Close Quarters from Kevin J Lim

[margitsw]

Hi,
Couldn't you place three coins in a triangle and another
three coins EXACTLY on top of the first three ?
Do we get into atomic impossibilities here ?
Is there a mathematical proof ?

[MacadamiamaN]

I don't think you understand the puzzle correctly - or maybe I don't understand what you're trying to say...
But, every coin has to touch EVERY OTHER COIN. In a triangle, (assuming you mean the quarters are standing on edge), every coin is touching every other coin. However, if you put another triangle ontop of that one (again, on edge - somehow balancing), the coins that are diagonal and opposite to eachother (on bottom/top triangles) would not be making contact. I don't know if I explained that well, I guess I need a picture.

[cha]

There may be two answers here, depending on whether we're talking about real coins or mathematical coins.

Another way of thinking about this may be to imagine two "stacks" of two coins each. If you move these stacks together, will the top coin of each stack touch the bottom coin of the other stack? Real coins don't work (try it).

But what about a mathematical coin, with perfectly straight edges... I would say that as the stacks moved closer together, the distances between all four coins (at the center intersection) would eventually reach zero, meaning that they are all touching. Add a third stack on the side, all coins are touching. Margitsw may have hit on something.

[Bicho the Inhaler]

I think if you look at the quarters as perfect cylinders, you have some quarters sharing single points of contact with others, and you have quarters intersecting through other intersections. On the atomic level (if you really want to go there), maybe there's a chance that there would be mutual contact between the six, but probably not, and you certainly couldn't guarantee it. So given some assumptions (that on the atomic level the quarters have the right imperfections), the scenario works, but for a given arrangement it would be very difficult to prove that those assumptions are met. With Kevin's five it is clear that all five have mutual contact.

At least that's my opinion.

[Bicho the Inhaler]

Err...yeah, what cha said. I just tried it and there's always a visible gap. I shouldn't have said "on the atomic level"; the imperfections in question actually need to be macroscopic. But yeah.

[margitsw]

This one is bugging me.
I've tried to get Kevins solution for 5 coins working
for Euro's (Being in Germany)
It does not work ! (Or I am not dexterous enough !)
The problem is in the coins dimensions (I think).
He worked with 2.4 cm x .15 cm
The Euro coins are all THICKER.
<http://www.euro.ecb.int/en/section/euro0/coins.html>
So, 50 cents - 2.425 x .238
1 Euro - 2.325 x .233
2 Euro - 2,575 x .220

Now this does not affect the bottom coin or the two sloping
vertical coins which can all be asumed to have negligable thickness
(Only the inside faces of these are relevant) BUT it DOES affect the
2 middle horizontal coins which changes the available angle of the
vertical coins.
Am I right ?
If so, is it a ratio of diameter/thickness thing or is only
the thickness a limiting factor ?
Any genius out there got the math for this ?

Is this analagous to the touching cigarettes problem ?
(I suppose you could consider a coin as an extremely fat and
extremely short cigarette)
The answer to the cigarettes problem was for a long time believed
to be 6. Then somebody came up with seven and I've seen somewhere
that it's now 8 (Blast, I wish I has saved the URL)

Comments anyone ?

[CrystyB]

i must confess i didn't understand your first reply here. What did you meant by triangle? If three flat pieces touching each other, i don't see how placing another on top of any triangle like that solves anything. Here's what i thought about it (grey means below):


As for the Euros, the thickness is not really that important. If they were heavy and fell apart i'd understand, but what exactly does thickness interfere with when you're reproducing the 'tower'?

[margitsw]

Hi Crysty,
Just swivel your top row 45 degrees.
You then have three piles of two coins.
I theorised that they all might be touching.

[margitsw]

Hi Crysty again,
Re the Euros :
As I said, I cannot get it to work.
I tried taping and glueing them.
Call the bottom coin number 1.
Call the two coins placed onto it 2 and 3.
Call the two sloping coins 4 and 5.
It should be obvious that changing the thickness
of coins 2 and 3 affects the possible angle of 4 and 5.
And that appears to be the problem.
Either I get coins 4 and 5 to meet at the top, in which
case one or the other does not touch coin 1; or ;
they both touch coin 1 and do not meet at the top.
There should be some math for this. Anyone ?

[margitsw]

I found some references to touching cigarettes problem :
http://mathworld.wolfram.com/Cylinder.html
and
http://www.ocf.berkeley.edu/~wwu/riddles/cigarettes.shtml

Hmmm

[margitsw]

So, somebody tell me why the 5 will work
for a given thickness !

[margitsw]

And if you habe not found it -
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027804466

[CrystyB]

Yes you're right. The 5 won't work from some given thickiness up. I'll be messing with the math for this. Thanks for explaining.

[margitsw]

Hi Crysty,
Hope that you are still working on this.
It's driving me mad.
I have NOT got the math background for this.
Maybe, we should raise this as a new problem ?

[CrystyB]

I have the math background for it. But like any other real-life applications, maths could get 'stuck' . I should solve a horribly-looking equation in x⁴ to get to the root of (roughly) 0.11 (IIRC) - meaning any cylinder that has a radius smaller than 9 times its height would not assemble like in the shown shape.

[margitsw]

Hi Crysty,
I give you a a kiss !
Not sure what this means ? Am I right
or is Kevins soltion wrong?

[CrystyB]

yeah it seems that you're right. With the given 2.4 and .15 cm dimensions, they would not fit!!

PS ~Places the kiss in his pocket~ Hey, maybe it will save my life too!

[This message has been edited by CrystyB (edited 09-10-2002 02:40 PM).]

[CrystyB]

Actually scratch that!

I'll just consider the most extreme case, where the coins barely just touch each other:



So h=xÖ3 and
r-Ö(r²-(r-x)²) = Ö(r²-(5-2x)²)
Working with x=x/r,
x⁴-2x³/3+5x²/9-x/3+1/36=0
Aprox. solutions: 0.0976 and 0.569 (these are approximations because solving the resolvent equations is murder)
So i didn't remember correctly. It was x/r, not h/r, and it wasn't 1/9~0.11, it was 1/11~0.9 ...

Anyway the maximum possible h/r ratio is a little over 0.169

[margitsw]

Woooh,
So max of our coins with ~ 2.4 cm is
1.2 * .169 = .2028
Which means that Kevin's coins work and Euro's do not.
Correct ?
I find this incredibly surprising and I only came across it
by physically trying it. The puzzle is five years old and I don't think
there has ever been a post about it.
OT - Crysty, What do you use to draw your diagrams and how do
you get them into the post ?

[CrystyB]

Click on the Pencil & Paper icon right above my post. You cannot alter my post (you lack the rights), but you can still see how i posted it. Through UBB Img tags.

Drawing is just a kids' play in Paint. I just save it as JPG, upload to my website and post it in here.

PS There's no need for you to manually add the ENTER character. The font isn't even monospace, if one ever considers justifying the text - that's just plain useless, as i found out on my first postings in here!

[This message has been edited by CrystyB (edited 09-11-2002 10:09 AM).]

[margitsw]

Hi Crysty,
Thanks.
For your perusal - Maybe a new puzzle ?
Re : http://www.greylabyrinth.com/Puzzles/puzzle036.htm
The touching pencils/cigarettes problem.
Ed Pegg over at mathpuzzle.com says there is a solution
for 8 if different lengths. He DIDN'T give me the solution.
Booo

[margitsw]

And I am going to put this as one of the puzzles that
anyone outside the USA cannot solve without resort to
either (a) a higher mathematics or (b) an empirical
solution.