Trisection: CR4 Challenge (04/15/08)

Let me clarify this, Let's say we start with a 45 degree angle. In the end you want to end up with two lines drawn at 15 and 30 degrees off one reference side. Is this correct?

Theoretially one could keep bisecting angles until we get close enough to the target, but I would assume this is not what you are thinking. In this case one could just keep using the same compass setting and drawing arcs to get the half point then draw a line, then repeat.

Also I am not sure what you mean by 5 feet. Could you express the desired resultant accuracy in terms on precentage of the orignal angle.

Hmm. It seems clarification is required.

1) The desired angle is required to be close in value to 1/3 of the original angle. As the location is unspecified you can place it wherever you wish (not necessarily sharing any line with the original).
[Nevertheless, all of my "low-operation-count" procedures to date have shared an edge and the vertex with the original angle, but I repeat that this is not a requirement]
2) An obtuse angle is strictly greater than 90^O and strictly less than 180^O
3) 5' = 5-arcminutes, or 1/12-degree. Apologies that the diagonalisation of the single prime got lost in translation.
4) The method of alternating bisections would be legitimate in principle - but it would take too many operations to attain this accuracy.

X = 1/4 + 1/16 + 1/64 + 1/256 +

= 1/4 + X/4

3X/4 = 1/4

X = 1/3

As you say, bisection would get there but, alas, too late. I just put in the series above because I like the way it adds up, but maybe there are other series that work better. It's a good job Fyz didn't ask for inches ! 1 Degree (^o) = 60 minutes ('), 1 minute = 60 seconds (").

Edit: dskktb posted his solution while I was working on this.

Here is a sequence I entered into AlgoSim (edited for spacing):

.25+.125 = 0.375
Ans-.0625 = 0.3125
Ans+.03125 = 0.34375
Ans-(.03125/2) = 0.328125
Ans+(.03125/4) = 0.3359375
Ans-(.03125/8) = 0.33203125
Ans+(.03125/16) = 0.333984375
Ans-(.03125/32) = 0.3330078125

Don't know how this applies. Kris got me to thinking, and I had to follow up on it. I'm beginning to suspect that an application of the tangent function would prove beneficial. I'll mull it over later, got other fish to fry right now.

Line AC ........................................... > 1

Cir centre any point on line.................. > 1

Arc through centre ............................ > 1

90° tri line OX, OY, OC........................ > 6

Arc OA ............................................ > 1

First cir R ......................................... > 1

Second cir R .................................... > 1

Line OB ............................................ > 2

Total > 14

Tri BOC = one third of obtuse tri AOC.

Regards JD.

Its wrong. must cut down on the port.

Regards JD.

well, take an angle, triple it, and reverse the process to find the original angle.

like squaring the circle, it cannot be done as long as you follow the restrictions on the gear you can use. To get to the minimum number of moves within that error limit needs some elegance.

Aha.
I imagine that you already know this, but "arbitrary" should be taken to mean imply that you know nothing about the angle except that it is obtuse - so you haven't created it by multiplying a known angle by 3.

While writing - I currently have a method that will do it using 15 operations (probably still not optimal) - but anything less than 18 will still answer the challenge

This method should be exact. I don't report all the passage because I'm busy at moment, but it is sure is possible to use only compass and straight-edge.

Given the angle BOA:

build the parallelogram BOAC;

Trace the diagonals;

Find the centre point D of OA;

Trace the lines DB and DC;

Find the point H as intersection between EF and GD

The angle HOA is exactly 1/3 of BOA.

This method was found starting from a general method to divide a segment in any part you desire:

Regards

The angle HOA is exactly 1/3 of BOA.

Why ?

You have successfully trisected the line - but that is not the same as trisecting the angle. You could try measuring the angle starting with an angle of (say) 120^O.

I chose 120^O because this
a) It is an example of an angle that cannot be trisected exactly using the methods specified, and
b) Unless you fortuitously have the trisected line in a suitable position, the error should be large enough to be visible.
[N.B.1 if you choose a position for the trisected line that gives a relatively small error at 1/3 of the angle, it will increase the error at 2/3 of the angle, showing that the "exact" angle is pure coincidence!
N.B.2 Even for bisection this method only works if you start with an isosceles triangle - and the reason it works here is that everything is symmetrical]

Looks like they add up to 18. Interesting, how did you come up with this little jem?

I don't know if I'm in the right track, but I've got an idea. I will elaborate it later if I find some time, but here it is for now:

At first, we draw a circle with its center at O and we get points A and B. Then we trisect segment AB (that's a simple Euclidian construction) and get rays Oz and Ow. Needless to say that Oz and Ow do not trisect the angle, but nevertheless, they form a good starting point for the iterative process that will follow.

Now, the iteration idea is that we take the bisector of Oz and Oy deriving a ray Oz' as well as the bisector of Ox and Ow, getting ray Ow'. Bisection is again a simple construction. We then go on and bisect Oz' with Oy to get Oz'', etc.etc. I have a feeling that this process converges, so if we continue like this we will be getting closer and closer to the solution. I may be wrong, but at first it seems as a quick and dirty engineering approach.

I can't reall follow your text. Are you alternately bisecting the angles:
OBC -> OD'
OAD' -> OC'
OBC' -> OD''
OAD''-> OC''
etc?

If so, it is better to perform the bisections before the trisection, because the bisections always halve the error, whereas the quality of the trisection improves when you reduce the initial angle.
However, I believe that the method using bisections followed by trisection of a chord requires 21 steps to achieve the specified accuracy (but it's more than possible I have missed some useful algorithms).

You are right to be confused, I made a typing mistake. What I meant:

"... we take the bisector of Oz and Oy deriving a ray Ow' as well as the bisector of Ox and Ow, getting ray Oz'..."

BTW, I used some trigonometry to derive the initial angle φ=zOw as a function of the whole thing, call it θ. I arrived to the formula: tan(φ/2) = 1/3 tan(θ/2)

For example, if θ=90^o then φ=38^o . Not exactly 30^o as it should be, but it is 4^o away the proper value on each side. After the first step of the iterative process, this error reduces to half, i.e. 2^o and after the second goes down to 1^o and so on. Problem is that for bigger angles, the initial error is higher, so you need more steps to reach the desired accuracy... I better take back the "quick and dirty" label I assigned to this methodology and leave only the "dirty" part!

I agree the formula for the trisection approximation. It would be better to use only one of the starting trisections OZ and OY (OZ), and successively halve the resultant 2/3 angles. The error halves at each step. However, it is much better to perform any bisections first and then trisect the final difference between bisections, as the error in this case falls by a factor of about eight for each bisection. (Even this requires a large number of operations if you use the method of trisecting a chord).

Well I was able with the help of CAD to prove that the meathod works and it took a total of 9 circles and 9 lines for a total of 18 marks. The results below show that it should work for any obtuse angle. The amazing thing though is that in the work, the accuracy is better than 1/60th of a degree, about +-0.0141 degrees which is better than the 1/12th that was asked.

I have numbered the order in which I drew the objects in the two pictures.

Steps 1 - 4: (4 Blue Circles)

1.Given an obtruse angle we draw any size circle with the compass from the apex of the two lines.

2 and 3.Keeping the compass the same size, draw again at the two points of intersection between the first circle and the two lines forming the obtuse angle. (2 circles)

4. Now again with the same size circle draw the center at the new intersection point between circles 2 and 3.

Steps 5-7: (3 green lines)

5. Draw a line from the apex of the obtuse angle and draw thru the intersection point of circles 2 and 3 and extend line till the edge of circle 4.

6. Draw a line from the intersection of circles 1 and 4 to the intersection of circles 3 and 1 and extend to the edge of circle 4.

7. Draw a line from the intersection of circles 4 and 1 to the intersection of circles 2 and 1 and extend to the dege of circle 4.

Steps 8 - 10: (Yellow circle and two yellow lines)

8. Draw a circle centered at the intersection of circle 4 and line 5 where the radius meets the intersecting points on circle 4 with lines 6 and 7.

9. Draw a line from the left intersection of circle 8 and circle 4 thru the intersection of circle 8 and line 5 and extend to line 6.

10. Draw a line from the right intersection of circle 8 and circle 4 thru the intersection of circle 8 and line 5 and extend to line 7.

Steps 11-13: (Magenta circle and two magenta lines)

11. Draw a circle centered at the intersection of circle 4 and line 5 simalar to circle 8, but with a larger radius where it intersects at lines (10 and 7) or (9 and 6).

12. Draw a line from the right intersection of circle 8 and line 10 on circle 4 thru the new intersection point of circle 11 and line 5 and extend to line 7.

13. Draw a line from the left intersection of circle 8 and line 9 on circle 4 thru the new intersection point of circle 11 and line 5 and extend to line 6.

Step 14-16: (3 red circles)

14. Draw a circle centered again a the intersection of circle 4 and line 5 where the radius goes thru the intersection of lines (12 and 6) or (13 and 7).

15. Draw the same size circle as 14 at the intersection of circle 4 and 3 not on the original obtuse angle.

16. Draw the same size circle as 14 at the intersection of circle 4 and 2 not on the original obtuse angle.

Steps 17-18: (black lines)

17. Draw a line from the apex of the obtuse angle tangent to the right side of circle 16, this is one trisection.

18. Draw a line from the apex of the obtuse angle tangent to the left side of circle 15, this is the other trisection.

I checked the dimensions in CAD for the angles only. The smaller trisection came out to 52.3193 degrees about. The origional arbitrary obtuse angle was 157 degrees. The trisect angle should be about 52.3333 repeating degrees. My solution uses 18 steps given the original angle, and straight ruler with no markings, and a compass where the size can be kept the same.

In the end, my solution is +-0.0141 degrees or about 1/70th of a degree or 0.846'.

In order to check the answer for a good solution, I drew two aqua colored lines in CAD at 52.3333 degrees and 104.6666 degrees for visual comparison.

Good work ! - I spent way too long thinking about this today.

It is good to see a division by three that is toleranced - although you haven't said whether it gives similar accuracy at smaller angles (e.g. 136^O, 120^O and 91^O). Also, you could omit some of the steps, because the challenge only requires a single angle of 1/3 the original size.

On the other hand, even if all the steps were acceptable (see next paragraph), the number of operations (as defined in the challenge) would be 30, not 18 (see final paragraph).

Euclid did not admit of drawing tangents without first defining the point on the circle through which the tangent passes. I should have made this clear in the original challenge, but overlooked it (in mitigation, I would claim that I had not expected it to be helpful). (N.B. the standard way to define the point is to draw a line from the centre of the relevant circle to the other point on the line, and draw an arc from the mid-point on this line that passes through the centre of the original circle)

Returning to number of operations: Every time you place the point of the compass, that is an operation (or step if you prefer). If you then adjust the radius of the compasses to pass through a specific point, that is another operation. Thus, your steps ##8,11,14 are each two steps. Similarly, according to the third bullet of the challenge, each of the lines in steps 5,6,7,9,10,12,13 is two operations. Using the standard technique, step 17 would have added 9 operations (step 18 is not necessary for the challenge as written, but could have been drawn with just a further four operations).

Regards

Fyz

Within the bounds of reasonable precision, it is possible to trisect any angle. The procedure is as follows.

!. Strike an arc from the vertex of the angle that crosses each leg. As large as possible is best.

1 placement, 2 strikes

2. Set the compass to a radius slightly more than 1/3 the arc length.

1 setting

3. Strike a secondary arc from the each of the two intersections of the primary arc and the angle legs.

2 placements, 2 strikes

4. Without changing the compass strike an arc from the intersection of the first leg and primary arc and second leg and primary arc. Without changing the compass strike an arc from the intersection of the first primary and secondary arcs in the direction of the other inter-section.

2 placements, 2 strikes

You will now have created a pair of parenthesis that bound the exact trisection of the angle. The exact trisection is one third of the distance from the most central arc strike and 2/3 of the way from the outer mark. If you underestimated the 1/3 radius the marks still bound the error and a simple estimate of the point result by eye will give a result so precise extremely good equipment would be required to measure the error.

5. Draw the line from the vertex to the trisection point.

2 placements, 1 strike.

However, if you need to be more precise simply repeat the above process on the acute angle of the error. This should result in a "estimate" that is good to a few arc seconds. depending on the quality of your equipment and the precision of your work. (If you are laying this out on flat ground the whole thing can be done with a string and two stakes.)

Since the exact trisection point is at 1/3 , 2/3 of the new acute arc and this ratio is exactly the same as the ratio of the first problem it is possible, in theory, to "prove" that this point can not be found exactly. However, as a practical matter it can be found easily and accurately as you mentioned.

Since calculus and the "law of infinitesimals" had not been formalized in Euclid's time the simple expedient of using the bounded error marks to more precisely estimate the exact 1/3 radius and rebounding the error by repeating step 4 with the new compass setting might not have occurred to him. (I would bet against that myself, Euclid was pretty bright...)

But, the methods of successive approximation, and also of repeated estimation which in have just outlined certainly were known to engineers of both past and present times. Euclid's fascination with minutia challenged peoples thinking for many centuries and almost certainly contributed to the creation of calculus and the development of modern computational algorithm's.

I do not know if my methods fit within your rules and I am also not certain my operations count is correct. Additionally, it is not possible to exactly quantify what the arc error of my method is so I may have missed something along the way.

While it would be possible to approximate the trisection by successive bisections chosen in the appropriate direction as a practical matter this is not a method a working engineer would typically choose. Interestingly enough if the obtuse angle in question is 360 degrees never changing the first compass radius produces an exact result.

Thank you for presenting this most interesting challenge.

Sincerely,

Mr. Gee

Perhaps this phrase in the challenge is relevant:
"and (finally) your draughtsman's estimation of angle size is so poor that initial guesses are completely pointless"

Unless you have an enormous piece of paper, visual estimates to 5' of arc would be quite difficult. But if you can provide bracketing within about 27 degrees the approximation using a trisected chord would give the specified theoretical accuracy.

Euclid's geometry was to a large extent a philosophical exercise, being concerned with the ideas of minimum assumptions and rigorous proof**, and his construction equipment was quite deliberately the least capable that could do the job (hence his collapsing compasses, which I haven't retained for this challenge - and incidentally consider most unlikely to have been a tool that was in practical use). But it appears that ideas of limits were already current; but perhaps problems of definition had not yet been sufficiently overcome for this to form part of Euclid's thesis. (Documentation shows that infinite sequences date to at least 4000 BC).
**Though he was by no means as rigorous as is generally believed.

Sace the wear and tear on your brain and penceils.

With origami, it is possible to trisect an angle without an actual straight edge, or compass.

Check out this link.

http://www.math.lsu.edu/~verrill/origami/trisect/

HTRN

You will have to institute rules and a suitable counting method. Rolling into three should be counted as a sequence of halving steps (equivalent to successive bisections to the same precision), as it is multiple successive approximations.

Fyz

I wasn't going to do this just yet, but how about a tomahawk;

It's no good as far as the challenge goes, but like the origami it's fun to know. As you can see, TD and TR do the same as the single height and double height creases made in the paper with origami

Second attempt.

Obtuse triangle AOB, scribe two half circles, one radius OA, and the other at mid point of AB at radius R1, scribe R1 again using centre point A, join points OC. Scribe radius R2 at intersection and at point B. Scribe a line through where radius's R2 intersect with line OE. angle EOB is approx 1/3 of obtuse triangle AOB.

There appears to be a small error of about 27' will have to do some calculation to check?

Regards JD.

Is this 17 operations?

I couldn't identify line OD. Is that an "exact" trisection for reference?

(As it doesn't meet the accuracy of the challenge, I haven't checked how the error varies with angle AOB - did you?)

Regards

Fyz

Yes OD is a reference, done with Auto-Cad, have done the maths and it does not meet the required accuracy. Looking at it with hind sight, R2 only dissects COB, therefore AOC would have needed to be more accurate. With AOB being 160° the error for AOC is around 2 1/2° and COB around 1 1/2°. I will have to check more carefully, third time lucky may be.

Regards JD

Even though 15 steps (maybe fewer) is possible, it's still a "hard" problem to do it in 18...

Above are obtuse triangles AOB. Top left 174°, top right 96°, and bottom 135°. The object being to find the worst case error of angle X. being 1/3 of angle AOB. E1 being the first error. Dissection of angle YOB being error 2 (E2). Then a dissection of AOZ, followed by dissection BOY etc.

FIND X.

(1) Find OY, by cosine rule, side R1=AW, and side AO and angle A+60°.

(2) Find X, by sine rule, (sineX)/R1 = (sineA+60)/YO.

(3) Error 1 = (E1) = AOB/3 - X.

(4) Error 2 = (AOB - E1)/2

(5) Error 3 = (AOB - E2)/2

(6) Etc.

Worst case, AOB = 135° = error 1 = 1.3°

Error 2 = 0.65

Error 3 = 0.325

Error 4 = 0.1625

Error 5 = 0.08125 (E5 x 60' = 4.8' minimum error. )

---------------------------------------------

Find X.

Line .......... AB ->1

Semi cir A to B -> 1

Find W ..........-> 3

Find Y ..........- > 3

.................Total 8

Error 2 ...........-> 3

Error 3 ..........-> 3

Error 4 ..........-> 3

Error 5 ..........-> 3

Total ...............20

Best I can do is twenty unless I'm counting wrong

Regards JD

Assuming that I've understood correctly, I have probably made a mistake with my trig, as I get a slightly larger error than yours: E2 = 1.44-degrees at 135-degrees, and 1.48-degrees at 155-degrees. That's not too significant, but the process seems to take rather more steps than you give:

1 Arc centre O to define A, B
2,3 Chord AB (all your lines have one alignment at each end)
4 Arc centre A
5 Arc centre B crosses previous arc at W
6,7 Line OW bisects O at C
8,9 Arc centre C (place point on C, and align scribe to pass through A)
10 Arc centre A to cross previous at Y
11,12 First approximate trisection line OY (crosses first arc at Y1)
13,14 Arcs centred at Y1 and B respectively to define Z
15,16 Trisector through OZ

I agree that subsequent bisections can each reuse one existing arc, so each takes three steps.

That makes 25 steps to reach E5, I think (one more if you make an approximate reset of the compasses from R1 to R2)

Here are the steps:

1. Draw the arc PQ
2. Construct the perpendicular MO to OQ at O
3. Set the compass to distance OQ and draw the arc ON up to the PQ arc. This produces a 60 degree angle NOQ at O, trisecting the right angle MOQ.
4. Bisect the angle MON, yielding a 15 degree angle MOX.
5. Set the compass to distance MX, then mark-off the arc segments MB, PC and CA.
6. Construct the straight line AB.
7. Tri-sect the line AB. (*)
8. Construct line OD.
9. The arc labeled ST will be approximately 1/3^rd the arc PQ, and thus the angle DON will be the tri-section of POQ.

(*) For small angles, the tangent of DOB is equivalent to the angle DOB in radians. For a 2 degree angle the difference amounts to 2.9 arc-seconds.

Note that if angle POM were less than 45 degrees, an extra step would be needed to tri-sect angle MOX to yield a 7.5 degree angle, which would then be used to mark-off the arc segments PC, CA and MB. Similarly, if angle POM were more than 60 degrees the 30-degree arc segment could be used to generate the arc segments PC, CA and MB.

Agree the precision - possibly overkill, as the spec is 5 arc-minute. But if we take just the first 6 stages of the construction shown on your drawing we have taken 17 operations - and we still have to trisect AOB.

Yeah,

When I looked at my steps and counted them your way, I knew it was too many steps. But I decided to post my solution anyway since I thought it was an interesting challenge.

It was proved years ago, of course, that angle trisection is impossible except for a few special cases like a right triangle. But trisecting an angle to a desired degree of accuracy is a very different problem and, I decided, worth spending some time thinking about.

At least I was able to prove a correlary of the problem (though this too was probably proved a long time ago) -- that the trisection of an obtuse angle can be reduced to the trisection of an accute angle, since the right-angle portion can itself be exactly trisected.

Sorry I misunderstood the intent.

In fact, the 'official' solution is similar to yours in that it comes down to creating an angle that is small enough that it can be trisected accurately enough. The final trisection uses a method that is equivalent to trisecting the arc - but in a way that requires far fewer steps.

In the end, I found two completely different solutions that take just 15 operations - there are probably more - and there may even be some that require even fewer operations.

One of these is a refinement of the 'official' solution. For larger angles it uses a slightly more accurate pair of first "approximations" than bisection, and then uses a slightly more accurate approximation to trisection than the one given (although it is very similar in most of the detail).
The other is based on Mark Stark's method. It uses a simple first approximation that is within 20-degrees of the actual trisection, and then a compact version of the Mark Stark refinement technique to provide the required angle.
Either of these methods would easily include acute angles. The restriction to obtuse was intended to allow the possibility of methods that didn't work for smaller angles (JDretired showed an interesting one, though it too is a lot of steps).

Personally, I prefer the 'refinement' of the official solution - even though the theoretical accuracy is not as good. The reason is that (for the same size of total drawing) constructional errors in Mark Stark's method give worst-case angular error that is amplified by about a factor of six relative to their effects in the 'refinement'.

I propose to delay posting these solutions for a few weeks to allow space for others to find better methods.

I'm not sure if my operations count is right, but I might have done this in 17 steps:

1. Draw line AB
2. Draw line AC
3. Draw line CB
4. Set compass to length of AC
5. Strike arc from C intersecting CB to find point a
6. Set compass to Ba
7. Strike intersecting arcs from B and a to find point b
8. Draw line Ab to find vertex D
9. Set compass to Da
10. Strike intersecting arcs from D and a to find point c
11. Draw line Ac to find vertex E
12. Set compass to DE
13. Strike intersecting arcs from D and E to find point d
14. Draw line Ad to find vertex F
15. Set compass to DF
16. Strike intersecting arcs from D and F to find point e
17. Draw line Ae to find vertex G

Angle of GAB measured by AutoCAD to 3 decimal places came to 40.399°.
If I'm off in my count or the margin of error, let me know. It's too late at night for me to convert from decimal to degree-minutes-seconds format.

Assuming that I have understood the construction: I think it only works for a very small range of angles - if my quick algebra is correct, this would be slightly larger (but not including) 144^O. On that basis, if you try this with 90^O I think you will find the error is greater than 7.5^O.

Regarding the steps that will be required:
As the initial angle is provided, you don't have to count construction of AB or AC. On the other hand, if you want AB=AC you have to draw a circle centre O.

Regarding counting: every point you align to counts as one, so:
. drawing a line through two points is two operations,
. placing the compass point is one operation, and
. adjusting the scribe to pass accurately through a point is another.
So - stages with number of operations in braces:
1 (0); 2 (0); 3 (3); 4 (already done); 5 (2); 6+7 (3); 8 (2), etc.

Haven't done my maths or count yet, but on the Auto-CAD it looks exact? scribe a line from the centre of a equilateral triangle to the inter section, and where it intersects the arc is one third of the obtuse angle?

Regards JD.

Count

Line AB ……………………………….. > 1

Cir C ………………………...………… > 2

Line OD ………….…………………….> 3

Arc E ……………….……………………> 4

Line WG ……….……………………….> 5

Cir F ……………………………………..> 6

Intersection H ………………………. >7

Cir J centre I ……….………………. > 8

Cir L centre K …….………………… > 9

Line OM …………….…………………. > 10

Cir N centre P .…………………….. > 11

Cir Q centre O …………………….. > 12

Line RS ……………..…………………. > 13

Line TH ………………………………... >14 Intersection U = 1/3 of arc E?

Maths to follow ( getting late, tomorrow). Sorry about the drawings got the wrong ink cartage.

Regards JD.

The basic construction is new to me, so it's of interest even though the construction takes rather more than 18 operations**. It is impossible for it to be exact (mathematically provable), but it may be very accurate. I'll come back to you on the size of the error if/when I can get my head around it.

**Your lines are constructed by aligning both position and angle (or we could say you are defining two points on each line). So they are each two operations.
Similarly, most of your circles require both the centre and the position of the scribe to be aligned - that too is two operations.
The sole exception ( I think) is circle L,, which looks like a single operation.
That would make this 27 operations - though I may have misinterpreted some part of the procedure.

This is the actual construct, I make it 12 but could be wrong?

Calculation references.

MATHS

Given. Obtuse angle AOB = 120°, Length OA = 100, Length OF = 50, Angle FAO = 30°.

(1) Find OK. Angle FOK = 90+60 = 150. FK = radius, 100 x Cos30 = 86.66

By Sine rule find angle FKO.

Sine150/FK = Sine(K)/50. Angle K = 16.7787°

Therefore angle KFH = 180-(150+16.7787) = 13.2213°

By Sine rule find length KO.

OK/ Sine13.2213 = 50/Sine16.7787. Length OK = 39.6141.

(2) Find ON.

Triangle height = OK x Cos30. ON = 2/3 of height.

ON = (39.6141x0.866x2)/3 = 22.8712.

(3) Find angle WND.

WND = Tan WD/( WF+FO+ON) = 43.3013/ (75+50+22.8712) = 16.3217°

(4) Find angle OPN

By Sine rule Sine(P)/ON = Sine(N)/AO.

OPN = (Sine16.3217x22.8712)/100 = 3.6852°

(5) Find angle POB.

Angle POH = 180-( 3.6852+16.3217) = 159.993°

Angle FOP = 180 - 159.993 = 20.007°

Angle POB = FOB - FOP. 60 - 20.007 = 39.993°

Error trisecting angle AOB = 40 - 39.993 = 0.007°. 0.4' = 25".

Accurate to within 25 seconds.

Regards JD.

Different construct which I believe to be more accurate and a smaller count? I hope. This is more accurate on the larger angles, which where getting close to the limit.

Regards JD.

Your claimed accuracy looks impressive; unlike Mark Stark's method** (which can be part of a sequence that reduces to 15 steps) you use no constructions that magnify small errors, which is a great improvement. I'll restart independent calculations of the first error because I mis-transcribed the original version

On the other hand, I think this one is 22 operations - it may be fewer if I have misinterpreted, so I'd appreciate your confirmation. (Did you see my previous attempts at clarification?)

Arc BCO ->2 (centre O to find centre for OCB, then draw OCB)
Arc AEB -> 2 (Centre O and adjust span to OB)
Line AB -> 2 (align at A and B)
Line FG -> 2 (align at O and F)
Arcs centre F, radius AF -> 2 (centre and adjust span to AF)
Arc centre B -> 1 (defines D)
Arc IJ -> 2 ((centre H and adjust span to OH)
Arc JH -> 1 (recentre on I)
Line JO -> 2 (align at J and O)
Line KB -> 2 (align at K and B)
Line ND -> 2 (align at N and D)
Line OP -> 2 (align at O and P)

**Kris (many thanks) provided a link that includes Mark Stark's basic method in earlier discussions.

Yes I do confirm your number of operations as 22, I now see my mistake of adding only 1 operation for a line that uses two points. Thank you, and Kris, for the link, have put same into favorites and will look closer at it. I had not seen the Mark Stark's solution prior to the above, when looking at it, it does seem have some similarities.

Regards JD.

Hi JD

Thank you for sharing this remarkably precise "single-step" approach - it's really interesting, even if it can't beat the number of operations that can be achieved using more obvious approaches.

After numerous false entries into the spreadsheet, I finally managed to confirm the accuracy at 120^O. So far as I can see, this family of methods is exact at 90^O and 180^O, and the version in post #38 give a theoretical peak error of just under 0.8' (at close to 143.1^O); that's more than a factor of 5 more precise than required by the challenge**. So I'm not quite certain what you mean in post #39 where you say "the larger angles, which where getting close to the limit".

Regards

Fyz

**Even near 53.13-degrees (i.e. before the method fails altogether), the error is just marginally over 0.5O (or 30').

Yes your correct the approach in #39 is not correct, and my statement is in error, use of a hand calculator and confusion does not give good results. Thank you for going to the trouble to check the accuracy, much appreciated. There is a small floating error in the method which I was trying to eliminate, plus reduce the count,but my efforts did not work.

Regards JD.

It's a pleasure to look at something that works this well. Although I found a few alternative approaches that would need the same number of operations, I haven't managed to reduce it below your 23 (I think) operations.

BTW, there are a couple of free spreadsheets on the web which should easily handle what is needed - "gnumerics" and OpenOffice "Calc" both seem quite good. Perhaps they are not quite as 'pretty' as EXCEL, but the interface is basically quite similar [including working in radians rather than degrees, so you would need to know that pi needs to be written as pi() ].

We obviously live in an irrational universe.There is no way to escape the irrationality of Pi in any construct, using current mathematical processes.Perhaps we need a different kind of math to measure circles,spheres, etc.? Obviously, a bubble is the most powerful computer in the universe.It can calculate Pi to the last place instantly when it forms.That makes it apparent that factors other than radius determine circumference.I do not claim to know what the factors are are, but they must exist, or a bubble could never completely form, it would always be ALMOST formed, chasing the elusive Pi out to infnity.An older sphere would be more complete than a younger one, etc.Perhaps our "straight" lines are actually slightly curved and we cannot detect it? Can a straight line not be defined as a segment of a circle with an Infinite Radius?

Beam me up Scotty.

Our work is finished here.!

This method uses 28 operations, but the accuracy is impressive (at least to me).

Given line AB:

1. Set compass to AB
2. Draw circle centered on A with radius of AB
3, 4. Draw AC
5. Draw line perpendicular to AB from A to circle to find point D (top quadrant).
6. Draw circle centered on D with radius of AB
7,8. Draw horizontal line through the 2 points where the circles intersect.

Bisect the right half of this line:
9. Set compass to half of horizontal line.
10. Draw circle at midpoint of line. (This is the left endpoint of this section. The reason for noting this will be seen in next paragraph.)
11. Draw circle centered on right endpoint.
12, 13. Draw vertical line from upper intersection point to horizontal line.

Repeat the bisection 3 times on right, then left, then right halves of previous bisections.

Angle from A to last bisection point is 40.023° or 40° 1' 22.8".

Notes:
I counted drawing the first vertical line to find point D as one operation because I would be aligning the squared end of my straightedge with AB, unless I'm allowed use of a drafting triangle as a straightedge. I am aligning with only one point to draw the line.

I counted the drawing of the other vertical lines as 2 operations because I have to align my straightedge with both intersection points because they are known points.

I'm not really concerned with the number of operations in light of the accuracy I've achieved.

Apparently, my thinking in post #15 paid off.

Hi Doug

I think you demonstrated the accuracy at 120^O. As the process is hard to analyse, can you indicate the errors at 90, 150 and 180?

Am I right that you used the right-angle at the tip of your straight edge? Unfortunately, Euclidian straight-edges do not have calibrated right-angles at their ends, so you need to use a couple of arcs* and a line to create a right-angle. (Even if you have a reference angle, I think that you would have to count two alignments - one for the point you take the line through, and a second to give you the direction)

Regards

Fyz

*One of which may often be a reuse

Well, as far as the vertical line from A to the top quadrant, to stay within the constraints, I'd have to do these steps:

1. With compass set to AB (already done), strike arcs on circle from B.
2. Extend AB to left quadrant.
3. With compass still set to AB, strike arcs on circle from left quadrant.
4. Draw line from left lower intersection to right lower intersection.
5. Set compass to length of this line.
6, 7. Draw circles centered on the endpoints of the line.
8, 9. Draw vertical line through the intersections of these circles.

As for other angles, that will take more drawing time. I'm sure the basic method is sound, the key issue is the placement of the horizontal line.

I have a feeling this thread is going to be one of those that lives on long past the posting of the official answer.

I'm probably being thick, but can you mark up which points and angles are which?

You ask, Fyz, and I deliver.

I have numbered the items according to the steps in which I created them. I'm also reposting the steps so you don't have to scroll back and forth.

1. With compass set to AB (already done), strike arcs on circle from B.
2. Extend AB to left quadrant.
3. With compass still set to AB, strike arcs on circle from left quadrant.
4. Draw line from left lower intersection to right lower intersection.
5. Set compass to length of this line.
6, 7. Draw circles centered on the endpoints of the line.
8, 9. Draw vertical line through the intersections of these circles to locate top quadrant.

Notes:
The horizontal line at the bottom of the main circle isn't entirely needed. I decided it would help in setting the compass and locating the center points of the circles. As it turns out, the line equals AB, but I had no way of knowing this beforehand.

The 2 arcs at the top of main circle are really needed either, but again, I had no way of knowing beforehand if I would use them.

Edit: I just noticed that I could have extended AC to intersect the circle at the right end of the horizontal line. A diagonal through A and the intersection of the upper right arc would have given the left end.

I still don't understand.

I assume you intend to trisecting the angle CAB - but I can't identify the angle that should be 1/3 of CAB. There are plenty of 60^O angles available (though the lines are not constructed), plus the right angle and CAB-90^O.

The illustration in post #73 is a clarification of the illustration in post #54, where I presented a Euclidian approach to locating the top quadrant point on the main circle, which is what I thought you wanted. Apparently, you want a clarification of the illustration in post #47. OK, I'll do that too.

BTW, I am working on an illustration of trisecting a 60° angle. I chose that angle because it was either the link in post #40 or some other comment somewhere said that 60° is particularly hard to trisect.

Thanks, you've clarified that. The reason I didn't understand was that I didn't follow back through the thread. And my original question was how you apparently counted the production of a right angle as a single operation.
Purely communications, I think. On which topic - realigning a straight-edge to pass through a point is normally counted as an operation, even if it was immediately previously aligned through that point; I think you could argue (with considerable justification) that this was not a separate alignment if you used pins for each placement and didn't need to move the pin for drawing purposes. But accepting that would save operations using other methods as well (typically at least two), so the goalposts would move along with the scoring method**.

Regarding specific angles. You are right that 60^O is often given as an example of an angle that is impossible to trisect using Euclidean methods. So far as I can make out the reason it is singled out is that it is regarded as a "commonplace" angle, and the measure (number of degrees) is so obviously divisible by three. Most trisection methods give a more accurate result for 60-degrees than (say) 120-degrees - though naturally it is possible to tailor a method for obtuse angles that will be less accurate at 60-degrees (the method presented by jdretired is one such).

**That's not a mixed metaphor - it's scrambled

I think I saw the angle trisected in Popular Science ages ago. All I remember was three equal circles on a common centerline, with the trisected angle within. The vertex may have been on the perimiter of the circle on the left, on the common centerline, and the trisected angles going to the point where a vertical centerline hit the circles. Just did a rough test on a piece of paper with some desk stuff, and it looked close. Very simple, too.

You might have seen an approximation, but not the real deal. It's a bit like trying to work back from Cos(3ß) = 4 cos³(ß)-3cos(ß). As Fyz mentioned, you end up trying to find a cube root which can't be done (in all cases) with euclidean tools.

< I shall hide until Fyz gives a better explanation !>

Physicist,

I guess I am having a hard time following your answer. If you would post a visual representation of the solution with references to each arc and line that would be great. I have a bunch of geometry teachers awaiting an answer. Thanks.

My tools won't talk to CR4, unfortunately. If you can recommend a free web-based drawing tool that I can learn quickly and will easily upload onto CR4, I'll have a go.

Regards

Fyz

I am not sure of an online tool, but I used AutoCAD to draw in, then printed to PDF which is free, then opened the PDF, took a snapshot, copied into a new bitmat image using paint and it worked well for me. I was able to use colors also, which help quite a bit and distinguishing different arcs. As far as a web-based drawing program, I do not know of any.

Sorry, incorrect description. I meant a package that is free to download. (It's just that I don't have an use for this type of drawing in my day-to-day work).

http://sketchup.google.com/download/

Physicist? try this link, I think it may work for you, it is a free sketch tool through google. What will they come up with next.

I also did my drawings in AutoCAD, then highlighted the parts I wanted to post, clipboarded into Paint, then clipboarded from Paint into Serif Photo Plus (free version), cropped to the size I wanted and exported to .jpg. When I tried clipboarding directly into Photo Plus, the lines came out too faint, so I would go into the drawing and convert everything to polylines and widen the lines. Clipboarding into and from Paint gives me a bitmap I can edit and the lines remain dark enough to see. Also I don't have to save the bitmaps or the .spp format image that Photo Plus creates.

As for free drawing programs, Kris had a thread in the General Discussion a few months ago about that.

...much to my shame, the drawing Fyz just posted at #68 (after some rapid learning of a drawing package) is far better than anything I've yet produced ! What did you use Fyz ?

Pure luck (I mistyped this as "puer" first time - equally appropriate).

I used the free version of CadStd, transferred the 'raster' to OpenOffice draw, and saved it as a .bmp file (ouch)

N.B. I tried with Google SketchUp initially - not my sort of thing at all.

A bit of advice, not just to you Fyz, but to anyone who hasn't yet posted a drawing in CR4, but is thinking of doing so: Try to set your background color to white to make things more readable in this forum. You can probably find it under Display Settings or Visual Styles under Options, Tools, or Format. I know that while drawing, a dark background sometimes works better, but a light background with dark lines makes for a better presentation in CR4.

I do intend to add some drawings later in answer to earlier requests by Fyz. It might take a while, especially if some storms mosey through my neighborhood and ask for my attention as a storm spotter.

Thanks - I agree entirely, but it took a while to find the invert background command in a new package - but I've got it now. But I haven't found out how to hide the paper edges yet...

(I don't expect I'll have time to fully learn this tool...)

Apologies that this is not pretty - but I hope this is sufficient to work with. I have numbered the constructions, but have not marked all the intersections, as these would be hard to see on this scale. Vertex names are in the text in case this helps. It's not identical to the official answer, but it is exactly equivalent. The reason for the difference is partly that the drawing would be even more cluttered, and also because this is slightly less prone to accumulating errors from the bisections.
There may be mistypes - please ask about specifics if still unclear.
The original angle is AOB.
First construction: Arc1, centre O, (arbitrary radius R1 to mark AB on rays of original angle at equal distances from O. (1 operation)
Second construction: Arc2, centre B, radius R1 (1 operation)
Third construction: Arc3, centre A, radius R1 (1 operation)
Arc2 and Arc3 cross at Z
Fourth construction: Line OZ (2 operations)
OZ is "First bisector". It bisects AOB, and crosses Arc1 at C
Fifth construction: Arc4, centre C, radius R1 (1 operation)
Arc2 and Arc4 cross at Y
Sixth construction: Line OY (2 operations)
OY is "Second Bisector". It bisects ZOB, and crosses Arc1 at D
Seventh construction: Arc5, centre D, radius R1 (1 operation)
Arc3 and Arc5 cross at X
Eighth construction: Line OX (2 operations)
OX is "Third Bisector". It bisects YOA, and crosses Arc1 at E
We now mark of three times the radius of Arc1 from O along the line OE.
Ninth construction: Arc6 centre E, radius R1 (1 operation)
Arc 6 crosses OE at W
Tenth construction: Arc7 centre W, radius R1 (1 operation)
Arc 6 crosses OE at V
We now measure CE
Eleventh construction: Arc8, centre C, to pass through E (2 operations) [Radius R2]
We now draw an arc this radius three times as far from O to create an angle that is approximately 1/3 of angle OCE. We use the same arc we used to mark point V to define the final position. This is not ideal, but it avoids extra operations. The precision is identical to using the method of trisecting an arc.
Twelfth construction: Arc9 centre V, radius R2 (1 operation)
Arc nine crosses Arc7 at T
Final operation: Draw OT
The angle AOT is approximately 1/3 of AOB.
AOT is built from AOE - COE/3** = 3.AOB/8 - AOB/8/3**

**This represents that the division by 3 is not exact.

OK, this will seem cumbersome and tedious to get through, but here's my attempt at trisecting a 60° angle. I broke the procedure down into discrete steps that may or may not equate to Euclidian operations. Part of this will be redundant, but it might help people understand my approach. It does involve a major modification of my procedure in post #47.

Phase 1 - Preparing for Bisections.

Draw AB.
Set compass to AB.
Draw circle 1 centered on A.
Strike arc on circle 1 above AB and centered on B.
Label this intersection as point C.
Draw AC. Angle CAB is 60°.
Extend AC to diameter of circle 1.
Label this end as point D.
Strike arc on circle 1 below AB centered on B.
Label this intersection as point E.
Draw line DE.
Set compass to DE if needed. (DE should = AB, so check first to see they are the same.)
Draw circle 2 centered on D.
Draw circle3 centered on E.
Draw vertical line through the intersections of circles 2 and 3 up to the top quadrant of circle 1.
Label the quadrant as point F.

Phase 2 - Bisecting to (almost) Trisect

Set compass to AF if needed. (AF should = DE or AB.)
Draw circle 4 centered on F.
Draw horizontal bisector 1 through the intersections of circles 1 and 4.
Label the intersection of bisector 1 with AF as point G.
Set compass to AG.
Draw circle 5 centered on A.
Draw circle 6 centered on G.
Draw bisector 2 through the intersections of circles 5 and 6 and over to circle 1.
Label the intersection of bisector 2 with AF as point H.
Set compass to AH.
Draw circle 7 centered on A.
Draw circle 8 centered on H.
Draw bisector 3 through the intersections of circles 7 and 8 and over to circle 1.
Label the intersection of bisector 3 with AF as point J.
Set compass to HJ.
Draw circle 9 centered on H.
Draw circle 10 centered on J.
Draw bisector 4 through intersections of circles 9 and 10 over to circle 1.
Label the intersection of bisector 4 and AF as point K.
Set the compass to JK.
Draw circle 11 centered on J.
Draw circle 12 centered on K.
Draw bisector 5 through intersections of circles 11 and 12 over to circle 1.
Label the intersection of circle 1 and bisector 5 as point L.
Draw AL.
Angle LAB = 20.106 degrees or 20` 6' 21.6".

Of course the specific procedure has changed, but the basic approach of alternating bisections appears to be sound. This will require me to experiment more, but not right away. I need a break from AutoCAD for a couple of days.

FWIW, this page has good info on these types of constructions beyond trisecting angles.

There's no question that alternating bisections halves the error each time. Choosing the ray closest to the trisection as the initial estimate gives an error of θ/(3.2^N), where N is the number of bisections. My preferred bisection method is to make the initial bisection by scribing circle AB, placing the compasses unadjusted centred on A and B successively to mark the bisection, and scribing the bisector to cross the arc AB at C. I would then adjust the compasses to AC, scribe new arcs centred on A and B. For each succeeding trisection I would use one of these arcs and an arc centred on the intersection of the preceding trisection with the arc AB. The only problem with this is that the reference arcs get rather crowded, so it can be worth adjusting the compass again - but in the end the crowding of the estimates will inevitably become problematical...

In ruminating on this, and trying different trisections in AutoCAD, a question has come to my mind: how can you know which angle you have drawn using Euclidian methods?

From what I can see, you can summarize Euclidian construction as drawing from at least one known point using either compass or straightedge, except for drawing a simple line of any length. Using compass and straightedge, you can locate points to draw right angles, plus angles of 60, 30 and 45 degrees without too much thought. You can bisect 30° to get 15°, and you can do the same the same to 45° to find 22.5°, etc. But let's say you want to draw an angle of 10°? or 20°? How would you know you have drawn that angle from the method you used to construct it? If you drew an angle at random, how would you know which angle you have drawn?

Of course, using CAD, a protractor, or distances and trigonometry will tell you the angle. But remember, we are talking Euclidian construction, using only the designated tools, and methods conforming to the given restrictions.

For whole numbers, it's possible to construct 3^o, and therefor any multiple of it. As far as the challenge goes, it's not necessary to know the size or the initial, or trisected angle ; the error in trisection can be worked out as a maximum value. If you wanted to draw 10^o, you'd have to draw 30^o and approximate a trisection (or use some other approximating method).

Obtuse triangle AOB

Arc BCO ->2 (centre to find centre for OCB,(X) draw OCB (W).

Arc AYB >2 (centre O and adjust span to OB. full circle)

Line AB ->2 (align A and B).

Line FG ->2 ( align O and F).

Arcs centre F, radius AF ->2 ( centre and adjust span to AF full circle).

Arc centre B ->1 (defines D).

Line AG ->2 (align A and G).

Line KM ->2 (align K and M).

Line ND ->2 (align N and D).

Line OP ->2 (align O and P).

Count 19? I thought it was less.

Regards JD

Interesting (and I agree the count). As the method is not immediately obvious to me, I'll ask rather than heading down blind alleys: - does this place point N identically to your previous method?

If so. it's economical for the accuracy - methods I now know can give:
Your method: 0.26'
19 operations: 0.14' **
18 operations: 0.55' **
17 operations: 0.77' **
16 operations: 1.3'
15 operations: 4.4'
14 operations: 6.2' (fails challenge on accuracy)

**These are based on the 16, 15 and 14 operation methods, with the first estimates improved by an additional bisection.

Will analyse it further, N is pretty close, but the theory is, can line AKG be curved to eliminate the error?

Regards JD.

I have come to the conclusion that the above does not in any way improve on what I have already done, the end.

Regards JD.

When I first proposed this challenge, I thought CR4 would do well to manage it in 21 operations. By the time it was accepted, 18 seemed more sensible. In the end, it seems it is possible to solve using 13 operations - until someone finds an even more efficient method (I doubt it will be me, however).
The proposed solution uses the method due to Mark Stark. It is by no means the most robust method (in terms of its sensitivity to drawing errors). But it is easily the most efficient in terms of operations - provided the operations are performed in the optimum order.
N.B. The next most efficient method I found requires 15 operations - but it does have the advantage of significantly lower sensitivity to drawing error - and doesn't present additional problems with acute angles.

In order to leave opportunity for any other obsessives to struggle, I don't plan to post this solution for a few weeks yet. But I can send it to anyone who sends a request via my CR4 mailbox.

N.B. If you provide Mark Stark's estimation method with an initial solution with an error ε (radians), the method will produce a new estimate with an error of about ε³/48 - (although, as I did, you may prefer to rewrite the formula in terms of the error rather than arbitrary intermediate stages). I used a construction that gave an initial estimate of the trisection that was accurate to within 20-degrees - that gives a next-stage error of 3.046 arc-minute.
I was also struck by the efficacy of the method at the following complexities (all using Mark Stark's method) interesting:
11 operations: 10.28 arc-minute (initial estimate is exactly 60^O)
16 operations: 0.381 arc-minute (initial estimate within 10^O)
17 operations: 0.000115 arc-seconds (starts from the 10.28 arc-minute error)

Hello Fyz,

Trisecting angles is not quite my cup of tea, but I went through your procedure as a matter of curiosity, and satisfied myself that the practical procedure works (for an angle which I measured as 104°).

I think a little re-wording is called for in the description of steps 14 onwards. Adjusting the compass in step 15 is only a radius setting (and no arc #8 need be drawn at that point?), while the last step should refer to arc #7 rather than arc #9 (?). Perhaps this is mere quibbling, unless I have misunderstood something. Hope I'm not repeating something that others may have pointed out, as I haven't read all the posts.

I have not tried to investigate whether the method works for acute angles, and if not why not. Maybe you have a ready answer to that.

I presume you would have employed suitable analytical or numerical methods to establish the accuracy of the method or maximum possible error over the whole range of obtuse angles. I am not however interested in that aspect, and it may be beyond my mathematical capabilities anyway.

My interest as an engineer is more oriented towards "accuracy of construction", whereas the values mentioned in your post 81 are no doubt based on "theoretically perfect" construction.

I have often had to resort to graphical procedures to deal with problems relating to beams, trusses, theory of machines, etc. Difficulties would arise whenever intersecting arcs cut each other at a shallow angle, because there could be an error in radius setting or compass point placement. A repeat construction could yield a substantially different result at times! When I could choose a radius freely, as in simple angle bisection, I would naturally select one which would allow the arcs to intersect close to a right angle.

In the trisection procedure, as the original angle approaches 180°, the very first practical bisection becomes increasingly error-prone. Suppose each point placement could be 'out' by say 0.2 mm absolute (or 0.2% of the initially selected radius) the cumulative error can become enormous. I certainly don't expect you to address this kind of problem in any detail, but if you have any qualitative insights or suggested references regarding constructional inaccuracies in general, I'd be grateful for your comments. I've never come across anything suitable, or never knew where to look, so I just tried to do my construction work each time with eyes open for possible errors. I don't think I've tried doing any accurate graphical work in the last two decades though.

Thanks, =TeeSquare=

Agreed. The 13-operation approach doesn't suffer from this, but is not brilliant at any angle, and fails completely at 60-degrees. The 15-operation approach is sound at all angles - and is reasonable for constructional sensitivity at all angles.

And the answer should have been... 10 - giving a maximum error of 2.6 arc-minutes. That is most unexpected (at least to me). The reason is that a minor modification to Mark Stark's method can increase its accuracy by a factor of approximately four. The construction is:

Provided with angle, vertex O
Operation 1: Draw circle to intersect the rays of the angle at A and B respectively
. all subsequent arcs will use the same radius as the circle
Operation 2: Draw arc passing through O to intersect the circle at E inside the angle AOB
. BOE = 60^O (not constructed) is used as the first estimate for the trisection.
. the error ε of this first estimate grows from zero for AOC=180^O to 30^O at AOC=90^O
Operations 3,4: Draw Chord AB, to intersect arc at C.
Operations 5,6: Draw line EC, cutting the circle again at G and extending outside the circle by at least one diameter.
Operation 7: Draw arc, centre G, intersecting EG at H outside EG
Operation 8: Draw arc, centre H, intersecting EH at K outside EH
Operation 9: Draw line KO, intersecting the circle inside the angle AOB at T
. TOB is the improved estimate of the trisection. Its maximum error for obtuse angles is 2.6 arc-minutes

Error calculation:
Following the provision of the (rather poor) initial estimate, the initial construction is the same as for Mark Stark's original, and provides an angle CEO (EO not drawn) that is 3/4 of the error ε in the initial estimate
. i.e. CEO = 3.ε/4 = 3.(AOB/3-EOB)/4.
. So we need a line that is ε/4 rotated w.r.t. EC. An approximation will be provided by KO.
Now consider the triangle GKO. Clearly, GOK = 180^O - CEO
So, by the cosine rule:
. OK = OG² + GK² - 2.OG.GK.cos(GOK) = r²(1² + 2² + 2.1.2.cos(CEO))
. = r²(5 + 4.cos(3.ε/4))
Now, by the sine rule:
. sin(GKO)/GO = sin(OGK)/OK, i.e.
. GKO = arcsin(sin(OGK)/(5 + 4.cos(3.ε/4)))
Which gives the error for the second estimate angle TOB as:
. arcsin(sin(OGK)/(5 + 4.cos(3.ε/4))) - 3.ε/4

Comment:
The improved accuracy provided by the slight reduction in the length OK from the 3.r value provided by Mark Stark's construction allows the challenge accuracy to be achieved with an initial estimate that has an error of up to 37.2-degrees. This makes it possible to use a fixed non-zero angle anywhere between -7.2 and 67.2 for the initial estimate. Keeping the compass radius fixed throughout the construction keeps the number of operations to a minimum.
. Note1: that, if a secoond stage of improvement is needed, the original Mark Stark method can be more economical than this modified - and given the extreme theoretical accuracy this would already provide, the factor of 256 accuracy improvement that this "improved" method would give is probably irrelevant.
. Note2: that the constructional error in the angle CEO becomes progressively more sensitive to alignment to the points C and E as the angle for trisection reduces. This particular method also fails completely at AOB=60^O. Thus, while I would a-priori have expected trisection of a small angle (to any given accuracy) to be less costly than a larger one, it appears that no comparably economical method exists that will cover the full range of acute angles - unless someone knows different.

What no other comments yet!

This is incredible. I checked the construction for 100^o and 170^o with results shown. 2 decimal places doesn't quite do them justice sorry.

33.36^o

56.67^o

I actually chose the obtuse angles because I assumed they would need the more refined construction - and I feared the descriptions "acute or obtuse", "non-reflex" or "internal angle" would further overcomplicate the reading of the problem.
So you can imagine why I still find it surprising that such economy is possible for obtuse angles - but not apparently for all acute ones.

Plus, I should have mentioned:
The third-order approximation
. ε' ≈ ε³/192 (angles in radians) is remarkably close throughout this range.
As the error in the initial estimate is ∏/3 - AOB/3**, that gives:
. Final error ≈(∏/3 - OAB/3)³/192 ≈ (∏ - AOB)³/5184**

i.e. maximum error in the initial estimate is 30^O (at AOB=90^O)
[Any chance of an improved symbol for pi?]

I must be an engineer by nature; I can't leave well enough alone. I took another stab (several, actually) at trisecting a 60° angle. Here's what I came up with as my best shot:

Draw AB
Set compass to AB
Draw circle 1 centered on A
Strike arcs on circle 1 from B
Label these intersections as C and D
Draw AC
Angle CAB = 60°
Draw CD
Draw circle 2 centered on C
Draw circle 3 centered on intersection of CD with circle 2
Label the intersections of circles of 2 & 3 as E and F
Draw EF
Draw AF
Label intersection of AF and circle 1 as G
Set compass to AG
Draw circle 4 centered on F
Label intersection of circle 1 and EF as H
Set compass to FH
Draw circle 5 centered on H
Draw vertical line through intersections of circles 4 & 5
Label the intersection of this line and AF as I
Set compass to FI
Draw circle 6 centered on I
Draw circle 7 centered on G
Draw line through intersections of lines 6 & 7
Label the intersection of this line with EF as J
Label the intersection of this line with AF as K
Set compass to JK
Draw circle 8 centered on J
Draw circle 9 centered on the left intersection of circle 8 and EF
Draw line through the intersections of circles 8 & 9
Label the intersection of this line with EF as L
Draw AL

Angle of LAB = 20°

Is this a perfect trisection? NO!

I did it in AutoCAD, but tried to stay within the constraints of drawing with compass and staightedge. Except for drawing AB, I only used points created (or, more accurately, marked) by something already drawn. Of course, AutoCAD positions elements of a drawing precisely on the those points. You can not count on that happening when drawing manually. The amount of error then depends on the skill of the person, and the quality of the instruments.

I am sure that Tee Square and some others here on CR4 could probably come within an error on 1' or less. I'm sorry I can't think of other names just now

I tried to follow the steps a few times but got a little (or a lot) confused along the way. I wasn't always sure which bisector or circle or arc was the right one. It would help me if each bisector, circle, or arc had a specific designation (most have already) and then these designations were used in following steps.

Perhaps I'll be able to work through this and post a graphical representation.

Thanks

If you have compass you can always mark the straight edge. And you do not need to mark too many divisions for this solution. After that the link below gives this answer. So easy with the internet but I hunted down this solution in High school, 40 years back after I was provoked by an uncle who was a very good Mathematician. It can also be done by the conchoid of Nicomedes.

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Trisecting_an_angle.html#s31

One of these days I will put in a problem, (the answer for which I do not know) he gave but expired before I could get his solution.

I wondered if the TRiSeK V3 (http://www.zolkorp/TRiSeK.html) could be simplified to be competitive, but the best I can do is 6 circles and 7 lines. I'm not sure how many steps that would count as, but it gets to within a hundredth of a degree for a 45 degree angle. Adding another 2 circles and 1 line gets to less than a thousandth. (see tha SIMPL8 on that page). Obviously, more steps would be added to do an obtuse angle.

A few questions about this thread.

Who's Mark Stark?

Why bother with minutes and radians?

I ran Mark Starks trisector with the refinement step.

Thats incredible!

It works all the way up to 180 and is accurate to 13 decimal places over most of the range.

Definitely has my trisectors beat for simplicity, and they only work up to 104. My latest one may only have a slite accuracy advantage. http://www.zolkorp.com/TRiSeK.html

I can't really tell though, since the computer measures the angle to be below perfect from one side and above from the other. But it measures Mark's consistently.