4 not so easy pieces?
I had thought of a problem that I don't have a clue how (nor likely the skills) to solve. I set about searching for an answer, etc. before I presented it here. While doing so, I came across a different challenge.
Given a circle and classic tools (straight edge, compass, and "pencil"); please divide the circle into four equal pieces. Child's play, right? But, here is the twist: The lines drawn to divide the circle must [SIZE="3"][B][COLOR="Blue"]not[/COLOR][/B][/SIZE] be radial. Have fun. 
[SPOILER] Does "lines drawn" include construction lines? Are curved boundaries "lines"? If so, how do you define "radial" with respect to a curve?
If you allow my interpretation, the construction is easy. First, locate four points on the circle such that they equally divide the circumference. (Construct two perpendicular diameters) Then, using in turn each of those points, create equilateral triangles which have a base which extends from the center of the circle to one of the designated points. Finally, using the top vertex of each triangle as a center, inscribe an arc to replace the base. This will form 4 congruent parts.[/SPOILER] 
[QUOTE=Wacky;88025][SPOILER] First divide the circle into the 4 parts by using perpendicular radial lines.
Then, using a convenient radius, for example the radius of the circle, replace each of the radial divisions with the arc of a circle. [/SPOILER][/QUOTE] [spoiler]While not 'radii' those lines are radial, more or less. For your solution a swastika would also work, however the construction would be harder.[/spoiler] 
Are temporary radial lines allowed for construction?

[QUOTE=Uncwilly;88026]While not 'radii'[/QUOTE]
Therefore you exclude all solutions that have 4 congruent parts 
[QUOTE=axn1;88027]Are temporary radial lines allowed for construction?[/QUOTE] Sure.
[Quote=Wacky]Therefore you exclude all solutions that have 4 congruent parts[/quote]In essence, yes. [URL="http://www.thefreedictionary.com/radial"]http://www.thefreedictionary.com/radial[/URL] Adj. Usage 1 (b & c, esp) See Stellate in the thesarus section. Obiously I want the uneasy. 
[spoiler] First create a "doughnut". The radius of the hole is half that of the circle. Thus, the hole has 1/4th of the area.
Next divide the remainder into 3 equal parts. The easiest way would be to use radial lines. But this is not allowed. Therefore simply rotate the outer endpoints a uniform distance along the circumference. These three parts will be congruent and each equal in area to the center circle.[/spoiler] 
Assume the initial circle is centered at (0,0) and has radius 1. Now draw a pair of circles of half that radius, one centered at (0, 1/2), the other at (1/2, 0). These intersect (in the sense of touching at a single point) at (0,0), and thus subdivide the large circle into 4 equalarea subsets, two of which are circles, two of which are not.
[quote="Wacky"]Next divide the remainder into 3 equal parts. The easiest way would be to use radial lines. But this is not allowed. Therefore simply rotate the outer endpoints a uniform distance along the circumference.[/quote] "Outer endpoints" of what? It seems to me a tricky problem might be to subdivide into 4 equalarea pieces, without having any of the resulting piece boundaries intersect the center of the original circle, and without ever placing the point of one's compass or ruler so it touches the center. I need to think about that a bit more... 
[QUOTE=ewmayer;88032]Assume the initial circle is centered at (0,0) ....two of which are not.[/QUOTE]A very good solution indeed, but not the one that I ran across. The one that I ran across does use the center point.
[QUOTE=Wacky;88031] Next divide the remainder into 3 equal parts. The easiest way would be to use radial lines. [/QUOTE]How do you divide it? Second, any rotated radial lines retains the radial aspect. The solution that I saw does not have any circles in the resultant shapes. 
[QUOTE=ewmayer;88032]Assume the initial circle is centered at (0,0) and has radius 1. Now draw a pair of circles of half that radius, one centered at (0, 1/2), the other at (1/2, 0). These intersect (in the sense of touching at a single point) at (0,0), and thus subdivide the large circle into 4 equalarea subsets, two of which are circles, two of which are not.
[/QUOTE] I would claim that this is a "radial" solution since it has point symmetry about the origin. 
[QUOTE=Uncwilly;88043]How do you divide it? Second, any rotated radial lines retains the radial aspect.[/QUOTE]
Let me describe the solution in polar coordinates. Assume that the circle is centered at (0,0) and has radius =2. Thus the area is 4 PI. The first part is bounded by a circle, also centered at (0,0) with radius =1 and area = PI. Now, draw the line segment from (r=1, Theta=0 degrees) to (r=2, Theta = 30 degrees) Also draw the line segment (r=1, Theta=120 degrees) to (r=2,Theta=150 degrees) And the segment (r=1, Theta=240 degrees) to (r=2, Theta=270 degrees) 
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