Suppose there are n bottles in the bottom layer. We will label the bottles in even layers (those containing n - 1 bottles) So for k < n let b(m, k) be the kth bottle from the left in layer 2m (because it will be useful later, we also include an imaginary layer 0 beneath layer 1). b(m, k) has two bottles resting on it from above, one on the left and one on the right. Let l(m, k) be the (directed) angle from the horizontal of the line joining the centre of b(m, k) to the centre of the bottle resting on it to the left. Define r(m, k) similarly on the right. Let a be the directed angle to the horizontal of the sides of the rack. Then:

l(m+1, k+1) = l(m, k) for k < n-1 by considering a couple of rhombi
r(m+1, k-1) = r(m, k) for k > 1 similarly
r(m+1, n-1) = 2a - l(m, n) by considering a rhombus and an isosceles triangle
l(m + 1, 1) = 2a - r(m, 1) similarly.
r(0, k) + l(0, k) = 0 since the bottom layer is horizontal.

We then deduce:
l(2n - 2, k) = l(2n - k - 1, 1)
= 2a - r(2n- k - 2, 1)
= 2a - r(n - k, n - 1)
= 2a - (2a - l(n - k - 1, n - 1))
= l(0, k)

Similarly, r(2n-2, k) = r(0, k). In particular, then, l(2n-2, k) + r(2n-2, k) = l(0, k) + r(0, k) = 0 so that layer 4n-3 is horizontal.

This is all much easier to see in a diagram, which you are better equipped than I to produce. Note that a similar argument gives the original result in the case that the sides are vertical, since then l(n-1, k) + r(n-1, k) = 2a - r(0, n - k) + 2a - l(0, n - k) = 4a - 0 = 0.

What is a pww?

Thankyou

sfwc
<><

https://www.cut-the-knot.org/Curriculum/Geometry/NBallsAtBottom.shtml
https://www.cut-the-knot.org/Curriculum/Geometry/RhombiRhombi.shtml

Having a center bottle ensures central symmetry of the arrangement with the top row being a reflection of the (horizontal) bottom row. The central symmetry may be around for a slanted rack, but my proof does not appear to work in this case.

First note that converse of the theorem at https://www.cut-the-knot.org/Curriculum/Geometry/NBallsAtBottom.shtml is true. I do not, of course, mean to say that if the central bottle is equidistant from the two base corner bottles then all the bottles in the bottom layer are collinear, since that is not true. However, it is true if we include the condition 'The heavy broken lines (make sure the Hint box is checked) formed by connecting the first and the last bottles of successive layers can be obtained from each other by reflection and translation' (I shall call this the reflection condition). In that case, if we do not hit problems filling in the bottles to complete the triangle, then we will end up with a flat base.

Now suppose we have some bottles arranged on the base of a (possibly badly broken) wine rack. Then we certainly build a triangular stack on them, and we know that the centre of the top bottle (which I call O) will be equidistant from those of the corner bottles, P on the left and Q on the right. Now construct an isosceles triangle OSP with OS = OP and SP parallel to the left hand side of the wine rack. Similarly construct OQR on the right. Then ORS is also isosceles.

Now add 2 chains of circles, one from O to R and one from O to S so that OSP and OQR both satisfy the reflection condition. Then since OPQ satisfies the reflection condition too, so does ORS. Now fill in these triangles with circles so as to get flat bases along QR, RS and SP. Then you have the arrangement of bottles in the first 2N-1 layers of the rack, so long as layer separation holds. So in particular, the bottles in layer 2N-1 are collinear. Furthermore, cyclicity of the resulting quadrilateral is trivial since PQRS is cyclic with centre O.

A picture is worth a thousand words, and I think a picture showing the division into 4 isosceles triangles would convey the ideas of this post more easily than the somewhat wordy Description above.

Thankyou

sfwc
<><

"Let a be the directed angle to the horizontal of the sides of the rack."

This makes a = 90°, does it not?

Thankyou

sfwc
<><

Yes, I realize that. I can't remember what jerked my mind to post this question. Perhaps it was a post effect of the local anesthesia adminestered by my dentist a short while beforehand.

However, I'd like you to think more on the problem. 4n-3 is not quite a generalization. For the straight rack, the number is 2n-1. Your number 4n-3 is twice the "straight" result minus the 2n-1^st layer, which would be counted twice - as the top of the first 2n-1 layers and the bottom of the next 2n-1 layers.

For the original theorem getting a derivation that ends with 2a would be quite sufficient. 4a is an overkill.

>For the original theorem getting a derivation that ends with
>2a would be quite sufficient. 4a is an overkill.
You are right that 2a was all I needed. But 4a was what the algebra spewed up.

Thankyou

sfwc
<><

Do you think this is true?

>You are right that 2a was all I needed. But 4a was what the
>algebra spewed up.

Something is bound to be true of layer 2n-1.

OK, I am starting to do what you say I am better equipped to do (than thinking :)

This is all words and ideas to me right now; I will be much more sure of myself when I have seen a diagram.

It would probably have been better for me to state the idea in terms of rhombi rather than circles in the first place since then it is more general, easier to see from a diagram and does not need the proviso that must be given in the wine bottle problem.

Thankyou

sfwc
<><

Thank you.

Thank you,
Alex

First, we must note that the arguments I gave 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 > 2 * pi / 3) or too sharp (base angle <pi/3), or if one of the isosceles triangles on the side has central angle < pi/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
(2) r(m, k) - l(m, k-1) is not in
(3) r(m+1, n-1) - l(m, n-1) < pi/3
(4) r(m, 1) - l(m+1, 1) < pi/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) >= pi/6. Similarly, l(m, n-1) <= 5 * pi/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) >= pi/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) <= pi/3. So we know that r(0, k) is in for all k and similarly l(0, k) is in <2 * pi/3, 5 * pi/6> for all k. It follows using the basic equalities that r(m, k) is in and l(m, k) in <2 * pi/3, 5 * pi/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 you have 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) < pi/3 for all k. So we know that l(m, k) - r(m', k') < 2 * pi/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) < pi - 2 * pi/3 = pi/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+1th 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 < pi/3
Now this is impossible, as is most easily seen in a diagram. Nevertheless, I shall give an algebraic proof. w.l.o.g. 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) < pi/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) - pi/3 rather than - l(n - K - 1, n-1)
l(K + 1, 0) = r(K, 0) + pi/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.

>In the slanted case, I think, the
>requirements may need to be more stringent.
You are right about this. The requirements may be worked out in a similar way, but they are not so elegant and so I have not included the working out here.

Thankyou

sfwc
<><

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 for a in {0, 2b, 2b - 2d, -2d} and all k1 != k2

2*r(k) - a in <0, pi/3> for a in {0, pi/3 + 2b, 2b - 2d, pi/3 - 2d} and all k.

These conditions may be simplified, and simplify 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 in my previous post.

A suggestion for an interactive proof without words:
Let one of the n-1 bottles in the second to bottom layer be chosen by the user (for example by clicking on it). Then add bold line segments from the centre of this bottle to those of the two on which it rests, and connect the centres of those bottles with a coloured segment. Then continue to add segments in the following manner:
Whenever you have a bold segment on one side of a rhombus (formed by centres of the bottles) make the opposite side also a bold segment, and the other two sides fainter segments. Whenever you have a bold segment terminating in a bottle touching a side of the rack, add in colour a segment joining the centre of that bottle to that of the bottle resting on the same side two layers up: This is the base of an isosceles triangle of which the bold segment was one of the sides. Draw a bold segment along the other side also. Eventually, you will have 2 adjacent bottles in the top layer (layer 2N-1) each of which has the endpoint of a bold segment as it's centre. Join these with a coloured line.

Of course, my choice of bold, fainter and coloured as the distinguishing features of the types of segements was arbitrary and could be changed. The diagram produced should make the structure of my algebraic framework completely clear. If it does not, you could add to one side a smaller diagram consisting of four isosceles triangles which are congruent to those with coloured bases in the main diagram, but have been translated so that their apices coincide at a point O, and so the coloured segments form a quadrilateral with sides parallel to the base, sides and lid of the wine rack. For the badly broken wine rack you may also wish to add a circle centred at O and passing through the vertices of this quadrilateral to demonstrate that it is cyclic.

Thankyou

sfwc
<><
Thankyou

sfwc
<><

Meanwhile, if you seek a problem to apply yourself to, I have posted a Sangaku to which I do not have yet a solution. Do not see a useful inversion and do not see a possibility to avoid lengthy calculations.

Also, almost a year ago, John Mason suggested there should be a way to generalize the "next step" algorithm from the common case to the 3-color case.

https://www.cut-the-knot.org/Curriculum/Combinatorics/Hanoi3c.shtml

This was almost a year ago. However I tried to do that, there would pop up several cases to consider. His solution (in a private communication) also has led to several cases of which some were missing. It's a nice problem that deserves a nice solution.

Thank you,
Alex

Call the circle centred at O S and that centred at P T. I shall use inversion. I shall denote the inverse of any object by the name of that object followed by a '.

First note that the problem is equivalent to showing that the circle with centre on the perpendicluar from B, and tangent to BE and to T is also tangent to S. I shall call this circle V.

Invert with respect to B. C'BE' will be similar to EBC and so isosceles. S' will be the circle on A'C' as diameter, and T' the tangent to it at A'. Let Q be the intersection of T' with E'B. Then A'QE' = pi/2 - QBA' = pi/2 - E'BC' = pi/2 - C'E'B = QE'A' so A'QE' is isosceles. Let P be the centre of V'. Since V' is tangent to both T' and BE', P lies on the bisector of A'QB. So PQB = A'QP = BPQ and so also BPQ is isosceles. Let O be the centre of C', and call the radius of this circle R. Call the radius of V' r.

There are right angles everywhere, so we may now indulge ourselves with a pythagorean orgy:
OP^2 = OB^2 + BP^2
= (R - A'B)^2 + BQ^2
= (R-r)^2 + BA'^2 + A'Q^2
= (R-r)^2 + r^2 + A'E'^2
= (R-r)^2 + r^2 + A'C'^2 - E'C'^2
= (R-r)^2 + r^2 + 4R^2 - BC'^2
= (R-r)^2 + r^2 + 4R^2 - (2R-r)^2
= R^2 + 2Rr + r^2
= (R + r)^2

So that V' is tangent to C' and so V is tangent to C as required.

>Also, almost a year ago, John Mason suggested there should
>be a way to generalize the "next step" algorithm from the
>common case to the 3-color case.
What is the "nest step" algorithm?

Thankyou

sfwc
<><

Thank you.

>What is the "next step" algorithm?

It's the algorithm that allows for the "Suggest a move" feature in

https://www.cut-the-knot.org/Curriculum/Combinatorics/TowerOfHanoi.shtml

Given a disk configuration, what is a desirable (or optimal) move?

>>What is the "next step" algorithm?
>
>It's the algorithm that allows for the "Suggest a move"
>feature in
>
>https://www.cut-the-knot.org/Curriculum/Combinatorics/TowerOfHanoi.shtml
>
>Given a disk configuration, what is a desirable (or optimal)
>move?

I've got a nice algorithm for a programmed solution for the 3-color Tower of Hanoi problem, similar to the recursive function you mention on the page for the one color Tower of Hanoi. Is this the kind of thing you're thinking of, or do you mean an algorithm that can start with an arbitrary position and make progress towards the solution? The algorithm I came up with works from the specified starting configuration (all disks separated into 3 monocolor stacks, one on each peg) to the specified final configuration (same three stacks, but rotated positionally among the 3 pegs). However, I don't think it can handle an arbitrary starting configuration. If this is of interest, I will post it in a new thread.

--Stuart Anderson

>I've got a nice algorithm for a programmed solution for the
>3-color Tower of Hanoi problem, similar to the recursive
>function you mention on the page for the one color Tower of
>Hanoi. Is this the kind of thing you're thinking of, or do
>you mean an algorithm that can start with an arbitrary
>position and make progress towards the solution?

It's not what I had in mind, but this may be a nice feature in itself. Let's see it.

Thank you,
Alex

As was expected, Nathan, the 'inversion wizard',solved the problem
nicely and quickly. Yet, I have a question.

The inversion trasformation was invented-introduced in
pre-WWI Europe sometime in 1850-1900, if I'm not mistaken.
Is it'safe to assume that some Japaneese (at least those
collegues of mine in the military tenderly called samurai)
had the advanced knowledge of Inversion in that lonely period
of Japan's history? Was inversion one of the methods
carved in wooden solutions? I doubt it, and do not have military
intelligence on the subject.

It looks more probable that only Euclid's
Elements (smuggled into Edo before the 200 troubled years started)
along with the widely publicized solutions to the earlier Sangaki
could possibly be used as tools for solving the newer ones.

The quest for a more reasonably appropriate solution should
continue, although inversion (and oak!) is nice.

If one constructs the other 3 similitude centers of the 3 circles,
the 6 of them will be on 4 lines, 3 points on each, making
a nice and complete quadrilateral. The center of the circle in
question in on the diagonals intersection of it. The perpendicular
from it onto the 3rd diagonal (the big diameter) makes equal
angles with the lines connecting the base of this perpendicular
with a pair of opposite vertices of the quadrilateral.
It is still far from a solution, but first step, maybe?
But now I have to question if Monge-D'Alambert theorem was
a known fact in 1600-1700 Japan?

What if some Sangaki (this is my plural for a singular Sangaku)
remained unsolved in Japan until the time when this
Japaneese 'iron curtain' rusted away and allowed on Sangaki front
some geometrical assistance from Western Europe?

Any views and/or ideas?

Salute,

Capt. Pestich

My understanding is that solutions were carved only in exceptional cases, if at all.

> The quest for a more reasonably appropriate solution
> should continue, although inversion (and oak!) is nice.

The quest should continue of course. We shall learn more from various solutions. But what or which will be more appropriate? What if there are two distinct solutions that do not use inversion?

> It is still far from a solution, but first step, maybe?

Who knows? Having a complete solution is by far more rewarding of course.

> What if some Sangaki (this is my plural for a singular
> Sangaku) remained unsolved in Japan until the time when
> this Japaneese 'iron curtain' rusted away and allowed on
> Sangaki front some geometrical assistance from Western
> Europe?

I believe every one who ever investigated the matter of Sangaku made an assumption that the authors would not hang the tablets without knowing a solution.

See also

https://www.cut-the-knot.org/Curriculum/Geometry/BottleProofs.shtml

Many thanks

Thankyou

sfwc
<><

It's relatively straightforward, so I took the liberty to put the intervals in. They were missing in two place: 2 intervals at the beginning and two towards the end. I just wanted to make sure that I got this right. So, do please have a look at the following:

https://www.cut-the-knot.org/Curriculum/Geometry/BottleProofs.shtml

Many thanks

Thankyou

sfwc
<><

Alex