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:
/ / \
/ / \
| /_______| 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:
F -- ___-- /--
_/_-- / \
___---/--__ / \
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.
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.