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401 unit diameter Circles in a 400 unit rectangle

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In a rectangle that's 2 by 200 units long, it's trivial to draw 400 non-overlapping unit-diameter circles.

But in the same rectangle, can you draw 401 circles?

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Posted · Report post

... related to hexagonal packing.

Or in the case of two rows, triangular packing.

Straightforward triangular packing gives a top row of 199.

We seek a top row of 201.

But we do get some headroom above the row of 199.

That provides some wiggle room, where the top row need not lie on a straight line.

And that allows some compaction of its length.

How much, I have no idea.

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Posted · Report post

I'm pretty sure you can fit way more than 401 circles in the rectangle.

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Posted · Report post

I'm pretty sure you can fit way more than 401 circles in the rectangle.

what would your array be?

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Posted · Report post

I'm pretty sure you can fit way more than 401 circles in the rectangle.

what would your array be?

Sorry. There was an error in my thinking. Is Bonanova right? I think so. I can't possibly think of another way of doing this.

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Posted · Report post

on the right track, but no sure answer yet

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Posted · Report post

post-53237-0-10519400-1373832928_thumb.j

my idea is to zigzag..but its too long

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attachicon.gifzigzag.JPG

my idea is to zigzag..but its too long

[spoiler=It has to be different in the following way]In your sketch, the top row circles are 1-1 with the bottom row.

It has to be the case that two extra top row circles appear.

That won't happen unless at some point (at least in two cases) a top row circle is directly above a bottom row circle. So the top row circles must move up and down in groups.

Start with a group of three top row circles nestled down in triangular packing position.

Then a group of three top row circles packed against the top of the box.

Then three more down, then three more up. Do that 66+ times, to the end of the box.

It seems like that permits the top row to be compacted in length.

That may be enough to fit two extra (199 = 2 = 201) top row balls.

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Posted · Report post

the key is that the OP doesn't say "fit 401 circles in a 200 x 2 box", it asks "

can you fit 401 circles?"

I don't feel I have a very solid proof. But: a rectangular packing clearly won't work, a hexagonal packing where circles are touching in lines parallel to the long axis of the box will never allow you to make a third row of circles (and I don't think it would allow enough free space on one side to somehow squeeze in another circle because rectangular packing is as squeezed into a corner as you can get), and the most efficient way of packing I can envision that will make a third row of circles is to take a slice of a hexagonal packing like so:

-----------------------

O_____O_____O_____O_

___O_____O_____O_____O

O_____O_____O_____O

-----------------------

Where the circles directly above and directly below each other are touching. In that case, each "repeating unit" of three circles would be separated from the previous one by sqrt(3) ~= 1.73, and you could only fit 115 of them in the 200 unit long rectangle to get about 345 circles in.

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Posted · Report post

Consider the drawing below.

The top (red) row of circles jog up and down in groups of three.

In 20 circles distance, the row has compacted about 1/7 diameter.

After 200 circles, the compaction would be of the order of 1.5 diameters.

See green circle and dimension lines on the right end.

post-1048-0-04445800-1373869887_thumb.gi

201 circles should fit across the top row.

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Posted · Report post

attachicon.gifzigzag.JPG

my idea is to zigzag..but its too long

[spoiler=It has to be different in the following way]In your sketch, the top row circles are 1-1 with the bottom row.

It has to be the case that two extra top row circles appear.

That won't happen unless at some point (at least in two cases) a top row circle is directly above a bottom row circle. So the top row circles must move up and down in groups.

Start with a group of three top row circles nestled down in triangular packing position.

Then a group of three top row circles packed against the top of the box.

Then three more down, then three more up. Do that 66+ times, to the end of the box.

It seems like that permits the top row to be compacted in length.

That may be enough to fit two extra (199 = 2 = 201) top row balls.

Diamond or zigzag (end)

post-53237-0-22871200-1373875015_thumb.j

the sides(cyan) of diamond is 1 unit

the center to center distance contraction (green) is 0.995

the gain in length (yellow) is (1-0.995)X199=0.995 unit

the extra unit circle (red) fits in..

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Posted (edited) · Report post

Consider the drawing below.

The top (red) row of circles jog up and down in groups of three.

In 20 circles distance, the row has compacted about 1/7 diameter.

After 200 circles, the compaction would be of the order of 1.5 diameters.

See green circle and dimension lines on the right end.

attachicon.gif401 circles.gif

201 circles should fit across the top row.

If you try to compact the top row like this, at some point during compaction the circles on the top row must lie directly above the circles on the bottom row at which point such compaction cannot continue. Or if you propose that the compaction traverses a zone where circle N-1 is just a tiny bit to the right of circle M-1 below it and circle N is just a tiny bit to the left of circle M below it, all without ever having the top row of circles exactly above the bottom row, then if the lower circles M-1 and M are touching then it would require you to somehow pack more efficiently in that block of four circles than rectangular packing.

Edit: although now that I understand TSLF's answer, I think that it should work in principle. Haven't verified the calculations myself yet though.

Edit2:

I'm coming up with the angle between (the long axis of the diamond) and (the axis of the rectangle) in the picture above being arcsin(1/sqrt(3)) ~= 35.264 degrees, so the angle between (a side of the diamond nearest the edge of the rectangle) and (the axis of the rectangle) would be arcsin(1/sqrt(3)) - 30 ~= 5.264. Then you can get the horizontal distance between two circles along the zigzag as

cos[ arcsin(1/sqrt(3)) - 30 ] ~= 0.99578

Still can't quite fit an extra circle in. But it is a really neat approach.

Edited by plasmid
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Posted · Report post

To fit 401 unit-diameter circles in a 2-by-200 rectangle, the construction is as follows: First set up the standard triangular packing with 399 circles: 200 along the bottom of the rectangle, and 199 resting in the gaps. Then, from the left, group the circles in sets of three. Observe that every second triad can be moved upward a small amount; go ahead and move each second triad upward until they just touch the top of the rectangle. This opens up small gaps between adjacent triads. Finally, compress all the triads horizontally to close these gaps. The compression is about 0.6%, which leaves just enough room at the edges of the rectangle to add circles 400 and 401.

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Posted (edited) · Report post

Really easy!!!

In a rectangle that's 2 by 200 units long, it's trivial to draw 400 non-overlapping unit-diameter circles.

But in the same rectangle, can you draw 401 circles?

And even more.

No one said that the 401 circles may not have a smaller diameter.

No one said that they may not ovelap.

Edited by harey
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Posted (edited) · Report post

I disagree with your first perverted twist of the rules harey.as i feel that the overlapping and unit circle (designated diameter) was implied from the previous sentence.

Edited by BMAD
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Posted · Report post

To fit 401 unit-diameter circles in a 2-by-200 rectangle, the construction is as follows: First set up the standard triangular packing with 399 circles: 200 along the bottom of the rectangle, and 199 resting in the gaps. Then, from the left, group the circles in sets of three. Observe that every second triad can be moved upward a small amount; go ahead and move each second triad upward until they just touch the top of the rectangle. This opens up small gaps between adjacent triads. Finally, compress all the triads horizontally to close these gaps. The compression is about 0.6%, which leaves just enough room at the edges of the rectangle to add circles 400 and 401.

Does these triads form an equilateral triangle?

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