the round table

we have a circular dining table made of marble which had come down to us as a family heirloom.we also have some beautiful bone-china saucers that i recently brought from japan.
diameter of our table top is 15 times the diameter of our saucers which are also circular.we would like to place the saucers on the table so that they neither overlap each other nor the edge of the table.how many can we place in this manner

i have no clue of how 2 go about doing this question.
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EngineerCommented:
One saucer can be placed at the centre.

Now, around it you can place 6 saucers, such that they touch each other. So, 3 times the diameter = 1+6 = 7 saucers. And, these can be surrounded by 12 saucers. So, for 5 times the diameter, 7+12 = 19 saucers. Similarly, you can continue the calculations to get the final answer as

1 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 763
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EngineerCommented:
Sorry.. it should be 1 + 6 + 12 +18 + 24 + 30 + 36 + 42 = 169
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Commented:
mgh_mgharish, this is not true.
If you order the saucers this way, they will be
1,6,12,18,24,30,36,42
and the sum is 169.
But I do not think that this is the best possible ordering. It has to be a lot better if the ordering starts from the table border toward the center.
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Commented:
Sorry, I had to refresh the page first, it was open several hours ago.
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EngineerCommented:
> It has to be a lot better if the ordering starts from the table border toward the center.

Let me know if you can come up with that !
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Commented:
They will be at least 43,37,31,25,18,12,6,1
This makes 173. But I'm not sure that this is the optimum too.
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EngineerCommented:
> They will be at least 43,37,31,25,18,12,6,1

How ?
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Commented:
This is the ordering if you start from the border of table toward the center and place the sauscers in the way that their centers be on a distances from the table's center equal to 14*r, 12*r and so on. (r is the radius of the saucer)
But I think the real answer is even bigger since they could be placed better - the inner circles could be a little more toward the table's border so the saucers in some of the inner circles could be a bigger number.
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EngineerCommented:
Given are the diameters, not radii
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Commented:
so what?
Do you find any difference? :))
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EngineerCommented:
The circumference at a radius of 7.25 would be 2*pi*7.25 = 14.5 pi = 45.553093477052001957708329057553

But it doesn't mean that it would be a better approach. In that case, following your order,

45 + 39 + 32 + 26 + 20 + 14 + 7 + 1 = 184
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Commented:
I've already shown you that it is a better approach. See several posts above.
And I do not understand anything of your last post.
It seems that you have not understand what I'm trying to say and post some random answers.
I cannot find any sense in your last post. Absolutely chaotic numbers.
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EngineerCommented:
But I still stick on to 169 :)
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EngineerCommented:
173 can't be placed.. Try it yourself
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Commented:
Ok, I say it is at least 173 but most probably a couple or even a dozen of saucers could be added.
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EngineerCommented:
I would certainly say that it is at the most 225 :)
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EngineerCommented:
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Commented:
This is quite obvious!
Although as we can see from your first posting it is not enough obvious for you. :)
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Commented:
You're going on with the chaos?
I do not see anything on this site that helps you.
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EngineerCommented:
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EngineerCommented:
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EngineerCommented:
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Commented:
so where does your confidence go? :))
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EngineerCommented:
I have gained enough confidence to say that the answer is 187 !
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Commented:
I don't think that you help shilpi84 a lot the way you behave, neither with this nor with the other questions.
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EngineerCommented:
http:M_3652891.html

We can talk about that !
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Commented:
Don't think this is urgent. Neither for me nor for anybody.
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EngineerCommented:
OK.. then if you want to have any off-topic comments, please post in a Community Support thread or contact me directly. You can click my name to get my email id
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Commented:
This one was not offtopic.
And you are the one that does not follow the administrative recomendation in the other question.
So don't play a smart guy. You doesn't seem very far from suspension.
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EngineerCommented:
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Commented:
"I don't think that you help shilpi84 a lot the way you behave" is absolutely related with the question.
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Commented:
187 is the best known.  There seems to be no proof yet that 188 is not possible.
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Commented:
Hi,

we start frim Center

1      1      0.5            ceneter Only 1
6      3      1.5            new 6 so total diameter will be 3 that is user and 1.5 radis used from table
12      5      2.5            new 12 so total diameter will be 5 that is user and 2.5 radis used from table
24      7      3.5            new 24 so total diameter will be 7 that is user and 3.5 radis used from table
36      9      4.5            new 36 so total diameter will be 9 that is user and 4.5 radis used from table
72      11      5.5            new 72 so total diameter will be 11 that is user and 5.5 radis used from table
144      13      6.5            new 144 so total diameter will be 13 that is user and 6.5 radis used from table
288      15      7.5            new 288 so total diameter will be 15 that is user and 7.5 radis used from table
------------------------------------
583

Total 583

Thanks
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Commented:
SystemExpert, if you have not read above, notice that since the diameter of the table is 15 times as much as the diameter of a saucer, using the formula for the circle surface - pi*r^2 we receive that the theoretical maximum is 225 (Saucer=pi*r^2, Table=pi*(15*r)^2=225*pi*r^2). So it cannot be 583 even theoretically.
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Commented:
I don't think there's anyway anyone expects you to get as high as 187 without googling 'circle packing'

I suggest you go for hexagonal packing, eg. as in mgh_mgharish's original post. Regarding SystemExpert's comment, For straight hexagonal packing, where there is a single circle in the center, the increase per layer should be linear, not exponential!
1     1
6     3
12   5
18   7
24   9
30  11
36  13
42  15
--------
169 is set as minimum.
Now how many more can you fit on the edge? if you actually had the dishes and table to work with, I'm sure you could fit at least 6 more.
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Commented:
jkmyoung, with this approach you cannot fit neither 6 nor even 1. It is just 169. You can try if you want.
I've already described an approach above that gives at least 173 and since I've used it roughly I've been convinced that even bigger number could be achieved.
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Commented:
the best usage of a circle space, is dividing it to hexagonals.

SpaceOfHexagonal = 6 * r^2*sqrt(3)/4
SpaceOfTable = R^2*PI = 225*r^2*PI
max hex over table = 225 *PI * 4 / 6 * sqrt(3) = 900*PI / 10.4 =~ 272

arround the table, some hex maybe overlapping,
max number of overlapping chinas, is 2*PI*R / 2*r = 6.28*15*r/2*r = 47

so, I prooved the number is between 272 to 225 (and so 225 IS the minimum)

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Commented:
Talmash, 225 is the maximum without overlapping.
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Commented:
6 * r^2*sqrt(3)/4 is smaller than r^2*PI  it would be hard to fit a circle into a smaller hexagon
SpaceOfHexagonal = 6 * r^2/sqrt(3)
max hex over table = 225 * PI * sqrt(3) / 6 =~ 204
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