trisection of the angle

The division of an angle (two line segments joined at one end) on paper into three equal angles. While it is possible to bisect an angle drawn on a piece of paper, using only a compass and a straightedge, trisection cannot be done except in very unusual circumstances (like, say, a 270-degree angle).

Well, okay, it can be done. You just need one more tool: a marked straightedge. That, and a little expertise in geometric theorems, so that you can follow the process.

Let's start by crawing a circle around the angle to be trisect, like so:

     ________ B
   --       /--
  /        /   \
 /        /     \
|        /_______| A
|       C        |
 \              /
  \            /
   --________--

∠BCA is the angle we want to trisect. We need to take our straightedge and mark on it the radius of the circle, the endpoints of segment AC or BC. Now: extend AC through the other side of the circle. Take your marked straightedge and position it so that one endpoint of the radius lies on AC extended, the other endpoint lies on the circle, and B lies somewhere else on the straightedge, like so:

                ________ B
           F  -- ___-- /--
            _/_--     /   \
      ___---/--__    /     \
   _--_____|_____--_/_______| A
 E        D|       C        |
            \              /
             \            /
              --________--

AE is the line extended from AC, and D is the point it crosses the circle. F lies on the circle, the length of EF is equal to that of AC, and EFB is a straight line (segment). Using a straightedge in this manner is illegal according to the ancient Greeks, but it does the trick: ∠FED is exactly one-third of ∠BCA.

Proof: Let ∠FED = q. Because CF is also a radius of the circle, |EF| = |CF| and EFC is an isosceles triangle. Therefore ∠FCD = q, and since the sum of the angles of any triangle equals 180 degrees, ∠EFC = 180-2q. Since EFB is a straight line (a 180-degree angle), ∠BFC = 180-∠EFC = 2q. Since CF and CB are both radii, CFB is also an icosceles triangle and ∠BFC = ∠ FBC = 2q, so ∠FCB = 180-∠BFC-∠FBC = 180-4q.

Now, DCA is also a straight line, so 180 = ∠DCF+∠FCB+∠BCA = q+(180-4q)+∠BCA. Doing the algebra yields ∠BCA = 3q = 3(∠FED), and we have our trisection.

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