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Bractals
Member since Jun-9-03
Aug-23-03, 06:01 PM (EST)
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"Divide Disk"
 
   Hi,

Here is a cute problem - I forget where I first saw it.

With straightedge and compass divide the unit disk into seven parts with equal areas and equal perimeters.

Bractals


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  Subject     Author     Message Date     ID  
  RE: Divide Disk Vladimir Aug-26-03 1
     RE: Divide Disk Bractals Aug-26-03 2
         RE: Divide Disk Vladimir Aug-26-03 3
             RE: Divide Disk Bractals Aug-26-03 4
                 RE: Divide Disk Vladimir Aug-27-03 5
                     RE: Divide Disk Bractals Aug-28-03 6
                     RE: Divide Disk sfwc Aug-29-03 7
                         RE: Divide Disk Vladimir Aug-29-03 8

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Vladimir
Member since Jun-22-03
Aug-26-03, 08:10 AM (EST)
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1. "RE: Divide Disk"
In response to message #0
 
   LAST EDITED ON Aug-30-03 AT 03:01 AM (EST)
 
The impossibility of squaring a circle puts severe restrictions on the possible dividing lines. The 7 areas cannot have the same shape, because a regular heptagon is also impossible to construct. Consequently, all differently shaped areas or their parts must be proportional to p. Similarly, all different dividing line segments must be proportional to p. It appers to me that the only possible shapes are circle sectors with straight segments replaced by circular archs (or shapes derived from such circle sectors), with all the corresponding circle centers inside the unit disk. The idea is to use a 6-fold symmetry - 6 equal shapes plus one different central shape.

Let's start with the area: Divide the unit circle into 6 equal sectors (by constructing a regular hexagon). Place another circle in the disk center. Select its radius R to ensure that the 7 shapes (the 6 sector parts with areas A6 and the central circle with area A1) have the same areas:

A6 = p/6·(1 - R2) = p/7
A1 = p·R2 = p/7
R = Ö7/7 » 0.3780

Let's see what the corresponding perimeters P6 and P1 are:

P6 = 2p/6·(1 + R) + 2·(1 - R)
P1 = 2p·R

Not only the perimeters P6 and P1 are not equal, a more serious defect is that P6 is not proportional to p. To correct this defect, replace the remaining straight segments by half-circles with radius r = (1 - R)/2. The areas do not change, but the perimeters P6 of the 6 sector parts increase to:

P6 = 2p/6·(1 + R) + 2p/2·(1 - R)

The perimeters P6 and P1 are still not equal, but at least they are now both proportional to p. Now increase the perimeter P1 without upsetting the area balance. Cut some area from each sector part, cut the same area from the central circle, and switch them. P1 cannot be increased without increasing P6 as well, but the sector part perimeter will increase by 2x the perimeter of the cut-out shape, while the central circle perimeter by 12x the perimeter of the cut-out shape. Since the perimeter of the cut-out shape must be proportional to p, make it a little circle. Let's calculate the little circle radius r:

2p/6·(1 + R) + 2p/2·(1 - R) + 2·2p·r = 2p·R + 12·2p·r

Solving for r:

10r = (1 + R)/6 + (1 - R)/2 - R = 2/3 - 4/3·R

r = (1 - 2R)/15 = (1 - 2·Ö7/7)/15 » 0.0163

All radii (R = Ö7/7), r = (1 - Ö7/7)/2, and r = (1 - 2·Ö7/7)/15) are constructible. Is it cheating to say that a sigle shape is made of 2 or more touching parts?


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Bractals
Member since Jun-9-03
Aug-26-03, 02:13 PM (EST)
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2. "RE: Divide Disk"
In response to message #1
 
   Hi Vladimir,

You lost me in the last section where you cut out the small circles. Is it possible to supply a rough sketch of your final configuration?

Bractals


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Vladimir
Member since Jun-22-03
Aug-26-03, 02:25 PM (EST)
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3. "RE: Divide Disk"
In response to message #2
 
   Hello Bractals,
Sorry, I forgot to include the sketch yesterday.
Regards, Vladimir

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Bractals
Member since Jun-9-03
Aug-26-03, 06:20 PM (EST)
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4. "RE: Divide Disk"
In response to message #3
 
   Hi Vladimir,

Your solution now gives me two. See the following figure for another.

Bractals


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Vladimir
Member since Jun-22-03
Aug-27-03, 06:46 AM (EST)
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5. "RE: Divide Disk"
In response to message #4
 
   LAST EDITED ON Aug-30-03 AT 03:01 AM (EST)
 
Hi Bractals,

You rascal, you knew this all the time.

I was just thinking about improving my construction: Make the central circle a little larger - its perimeter will not increase much. Balance the areas and perimeters simultaneously by subtracting 6 little circles from the central circle (rather than adding/subtracting 12 little circles). Denote the central circle radius R and the little circle radius r. Denote the areas and perimeters as before (A1, A6, P1, P6):

A1 = p·(R2 - 6r2)

A6 = p·(1/6 - R2/6 + r2)

P1 = 2p·(R + 6r)

P6 = 2p·(1/6 + R/6 + (1 - R)/2 + r)

Equal areas A1 = A6 and equal perimeters P1 = P6 yield 2 equations for 2 unknowns R and r:

R2 - 6r2 = 1/6 - R2/6 + r2
R + 6r = 1/6 + R/6 + (1 - R)/2 + r

7/6·R2 - 7r2 = 1/6
5r = 2/3 - 4/3·R

7R2 - 42r2 - 1 = 0
r = 2/15·(1 - 1R)

These equations look quite innocent, but the solution is ugly. Substitute the 2nd equation into the 1st one:

7R2 - 42·4/225·(1 - 4R - 4R2) - 1 = 0
75·7R2 - 42·4 + 42·16R - 42·16R2) - 75 = 0
301R2 + 224R - 131 = 0

R = {-224 ± Ö(2242 + 4·301·131)}/(2·301) = (-112 ± Ö51975)/301

Only one root is positive (the one with the plus sign):

R = (-112 + 15·Ö231)/301 » 0.3853
R = (70 - 4·Ö231)/301 » 0.0306

The sketch looks a little better. Both radii (R and r) are constructible, but I would not want to be the one to perform the construction.

Can you divide a unit disk into 2 similar parts not symmetrical with respect to the disk center? Just divide, not necessarily with a straightedge and compass.

Regards, Vladimir

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Bractals
Member since Jun-9-03
Aug-28-03, 03:36 PM (EST)
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6. "RE: Divide Disk"
In response to message #5
 
   Hi Vladimir,

This makes three solutions to the problem and I think you can get another one if you add the six little circles instead of subtracting them from the central circle.

I have been thinking about your problem of dividing the disk into two similar parts. Every time I try the result is symmetrical about the disk center. I will keep trying.

Bractals


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sfwc
Member since Jun-19-03
Aug-29-03, 02:53 PM (EST)
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7. "RE: Divide Disk"
In response to message #5
 
   It is not specified in the question whether the open or closed unit disk is to be dissected.

In fact, the answers are different for these two cases.
In the following, by f^n(x) I men the function f applied to x n times, not the nth power of f(x). Similarly, by f^-1 I mean the inverse function of f, and by f^-n(x) I mean (f^-1)^n(x).

case 1: the open unit disk.
This may be dissected
Construction:
Denote the disk by U the point (0, 1) by p, and the function dilating the plane by a factor of two about P by f. For any x in U, |x-p| > 0, |f(x)-p| = 2 * |x-p|. By induction |f^n(x)-p| = 2^n * |x-p| for any positive integer n. So for each x in U there exists a positive integer m such that f^m(x) is not in U. For any x in U, let g(x) to be the least positive integer n such that f^n(x) is not in U. Let the set S_0 be the set of all x in U with g(x) ~ 0 mod 2, and S_1 the set of all x in U with g(x) ~ 1 mod to. Then {S_0, S_1} is a partition of the unit disk into two similar sets: It is easy to show that S_0 = f(S_1), using the fact that f(U) entirely contains U.

case 2: the closed unit disk.
This may not be dissected into two similar sets (note that this includes pairs of sets similar about the origin).
Outline proof: by contradiction
Let U be the closed unit disk, and let S_O and S_1 be sets, and f a similarity of the plane, centred at p, such that {S_0, S_1} is a partition of U and S_0 = f(S_1).

Claim 1: p is not contained in U
Proof: contradiction
Assume p to be contained in U. Since p = f(p), p is in S_0 iff p is in S_1. But p is required to lie in exactly one of these two sets, which is the required contradiction.

For similar reasons, f may not be the identity function.
Let U_1 = f(U) and U_-1 = f^-1(U). Let R_-1 = U - U_-1 and R_1 = U - U_1. Let R be the intersection of R_-1 and R_1.

Claim 2: R is non-empty
Proof: In three cases
case 1: U does not intersect U_-1
then U_1 does not intersect U, and so R_-1 and R_1 are both equal to U, as is R. Hence R is non-empty, as required.
case 2: U intersects U_-1 in exactly 1 point.
then U_1 intersects U in exactly 1 point. R is U minus those two points, and so is again non-empty, as required.
case 3: U intersects U_-1 in at least two points.
Let C_0 be the boundary of U, C_-1 the boundary of U_-1 and C_1 the boundary of U_1. Using claim 1, it may be seen that C_-1 and C_0 intersect at exactly two points, which I will call x and y. Indeed, the boundary of the intersection of U_-1 and U consists of x, y and the two open arcs A_-1 of C_-1 and A_0 of C_0. Some elementary geometry shows that the angles subtended by these arcs at the centres of their respective circles sum to less than 2 pi. Two arcs of C_0 are not in R: The first, A_0, is not contained in R_-1. The second, f(A_-1), is not contained in R_1. But C_0 subtends an angle of 2 pi at 0, whereas the sum of the angles subtended by these two arcs is less than 2 pi. So at least one point of C_0 is in neither arc, and so is contained in R. Hence R is non-empty, as required.

Claim 3: Any point in R_-1 must be contained in S_1
Proof: contradiction
Let x be contained in R_-1 but not S_1. Then x is contained in S_0. This implies f(x) is contained in S_1, which is contained in U. But f(R_-1) = f(U-U_-1) = f(U) - f(U_-1) = U_1 - U, which has an empty intersection with U. However, x is contained in this empty intersection! This is a contradiction, as required.

For similar reasons, R_1 is contained in S_0
So R is contained in both S_0 and S_1 and is nonempty
Hence {S_0, S_1} is not a partition of U, which is the original desired contradiction.


Note that the problem becomes simpler (and has a different answer) if we redefine a dissection as follows: a pair of open sets S_0 and S_1 such that:
1: They have empty intersection
2: The union of their closures is the closed unit disk

This only allows for dissections of closed figures. Now let S be any closed convex subset of the plane, and let f be any similarity, centre p, of the plane mapping S onto a subset of itself. Let n(x) (for x in S - {p}) be the least positive integer such that f^-n(x) is on the boundary of or outside S.
Let g(x) = -1 if f^(-n(x))(x) is on the boundary or if x = p, and be congruent to n(x) mod 2 otherwise. If we then define S_-1, S_0 and S_1 in the natural way, we have that (S_0, S_1) is a dissection of S
Here S could be U, the closed unit disk.

Thankyou

sfwc
<><


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Vladimir
Member since Jun-22-03
Aug-29-03, 08:41 PM (EST)
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8. "RE: Divide Disk"
In response to message #7
 
   Very nice proof. I completely missed the fact that the unit disk must be open, othervise, your construction is exactly what I had in mind. Except I do not dilate the plane about the point P º (1, 0) by a factor of 2, but by any factor a > 1. For the drawing, a = 5/4. Or I might say that I contract the plane by any factor 1/a < 1 and for the drawing, 1/a = 4/5.

Regards, Vladimir

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