"Automatic trisection" of an angle

[Turtle]

Sorry, I hadn't followed your exchanges quite exactly, so in the light of this subtlety let me put it as follows: I wouldn't require this guy to spell it out in detail. If a student uses something already known, in spelling out a method, one may consider it an independent thing. Unless of course it is an actual execution that is required. When the aim is to argue a point, one can certainly use standard things. For instance, a student won't always demonstrate Pythagoras' theorem each time it comes into demostrating something else. Even less does a researcher do so when publishing a novel finding.

 

Your construction has no error. It can be done more simply though; once the segment of length two is done there's your hypotenuse and, without changing the compass, you can make the intersection between arcs from the midpoint and one end.

 

acknowledged that your leniency exceeds mine. :) on that point, how about my taking off points for his use of graph paper? :omg: :lol: also, while i let it go before, now i'm taking off 2 points for his declaring point E and then not using it.

 

roger the simpler construction. i did not set the compass to 2 for that, only an arbitray measure >1.

 

do you also agree with me that diagram 1, or any drawing of a trisection, requires points and lines inside the angle to be trisected that correspond to a trisection? after all, a construction is a drawing, isn't it? if you askered a student to do a construction & got no drawering, would you accept that as a proper response?

I'm not sure if I read your intent. Do you mean he should trace a parallel to his line AC through H and then remaining line through H? As far goes proving his point, this is superfluous, I rest that his real shortcoming is that he would need to give a construction for the reverse of what he does.

 

well maybe, if that parallel line is a trisector of IHJ. i eye-balled & drew what i meant in second attachment to post #13. i have it below again; the green lines are what i want/expect for/from a trisection.

 

i see in reviewing the thread that craig in post#3 cited the same fault with one of the original poster's drawings.

 

[Qfwfq]

taking off points for his use of graph paper?

I agree that doesn't count as a SE&C construction, but I hadn't really inspected that detail.

[Turtle]

... As far goes proving his point, this is superfluous, I rest that his real shortcoming is that he would need to give a construction for the reverse of what he does.

 

i agree. i made a drawering to start with.

 

assignment: [edit: using only compass & straight-edge]on the below attached construction, trisect angle IHJ.

Edited November 1, 2011 by Turtle

[Jayeskay]

......I rest that his real shortcoming is that he would need to give a construction for the reverse of what he does.

 

 

See Diagram-3: http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm.

[Qfwfq]

See Diagram-3: http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm.

Show us how one can START WITH OMN and arrive at OCN s'il vous plait.

[Jayeskay]

Show us how one can START WITH OMN and arrive at OCN s'il vous plait.

 

 

As per the general impossibility rule, you cannot do it that way, hence my proposed method.

[Qfwfq]

you cannot do it that way, hence my proposed method.

Well, OK bit I don't see the point in proposing such a complicated thing when there is a way of tripling angles which is not only simpler, but also better suited to the aim of trisecting by successive approximations.

[Turtle]

As per the general impossibility rule, you cannot do it that way, hence my proposed method.

 

well, on your page you make it clear that you hold that your method does call into question the impossibility rule. we can't have our cake & eat it too. :cake: :naughty:

 

Approaching the Impossible

It is geometrically given as impossible to trisect the general angle using only an unmarked straightedge and compass, as shown at en.wikipedia.org/wiki/Angle_trisection and many other sites.

 

The construction procedure of this proposed general trisection is to locate the given angle to be trisected within an existing automatic trisecting model.

 

Should the angle located within the automatic trisecting model be geometrically judged to be equal to the given angle to be trisected, the result will be a definitive solution to the trisection problem.

 

This will raise an interesting question as to a true possible constructible general trisection and the above-mentioned accepted general trisection impossibility rule.

 

If the proposed automatic trisecting model be judged as geometrically correct, an element of uncertainty will exist between this correct general trisecting model and the general trisection impossibility rule.

 

...

[phillip1882]

geometry.bmp

given: two random lines AD, BC forming two angles; construct the angle trisector of angle AEC.

step 1:

construct a right triangle such that the hypotenuse is twice that of the shortest side.

(turtle already did this, never the less I'll present here.)

construct a line perpendicular to AD, qr.

construct two points equidistant from the intersection of AD and qr. (in the diagram above, i used point E as one of the two points.)

using the distance from the two points, construct a point q from one of the two points on line qr.

draw line pq.

bicect line pq, s.

from point s using the distance from point s to point p, draw arc, rp.

step 2:

bicect angle BED, which is equal to AEC

step 3:

now the "tricky" part. (read impossible)

translate angle BEF to the pqr triangle, such that:

a: line EF goes though point q.

b: the vertex of BEF is on the arc.

c: line EB is equal in length to line sE.

you can meet any two of these requirements, but not all 3. if you could then you would have succeeded.

[CraigD]

Welcome to hypography, Jay! :) Thanks for the interesting posts. :thumbs_up

 

As Qfwfq and Turtle have already pointed out, the construction you’ve sketched isn’t a trisection of a given angle – that is, given an angle a, constructing an angle [imath]b=\frac{a}{3}[/imath] – which is known (proven) to be impossible, but a “tripling of a given angle” – given a, construct b=3a – which can easily be shown possible. Here’s a simple construction that does:

 

I puzzled for a while over your unusuall approach, realizing finally that it doesn’t work – that is, given a, it constructs, in general [imath]b \not= 3a[/imath]. Though I didn’t try proving this by construction, here’s a counter-case proof using trig functions and a simplified version of your construction sketch

Picking convenient units, we set the coordinates of the labeled points C=(0,0), D=(2,0), A=(4,0), J=(5,0).

Because H is a point on a circle radius 2 centered at D, its coordinates are [imath]\left(2+\sqrt{4-y^2} , y \right)[/imath].

Given y, then, angle [imath]a = \arctan\left( \frac{y}{2+\sqrt{4-y^2}}\right)[/imath], and [imath]b-a = \arctan\left(\frac{y}{5 -2 -\sqrt{4-y^2}} \right)[/imath]

Evaluating for y=1 gives [imath]a \dot= 0.2618[/imath], [imath]b-a \dot= 0.6678[/imath], [imath]b \dot= 0.9296 \not= 3a[/imath]

 

Though in general [imath]b \not= 3a[/imath], an interesting question is: for what exact value of a does b=3a :QuestionM Some quick calculating shows it to be approximately 1.325.

 

Correction: as Qfwfq noted, the construction actually does work. [imath]J=\left( 2+2\sqrt{4-y^2} ,0 \right)[/imath], so [imath]b-a = \arctan\left(\frac{y}{\sqrt{4-y^2}} \right)[/imath].

Evaluating for y=1 gives [imath]a \dot= 0.2618[/imath], [imath]b-a \dot= 0.5236[/imath], [imath]b \dot= 3a[/imath]. Evaluating for any [imath]0 < y < 2[/imath] gives the same [imath]b \dot= 3a[/imath].

 

I’ve not been able to make any headway on proving this, though. Any help would be appreciated. :shrug:

Edited November 6, 2011 by CraigD
Major correction

[Qfwfq]

Picking convenient units, we set the coordinates of the labeled points C=(0,0), D=(2,0), A=(4,0), J=(5,0).

Given the choice for C, D and A I don't get why J should be that. Indeed, it could only be so for [imath]y=\frac{\sqrt{7}}{2}[/imath] unless CD, DH and HJ have different lengths.

 

As I had acknowledged, he does triple the angle, except that for trisecting by successive approximations it is better to use the method shown in your first drawing.

[CraigD]

Given the choice for C, D and A I don't get why J should be that. Indeed, it could only be so for [imath]y=\frac{\sqrt{7}}{2}[/imath] unless CD, DH and HJ have different lengths.

You don't get it 'cause I erred. I miss-read Jay's description, where it says "HJ = HD ...", thinking I'd read somewhere "2AJ = AD" :doh:

 

Thanks for the catch. I'll rework my look at it. This construction's much more fun if it actually works. :)

[Jayeskay]

You don't get it 'cause I erred. I miss-read Jay's description, where it says "HJ = HD ...", thinking I'd read somewhere "2AJ = AD" :doh:

 

Thanks for the catch. I'll rework my look at it. This construction's much more fun if it actually works. :)

 

Have edited the link < http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm > to be hopefully more informative and clearer, with the longish absence due to an editing technical glitch.

 

Jayeskay.

[Guest MacPhee]

I've read many of the posts on this long and very interesting thread. They are ingenious and illuminating, and increase my respect for the intellectual power of posters on this site.

 

I can only offer a simple take on the matter, which is this:

 

You can't trisect anything using just a pair of compasses. Because the compasses have only got 2 points, one on each leg.

 

These 2 legs can bisect things, into any number of fractions: 1/2, 1/4, 1/8, 1/16, 1/32... and so on ad infinitum. An endless supply of fractions - all with an even numbered denominator.

 

But trisection, requires an odd-numbered denominator to be generated - specifically, 3.

 

Surely it's mathematically not possible for 3, or any odd number,to be generated by any combination of even numbers - which are all the compasses can make?

[phillip1882]

well, its quite possible to trisect a length of line, or any other odd number of divisions.

[Turtle]

I've read many of the posts on this long and very interesting thread. They are ingenious and illuminating, and increase my respect for the intellectual power of posters on this site.

 

I can only offer a simple take on the matter, which is this:

 

You can't trisect anything using just a pair of compasses. Because the compasses have only got 2 points, one on each leg.

 

...

 

using "compasses" to refer to one instrument is grammatically incorrect. it is not the same as "scissors".

 

inasmuch as no one has bothered to post the proof that trisecting an angle with a comapass & straightedge is impossible, i will.

 

Angle trisection

 

Proof of impossibility

 

The geometric problem of angle trisection can be related to algebra—specifically, to the problem of finding the roots of a cubic polynomial—since by the triple-angle formula, cos(3θ) = 4cos 3(θ) − 3cos(θ).

 

One can show that any number constructible in one step from a field K is a solution of a second-order polynomial. Note also that π / 3 radians (60 degrees, written 60°) is constructible. We now show that it is impossible to construct a 20° angle; this implies that a 60° angle cannot be trisected, and thus that an arbitrary angle cannot be trisected.

 

Denote the set of rational numbers by Q. If 60° could be trisected, the minimal polynomial of cos(20°) over by Q would be of second order. Now let y = cos(20°).

 

Note that cos(60°) = cos(π / 3) = 1 / 2. Then by the triple-angle formula, cos(π / 3) = 1 / 2 = 4y3 − 3y and so 4y3 − 3y − 1 / 2 = 0. Thus 8y3 − 6y − 1 = 0, or equivalently (2y)3 − 3(2y) − 1 = 0. Now substitute x = 2y, so that x3 − 3x − 1 = 0. Let p(x) = x3 − 3x − 1.

 

The minimal polynomial for x (hence cos(20°)) is a factor of p(x). Because p(x) is degree 3, if it is reducible over by Q then it has a rational root. By the rational root theorem, this root must be 1 or −1, but both are clearly not roots. Therefore p(x) is irreducible over by Q, and the minimal polynomial for cos(20°) is of degree 3.

 

So an angle of 60° = (1/3)π radians cannot be trisected.

 

Many people (who presumably are unaware of the above result, misunderstand it, or incorrectly reject it) have proposed methods of trisecting the general angle. Some of these methods provide reasonable approximations; others (some of which are mentioned below) involve tools not permitted in the classical problem. The mathematician Underwood Dudley has detailed some of these failed attempts in his book The Trisectors.[2] ...

[CraigD]

You can't trisect anything using just a pair of compasses. Because the compasses have only got 2 points, one on each leg.

well, its quite possible to trisect a length of line, or any other odd number of divisions.

Good counterexample to MacPhee’s claim, Phillip :thumbs_up Clearly, line segments can be divided into 3 equal length subsegments, or any other number of divisions, even or odd.

 

For example, though not as terse and pretty as this animated construction at wikipedia, this construction, which I thought up just now, trisects, and can be easily generalized to n-sect, line segment AB:

1. Draw half-line AC terminating at A non-coincidental with AB
   
2. Draw circle centered at A passing through B intersecting AC at D
   
3. Draw circle centered at D passing through A intersecting AC at E
   
4. Draw circle centered at E passing through D intersecting AC at F
   
5. Draw line BG perpendicular to AB
   
6. Draw circle centered at A passing through F intersecting BG at H and I
   
7. Draw half-line AH terminating at A
   
8. Draw half-line AI terminating at A
   
9. Draw circle centered at A passing through E intersecting BG at J and K
   
10. Draw circle centered at A passing through D intersecting BG at L and M
    
11. Draw line LM intersecting AB at N
    
12. Draw line JK intersecting AB at O
    

Points N and O trisect line segment AB.

 

MacPhee’s claim got me to thinking, though, and an internet search didn’t give me an easy answer to the question: is it possible, or proven impossible, to pentasect (5), septasect (7), etc an arbitrary angle by construction :QuestionM: