Russian Beauty and 1729
The two legs of a compass are located at 2 lattice points(points with integer co-ordinates)in the co-ordinate plane drawn on a (One corner of the paper is at (0,0) and the diagonally other corner is at (1729,1729))sheet of paper.The distance between 2 legs can not be changed.It is allowed to fix one of the legs, and move the other leg to any other lattice point.Determine the number of initial lattice points (where one of the legs of the compass was initially placed) for which you can switch the position of 2 legs after a finite number of steps.
The answer is 0.
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1 solution
Call a lattice point even if the sum of its co-ordinate is even,and call it odd otherwise.Call one of the legs first,and the other second,Then every time a leg is moved from an even point to an even point,or from an odd point to an odd point.(Why?...Use pythagorus dist formula....Dist between legs is invariant).If in the beginning one of the legs is at an even poit,and the other - at an odd point,the legs can not switch.If both legs are at even points or both at odd points,consider a new co-ordinate system on the sheet of paper,with unit length twice as large as unit length iin the original coordinate system,and so that the two legs are located at lattice points of the new co-ordinate system by appropriate shifting of origin...continue this process and complete the sum....THIS IS A SLIGHT MODIFICATION OF RUSSIAN MATH OLYMPIAD PROBLEM 1998....THESE ARE FAMOUS AS RUSSIAN-STYLE PROBLEMS...AND ARE JUST SO BEAUTIFUL!!!!!!HOPE U ENJOYED SOLVING OR TRYING THE SUM...:)