It is clear that the largest semicircle will touch the sides of the square at both ends of its diameter, and will also be tangent to the perimeter.
One obvious solution is a semicircle whose diameter coincides with one side of the square.
Such a semicircle will have radius = 1/2 and area = (1/2)2/2 = /8 0.3927.
Can we do better?
Consider a semicircle whose diameter endpoints touch two adjacent sides of the square. It is intuitively obvious that such a semicircle of maximal area will be tangent to both of the other sides of the square, but see the remarks below for a more rigorous justification.
Since the figure is symmetrical in the diagonal BD, QPB = 45°.
Consider the point X on AD at which the semicircle is tangent to AD. A line extended from X that is perpendicular to the tangent will be parallel to AB, and will also pass through the middle of the semicircle diameter. Let the line meet BC at Y.
OY = r cos 45° = r/.
Hence 1 = AB = r + r/ = r(1 + 1/).
Thus r = 1/(1 + 1/).
Rationalizing the denominator, we obtain r = 2 − and area = r2/2 = (3 − 2).
Thus, the area of the largest semicircle that can be inscribed in the unit square is (3 − 2) 0.539.
Although it is intuitively clear that the maximal inscribed semicircle will be tangent to two sides of the square, we can vary t = QPB. If t 45°, the above argument holds, and r = 1/(1 + cos t) is maximized when t = 45°.
Can we be sure that both diameter endpoints of the maximal inscribed semicircle touch the perimeter of the square?
Does the above semicircle, when extended to the full circle, pass through vertex B?
Source: Inspired by a puzzle posted by Mark Nandor