20060927, 19:24  #1 
6809 > 6502
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Aug 2003
101×103 Posts
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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. 
20060927, 19:49  #2 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} 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 20060927 at 20:08 
20060927, 19:59  #3  
6809 > 6502
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Aug 2003
101×103 Posts
10891_{10} Posts 
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20060927, 20:05  #4 
Jun 2003
2^{2}·3^{2}·151 Posts 
Are temporary radial lines allowed for construction?

20060927, 20:13  #5 
Jun 2003
The Texas Hill Country
441_{16} Posts 

20060927, 20:38  #6  
6809 > 6502
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Aug 2003
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Sure.
Quote:
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 20060927 at 20:50 Reason: why not? 

20060927, 20:58  #7 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} 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. 
20060927, 21:21  #8  
∂^{2}ω=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 equalarea subsets, two of which are circles, two of which are not.
Quote:
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... 

20060927, 22:24  #9  
6809 > 6502
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The solution that I saw does not have any circles in the resultant shapes. Last fiddled with by Wacky on 20060927 at 23:22 Reason: Speeelling typos 

20060927, 23:09  #10  
Jun 2003
The Texas Hill Country
10001000001_{2} Posts 
Quote:
Last fiddled with by Wacky on 20060927 at 23:21 

20060927, 23:21  #11  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
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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