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Geometry ProblemClick the figure below to see the complete problem 605.
Let 2a, 2b be the sides of the squares (a>b)Note a-b = 4 and (2a-x)^2 + a^2 = r^2 = (2b+x)^2 + b^25(a^2-b^2)= 4(a+b)x,Hence x = 5(a-b)/4 = 5
http://img836.imageshack.us/img836/732/problem605.pngLet R= radius of the circleNote that MN=2.LN and CF=2.CLSo 3 points F, L, M are collinearDistances LN and LC are the solutions of system equations:1. equation y=2.t represent line FM2. equation t^2+(y+x)^2=R^2 represent circle radius R, center O2 positive solutions are LN= abs(t1)=1/5(-2x+SQRT(5R^2-x^2))And LC=abs(t2)=1/5(2x+SQRT(5R^2-x^2))DB=LC-LN=4/5.x=4So x=5Peter Tran