Plates that cover a table

[bonanova]

In our last table and plates puzzle, we filled a circular table with plates that touched without overlap.

Now we want more plates. Read on ...

I've just placed 12 [circular] plates on a rectangular table.

They don't touch each other; but they're close enough that you can't add a plate without overlap,

Edit: even tho it overhangs the edge, so long as its center is over the table.

The plates are now removed and you are asked to completely cover the table with plates.

Edit: cover means you can't see any part of the table.

Twelve won't do it, so you'll have to order some more.

But they're expensive, so you don't want to order more than are needed.

How many plates are needed to ensure the table can be completely covered?

Obviously we now allow overlap and overhang, so long as the center of the plate is on the table.

Edited February 3, 2009 by bonanova
Clarify overhang, overlap and cover.

[HoustonHokie]

I've just placed 12 [circular] plates on a rectangular table.

Rectangle or square? If rectangle, I might need another dimension to get the answer. If square,

12² + 13² = 313 will definitely cover the table. I don't think you can be more efficient than that, but it might be possible.

[bonanova]

Rectangle or square? If rectangle, I might need another dimension to get the answer. If square,

12² + 13² = 313 will definitely cover the table. I don't think you can be more efficient than that, but it might be possible.

Rectangle.

I bet you can prove that many fewer will do the job.

[Prof. Templeton]

Rectangle or square? If rectangle, I might need another dimension to get the answer. If square,

12² + 13² = 313 will definitely cover the table. I don't think you can be more efficient than that, but it might be possible.

That seems like a lot.

114 would cover it, maybe less.

[bonanova]

Coffee and donut next visit to NY for the lowest provable number.

It's the proof that's so cool.

We're still high on the number, btw.

[HoustonHokie]

Coffee and donut next visit to NY for the lowest provable number.

It's the proof that's so cool.

We're still high on the number, btw.

Re-reading the question, I see that 12 plates already cover the table... I was thinking this was like the last one, and the 12 went along one axis. In that case, I can do it with...

32. It depends a little on how the rectangle is formed, I think.

If D is the diameter of a plate, I can get 32 plates if the rectangle's dimensions are 3D x 4D.

If the dimensions are 2D x 6D, I have to go up to 33 plates.

If the dimensions are 1D x 12D, I get 38 plates.

Donuts?? Or am I still too high?

edit: wait, reading the puzzle again, I think I see what you are looking for is a general solution no matter the what the dimensions are. I'll keep looking

Edited February 2, 2009 by HoustonHokie

[Guest]

I have only one question.

What kind of plate are we talking about here? There's different sizes for dinner, salad, lunch, dessert, bread....

and no, I'm leaving the platters out. Those things are oval, and most plates are round/circular.

[Prof. Templeton]

Re-reading the question, I see that 12 plates already cover the table... I was thinking this was like the last one, and the 12 went along one axis. In that case, I can do it with...

32. It depends a little on how the rectangle is formed, I think.

If D is the diameter of a plate, I can get 32 plates if the rectangle's dimensions are 3D x 4D.

If the dimensions are 2D x 6D, I have to go up to 33 plates.

If the dimensions are 1D x 12D, I get 38 plates.

Donuts?? Or am I still too high?

edit: wait, reading the puzzle again, I think I see what you are looking for is a general solution no matter the what the dimensions are. I'll keep looking

is bigger then 3D x 4D. The OP says that twelve plate are on the table and you can't fit anymore on without over-lapping. I had originally thought that meant if the plates were d=1 then the table was just under 9 by 7, but then I realized this would leave room for plates placed in the diagonal between the others. So the table can be just under 4d + 5(sqrt2-1) by 3d + 4(sqrt2-1) or 6.07 by 4.65. Since thats the case, 7 plates (formed like a flower) put in place of each single plate would certainly cover the table. 84 plates, but still it's probably less then that.

[bonanova]

What kind of plate are we talking about here?

Circular, flat.

[bonanova]

Clarification...

By "cover" we mean that the table does not show at all.

Overlap therefore is not only allowed, it's needed.

[Guest]

Circular, flat.

You obviously don't know your plates. All of the plates I've named can be the ones like you've described. I was wondering what kind of plate so I know which "standard" to use in my measurements.

[Prof. Templeton]

60 would be the least.

[bonanova]

You obviously don't know your plates.

All of the plates I've named can be the ones like you've described.

I was wondering what kind of plate so I know which "standard" to use in my measurements.

You're probably right about my knowledge of plates.

If it will help, I'll say they are dinner plates.

But if all the plates you named can be like what I described,

then it would be fair to assume the plates are of any of those types.

You know what? Now that I'm thinking about food, let's say they are dessert plates.

I suppose it would help to know what type of table [kitchen, dining room, banquet, TV tray] is involved, also.

But I'll leave that as part of the puzzle.

Cute avitar, btw.

[bonanova]

60 would be the least.

Prof T has come in with the lowest estimate to date.

Can anyone prove that fewer plates will do the job?

[Guest]

You're probably right about my knowledge of plates.

If it will help, I'll say they are dinner plates.

But if all the plates you named can be like what I described,

then it would be fair to assume the plates are of any of those types.

You know what? Now that I'm thinking about food, let's say they are dessert plates.

I suppose it would help to know what type of table [kitchen, dining room, banquet, TV tray] is involved, also.

But I'll leave that as part of the puzzle.

Cute avitar, btw.

...and we hear the sound of shredding paper as everyone rips up their notes before reading this. Now that you've decided, it leaves me out of the competition as math is not my usual forté.

Re: avatar - that's from Witch Hunter Robin, an Anime (Japanese Animation.) Interesting series....

[bonanova]

...and we hear the sound of shredding paper as everyone rips up their notes before reading this.

Now that you've decided, it leaves me out of the competition as math is not my usual forté.

Re: avatar - that's from Witch Hunter Robin, an Anime (Japanese Animation.) Interesting series....

Your disqualification might be premature. Give it a little more thought.

The dessert plate thing was triggered by some great strawberry short cake I had yesterday during the super bowl....

[Guest]

x x x x x

x - - - x

x x x x x

Answer is 3

Edited February 2, 2009 by Embrach

[Guest]

I can't beat Prof T

on the largest table that meets your specs, it takes me 63 plates to ensure coverage.

I'd "show my work", but my scribbles aren't legible enough to type right now. Perhaps tomorrow, if needed.

[bonanova]

x x x x x

x - - - x

x x x x x

Answer is 3

Nope. I see your approach, but recall we have to completely cover the table.

Because the plates are circular, that means they will have to overlap.

And we need a proof of sorts, and spoilers, please.

[Prof. Templeton]

of 60 plates, but there seems to be too much overlap still so there must be a more efficient solution.

[bonanova]

I can't beat Prof T

on the largest table that meets your specs, it takes me 63 plates to ensure coverage.

I'd "show my work", but my scribbles aren't legible enough to type right now. Perhaps tomorrow, if needed.

Well you [both] are clearly in the ballpark.

So let's take that number for now:

1. How do we know that number is certain to cover the table.
   
2. How do we know a lower number might not cover the table.
   

[Guest]

I don't have a solid proof yet, but

48 plates

[HoustonHokie]

is bigger then 3D x 4D. The OP says that twelve plate are on the table and you can't fit anymore on without over-lapping. I had originally thought that meant if the plates were d=1 then the table was just under 9 by 7, but then I realized this would leave room for plates placed in the diagonal between the others. So the table can be just under 4d + 5(sqrt2-1) by 3d + 4(sqrt2-1) or 6.07 by 4.65. Since thats the case, 7 plates (formed like a flower) put in place of each single plate would certainly cover the table. 84 plates, but still it's probably less then that.

I'm questioning your dimensions for the table somewhat. I think I can place 12 plates (dia=1) on a table "just smaller" than 6.5 x 5.33 such that you can't put another plate without overlap, which makes the table a little bigger than what you had. I've also come up with 2 other arrangements of 12.5 x 2.73 and 18.5 x 1.87. They all have similar areas (34.6, 34.1, 34.6). I think these arrangements - particularly the 6.5 x 5.33 - maximize the size of the table we are to cover with plates. If that's right, then we can move on to the minimum number of plates required for complete coverage. You might consider this the first step in the proof we're seeking.

[Prof. Templeton]

I don't have a solid proof yet, but

48 plates

I think the answer

somewhere between 48 and 60. Here's a picture of the optimal covering for four plates. The grey square is sqrt2 by sqrt2 (d=1). Layed out in a rectangle of 4 by 3 squares we could cover a table 5.657 by 4.243. But I think this table could be a little over that and still fit the description of the OP. Of course the OP leaves the size of the table up to us, so maybe we need to find the optimal sized table first.

Just saw HH's post. I was thinking of plates lined up in neat rows and columns so that you couldn't put a plate in between them diagonally, maybe I'll make a picture.

[Prime]

I have a hypothesis about the worst case scenario

I think the worst case scenario could be the case where the table is exactly one plate diameter wide and just a touch less than 13 plates long.

We could use a rectangle inscribed into a circle of diameter D with the sides (1/2)D and (3^1/2/2)D. Of those rectangles, we need exactly 2 to cover the width of the table (using rectangles' 1/2 sides) and just a touch over 15 lengthwise. So 31 plates would be enough.

I suspect, other cases where the table shape tends more towards square, could require less plates.

On the other hand, the largest area rectangle inscribed into a circle has pi/2 times smaller area than the circle. Multiplying 12 by that factor, we find that we could not get away with less than 19 whole plates.

[Guest]

I think the answer

somewhere between 48 and 60. Here's a picture of the optimal covering for four plates. The grey square is sqrt2 by sqrt2 (d=1). Layed out in a rectangle of 4 by 3 squares we could cover a table 5.657 by 4.243. But I think this table could be a little over that and still fit the description of the OP. Of course the OP leaves the size of the table up to us, so maybe we need to find the optimal sized table first.

Just saw HH's post. I was thinking of plates lined up in neat rows and columns so that you couldn't put a plate in between them diagonally, maybe I'll make a picture.

I would argue that that is the optimal size of the table provided that overhang is allowed, which HoustonHokie does not seem to account for.