radius = √40, circumference = 2√40 π Solution Consider the arc of the circle that lies in a particular black square. Its endpoints must be vertices of the square (otherwise the continuation of the arc would enter a white square). Suppose the two vertices are opposite (A and B in the diagram below). Then the two vertices of the next square along the circle cannot be opposite or the circle would pass through three collinear points (so they must be B and C in the diagram). Hence the center lies on the the perpendicular bisectors of AB and BC, so it must be O. That works, as shown in the diagram above. If no square on the circle has opposite vertices on the circle, then it is easily seen that the circle must pass through the four vertices of a square and so is smaller than the case above. (Of course, if the circle is contained in a single square, then it is smaller still.) 3rd Swedish 1963 © John Scholes jscholes@kalva.demon.co.uk 30 December 2003 Last corrected/updated 30 Dec 03
3rd Swedish 1963 Problem 2 The squares of a chessboard have side 4. What is the circumference of the largest circle that can be drawn entirely on the black squares of the board?