tag:blogger.com,1999:blog-4115025577315673827.post5848497227356386010..comments2019-10-08T12:02:40.465+05:30Comments on CSE Blog - quant, math, computer science puzzles: Chessboard CircleUnknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4115025577315673827.post-8399098473576231602010-01-10T16:12:10.266+05:302010-01-10T16:12:10.266+05:30nice explanation.. I actually got the solution by ...nice explanation.. I actually got the solution by "drawing" circles on a chessboard in bitmap. :) ThanxPratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-81209127148026333272010-01-10T08:05:33.145+05:302010-01-10T08:05:33.145+05:30for the circumference to lie entirely on the black...for the circumference to lie entirely on the black squares, it should not intersect any edge of the squares at points other than the vertices. so the question changes to 'what is the radius of the largest circle that can be drawn on a chessboard such that it never intersects any edge of the squares at points other than the vertices'. <br />we know that three non-collinear points in a 2-D space determine a unique circle. Since the circumference lies entirely on black squares, if one black square is containing the circumference, one of its diagonally adjacent black square has to contain it as well (excluding the case where the circle completely lies inside one black square, because that is clearly not the largest radius case). <br />now we have seven points (of the two black squares combines) out of which we have to choose 3 non-collinear ones such that the radius of the circle so formed is the largest. the common point is always included. for the remaining two points we consider the following cases:<br />1)point adjacent to common point in both black squares<br />2)point diagonally opposite to common point in both black squares<br />3)point adjacent to common point in one black square and diagonally opposite to it in another.<br />in case 2 and a subcase of 1, a circle is not formed because the points are collinear. out of the remaining subcases of 1 and case 3, it is easily seen that the radius is largest in case 3.<br />this take cares of a quarter of the circumference. the symmetry of the circle will make sure that the other 3 quarters also lie entirely on black squares.<br />By simple geometry (perpendicular bisector, pythagoras) we can prove that the radius of the circle so obtained is sqrt(10) inches.suyashhttps://www.blogger.com/profile/09632474759324558691noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-4563057538534658492010-01-10T00:46:29.183+05:302010-01-10T00:46:29.183+05:30yes suyash.. correct answer..
can someone prove th...yes suyash.. correct answer..<br />can someone prove that this is the best one can do :)Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-9998222665469154432010-01-09T21:57:27.722+05:302010-01-09T21:57:27.722+05:30is the answer sqrt(10) inches?is the answer sqrt(10) inches?suyashhttps://www.blogger.com/profile/09632474759324558691noreply@blogger.com