[QUOTE=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 :coffee:[/QUOTE]

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.

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

sin 3° = [tex]\large \frac {\sqrt {8-\sqrt{3}-\sqrt{15}-\sqrt{10-2*\sqrt{5}}}}{4}[/tex]

cos 3° = [tex]\large \frac {\sqrt {8+\sqrt{3}+\sqrt{15}+\sqrt{10-2*\sqrt{5}}}}{4}[/tex]

In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.

In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.

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

[tex]\large \frac {\sqrt{15 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} + \frac {\sqrt{2(34+6\sqrt{17}+\sqrt{2(17-\sqrt{17})} - \sqrt{34(17-\sqrt{17})} + 8\sqrt{2(17+\sqrt{17})}})}{2}}}{4}[/tex]

[QUOTE=alpertron]In general, the only angles that we can construct are multiple of [tex]\frac {3}{17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number.[/QUOTE]

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 [tex]\frac {3}{2^n*17*257*65537}[/tex] degrees, provided that there is not an unknown Fermat prime number

Trisecting an angle
 

[QUOTE=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.[/QUOTE]
:surrender 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. :unsure:
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 :coffee:

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°.

Trisecting an angle
 

: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? :confused:
Mally :coffee:

The heptadecagon can be constructed, see for example this [URL=http://en.wikipedia.org/wiki/Heptadecagon]Wikipedia article[/URL].

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)

Trisecting an angle
 

[QUOTE=alpertron]The heptadecagon can be constructed, see for example this [URL=http://en.wikipedia.org/wiki/Heptadecagon]Wikipedia article[/URL].
:bow:
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 :coffee:

[QUOTE=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 :coffee:[/QUOTE]

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.

[QUOTE=R.D. Silverman]Sigh. Where is Underwood Dudley? We need him.[/QUOTE]

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