# 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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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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The answer is $\sqrt {10}$ inches, right? What is there to debate? –  TonyK Sep 26 '11 at 22:56