=(2*2/2-∏*2? /8)+(∏*(2? +2? )/8-2*2/2)=∏/2

2.p has not changed, as shown in Figure 2:

When rotating, point M moves from B to A, and point N moves from C to B, which is synchronous with θ, that is, AN=BM, then MB+NB=2.

Similarly, as shown in figure 1:

When rotating, point m moves from b to A', and point n moves from c to B', which is synchronous with θ, that is, concentric rotation, then MN= constant.

P=2+ constant = constant

(Maybe it can be calculated by trigonometric function and cosine theorem. )

3. Suppose the area of △OMN is y and AM=x, then the sum of the areas of the four triangles is:

y+ 1/2(2 * x+(2-x)* x+(2-x)* 2)= 2 * 2

After finishing: y= 1/2(x- 1)? +3/2

Obviously, when x= 1, y is the smallest, so the radius △BMN of the inscribed circle at the isosceles right angle is:

(√2/2)(tg∏/8)

^_^