20060927, 23:24  #12  
∂^{2}ω=0
Sep 2002
República de California
263F_{16} Posts 
Quote:
So this begs the question: is a solution possible (again via classical ruler & compass) which contains more than 2 kinds of pieces? (All the ones proffered so far have either 4 identical pieces, 3 of one kind and a fourth that is different, or 2 pair of congruent pieces). Last fiddled with by ewmayer on 20060927 at 23:25 

20060927, 23:30  #13  
6809 > 6502
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Aug 2003
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Quote:
What about something like this? (Only a quick illustration, not to scale.) 

20060927, 23:50  #14 
∂^{2}ω=0
Sep 2002
República de California
9,791 Posts 
OK, it's also straightforward to construct a solution consisting of 4 nested circles  for simplicity let's have them all share a common center at the origin, outermost circle has radius 1 and area A = pi:
1) Innermost circle (circle 1) has radius 1/2 and area A/4, construct via straightforward bisection of radius of the largest circle. 2) Circle 2 has radius 1/sqrt(2), construct by e.g. inscribing the largest circle in a 2x2 square (say with horizontal top and bottom sides for simplicity), quartering the square, drawing crossing diagonals of one of the 1x1 squares, and drawing a line segment from the center of the 1x1 square perpendicularly to (say) the xaxis: this line segment will intersect the axis at x = 1/sqrt(2), the circle intersecting the same point has area 1/2 the largest circle, hence the annular region between it and circle 1 has area A/4; 3) Circle 3 will similarly be such that the annular region between it and circle 2 has area A/4, thus circle 3 must have radius sqrt(3)/2. We can construct this circle by inscribing an equilateral triangle with apex at the common center and symmetric about (say) the xaxis; the base of the triangle will intersect the xaxis at x= sqrt(3)/2; draw a circle intersecting that point, and we're done. Last fiddled with by ewmayer on 20060927 at 23:55 
20060927, 23:53  #15  
∂^{2}ω=0
Sep 2002
República de California
9,791 Posts 
Quote:
But note that that solution, even though it lacks 4fold symmetry, does involve an explicit diameter of the circle, so it all comes down to what one considers a "radial" solution. 

20060928, 08:42  #16  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2^{2}·2,531 Posts 
Quote:
Ernst's solution is divinding the disk into four portions. The question asked for the circle to be so divided. It's a matter of dimensionality  a circle is onedimensional whereas the disk is twodimensional. I spent a minute or two last night thinking about how to divide the circle into four parts without any of the construction lines being radii of the circle. Didn't quite manage it. Paul 

20060928, 09:47  #17  
Nov 2003
2^{6}·113 Posts 
Quote:


20060928, 14:06  #18  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
278C_{16} Posts 
Quote:
An example of cheating is to draw two circles of equal radius. Subdivide one of them into four pieces using radial lines. This is very easy (for instance, draw a diameter and then erect the perpendicular through the centre). When that has been done, set the compass to the length between adjacent subdividing points and then mark off the divisions on the other circle. The latter is then subdivided but no radii of that circle have been used in the construction. Paul 

20060928, 14:25  #19  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
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10011110001100_{2} Posts 
Quote:
If you wish to drop the continuity requirement, fine. They then need not be congruent. You could, for instance, divide the circle into 24 portions in the obvious manner (split into six then bisect twice) and create one arc of length six, two of length 3, and four each of length 1 and 2. Then the four portions are (6), (3+2+1), (2+2+2) and (3+1+1+1). This may make it clearer: AAAAAA D BBB CC DDD BB CC D B CC D where I've unwrapped the circle into a line (and so the final D is really adjacent to the initial A), identified the four portions by letters A B C and D, and put a space at the divisions of the circle. Each of A, B, C and D constitute one quarter of the circle. None of them is congruent to any other. Unfortunately, only one of them (A) is continuous. Paul 

20060928, 14:58  #20  
Nov 2003
2^{6}·113 Posts 
Quote:
construction line. Instead, draw a line segment outside of the circle whose length equals the radius. (assume r = 1 wlog) This a line segment, but it is not a radius. Now construct a square whose sides equal the length of the radius. Do this using the external line segment. Now, adjust the compass to a width equal to the length of a DIAGONAL of that square. Place one end of the compass on the circle. Draw another circle. It will intersect the first circle. Now place the compass at the intersection point and repeat. You get 4 points on the circle, each distance sqrt(2) from its nearest neighbors. QED One only need construct a line segment of length sqrt(2). This is trivial. But as I said, 5 pieces are even easier. 

20060928, 19:39  #21 
6809 > 6502
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Aug 2003
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20060928, 19:46  #22 
Bamboozled!
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May 2003
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2^{2}·2,531 Posts 

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