"The Round Table"

[zenith20]

"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 fifteen 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?

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solution: 187
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could anybody help me understand how this 187 is obtained?

 

[wjm11]

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 fifteen 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?

solution: 187

http://mathworld.wolfram.com/CirclePacking.html

 

[zenith20]

Dear Wim11, thanks for the article, new concepts for me, but i still can't solve the puzzle
any explanations for this 187? (solution by shakuntala Devi)

 

[soroban]

Hello, zenith20!

I don't agree with their answer . . .

"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 fifteen 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?

Solution: 187 .??

The most efficient arrangement is a "honeycomb" (hexagonal) packing.

. . \(\displaystyle \begin{array}{c}\bigcirc \\ [-2mm] \bigcirc\quad \bigcirc \\ [-2mm] \bigcirc \\ [-2mm] \bigcirc\quad \bigcirc\\ [-2mm] \bigcirc\end{array}\)

There is one saucer in the center, surrounded a "ring" by six saucers.



The next ring contains 12 saucers.

. . \(\displaystyle \begin{array}{c} \bigcirc \\[-2mm] \bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc\quad\bigcirc \\ [-2mm] \bigcirc\quad\bigcirc \\ [-2mm] \bigcirc \end{array}\)

And the next has 24 saucers, followed by 48 saucers, and so on.



With an arrangement with five rings,
. . there are: .\(\displaystyle 1 + 6 + 12 + 24 + 48 + 96 \:=\:187\) saucers.



I'm puzzled . . .

The table is 15 times the size of a saucer.
There should be 7 possible rings.

Perhaps part of the 7th ring extends beyond the table's edge?
Then there should be at least 6 rings.

Can anyone explain the discrepancy?

 

[zenith20]

Perfect as usual

Thank you Soroban

 

[zenith20]

Dear Soroban,

i found this scanned part of the book (Puzzles to puzzle you), i cannot get the whole.
esp the part (3x6), and it seems it's about 8 rings then, yes?

could you please explain it more for me then?

Thank you in advance