mersenneforum.org 4 not so easy pieces?
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 2006-09-27, 19:24 #1 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 10,891 Posts 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 not be radial. Have fun.
 2006-09-27, 19:49 #2 Wacky     Jun 2003 The Texas Hill Country 32×112 Posts 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. Last fiddled with by Wacky on 2006-09-27 at 20:08
2006-09-27, 19:59   #3
Uncwilly
6809 > 6502

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1089110 Posts

Quote:
 Originally Posted by Wacky 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.
While not 'radii' those lines are radial, more or less. For your solution a swastika would also work, however the construction would be harder.

 2006-09-27, 20:05 #4 axn     Jun 2003 22·32·151 Posts Are temporary radial lines allowed for construction?
2006-09-27, 20:13   #5
Wacky

Jun 2003
The Texas Hill Country

44116 Posts

Quote:
 Originally Posted by Uncwilly While not 'radii'
Therefore you exclude all solutions that have 4 congruent parts

2006-09-27, 20:38   #6
Uncwilly
6809 > 6502

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Aug 2003
101×103 Posts

10,891 Posts

Quote:
 Originally Posted by axn1 Are temporary radial lines allowed for construction?
Sure.
Quote:
 Originally Posted by Wacky Therefore you exclude all solutions that have 4 congruent parts
In essence, yes.
http://www.thefreedictionary.com/radial Adj. Usage 1 (b & c, esp) See Stellate in the thesarus section.

Obiously I want the uneasy.

Last fiddled with by Uncwilly on 2006-09-27 at 20:50 Reason: why not?

 2006-09-27, 20:58 #7 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts 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.
2006-09-27, 21:21   #8
ewmayer
2ω=0

Sep 2002
República de California

5·2,351 Posts

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 equal-area subsets, two of which are circles, two of which are not.

Quote:
 Originally Posted by 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.
"Outer endpoints" of what?

It seems to me a tricky problem might be to subdivide into 4 equal-area 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...

2006-09-27, 22:24   #9
Uncwilly
6809 > 6502

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10,891 Posts

Quote:
 Originally Posted by ewmayer Assume the initial circle is centered at (0,0) ....two of which are not.
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:
 Originally Posted by Wacky Next divide the remainder into 3 equal parts. The easiest way would be to use radial lines.
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.

Last fiddled with by Wacky on 2006-09-27 at 23:22 Reason: Speeelling typos

2006-09-27, 23:09   #10
Wacky

Jun 2003
The Texas Hill Country

100010000012 Posts

Quote:
 Originally Posted by ewmayer 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 equal-area subsets, two of which are circles, two of which are not.
I would claim that this is a "radial" solution since it has point symmetry about the origin.

Last fiddled with by Wacky on 2006-09-27 at 23:21

2006-09-27, 23:21   #11
Wacky

Jun 2003
The Texas Hill Country

32·112 Posts

Quote:
 Originally Posted by Uncwilly How do you divide it? Second, any rotated radial lines retains the radial aspect.
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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