Finding diameter of the largest circle

What is the diameter of the largest circle that can be drawn on a chessboard so that its entire circumference gets covered by the black squares and no part of the circumference falls on any white space, given that the chessboard has black and white squares of size one inch.

I am looking forward to some ideas/hints for this problem.

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There are two solutions.

1. The circle covering four squares (arranged in a diamond) with $$R=\sqrt{0.5^2+0.5^2}=\sqrt{2}/2$$
2. The larger circle covering eight squares (also in a diamond, 3 per side) with $$R=\sqrt{1.5^2+0.5^2}=\sqrt{10}/2$$

It helps to draw a diagram like this:

or this:

I don't think any bigger would fit. But how do you prove this statement???

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I think the later picture is the correct representation. –  Quixotic Dec 8 '11 at 21:02

Well I thought of a kind of a pseudo-code to figure it out.

First of all, if it covers only black squares, it is obvious that it has to pass through corners of the squares, otherwise it would necessarily pass through white squares.

Any point on a circle satisfies the relation $x^2 + y^2 = R^2$, where $x$ and $y$ are the horizontal and vertical distance of the point from the circle's center respectively.

If you look at a chessboard, you'll realize that the center of the circle you're looking for must be in the middle of a field (not on a corner), because you'll only be able to draw a circle through the corners where the black squares are neighbouring in the direction SW-NE on the left side, and NW-SE on the right side of the circle.

Having the problem set up in this fashion, you can now check whether the corners that are $0.5$ square away, $1.5$ square away, etc. satisfy the above relation.

Taking a quick look at a chessboard with this in mind, I would say that the largest circle you can draw is from the center of a black square passing through the corners of the farther edge of the neighbouring white square, i.e. with $R = \sqrt{5/2}$.

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I would stare at a chessboard and try to imagine where you could put a circle to meet the requirement. Of course, you can draw one within a black square, diameter one inch, but you can do better. The solution I am thinking of is subject to a bit of debate.

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The answer is $\sqrt {10}$ inches, right? What is there to debate? –  TonyK Sep 26 '11 at 22:56
@TonyK: How the intersections of the grid lines are colored. –  Brian M. Scott Sep 26 '11 at 23:20
Chessboards don't need grid lines! And obviously we have to assume that our particular chessboard doesn't have them, otherwise the question has no substance. Oh dang -- there I go, debating again. –  TonyK Sep 26 '11 at 23:26
Even without grid lines, the coloring of the lattice points is open to debate. I think they are supposed to be black in this question. –  Ross Millikan Sep 26 '11 at 23:35
@TonyK: Not quite right. –  Ross Millikan Sep 27 '11 at 3:01

Any circle with a diameter larger than 1 will have to enter and exit each square through a corner. There are, up to symmetry, only two ways three consecutive corners can be positioned: adjacent-adjacent or adjacent-diagonal. Because three points fix a circle there are no more than 2 possible diameters.

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Refer to this link for more information about the proof (or discussion rather). I find it really useful when trying to prove it on my own too. :)

Take note that the dimensions of each squares of the chessboard in the link are 2 inches by 2 inches, instead of 1 inch by 1 inch. But the same principle applies.

P.S. I wanted to comment under the post (because this post doesn't really count as "my" answer) but couldn't do so as a message saying "You need at least 50 reputation" to comment popped up.

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