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?

If we say the radius of the circle lies on point (R, R), then the far right point of the circle lies on (2R, R).

This point is part of the line x = 2y, which also includes the points (2, 1) and (0, 0).

If you draw the figure, then, it's clear that the point (2, 1) lies somewhere on the lower left arc of the circle. (I guess realizing this isn't essential to solving the problem, but it's what I did first, anyway.)

The distance between (R, R) and (2, 1) is √(2Rē - 6R + 5), which is also the distance of the radius itself. Just trying different integers for R, I got an answer of R = 5.

Nice problem!

Posted by Jyqm on 2006-11-25 07:38:43