Trisecting an angle

[R.D. Silverman]

Originally Posted by mfgoode

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

This is trivial. Note that 4*27 = 108 and that 108 - 90 = 18.

We can construct a 108 degree angle because it is the interior angle
of a regular pentagon and 2^2^1 + 1 = 5 is prime. We can construct a
27 degree angle by bisecting it twice. We can construct an 18 degree
angle by subtracting 90 from 108. Thus, we can construct a 9 degree
angle.

[alpertron]

All angles that are multiple of 3 degrees can be constructed since:

sin 3° =

cos 3° =

In general, the only angles that we can construct are multiple of degrees, provided that there is not an unknown Fermat prime number.

[alpertron]

In general, the only angles that we can construct are multiple of degrees, provided that there is not an unknown Fermat prime number.

For example, the cosine of 360/17 degrees is:

[alpertron]

Originally Posted by alpertron

In general, the only angles that we can construct are multiple of degrees, provided that there is not an unknown Fermat prime number.

The paragraph quoted above is incorrect, but I cannot edit my own messages, so I write the corrected paragraph below:

In general, the only angles that we can construct are multiple of degrees, provided that there is not an unknown Fermat prime number

[mfgoode]

Originally Posted by R.D. Silverman

This is trivial. Note that 4*27 = 108 and that 108 - 90 = 18.

We can construct a 108 degree angle because it is the interior angle
of a regular pentagon and 2^2^1 + 1 = 5 is prime. We can construct a
27 degree angle by bisecting it twice. We can construct an 18 degree
angle by subtracting 90 from 108. Thus, we can construct a 9 degree
angle.

Thank you Bob for the excellent explanation for not only the 27*angle but also the 9* angle.
Gauss aged 17 yrs. investigated the constructability of regular "p-gons" (polygons with p sides) where p is a prime number. He derived that only if p is a prime "Fermat number" that this is possible.
p = 2^ 2^n +1

Now the first Fermat numbers are 5 , 17 , 257 , 65537 . Euler factorised F(5) and hence proved that it is composite. No further F(n) primes have been found for n>4 so far. May I say that this is also a challenge for GIMPS?
I would include the prime 3 for n=0 . Why has this prime not been included?

Alperton: Thank you for expanding my horizon on this topic. I need to study it further before I can comment on it.
As you say that all multiples of 3/expression are the number of angles that can be constructed then knowing the 9* angle can we construct a 3* angle?
Mally

[alpertron]

In order to actually construct the 3 degree angle, you can start from the well known pentagon and hexagon constructions. These have angles of 360°/5 = 72° and 360°/6 = 60° respectively. Subtract both angles and you get 12°. Bisect it twice and finally you get 12°/2^2 = 3°.

[mfgoode]

:surprised Thank you Alperton. You have to spoon feed me!
Please explain the denominator of 3 in your previous post. I recognise the \Fermat numbers in it but I would imagine the angles will either be huge or very tiny.
Can a relationship be worked out for the number of sides in a regular polygon and the number of angles that can be constructed by unmarked ruler and compass?
Mally

[alpertron]

The heptadecagon can be constructed, see for example this Wikipedia article.

The angle between two succesive vertex is 360°/17.

Using the 3° angle constructed in my last post, just start drawing a 120-gon, where one of its vertex is a vertex of the 17-gon. You can draw other 16 120-gons using the other 16 vertex of the 17-gon.

Finally you have 120*17 = 2040 points in the circle. The angle between these points is 3°/17.

Since the 257- and 65537-gons are also constructible, using the same procedure you can construct an angle of 3°/(17*257*65537). Finally you can bisect this angle n times to have an angle of 3°/(2^n*17*257*65537)

[mfgoode]

[QUOTE=alpertron]The heptadecagon can be constructed, see for example this Wikipedia article.

Thank you for the URL. It was most enlightening. I knew it was possible according to Gauss but did not have a method. But it seems very difficult to divide into 17 parts as The stone mason hired to erect it on Gauss' tomb refused as it was too complicated.
I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.
Mally

[R.D. Silverman]

Originally Posted by mfgoode

I will try and tabulate the different angles starting from 3* angles up to 90* angles which can be constructed geometrically. Do you by chance have a ready reckoner for these constructions? It will be valuable information for the forum.
Mally

I see nothing wrong with your doing so for your own education and
entertainment. In fact, I applaud such efforts.

However:

(1) Unless you restrict to angles that are an integral number of degrees, your
list will be infinite. (e.g. bisecting a 45 degree angle etc.)

(2) Caiming that the result will be "valuable information" is ludicrous beyond
words. This information is *useless*. And it is easily reconstructed.

[fatphil]

Originally Posted by R.D. Silverman

Sigh. Where is Underwood Dudley? We need him.

I had to google for him (which lead immediately to wikipedia).
Author of /The Trisectors/, it appears.

Thanks Bob - my Christmas pressy wish list has grown by one!

Phil