Trisecting an angle

[Fusion_power]

One of the oldest of the impossibles is that it is impossible to trisect an angle using only a compass and straight edge. It is very easy to bisect an angle. It is possible to trisect an exact 60 degree angle. (I've done it, its easy) So would someone care to take a stab at explaining why you can't trisect angles other than exact multiples of 60 degrees?



Fusion

[Kevin]

Here's what it says in one of my books......

1. If we can trisect any angle, then we can trisect a 60 degree angle.
2. If we can trisect a 60 degree angle, trig shows us that x^3-3x-1 must have a constructible solution.
3. If the cubic had a constructible solution, then it must have a rational solution.
4. Assume solution c/d where c and d have no common factors.
5. Plugging in and multiplying by d^3 gives us c^3-3cd^2-d^3=0.
6. Rewriting this one way, c(c^2-3d^2)=d^3, so c is a factor of d^3, but since c and d have no common factors, c=+-1.
7. Rewriting another way, we get d(3cd+d^2)=c^3, so similarly d=+-1.
8. We conclude that c/d must be +-1, and plugging both into the cubic fail to solve the equation.
9. Since all steps up to the beginning are (supposedly) valid, we must have erred in out assumption that we can trisect any angle.

So i guess you're thing about trisecting a 60 degree angle must not work. :?

[Jwb52z]

He's talking about trisecting a 180 degree angle in exact 60 degree segments, not a 60 degree angle.

[Kevin]

First of all, that's only 1 case. Secondly, in standard geometry angles are only defined 0<theta<180 non-inclusive. A "straight angle" of 180 degrees doesn't qualify as a real angle. Neither does a 0 degree angle (which I bet I can trisect ;) ).

[Fusion_power]

Its been a lot of years since I did this but I was able to trisect a 60 degree angle. Off the top of my head though, I don't remember the exact steps. I do remember that I had to convert one of the arms to a straight line in the process. I don't think it would be hard to do again. It takes advantage of dividing a circle into six equal sections when the compass is first used to draw a circle, then the point is placed on the perimeter and an arc is drawn from the perimeter on one side through the center and to the perimeter on the other side.

Surprisingly, most people approach this problem from the wrong perspective. Draw an acute angle. Then take a compass with the point on the vertex and draw an arc connecting the sides of the angle. The problem is to trisect the arc.

Fusion

[nucleon]

I think I know the answer, but I can't prove it :)

Draw angle @ 60deg, mark vertex as A, fix compass to a length 'x', mark off B and C. Connect B & C. Now we have equalateral triangle ABC.

Bisect line BC. The mid point of BC becomes the centre with circle of radius BC/2. With compass mark point D on the external arc from B, and similarly create point E, with compass on C.

Draw AE, and AD. AE and AD trisect BAC. Angle DBA = Angle EAC regardless of the value of angle BAC, but only trisects BAC when BAC = 60deg.

-- Craig

[Wynand]

You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter.
Now you bisect the angle, as usual. Now you extend the bisect line far
enough to draw an exact invert of the angle. Now you bisect one of the
arms of the inverted angle. If you draw a line back to the initial vertex
of the initial angle, it will be exatly one third of the angle.
I can include a graphic example, but don't know how to do it.
Maybe you can tell me how ?
Wynand

[ixfd64]

This might help.

[R.D. Silverman]

Originally Posted by Wynand

You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter.
Now you bisect the angle, as usual. Now you extend the bisect line far
enough to draw an exact invert of the angle. Now you bisect one of the
arms of the inverted angle. If you draw a line back to the initial vertex
of the initial angle, it will be exatly one third of the angle.
I can include a graphic example, but don't know how to do it.
Maybe you can tell me how ?
Wynand

Sigh. Where is Underwood Dudley? We need him.

[cheesehead]

Originally Posted by Wynand

You draw an acute angle, any angle 10 degrees, 20, 35 , doesn't matter. Now you bisect the angle, as usual. Now you extend the bisect line far enough to draw an exact invert of the angle. Now you bisect one of the arms of the inverted angle. If you draw a line back to the initial vertex of the initial angle, it will be exatly one third of the angle.

No, it won't.

[mfgoode]

Originally Posted by Kevin

Here's what it says in one of my books......

1. If we can trisect any angle, then we can trisect a 60 degree angle.
2. If we can trisect a 60 degree angle, trig shows us that x^3-3x-1 must have a constructable solution.
3. If the cubic had a constructable solution, then it must have a rational solution.
4. Assume solution c/d where c and d have no common factors.
5. Plugging in and multiplying by d^3 gives us c^3-3cd^2-d^3=0.
6. Rewriting this one way, c(c^2-3d^2)=d^3, so c is a factor of d^3, but since c and d have no common factors, c=+-1.
7. Rewriting another way, we get d(3cd+d^2)=c^3, so similarly d=+-1.
8. We conclude that c/d must be +-1, and plugging both into the cubic fail to solve the equation.
9. Since all steps up to the beginning are (supposedly) valid, we must have erred in out assumption that we can trisect any angle.

So i guess you're thing about trisecting a 60 degree angle must not work. :?

Kevin your book is right. Note the ruler must be unmarked.
Trisection of an angle in general is impossible
.
The multiples of 60* can be trisected. 90* and multiples also can be trisected.
0* and 180* are very much angles.
Your cubic stems from the formula that cos theta (\0\) = g say and follows from the trig formula. in terms of \0\* divided by 3
Your proof stands for 60* and it is sufficient to exhibit only one angle that cannot be trisected, since, a valid 'general method' would have to cover every single example

Hey the URL given by ixfd64 is a good one. It says angle 27* can be trisected under the usual conditions. I have put on my thinking cap!
Mally