Bottles in a Wine Rack:
Proofs and Generalizations

By N. Bowler

First, we must note that the arguments I gave earlier are better suited to rhombi and indeed will always work if the construction is performed using rhombi rather than circles.

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 (base angle > 2p/3) or too sharp (base angle < p/3), or if one of the isosceles triangles on the side has central angle < p/3.

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 l(m+1, 1) = -r(m, 1) gives the equation r(m, 1) ≥ π>/6. Similarly, l(m, n-1) ≤ 5π>/6. Since this is to be true for all m, using the known equations we deduce that for all k we have r(0, k) ≥ π>/6. That is, any pair of adjacent circles in the base layer are to have centres at most sqrt(3) apart, where the circles are considered to all have unit diameter.

I shall now show that this is a sufficient condition. Since circles in the base layer cannot overlap, we have the additional condition that r(0, k) ≤ π>/3. So we know that r(0, k) is in for all k and similarly l(0, k) is in (the interval) (2π>/3, 5π>/6) for all k. It follows using the basic equalities that r(m, k) is in and l(m, k) in (2π>/3, 5π>/6) for all m and k, and simple checking now shows that none of the problem cases listed above can occur.

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(m, k) = r(0, k') and l(m, k) = -r(0, k''). We also know that the circles in the base layer do not overlap so r(0, k) < π>/3 for all k. So we know that l(m, k) - r(m', k') < 2π>/3 for all m, k, m' and k'. Thus none of the rhombi is too sharp, and any problem rhombus must be too flat. For there to be such a rhombus we must have

r(0, k₁) + r(0, k₂) < π> - 2π>/3 = π>/3,

for some k₁ and k₂.

Now assume for a contradiction that k₁ ≠ k₂. Let d(k) be the distance from the centre of the kth circle to that of the (k + 1)^st circle in the bottom row. Thus d(k) = 2·cos(r(0, k)). Also, d(0, k) ≥ 1 for all k and:

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 r₁ < r₂. For fixed R = r₁ + r₂, the rate of change of C = cos(r₁) + cos(r₂) with respect to r₁ is -sin(r₁) + sin(r₂) > 0. So C is minimised when r₁ is 0. But then C = cos(R) + 1 > 3/2 which is the desired contradiction.

So in fact k₁ = k₂, and problems arise iff there is some (necessarily unique) K with r(0, K) < π>/6. In this case, we have one problem of type (3) since

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(n - K - 1, n-1),
l(K + 1, 0) = r(K, 0) + π>/3 rather than -r(K, 0),
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 (n - K)^th and (n + 1 - K)^th bottles in the top layer will be higher than the others and level with one another, as conjectured. It also shows that all the bottles to the left of these two will be level with one another, and similarly for those on the right. However, the height of the bottles on the left may differ from that of those on the right. These extra predictions are easily seen to be true upon experiment.

(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.)

If you are reading this, your browser is not set to run Java applets. Try IE11 or Safari and declare the site https://www.cut-the-knot.org as trusted in the Java setup.

Bottles in a Wine Rack: Proofs and Generalizations

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 r(k) = r(0, k) for any k. Recall that d(k) = 2·cos(r(k)) so that these are natural quantities to work with.

It is then necessary and sufficient to have:

r(k₁) + r(k₂) - a in [π/3, 2π/3], for a in {0, 2b, 2b - 2d, -2d} and all k₁ ≠ k₂

2·r(k) - a in [0, π/3], for a in {0, π/3 + 2b, 2b - 2d, π/3 - 2d} and all k.

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.

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