(idea) by m_turner Sun Dec 23 2001 at 2:52:38
The classic problem of antiquity of trisecting an angle with only a straight edge and a compass is still impossible (the other two classic problems are construct a cube with twice the volume of a given cube (doubling the cube), and construct a square with the same area as a given circle (squaring the circle)). There are special cases where with common sense (and a straight edge and a compass as the only tools) it becomes possible.

270 degrees
This shape looks like an 'L' and is the outside angle of a right angle. Trisecting this angle is trivial: 
   A
     +--------->X
     |_|
     |  90
     |
     |
     V
     Y
To trisect angle 'A':
  1. continue the line 'X' to the left
  2. continue the line 'Y' up. 
     ^
     |
     |
   A |
<----+--------->X
     |_|
     |  90
     |
     |
     V
     Y
Each of the angles formed by the lines is 90 degrees, one third of 270 degrees.

180 degrees (a line)
Described above.

The trisecting of an angle is equivalent to solving a cubic equation. Using an unmarked ruler and compass there is a limited set of equations that may be solved - and these are all quadratic equations or reducable to quadratics. It was proven in the 19th century that this was impossible to trisect an angle with only a straight edge and compass.

The key to realizing the why the 270 degree, 180 and 90 degree (not shown) trisections work is that the angles which result (90 degrees, 60 degrees, and 30 degrees) are able to be constructed simply with a straight edge and compass. However, it is not possible to construct a 20 degree angle, and thus it is likewise impossible to trisect a 60 degree angle (this was proven by Pierre Laurant Wantzel (1814 - 1848) in 1836).

To trisect a 60 degree angle, this is the equivalent of constructing a 20 degree angle from scratch - which is the same as constructing a line of cos(20).

The cosine of 60 is 1/2 thus, cos(60) = cos(20 + 20 + 20) = 1/2.

This becomes:
4cos3(20) - 3cos(20) = 1/2
assigning x = cos(20), multiplying each side by 2, and moving the 1 to the left hand side, we get
8x3 - 6x - 1 = 0

This cubic equation cannot be reduced any further and does not have an equivalent solution as a quadratic.

http://www.cs.unb.ca/~alopez-o/math-faq/mathtext/node28.html http://mathcircle.berkeley.edu/BMC3/construct/node12.html

I like it!