# A Nice Riddle

Discussion in 'Physics & Math' started by §outh§tar, Aug 6, 2009.

1. ### CptBorkValued Senior Member

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If you're saying that calculus and other elementary techniques won't work, are you still standing by the claim that this is a high school problem? I have a couple of schemes I was thinking of, but I don't want to bother working out the details if it's going to turn out to be some sort of abstract algebraic or topological solution. I'm still assuming this is a high school-level problem that just requires some straight up ingenuity, but please don't hesitate to let us know if high school math won't suffice.

3. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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I took a look at the portion of the solution that Google provides, but didn't get much out of it. It makes it sound complicated, since there's a bunch of citations and references to historical work, but I don't get any feel for how the solution would operate.

5. ### §outh§taris feeling causticRegistered Senior Member

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Well the person who suggested it told me to come up with a clever solution and so my mind traced out the same ideas you all have explored. But high school math has nothing to do with this.

7. ### CptBorkValued Senior Member

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Ok, I'll think about it some more later tonight perhaps. But you did mention in the OP that this problem was "for high school students". Now I happen to have met a high school student or two who had decided to go WELL beyond the curriculum and could have practically obtained a bachellor's within the year if they wanted (how many high school students know about matrix diagonalization, differential equations, and residue calculus?). So was that an error in transcription, or is it meant for high school students who decide on their own to study some real math for a change, or is it something else?

8. ### §outh§taris feeling causticRegistered Senior Member

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Unless there's a marvelous solution, the proof I saw isn't appropriate for most high school students. A prepared one may understand it, but come up with it on her own.. probably not. It's just that you'd have to be really special to spontaneously come up with the given solution without working on similar problems.

9. ### iceauraValued Senior Member

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In the perpendicular line tactic, the distance from the point to the circumference decreases monotonically, and the lengths of the successive line segments in D decrease monotonically as well - in other words, the choice of point cannot "bounce" more than so far, and that distance monotonically decreases with each choice.

That is, since the spiral objection vanishes in the fact that a greater angle line segment change implies a greater reduction in line segment length (point chosen closer to the circumference); and since the positive to negative bounce vanishes in the decreasing line segment length, alpha numeric's proof holds?

10. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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No, przyk showed in post 16 why this doesn't work. It's true that, on a given step, the bigger the angular jump point makes, the bigger the increase in radius. But the kicker is that point can instead take many very small steps, and in that way attain arbitrary angular movement with arbitrarily small increase in radius.

Seen another way, we know that the distances between successive points will monoticially decrease to zero under the perendicular line strategy. However, they must decay sufficiently rapidly in order for that to imply convergence of the sequence of points: if they shrink at a rate of 1/n, for example, the sequence will not converge. And this is essentially what the counterstrategy proposed by przyk in post 16 consists of.

11. ### iceauraValued Senior Member

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Deleted to ponder. My intuition is boggling at generating an unbounded angular step sequence without producing an unbounded radial increase, in bounded ratio - I keep thinking about x/sinx, bounded approaching 0.

Last edited: Aug 15, 2009
12. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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Right, the points can't be arbitrarily close: the distance between successive points needs to shrink as 1/n (where n is the number of turns). If it shrinks too much more quickly or slowly than that, convergence should result (to an interior point in the former case, and to the boundary in the latter).

Only if you try to get the angular motion all in one turn. If you're content to achieve the motion over many turns, then you can do it with arbitrarily small increase in radius.

At least I think that's what you're referring to... where did you get the sinx/x from?

Under przyk's strategy, the radius keeps increasing forever, asymptoting towards 1, as I understand it.

Edit: just saw your edit, but figured I'd leave this as food for thought anyhow.

13. ### przyksquishyValued Senior Member

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In case there's any confusion, I didn't actually explicitly post a full divergent strategy in the sense of specifying $\varepsilon_{k}$ for all $k$. What I showed was that, at any stage in the game, Point could choose a sequence of equally sized[sup]*[/sup] steps in order to increase his angle by as much as he wants while increasing his radius as little as he wants by pre-selecting $\tilde{\varepsilon}$ and a sufficiently large $n$. The point is that, no matter how close he is to the outer edge, he's never bound within some particular angular range.

[sup]*[/sup]technically in a multiplicative sense since $| x_{k+1} \,-\, x_{k} | \,=\, \varepsilon_{k} \, | x_{k} |$.

14. ### temurman of no wordsRegistered Senior Member

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Qualitatively, for this "perpendicular" strategy radial increment behaves like the square of the angular increment, so for instance when the radial increment is like 1/n^2, which is ok for the sequence to stay in the disk, the angular increment is like 1/n which induces divergence.

15. ### iceauraValued Senior Member

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30,994
Thanks - that's what I was muddling over.