A circle of unit radius is completely covered by three identical squares. What is the smallest size of the squares?

the 'non overlap solution' was very close, just the math that was wrong.
place two squares side by side with the centre of the circle at some point on their edges. the whole circle will not be covered by these two, only part of it. the other square should be placed on top of these two squares to cover the part of the circle poking out.

take a right angled triangle with one point at the centre of the circle(o), the other at the point where all three squares meet(m) and the other on the edge of the circle(e), found along the square edge. the right angled triangle this results in(ome) gives the following relationship where om=b me=a oe=1
2a=b+1
or 2(√(1-b^2)) = b+1
which ends in a quadratic 4b^2 + b -3 = 0
giving two options for b, one negative, and the other resulting in a square of side 7/8, or 1.75.

Posted by ally on 2005-10-29 11:55:59