Bottles in a Wine Rack: Proofs and Generalizations
By N. Bowler
Now if circles are drawn with centres on the corners of the rhombi and with radius half the side of a rhombus, and the circles produced like this do not overlap, this will give a suitable stack of circles. Indeed, if there is a stack satisfying layer separation (which I think is certainly a good condition to start from) then it will be generated in this way. So we have a problem iff the circles would overlap, that is iff one of the rhombi is too flat
So using the notation developed in my previous post, there will be a problem iff one of the following occurs for some m and k:
|(1)||r(m, k) - l(m, k) is not in [π>/3, 2π>/3]|
|(2)||r(m, k) - l(m, k-1) is not in [π>/3, 2π>/3]|
|(3)||r(m+1, n-1) - l(m, n-1) < π>/3|
|(4)||r(m, 1) - l(m+1, 1) < π>/3.|
First let us consider the case that the sides of the wine rack are vertical.
Then condition (4), together with the equation
I shall now show that this is a sufficient condition. Since circles in the base layer cannot overlap, we have the additional condition that
To summarise: If the sides are vertical, to avoid problems it is necessary and sufficient to ensure that for any pair of adjacent circles in the bottom layer the distance between the centres is at most sqrt(3).
However, as was pointed out, there is still some good behaviour even in the problem cases. Indeed, we have the following conjecture:
|If this happens, the very top bottles (it appears there are always two of them) are still horizontal,|
which I shall now prove.
In order to do this we must analyse the possible types of bad behaviour. We know by the equalities mentioned in my earlier post that for all m and k there exist k' and k'' such that
r(0, k1) + r(0, k2) < π> - 2π>/3 = π>/3,
for some k1 and k2.
Now assume for a contradiction that
n ≥ Sum(k = 1..n-1, d(k)) ≥ n - 3 + d(k1) + d(k2)
so letting r1 = r(0, k1) and r2 = r(0, k2) we have:
3/2 ≥ cos(r1) + cos(r2) but r1 + r2 < π>/3
Now this is impossible, as is most easily seen in a diagram. Nevertheless, I shall give an algebraic proof. Without loss of generality, assume
So in fact k1 = k2, and problems arise iff there is some (necessarily unique) K with
r(n - K, n-1) = -l(n - 1 - K, n-1) = r(0, K)
and one of type (4) since
r(K, 0) = -l(K + 1, 0) = r(0, K).
In fact, if the bottles are placed in the manner employed by the applet you wrote, and we then consider the values this gives us for the l and r functions, we obtain the same equalities as before except that:
r(n-K, n-1) = l(n - K - 1, n-1) - π>/3 rather than
l(K + 1, 0) = r(K, 0) + π>/3 rather than
and r(n- K + 1, n-1) < - l(n - K, n-1) rather than equality
and l(K + 2, 0) > -r(K + 1, 0) rather than equality.
Working out the consequences of this for the top layer in the same way as before shows that the
(The applet below helps visualize the geometric properties underlying the foregoing algebraic derivations. The inner circles at the bottom can be dragged left or right and the ones in the next row - those with the red center - can be clicked upon. See what happens.)
|What if applet does not run?|
The proofs and derivations of the results below follow exactly the same pattern as the above, with only slight modifications of the dull algebra. So I shall simply mention the results without proof.
Bottles in a badly broken wine rack:
Suppose that the wine rack is so broken that the sides, although straight lines, are both slanted, and are not necessarily slanted to the same angle. Then if we have good behaviour (which I will define better below, and which is almost equivalent to layer separation) up to row 2N-1, the bottles in row 2N-1 will be collinear, so that we may add a lid which is tangent to them all. Furthermore, the lid, the base and the two sides of the rack will form the sides of a cyclic quadrilateral.
The nature of good behaviour
First I shall set up some convenient notation. Let
It is then necessary and sufficient to have:
r(k1) + r(k2) - a in [π/3, 2π/3], for a in
2·r(k) - a in [0, π/3], for a in
These conditions may be simplified, but have to be simplified differently in three different cases; that the sides both slope inwards or both slope outwards or one slopes in and one out. Note that if the sides are vertical, they simplify to precisely the condition given previously.
- A Circle-Stacking Theorem
- A Property of Rhombi
- Bottles in a Slanted Rack
- Bottles in a Wine Rack
- More Bottles in a Wine Rack
- Proofs and Generalizations
Copyright © 1996-2017 Alexander Bogomolny