## Coins on a Jar

Consider a round jar with no lid. The radius of the opening in the jar is 10. A coin (with a radius R < 10) can be placed flat on the edge of the jar without falling. The coin can be placed such that its center is on the rim of the jar, or it can be moved toward the center of the jar.
1. What is R, when the coin is as close to the center of the jar as possible without falling in, and the inside edge of the coin is at the center of the jar's opening?
A number of identical coins are placed around the rim of the jar, each touching a coin on either side.
1. What are the possible values of R if the center of each coin is on the rim of the jar?
2. What are the possible values of R if each coin is as close to the center of the jar as possible?
3. Are there any size coins which can be placed in both of the previous two arrangements (with a different number of coins in each)?

Source: Original.

Solutions were received from Joseph DeVincentis, G. Oliver Stone, and Philippe Fondanaiche. Oliver's solutions:
```1.  When the coin is as close to the center as possible, the endpoints
of a diameter of the coin must be touching the jar rim.  There is a
right triangle then from the center of the jar to the center of the coin
to one of the end points where the jar rim and coin touch with sides
Rcoin, Rcoin, and Rjar.  Thus the radius of a coin in this configuration
is Rjar/sqrt(2).  If Rjar is 10, Rcoin is 5*sqrt(2).

2. If N coins are placed around the jar with their centers on the rim
and tangent to each other, the angle made by the two jar radii tangent
to one coin must be 360/N.  The angle made by one of these radii and a
jar radius drawn to the center of the coin will be half this, 180/N.
The line from the center of the coin to the tangency point (Rcoin) is
just the sine of this angle times the jar radius.  So Rcoin is any value
that satisifies Rcoin = Rjar * sin (180/N) for any whole number N (>1).

2        10
3        8.66
4        7.07
5        5.88
6        5
7        4.34
8        3.83
9        3.42
10        3.09

3.  This is this same as 2) except the coin centers are on an inner
circle with radius Jinner.  Rcoin = RJinner * sin (180/N).  The right
triangle from the center of the jar to the center of the coin to the
point where the edges of jar and coin intersect has sides of lengths
RJinner, Rcoin, Rjar.  So RJinner is sqrt (Rjar^2 - Rcoin^2).  Solving
gives Rcoin = Rjar/sqrt(1+1/sin(180/N)).

2        7.07
3        6.55
4        5.77
5        5.07
6        4.47
7        3.98
8        3.57
9        3.24
10        2.95

4)  Solving these two equations together and trying the first few values
shows that the only radius that works for both part 2 and 3 is achieved
when you have 2 coins in the part 3 arrangement, or 4 coins the part 2
arrangement.  For Rjar = 10, this is 5*sqrt(2) or 7.07.
```

```This asks: Can we find different values of N for the answers to parts
2 and 3 such that the radii R are equal?  Substituing M for N in part 3,
we need:

10 sin(pi/N) = 10 sin(pi/M) / sqrt(1 + sin^2(pi/M))

or

sqrt(1 + sin^2(pi/M)) = sin(pi/M) / sin(pi/N).

for M, N integers >= 3 (or >= 2 if you accept the degenerate case).

Squaring both sides, we get

1 + sin^2(pi/M) = sin^2(pi/M) / sin^2(pi/N).

or, dividing by sin^2(pi/M):

1 + 1/sin^2(pi/M) = 1/sin^2(pi/N)

We need two different values of X for which the values of 1/sin^2(pi/X)
differ by 1.  By inspection, the only possible answers are X=2 and 4;
for higher values of X, the values of 1/sin^2(pi/X) differ by more than 1
for consecutive values of X.

The coins with R = 5 sqrt(2) fit 4 around the jar when they are placed
eith their centers on the rim, and 2 across the jar when they are placed
with the centers in as close as possible to the center without falling in.
```

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