Red and Blue squares

[bonanova]

A table top is tiled with red unit squares, none of them overlapping.

A blue unit square is laid on top.

What is the maximum number of red squares the blue square can touch overlap for some non-zero area?

Edited August 6, 2012 by bonanova
Clarify touch.

[Pickett]

is 5:

...I'm seeing if more can be done...I mean 9 can be done if you count edges touching...but since there is "no overlap" I don't know if that would count as "touching" :c)

Edited August 6, 2012 by Pickett

[unrealdon]

9

[Pickett]

is 5:

...I'm seeing if more can be done...I mean 9 can be done if you count edges touching...but since there is "no overlap" I don't know if that would count as "touching" :c)

I got 6...It may not look accurate, but that's due to my horrible paint skills...I actually laid this out with paper and it works...

Edited August 6, 2012 by Pickett

[bonanova]

Good going, but so far no one has the answer.

Clarification about overlap and touching. Apologies to unrealdon.

I used "touch" for red-blue contact, when I meant to say overlap.

I'll edit the OP accordingly.

1. Touching at the edges does not count as overlap.
   
2. The red squares cover the table without overlap.
   
3. The blue square must overlap a portion of some number of red squares.
   That is, blue and red can't just touch at their edges.
   

[curr3nt]

I can't get more than six...

[plainglazed]

[bonanova]

Bingo. Nicely done.

Now we'll just put red squares anywhere on the table, (i.e. let them move apart,) but not overlapping each other.

Can the blue square touch (overlap) even more red squares?

This is counterintuitive, which makes it a good puzzle.

[plainglazed]

had originally assumed the table would not necessarily have to be fully tiled but even then could not beat seven. have not done the math yet but this lays out to work:

remain uncertain one can do better than seven. tho of course, am still trying. especially so due to your counterintuitive comment

[bonanova]

That'll do it. Nice.

[TheChad08]

had originally assumed the table would not necessarily have to be fully tiled but even then could not beat seven. have not done the math yet but this lays out to work:

remain uncertain one can do better than seven. tho of course, am still trying. especially so due to your counterintuitive comment

Umm... I count 8 in that picture.