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

(In reply to re: overlap solution by Jer)

In the scenario presented, one of the diagonals of each square must lie along an extended diameter of the circle.  The corner that lies within the circle must be far enough back so that its 90-degree angle subtends 120 degrees of arc on the circle.  The points of intersection of the two sides with the circle form a chord of length sqrt(3). As a triangle formed from the inner vertex of the square, the intesection of a side of the square with the circle and the midpoint of the chord is an isosceles right triangle, the distance from the chord's intersection with the diagonal of the square (coincident with the diameter of the circle, as mentioned), is also sqrt(3)/2.  Meanwhile the distance from the center of the circle to the chord's intersection with the diameter is 1/2, so the inner corner of the square is sqrt(3)/2 - 1/2 units back from the center of the circle.

The sides of the square must extend farther out than merely to the circle--far enough so that the other two sides will be tangent to the circle.  The distance from the center of the circle to the point of tangency is, of course, 1.  A rectangle can be formed from the center of the circle to the point of tangency, to one corner of the square that's off the axis we've drawn (i.e., not on the diagonal containing the internal corner of the square), and completed along an edge of the square near the inner corner of the square.  That marks off a length of 1 on that edge of the square, and the rectangle leaves off a small isosceles right triangle, whose hypotenuse is center of the circle to the inner corner of the square.  The side of that triangle is therefore (sqrt(3)/2 - 1/2) / sqrt(2).  Together with the portion of length 1, that makes up the whole side of the square: 1+ (sqrt(3)/2-1/2)/sqrt(2) = 1.25881904510252.

Posted by Charlie on 2005-10-21 17:08:12