB so that AB = BD/3. Then we draw AC perpendicular to BD (we may already have this drawn while graphing the curve), giving us C. We draw CE perpendicular to AC, and where it crosses the right part of the curve is the trisction point E. So we draw BE.

This one may be a hyperbola. The two curves on the left side of the diagram may be related hyperbolas? I'll have to figure out the equation, in order to tell.

anotherthe curve4. Another: Here is another curve. We want to trisect angle AOB (the entire curve is shown on the right; I think I've seen him before). We fit OA into the curve as shown, and draw circle O with radius OA. Then we find point C where the small loop of the curve intersects OB, drawing a circle with center B and radius BC. This circle intersects circle O at point D, within the angle, which is a trisection point.

a useless curve5. A useless curve: Here is a complicated and useless curve for trisecting angles. For angles less than 120 degrees, we use the branch of the curve shown. We want to trisect angle AOB, we fit our angle into the diagram, which should already contain segment AO, and the circle O. The extended segment OB intersects the correct branch of the curve at C. Line AC intersects the circle at the intersection point D. This curve is almost unusable for most angles, as line AC is almost parallel to the circle for small angles, making the intersection point hard to choose, and point C is way out there (off the chart) for most other angles.

Sometimes I prefer the useless methods.

another6. Another complicated curve: This one is based on the previous one, but is more useful. We want to trisect angle AOB. We will use the left part of the curve which looks like a hyperbola, and the circle O with radius OA/2. Segment OB intersects the curve at C. We draw segment AC, which intersects the circle at D, which is the trisection point.

This graph is the previous graph, but with a smaller circle, which makes AC cross the circle at a better angle.

useless7. Another useless one: We trisect angle AOB within circle O by extending OB to C on the correct branch of the curve. We draw CD perpendicular to OA, intersecting the circle at D, which is the trisection point. This one is useless for exactly the same reasons as is method #5.

a better one8. A better one: Here we use the branch of the curve that is concave to the left. We draw circle O, as shown. We are trying to trisect angle AOB, and we label B as the point on OB that intersects the correct branch of the curve. We draw circle B with radius BO. This circle intersects the original circle O at C within the angle, which is a trisection point.

How did I come up with all of these interesting curves? Well, I just drew a bunch of loci (paths of points) which involved trisections in various ways. Sometimes I came up with a well-known curve (often a limacon), and at other times I discovered new curves.

yet another curveLet me give an example. In method #2 above, which used this diagram. I drew the angle ABF (I'll define point F later), and then tripled it to get angle ABC. Then I drew the rhombus, and trisected diagonal AC as shown. Drawing segment ED gave me point F. I then varied angle ABF and drew the path of point F, which gave me the curve. This was fairly easy to do, using Cinderella. Then I did a screen capture with Paint Shop Pro, and cropped the picture, added the labels, and saved it as a gif image with a transparent background.

Many of the images on this page, not just in this section, were drawn in a similar way.

Part VII - "Cheating" (choosing an arbitrary point)

When proving things, or constructing things, it is sometimes useful to "choose an arbitrary point," perhaps on a line or between two points. This is perfectly valid. If we "draw an arbitrary line" through a point, we are essentially choosing an arbitrary point, and then drawing the line through the two points. Some trisections involve arbitrary points, and these arbitrary points are usually chosen so well that the trisection comes out looking good. Other choices for these points will probably not look very good at all. In that case, "choose an arbitrary point" is a scam (the scammer may be scamming his/herself, so don't blame him/her). You should try moving the arbitrary point, and see what happens to the trisection. Rarely, the choice of the arbitrary point may have nothing whatsoever to do with the trisection; it may just be part of the smoke screen.

This method of cheating amounts to drawing the diagram just right (a Baby Bear diagram) so that the trisection comes out almost perfectly. There is probably one perfect point (or angle or length) out there, which you cannot locate with compasses and straightedge (by the way), which will perfectly trisect your angle. Get close to that point, and you will not be able to tell that you have not trisected the angle. Miss it by a mile, and the trisection falls apart. These methods can get fairly amusing. Here are a few of those: [under construction]

56.60 degreesa smaller angle1. A 56.60 degree angle: Here we trisect side AO giving us C, so AC = AO/3. And we bisect angle A, the bisector intersects segment BC at D, the trisector. This method is fine tuned for an angle of about 56.60 degrees. For any other angle (including small angles), the point D is way off. There is an angle here (about 56.60 degrees) which can be exactly trisected with this method, but we don't know what it is exactly. Above right, I have used this method to "trisect" a smaller angle, and the trisection ends up being larger than a bisection.

LJRH methodmake BC bigger2. LJRH method: This one is found in The Trisection Problem, by Robert C. Yates. He says that this appeared in a science periodical, and was invented by L. J. R. H. We are trying to trisect angle AOB. We draw the arc AB with center O. Then we draw arc AO with center B, and choose an arbitrary point N on that arc. Then, with center N, we draw the trisected arc BT of "any" length (actually we trisect it by just drawing the circle, then making three arcs BC, CD, and DT with equal chords of any length BC). We bisect the arcs AB (at Q) and BT (at P). Then we draw lines AT and PQ, which intersect at some point K. Draw lines KD and KC, which intersect arc AB at the trisection points E and F. This looks very good for most positions of N and relatively small lengths BC (within a fraction of a degree). When BC gets longer, then the "trisection" gets really bad; see the diagram above right. For some angles, it is even possible that line KD could miss circle O completely, and then where would that trisection point be?

If this method specified where N should be, and how long BC should be, then it would be a good trisection attempt for angles less than 90 degrees.

Part VIII - Comments

I didn't mean to become a trisection buster (or debunker). I was willing to point out obvious errors. But some of the methods sent to me were complicated or poorly described. How do you bust a construction that you cannot understand? Well, this is now one of my really fun projects. Having busted some of the more complicated methods above, I find that I enjoy this. And I am certainly improving my trigonometry skills.

There is a joke that there are two types of people, those who divide people into two types, and those who do not. Well, there are several different types of people who choose to believe that trisections (using the ancient tools) are possible:

  1. The first kind are mathematically illiterate, and seem unsure if 2+2 always equals 4. Most of humankind, mostly nonscientists and nonengineers, are similarly mathematically illiterate. I told one of these people that simple trigonometry shows that his trisection of 60 degrees made one of the angles 19.07 degrees, and he informed me that the trisection was a geometry problem, not a trigonometry problem, a response that left me speechless (for a moment).
  2. A second kind of people are mathematically adept (algebraically and geometrically) to some extent, but seem to think that an answer that is close to being right is right. They measure two angles and think that they have the same measure, and they don't bother to prove that they have the same measure. There are many such people.
  3. A third kind are also adept at mathematics, and think that they have trisected an angle, but have made some sort of mistake; and then they find out that it is supposed to be impossible. These people react with indignation when told they did not really trisect an angle. Instead of trying to find their errors, which would be very educational, they are angry at me, as if it is my fault.
  4. Is there a fourth kind of person, who has actually trisected an arbitrary angle, with the appropriate tools? Well, there is apparently a mathematical proof that such a person cannot exist. A mathematical proof meets a very high standard of proof.

construct a square rootIn the excellent book Eureka! (Math fun from many angles) by David B. Lewis, Mr. Lewis says that one of the reasons that you cannot square the circle, is that you cannot construct the square root of an irrational number. This is not true. Here is a construction that produces the square root of any length x. The two lines are perpendicular. If that length is irrational (like the square root of 2) then the length shown is the square root of it (fourth root of 2, in this case where x is the square root of 2). It is impossible to construct a line segment equal to pi. If you could, then you could take its square root.

Of course some angles (90 degrees, for example) can be trisected. I have a truly remarkable trisection of a 0 degree angle, which can be done without any tools, whatsoever. There is the stange case of the angle 3pi/7 (540/7 degrees). This angle cannot be constructed. But (if you managed to miraculously have it before you) it can be trisected (Honsberger 1991).

Conway's test: John Horton Conway suggests that a trisection method be tested by seeing what happens if you double the angle being trisected. If you really have a true trisection method, then doubling the angle will double the trisection. In other words, we use our method on angle A, and get some angle B (hopefully A/3). Then if we use our method on 2A, then our trisection should be 2B. If we get some other angle than 2B, then our method doesn't work. Of course we are very suspicious of methods which only seem to work for small angles (or for some other range of angles).

An analogy:

Here is an analogy which may put the above into perspective for you:

We have a line segment, and our construction tool is a rubber band with a point in the middle. All we can do with this tool is bisect a segment. Both ends must be on the line, and so, with this tool, we cannot construct any point outside the line.

Let's say we want to trisect the line segment (or any smaller sub-segment of the segment). All we can do is repeatedly bisect segments. If we are allowed to do this infinitely many times, then we can easily trisect the line segment (1/3 = 1/4 + 1/16 + 1/64 + 1/256 + ...). But no finite sum will ever do the trick. You can do it to any accuracy you want (certainly more accurate than your tools), but you can never be exact. There is no sum of these fractions (1/2, 1/4, 1/8, ...) which can possibly add up to 1/3, without using infinitely many of them.

The situation with straightedge and compasses is more complicated because those tools can easily produce square roots. But they can't do cube roots and the roots to other cubic equations. And they can't do many other lengths.

Most of the above diagrams were drawn using Cinderella and Paint Shop Pro.

Return to my Mathematics pages
Go to my home page