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LostInParadise's avatar

Can you solve this Russian nested doll problem?

Asked by LostInParadise (28385points) August 14th, 2012

Nested dolls are a traditional Russian toy and have common metaphorical use. This problem highlights an interesting property that they have due to an important mathematical result.

Suppose that a given set of Russian dolls are all the same shape, though different in size, and have identical height ratios between top and bottom parts. The bases of these dolls are a bit thicker than the rest of them. When the tops are all removed, as well as the innermost doll (which traditionally is all one piece), the edges of the bottoms are all flush with one another. In order for this to hold, what can we conclude about the relative heights of the tops and bottoms of each doll?

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17 Answers

zenvelo's avatar

Each individual piece (either top or bottom) differs in height from the next smallest or next largest by the thickness of the wood.

Sunny2's avatar

My brain can’t begin to think about that. I’m impressed that yours can.

fremen_warrior's avatar

Stop making me feel… uh… what is the word.. ugh… un-smart. Yeah!

ccrow's avatar

um… the tops are higher than the bottoms when they are assembled??

LostInParadise's avatar

This may not have been a good problem to give. I should have given the the math and illustrated it, and even then it may not have been that clear.

Let me explain what I was thinking of. The math result is the Brower Fixed Point Theorem, which says that if you map an object into itself continously then at least one point of the map must corrspond to the object. For example, if you take a cup of coffee and swirl it around, there is at least one point in the coffee that is in its original location. In terms of the dolls, if you think of a doll nested in another as a mapping of the original into itself, then there is one point of the outer doll that lies on top of the corrponding point of the inner one.

When all the dolls are of uniform thickness, they have the same geometric center, like a set of concentric spheres or cylinders. If the tops and bottoms are the same height then the line between them encircles the common center and the edges are all flush.

If the base is made thicker than the rest, the dolls will no longer have the same geometric center. The neat thing is that they still all share a corresponding point. By Brower’s Theorem, the frist and second share a common point as do the second and the third. Considering the second and the third together is just a scaled down version of looking at the first and second together. The point shared by the second and third is the same as that shared by the first and second. That is, alll three share a common corresponding point and the same argument holds for all the rest. The common point will be above the geometric center so the top must have a smaller height in order for the line between the two parts to encircle the common point.

I hope this makes sense.

zenvelo's avatar

@LostInParadise So you just described the rotational axis of the dolls.That has nothing to do with the relative heights.

So what was your point in posing this conundrum?

LostInParadise's avatar

I originally started thinking about the one dimensional version of this problem, which is much easier to explain. Suppose you have a one foot ruler. Make a copy of the ruler that is reduced in size and place it alongside the original. The new ruler exactly fits between 2 and 10 inches on the original. The two rulers coincide at the 6 inch mark. Now repeat the process. Create a third ruler, which fits between 2 and 10 inches on the second one. Rulers two and three coincide at the 6 inch mark and so all three rulers do, which should be fairly obvious since each ruler is centered with respect to the previous one. We could continue this process and all the rulers will coincide at the 6 inch mark. Now let’s break each ruler into two parts in the same way. Unless we do this at the 6 inch mark, the edges of the rulers will not be flush with one another. This is the analog of the way that the Russian nesting dolls are ordinarily set up.

Now suppose we take this same set of rulers (before they were broken) and line them all up between the 3 and 11 inch marks. It should be obvious, without doing the algebra, that they will all have common corresponding values at some place past the 6 inch mark. Again, we can only cut the rulers at the place where they coincide in order to make their edges flush with one another, meaning that the top piece is shorter than the bottom one. My 3 dimensional analogy was to thicken the base of the dolls to force their corresponding point to be above the geometric center.

fremen_warrior's avatar

@LostInParadise seriously, what is the point of this question, in lay terms?

erichw1504's avatar

In the words of @fremen_warrior I am too un-smart for this. Sorry.

LostInParadise's avatar

@fremen_warrior , The point was to explore the interesting (to me at least) fact that when you nest objects, they all share a common corresponding point, and to see how it is necessary to know what this common point is if you want to cut them into two parts such that the edges are all flush with one another.

__Life without geometry is pointless.__

fremen_warrior's avatar

@LostInParadise given, fluther is a community of science/math/it majors you are right I too am stumped nobody has yet even attempted to solve this here. I wonder why…

LostInParadise's avatar

Maybe a picture will work better than words. This crude drawing represents four rulers with each one reduced in size and placed between 3 and 11 of the previous one. The marks at 9, colored red, all line up. If the rulers were all cut at the 9 mark, the edges of the two halves would line up and clearly this would not hold for at any other place.

zenvelo's avatar

@LostInParadise It’s not that we don’t understand, but we’re all wondering why you posted this? There hasn’t been any exploration of anything here, and my original observation, which I believe is true, has not yet been addressed.

LostInParadise's avatar

If this is all transparently obvious to you then I have to give you credit. It was not immediately obvious to me, so I thought I would share what I had figured out. I hope that it is clear that the division point between top and bottom has to line up with the common point in order for all the sides to line up. Again, if you found this blatantly obvious, I tip my hat to you.

gasman's avatar

I found this to be a delightful and non-trivial mathematical puzzle. I’m a fan of such puzzles but don’t recall ever seeing this one before. Here’s my approach:

Each doll has a cup-shaped bottom half. Let the outermost of these cups have height h (measured from bottom of base to the top rim) and base thickness b (measured from bottom to top surface of the base). Each cup/doll is reduced in size from the previous by some fixed ratio r , 0<r<1. These 3 parameters (h, b, r) determine the whole setup.

Consider just the first two (outermost) dolls.

The second cup rests on the first cup’s base. The distance from the outer cup’s base to its top is (h – b). This must equal the total height of the inner cup (so their tops line up), which is r times h. [Sure wish I could post a diagram.]

rh = h – b
From this we can derive various relations:
r = 1 – b/h
h = b / (1-r)
b = h (1-r)
Given any two parameters, you calculate the third one using one of these formulas. This shows how they are constrained by conditions of the puzzle.

Example 1: Give a base thickness 10% of the height (of the bottom cup), then since b/h = 0.1 then r = 0.9. Each doll must be 90% the size of its predecessor. Example 2. If each doll is 80% smaller than the last, then the base must be 20% the height.

Alternate approach using infinite dolls instead of just two:

Bases of all the nested cups stack up on each other, forming a geometric series whose sum converges to some particular height. The final height of all the stacked bases will be:

S = b + rb + r^2b + r^3b + ...
S = b (1 + r + r^2 + ...)
S = b (1 / (1-r)) {applying formula for sum of geometric series}
S = b / (1-r)
But wait! The cups eventually get quite tiny, yet their tops all have to reach the top of the outermost cup. The only way this can happen is if the top of the innermost cup’s base (on which successive dolls rest) approaches the top rim. In other words, the sum of base thicknesses must converge to a total height equal to h, the height of the outer cup. S = h.

So if S = b / (1-r) and S = h then
h = b / (1-r)
...which is exactly what was already worked out above by considering just two dolls.

@LostInParadise: In order for this to hold, what can we conclude about the relative heights of the tops and bottoms of each doll? Do you mean the relative heights of the bottom and its base? The doll tops, it would seem, are an irrelevant red herring. Since they need not have any appreciable thickness, their proportions are not constrained as with the bottoms, so long as they decrease in size for nesting.

LostInParadise's avatar

We are looking at the problem from different points of view and maybe that is what got people confused. I was thinking of what would happen if the total height is kept constant and the thickness of the base is varied, that is, varying the ratio of base to total height. In the ruler example, the base could be considered as being 3 inches and the total size would be 12 inches.

Using my interpretation, there is still a geometric series relationship.
Let b=base thickness of largest ruler
r=ratio of ruler size to next larger one
Let’s do it for the first two rulers. To find the common point x, x-b is the distance to the common point along the smaller using the measure of the larger one. This distance in terms of the smaller ruler on its scale is (x-b)/r. (x-b)/r = x or x=b/(1-r)

If the total height is kept constant then the ratio of common point to height increases directly with b, so that the top portion keeps getting smaller relative to the bottom one.

The starting point of the nth ruler in terms of the first one is the geometric series b+br + ...+br^(n-1), which again converges to b/(1 – r), the common point.

The geometric series is fun to play with because there are so many ways of deriving it. I have listed a few on my Web site

LostInParadise's avatar

One final comment and a result that I find a bit counterintuitive. The problem originally mentioned the thickness at the base compared to the thickness elsewhere. To simplify things, we can just look at b= thickness at the bottom and t = thickness at the top, which would also include clearance between the two dolls. Since everything depends on ratios of distances, we can take the total height to be 1.

The size of the second doll would then be 1 – b – t, and this is also equal to the size ratio r.. r = 1 – b – t.

We found that the common distance is b/(1 – r) = b/(b + t). The only thing that matters is the relative thicknesses at the top and bottom. The distance between them is irrelevant. In the usual case, the thicknesses are uniform, but t > b because we need some clearance. This would predict that the top piece should be slightly larger than the bottom one.

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