Jump to content
BrainDen.com - Brain Teasers
  • 0

Last Puzzle of 2014 - Plates on a table


bonanova
 Share

Question

I may have posted this delightful puzzle already, but search didn't find it for me.

Hopefully it is new to most of our current puzzle solvers:

 

I've just placed 13 non-touching plates on a table in such a way that a 14th non-touching plate cannot be added.

 

[1] If the table is so large that a larger table would permit a 14th non-touching plate,

what is the smallest number of additional plates that will ensure that the table is completely covered?

 

[2] If the table is so small that a smaller table cannot contain 13 non-touching plates,

additional plates that are fewer in number than in case [1] can cover the table.

What is that number?

 

Assume:

  1. the plates are identical, thin circular disks
  2. the table is rectangular, aspect ratio unknown
  3. the center of every plate lies within the table's perimeter (it will not fall)
  4. "completely covered" means that looking from above, the table is completely obscured
Link to comment
Share on other sites

13 answers to this question

Recommended Posts

  • 0

I feel like a tiling of regular octagons could represent all of the following: the least efficient packing of non-touching plates, the most efficient packing of non-touching plates, and the most efficient placement of plates that completely cover the table. In the first case, each octagon has a circumradius equal to the diameter of a plate. In the second case, each octagon has an inradius equal to the radius of a plate. And in the last case, each octagon has a circumradius equal to the radius of a plate.

 

 

The answer to [1] would be the number of smalloctagons that fit into the area that exactly (almost) 14 large octagons fit.

The answer to [2] would be the number of small octagons that fit into the area that exactly (barely) 13 medium octagons fit.

 

Am I moving in the right direction?

Edited by gavinksong
Link to comment
Share on other sites

  • 0

Since the circumradius of the large octagons is twice that of the small octagons, The large octagons must occupy an area four times larger than the small octagons do. So I estimate that the smallest number of additional plates required to cover the table will be around 40.

Edited by gavinksong
Link to comment
Share on other sites

  • 0

I feel like a tiling of regular octagons could represent all of the following: the least efficient packing of non-touching plates, the most efficient packing of non-touching plates, and the most efficient placement of plates that completely cover the table. In the first case, each octagon has a circumradius equal to the diameter of a plate. In the second case, each octagon has an inradius equal to the radius of a plate. And in the last case, each octagon has a circumradius equal to the radius of a plate.

 

 

The answer to [1] would be the number of smalloctagons that fit into the area that exactly (almost) 14 large octagons fit.

The answer to [2] would be the number of small octagons that fit into the area that exactly (barely) 13 medium octagons fit.

 

Am I moving in the right direction?

 

 

Since the circumradius of the large octagons is twice that of the small octagons, The large octagons must occupy an area four times larger than the small octagons do. So I estimate that the smallest number of additional plates required to cover the table will be around 40.

 

I meant to say hexagons.

Link to comment
Share on other sites

  • 0

I think it is 48 additional plates. I could be wrong. I didn't arrive at this number rigorously. I just counted up the edges in a hexagonal tiling of a rectangular area that contained 13 centers, excluding the ones on the perimeter.

 

[2] is less nice.

 

 

If we consider the radius of a plate to be 1, then the dimensions of the table should be 4 by 12/sqrt(3).

The number of plates required to cover the table should be equal to the number of hexagons with a inradius of 1 we can fit on a 8/sqrt(3) by 24 table. With a good amount of pseudomathematics, I arrived at 35 total plates, or 22 additional plates.

Edited by gavinksong
Link to comment
Share on other sites

  • 0

 

I think it is 48 additional plates. I could be wrong. I didn't arrive at this number rigorously. I just counted up the edges in a hexagonal tiling of a rectangular area that contained 13 centers, excluding the ones on the perimeter.

 

[2] is less nice.

 

 

If we consider the radius of a plate to be 1, then the dimensions of the table should be 4 by 12/sqrt(3).

The number of plates required to cover the table should be equal to the number of hexagons with a inradius of 1 we can fit on a 8/sqrt(3) by 24 table. With a good amount of pseudomathematics, I arrived at 35 total plates, or 22 additional plates.

 

 

The number of plates required to cover the table should be equal to the number of hexagons with a inradius of 1 we need to cover a 8/sqrt(3) by 8 table. We actually only need 14 total plates, or 1 additional plate. This is a strange result.

Edited by gavinksong
Link to comment
Share on other sites

  • 0

Okay, I know I've made a lot of posts, but here's my final answer.

I realized I made a mistake on the first problem, or at least the first problem as I understand it.

I used hexagonal tilings to model the plate arrangements.

I assume that the plates can be rearranged after adding new plates to cover the table.

[1] requires 25 additional plates.

[2] requires 1 additional plate.

Edited by gavinksong
Link to comment
Share on other sites

  • 0

The aspect ratio is not given. How can we be sure of covering the table?

 

Note that doubling the diameter of the plates does it

 

Oh..

 

Six additional small hexagons are needed to cover one big hexagon, but adjacent big hexagons can share a small hexagon. The case where the fewest small hexagons are shared are when the big hexagons are in a line. So six times thirteen small plates minus twelve shared small plates equals sixty-six additional plates required for [1]?

 

Hmm...

Edited by gavinksong
Link to comment
Share on other sites

  • 0

Double the diameter of the plates, thus covering the table.
Shrink everything 50%, thus creating a half-size table covered with full-size plates.

Clone three times, and arrange everything in a 2x2 array.
This makes a full-size table covered with full size plates (4x13=52 of them.)

 

Answer: you need 39 additional plates to ensure a covered table.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...