



 
Hi Jamie. I drew a diagram of half of the square and labeled some points.
Let r be the radius of the circles and x the distance from S vertex to the circle. ABC is a right triangle and BC = CA = 2r so from Pythagoras' theorem AB = 2√2r. Thus QR = 2√2r and hence
Hence
Triangle PSC is a right triangle and hence
so
Substitute this value for x into equation (1), simplify to obtain a quadratic in r and solve. You will obtain two values for r so make sure you choose the correct one. Hope this helps,  


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