You draw a square. Then, you draw the largest possible circle, tangent to all four sides. If the lower left corner of the square is at (0,0), and the sides of the square are parallel to the coordinate axis, the point (2,1) is on the circle, within the square.
What is the radius of the circle?
(In reply to
Solution by Jyqm)
I think you'll find that R = sqrt(2R^2 - 6r + 5) has two real solutions.
Simplifying gives (R-1)*(R-5) = 0.
So R = 1 when the point is on the lower right side of the circle.
re: Solution (pt. 2) (spolier)
