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A circle in a square hole.


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  • Solution

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.

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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.

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