-- regions in R^2 determined by n lines
quasi
of which are parallel, and no 3 of which have a common point. Of the
bounded regions determined by S (more precisely -- the areas of the
bounded connected components of R^2\S, consider the ratio of the area
of the region with greatest area to that of the least. Let r(n) be the
least such ratio for all possible choices of S.
Clearly r(3) = 1.
Question:
What is the value of r(4)?
quasi
ge...@math.mq.edu.au
I asked a couple of related questions at a conference a few years
ago, I'm not sure whether I ever posted them here:
If you cut a circle into the maximum number of pieces with n straight
lines, how large can you make the smallest piece?
How many equal pieces can you cut a circle into with n straight
lines?
For the second question, there are a couple of ways to cut the circle
into 2 n equal pieces, but is there any n for which you can do
better?
--
GM
Tim Little
> consider the ratio of the area of the region with greatest area to
> that of the least. Let r(n) be the least such ratio for all possible
> choices of S.
> What is the value of r(4)?
r(4) = 1, also.
Example: let S_1 and S_2 be x and y axes. Let S_3 be a line passing
through (3,3) and (0,2), while S_4 passes through (3,3) and (2,0).
These divide the plane into three finite regions, all of which have
area 6.
- Tim
quasi
wrote:
Cool.
Is r(5) also equal to 1?
quasi
quasi
>On Mar 8, 1:42 pm, quasi <qu...@null.set> wrote:
>> Let n in N, n > 2, and let S be a set of n distinct lines in R^2, no 2
>> of which are parallel, and no 3 of which have a common point. Of the
>> bounded regions determined by S (more precisely -- the areas of the
>> bounded connected components of R^2\S, consider the ratio of the area
>> of the region with greatest area to that of the least. Let r(n) be the
>> least such ratio for all possible choices of S.
>>
>> Clearly r(3) = 1.
>>
>> Question:
>>
>> What is the value of r(4)?
>
>I asked a couple of related questions at a conference a few years
>ago, I'm not sure whether I ever posted them here:
>
>If you cut a circle into the maximum number of pieces with n straight
>lines, how large can you make the smallest piece?
Yes, that's similar in spirit to the question I asked.
>How many equal pieces can you cut a circle into with n straight
>lines?
>
>For the second question, there are a couple of ways to cut the circle
>into 2 n equal pieces, but is there any n for which you can do
>better?
Interesting question.
quasi
Tim Little
> Is r(5) also equal to 1?
No. A suitable choice of intersection labelling and affine
transformation can reduce the cases to one parametrized diagram.
S_1 and S_2 are coordinate axes, the axis intersections are S_3 at
(0,1) & (x,0), S_4 at (1,0) & (0,y), with S_5 intersecting at some
(w,0) and (0,z) with 1 < y, 1 < z, 1 < x, 1 < w, and z < w y. This
splits into three cases based on whether zero, one or both of y < z
and x < w hold.
Trying to equate the areas yields an inconsistent set of equations in
both cases, though it is close: 5 regions can have equal area.
The various boundaries of area ratio inequalities could be explored
for the ratio minimum, but that seems likely to be very messy, and
possibly unenlightening.
- Tim
quasi
wrote:
>On 2009-03-08, quasi <qu...@null.set> wrote:
>> Is r(5) also equal to 1?
>
>No. A suitable choice of intersection labelling and affine
>transformation can reduce the cases to one parametrized diagram.
>
>S_1 and S_2 are coordinate axes, the axis intersections are S_3 at
>(0,1) & (x,0), S_4 at (1,0) & (0,y), with S_5 intersecting at some
>(w,0) and (0,z) with 1 < y, 1 < z, 1 < x, 1 < w, and z < w y. This
>splits into three cases based on whether zero, one or both of y < z
>and x < w hold.
>
>Trying to equate the areas yields an inconsistent set of equations in
>both cases, though it is close: 5 regions can have equal area.
I expected that r(4) would be greater than 1.
>The various boundaries of area ratio inequalities could be explored
>for the ratio minimum, but that seems likely to be very messy, and
>possibly unenlightening.
Well, I was curious as to what an optimal configuration would look
like.
quasi
quasi
>On 08 Mar 2009 23:21:21 GMT, Tim Little <t...@little-possums.net>
>wrote:
>
>>On 2009-03-08, quasi <qu...@null.set> wrote:
>>> Is r(5) also equal to 1?
>>
>>No. A suitable choice of intersection labelling and affine
>>transformation can reduce the cases to one parametrized diagram.
>>
>>S_1 and S_2 are coordinate axes, the axis intersections are S_3 at
>>(0,1) & (x,0), S_4 at (1,0) & (0,y), with S_5 intersecting at some
>>(w,0) and (0,z) with 1 < y, 1 < z, 1 < x, 1 < w, and z < w y. This
>>splits into three cases based on whether zero, one or both of y < z
>>and x < w hold.
>>
>>Trying to equate the areas yields an inconsistent set of equations in
>>both cases, though it is close: 5 regions can have equal area.
>
>I expected that r(4) would be greater than 1.
I meant r(5).
Tim Little
> Well, I was curious as to what an optimal configuration would look
> like.
As a heuristic guide, it looks like it should be achieved with the
configuration of lines and labelling from my previous post, with x=y
and w=z, with 1 < w < x. In that case all the areas are in the first
quadrant.
As a guide, one 5-areas-equal configuration for that case has x = 5
and w = 5 - 2 sqrt(5/3). There are 5 equal areas of size 5/6, and the
remaining one has area 40/3 - 10 sqrt(5/3), about 51% of the others.
More generally there are 4 non-congruent regions in that layout, with
areas (unless I've messed up the algebra):
A0 = x / (x+1)
A1 = (x-w)^2 / 2(x-1)
A2 = x(x-1) / 2(x+1) - A1
A3 = w^2 / 2 - A0 - 2 A2
Various values for x and w will order these by size, with the ratio to
be minimized corresponding to a boundary or interior condition of one
of up to 24 orderings. Or it may turn out that my guess that some
values within this general class of configurations is optimal may be
incorrect, in which case some different set of expressions for areas
will apply.
Good luck!
- Tim
David W. Cantrell
One can do better. I have a configuration showing that
r(5) <= (sqrt(17) + 1)/4 = 1.2807...
If nobody finds anything better in the next day or so, I'll post a link to
my figure.
David
David W. Cantrell
OK, here's my figure:
<http://i403.photobucket.com/albums/pp113/DWCantrell_photos/arearatio5.gif>
Yellow to cyan area ratio = (sqrt(17) + 1)/4
David
quasi
<DWCan...@sigmaxi.net> wrote:
Very nice.
Believe it or not, I was just about to reply to your previous post to
say that I had matched your ratio exactly,
In fact, it's the same configuration.
It took some work (with Maple) to find it.
I think that ratio is probably best possible.
quasi
Rainer Rosenthal
> Believe it or not, ...
Well, I believe you, quasi ;-)
Cheers,
Rainer