A circle in a square hole.

[BMAD]

Given 120 unit squares arbitrarily situated in the 20x25 rectangle. 

Prove that You can place a circle with the unit diameter without intersecting any of the squares.

[bonanova]

Drawing this from the puzzle.

 

Partition the 20 (wide) x 25 (high) rectangle with a unit square grid.

 

Place the 120 unit squares on the intersections of

columns 1 3 5 7 9 11 13 15 17 19 and

rows 1 3 5 7 9 11 13 15 17 19 21 23.

 

Double the width and height of the squares, keeping the lower-ranked corner fixed.

The rectangle is now covered completely except for the 25th row - a 1" strip.

A unit circle can be inserted anywhere along that strip.

 

Why double the width and height of the squares?

 

If doubling the linear dimension of a collection of objects creates a covering,

then a similar object cannot be placed among the original array

without touching one of them. Any area left uncovered after the doubling

provides a location where a similar object could be placed within the

initial-size array without touching any of them.

 

Thus,

1. If we had been asked to place a unit square on the original array without
   touching any of the others, we could do it, but only by placing it
   along the 25th row.
   
   The fact that doubling the squares covers the first 24 rows without
   overlap proves their placement is optimal with regard to the problem
   to be solved.
   
    
2. Since we are asked to place a unit-diameter circle, the task is easier.
   Suppose we color the original partitioning of the rectangle red and black,
   and say the 1" squares had been placed on red squares. 
   
   Then a unit-diameter circle could be placed on any unoccupied red square
   without even touching any of the squares.

A unit-diameter circle can therefore be placed anywhere on the 25th row,

or on any unoccupied red square, without touching (much less intersecting)
any of the 1" squares.

 

QED.

[phil1882]

in a 20x25 rectangle, 500 unit squares can be fitted.

spacing the 120 squares out such that slightly less than a unit square is between them, gives roughly 240 area that the circle cannot be.

rotating the squares in an alternating pattern of diamond and square gives roughly 324 area that the circle cannot enter.

moving the squares out of sync such that they are in a somewhat hexagonal pattern maximizes the area that the circle cannot be,  giving roughy 411 area the circle connot enter. this still leaves plenty of space for the circle, even in the worst case.

[bonanova]

@Phil, I am thinking along these same lines.

The task I'm working on is how to determine a provable worst case for the squares placement.

It's the nature of such a proof that is puzzling me for the moment.

[Barcallica]

Prove for 5 squares and 4x5 rectangle, and multiply by 5. Will it work?

[bonanova]

I thought of that, too. But it scales down to 120/25 = 4.8 squares. Not quite 5.

So I don't think it would apply, unless that turns out to be harder than the original problem.