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2006-09-27, 23:24   #12
ewmayer
2ω=0

Sep 2002
República de California

263F16 Posts

Quote:
 Originally Posted by Wacky I would claim that this is a "radial" solution sine it has radial symmetry.
Well, it is indeed symmetric about e.g. the x and y-axes, but I would counter that of all the solutions offered so far it is the "least radial" since it has the lowest symmetry degree with respect to rotations in the plane (twofold, rather than 3 or 4-fold).

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 2006-09-27 at 23:25

2006-09-27, 23:30   #13
Uncwilly
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Quote:
 Originally Posted by Wacky I would claim that this is a "radial" solution since it has point symmetry about the origin.
But, the lines are not.

What about something like this? (Only a quick illustration, not to scale.)
Attached Thumbnails

 2006-09-27, 23:50 #14 ewmayer ∂2ω=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 x-axis: 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 x-axis; the base of the triangle will intersect the x-axis at |x|= sqrt(3)/2; draw a circle intersecting that point, and we're done. Last fiddled with by ewmayer on 2006-09-27 at 23:55
2006-09-27, 23:53   #15
ewmayer
2ω=0

Sep 2002
República de California

9,791 Posts

Quote:
 Originally Posted by Uncwilly What about something like this? (Only a quick illustration, not to scale.)
You just need to figure out what the angles in the figure are (I don't have time at the moment - have to get some actual work done, annoyingly enough) such that the areas come out right and whether those angles are constructible via rule and compass.

But note that that solution, even though it lacks 4-fold symmetry, does involve an explicit diameter of the circle, so it all comes down to what one considers a "radial" solution.

2006-09-28, 08:42   #16
xilman
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May 2003
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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.
It appears to me that the solutions proposed so far are not answering the question as asked.

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 one-dimensional whereas the disk is two-dimensional.

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

2006-09-28, 09:47   #17
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by xilman It appears to me that the solutions proposed so far are not answering the question as asked. 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 one-dimensional whereas the disk is two-dimensional. 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
5 pieces are a lot easier.

2006-09-28, 14:06   #18
xilman
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May 2003
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Quote:
 Originally Posted by xilman 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.
Clarification: I didn't quite manage it without cheating.

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 sub-dividing 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

2006-09-28, 14:25   #19
xilman
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Quote:
 Originally Posted by Uncwilly Sure. 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.
If you wish the portions of the circle to be continuous, they are necessarily congruent. Portions of a circle are arcs. All the portions must be of equal size, by the statement of the problem, and arcs of equal size are congruent to each other. They all subtend the same angle, a right-angle, at the centre of the circle.

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

2006-09-28, 14:58   #20
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by xilman It appears to me that the solutions proposed so far are not answering the question as asked. 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 one-dimensional whereas the disk is two-dimensional. 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
Dividing the circle into 4 parts is easy without using any radii as a
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

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.

2006-09-28, 19:39   #21
Uncwilly
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Quote:
 Originally Posted by xilman Ernst's solution is divinding the disk into four portions. The question asked for the circle to be so divided.
By circle, I meant the area encompassed by the circle, the disk.

2006-09-28, 19:46   #22
xilman
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May 2003
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Quote:
 Originally Posted by Uncwilly By circle, I meant the area encompassed by the circle, the disk.
Ah. In that case you should have said so.

Bob and I were solving the question you asked, not the question you intended to ask.

Paul

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