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bonanova

How many squares?

Question

I randomly drew some squares on a sheet of paper and colored them red. Then I drew a gray square of equal size and counted the number of red squares it touched. Not very many. I forget the actual number, might have been 4 or 5. But it made me wonder: What is the largest number of red squares that a single gray square can touch?

The squares are all of equal size and none of the red squares touch each other.

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16 answers to this question

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Thanks for the Christmas tree idea! Changed the red to green. Hope my sketch is close enough! Don't know how to put an image into the spoiler section only - Sorry for that glitch.

Spoiler

image.png.b872db2940b9087f3440f944f8ca964e.png

bona_gold_star.gif

Edited by bonanova
spoiler and star added

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Dang, I blew my spoiler again...I’ll be back. I’ve got an answer...sigh

Edited by CaptainEd
Spoiler didn’t hide my answer

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Assuming the gray square can only touch each red square corner-to-corner or corner-to-edge, my answer is ...
Spoiler

Six

 

Edited by bonanova
Edit to provide a spoiler

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3 hours ago, CaptainEd said:

“Touch” elucidation question: Does the gray square overlap red squares? Or can only edges overlap? Or can they only share single points?

Gray squares overlap red squares but the overlap need not involve a corner (e.g. aligned centers and rotated 45 degrees.) Overlap can be any positive amount, from full overlap to a small portion of a corner, but positive - not zero.

And of course there is no Red/Red overlap; red squares do not touch.

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2 hours ago, CaptainEd said:

Assuming the gray square can only touch each red square corner-to-corner or corner-to-edge, my answer is ...
  Reveal hidden contents

 

More red squares can be overlapped.

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How about

Spoiler

Seven? something like this: (pardon my poor paint skills):

squares.png.85058f43c66364f8e31fe1ae5378acac.png

 

Edited by Pickett
Had nested spoilers for no reason...removed those...

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Captain Ed had the answer a red beneath the gray but rotated 60 degrees; another red a top of gray and rotated 30 degrees.  Then you have one red at each of the four corners,which totals 6

new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd beneath the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray.  You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray  at the midpoint of the grays' bottom side.                                                                              If the red over gray is considered to be touching  the red beneath the gray then the answer is  7

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Tried to edit my previous answer and lost everything .  Correction ...Red sq "beneath "gray and rotated 60 degrees; Red sq "a top" gray sq and rotated 30 degrees

22 minutes ago, Donald Cartmill said:

Captain Ed had the answer a red beneath the gray but rotated 60 degrees; another red a top of gray and rotated 30 degrees.  Then you have one red at each of the four corners,which totals 6

new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd above the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray.  You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray  at the midpoint of the grays' bottom side.                                                                              If the red over gray is considered to be touching  the red beneath the gray then the answer is  7

 

25 minutes ago, Donald Cartmill said:

 

26 minutes ago, Donald Cartmill said:

 

new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd above the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray.  You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray  at the midpoint of the grays' bottom side.                                                                              If the red over gray is considered to be touching  the red beneath the gray then the answer is  7

new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd above the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray.  You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray  at the midpoint of the grays' bottom side.                                                                              If the red over gray is considered to be touching  the red beneath the gray then the answer is  7

 

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On 12/15/2017 at 8:10 AM, Pickett said:

How about

  Hide contents

Seven? something like this: (pardon my poor paint skills):

squares.png.85058f43c66364f8e31fe1ae5378acac.png

 

Best answer so far. Can you do exactly one (1) better?.

22 hours ago, BMAD said:

how are we defining touching?

Touching refers to red squares, which do not touch: Every point in the plane belongs to at most one red square.

 

Donald Cartmill:

Spoiler

new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd above the gray but rotated 30 degrees. Red squares cannot have points in common. 4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray.  You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray  at the midpoint of the grays' bottom side.                                                                              If the red over gray is considered to be touching (yes) the red beneath the gray then the answer is  7

Can you do exactly one (1) better?.

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To clarify: Picket's configuration has yielded the best result so far, but it is possible to do better.

Puzzle still unsolved.

 

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So, how can one square touch more than seven other non-touching squares?

Clue

Spoiler

If the red squares were green, the configuration might look a little like a Christmas tree!
Can you find the answer in time for the celebration?

 

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@CaptainEd Thanks.

I read somewhere (probably in a Martin Gardner book) that seven was the commonly accepted result (as given by @Pickett); this eight solution is a relatively recent discovery and published only incidentally in regard to a separate problem. A third party then recognized its relevance here. Surprising, because results of this type are things we imagine were settled centuries ago. Happy to share it here.

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