Idea: there is only one small square that will meet the constraints of the problem, so it needs only be constructed to produce the desired answer.
Divide the 5*5 square ABCD into 25 little 1*1 squares (1,1) to (5,5) and construct semidiagonal DO from D to the centre O of the circumcircle, S. DO is also a diagonal of (1,1), (2,2) and (3,3), the last of these being centred on O. Construct the other diagonal of (1,1) and extend it to intersect S at H. Do the same with little square (2,2) to intersect CD at F. Lastly, at the common vertex of (1,1) and (2,2) construct a line, parallel to the two diagonals just drawn, intersecting CD at E and S at G. Clearly quadrilateral EFGH is a square, and is of the same size as (3,3), so its area is exactly 1.
