How many coins can be put on a round table? having; coins=1cm &
table=1 m diameter. I need a clear proof, not only a numerical
answer...

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Request for Question Clarification by mathtalk-ga on 31 Mar 2003 10:14 PST

Hi, haluk-ga:

The optimal packing for circles in the plane is known to be the
hexagonal packing.  However this is optimality is only proven for the
full plane (greatest density) and may be false in bounded regions (see
the reference cited by justaskscott-ga).  Given the relatively large
size of the circular table top in relation to the coins in your
problem, using the hexagonal packing should provide a fairly tight
lower bound on the number of coins that can be packed there.

Given the price offered for this question, I'm wondering if that
approach would provide you with an acceptable response?

regards, mathtalk-ga

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Clarification of Question by haluk-ga on 04 Apr 2003 05:46 PST

solved the problem by using hexagonal packing here is the "Table -
Coin" Solution

Any plane can be covered by using adjacent regular hexagons. If we
replace each hexagon by a coin we can find the maximum number of coins
we can place on the table.

Area of the regular hexagon = SQRT(3)/2 . Number of hexagons can be
found by dividing the area of the table by this number (the area of
the hexagon).

Number of hexagons=Pi*r²/.866025403...=9068.99
Since the number of coins has to be a whole number we take =9068
In this case some of the coins will be cut by  the circumference. If
the diameter of the table was 99.5 cm some of the coins would fall;
and if it was 100.5 cm we would have more coins. If we find the number
of hexagons in that 1 cm ring this will be the number of coins cut by
the circumference.

Area of the ring=Pi(D²-d²)/4=Pi*(100.5²-99.5²)/4=157.079 cm²
Number of hexagons in the ring = Area of the ring/SQRT(3)/2 = 181,37 
(= 181 since no. of coins has to be a whole number)

The number of coins cut by the circumference of the 100 cm diameter
table is 181.

Therfore the maximum number of coins which can be placed on the table
is:
9068-181=8887

Please note that 8887 is a prime number (in 6p+1 form, where p is also
a prime) and 1coin is on the center of the table.

Haluk Somer

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Request for Question Clarification by mathtalk-ga on 06 Apr 2003 14:26 PDT

Hi, haluk-ga:

I'm not entirely confident in how you calculated the number of coins,
but I agree that with a coin placed in the center of the table and
other coins packed hexagonally around it, there would be 8887 coins
within the margins of the table.

On the other hand I think it is possible to improve the count slightly
by varying the location of the coins in relation to the center of the
table.  If you would be interested in having the details posted as an
Answer, let me know.  Otherwise I'd be happy to provide a summary as a
Comment.

regards, mathtalk-ga

One would assume that this would be fairly straightforward ... but
apparently, one would be wrong in that assumption.  See, for example:

"Pennies in a Tray" (Ivars Peterson's MathLand) (November 25, 1996)
The Mathematical Association of America
http://www.maa.org/mathland/mathland_11_25.html

Are the coins allowed to hang over the edge of the table?  That is,
does the entire coin have to be on the table, or just the center of
the coin?

The coins are not allowed to hang over the edge of the table; they are
at least tengent to the edge.