(6) As can be seen from the figure, the bottom surface of the triangular prism is an isosceles Rt△, the length of its hypotenuse is 1+ 1=2, and the center of the orthographic (overlooking) circle inscribed with the spherical surface is at the bottom surface of the isosceles Rt△, that is, the hypotenuse center line, with the length = 1, because the hypotenuse center line of the isosceles Rt△. The straight line connecting the center of a square and the vertex of a right angle is a diagonal of the square, and the distance from the center to the tangent point is two adjacent sides of the square, which is equal to √2- 1, so the diagonal length of the square is √2(√2- 1)=2-√2, and the radius of the circle is = midline length-diagonal length =1-.

(7)S=S+ 1/i, that is, sum: s = 1/2+1/4+1/6+...+12016. It can be seen that when i=20 15, there is the last cycle. When i=20 16≥20 15, the cycle ends, so i≤20 15. Choose D.

(8)y？ The focus of 4x is x= 1/2×(4/2)= 1, that is, (1, 0), so the hyperbola c= 1. Hyperbolic c again? =a？ +b？ =m+n= 1, n= 1-m, then e? =c？ /m= 1/m=2？ That is, m= 1/4, then n=3/4, mn=3/ 16? Choose a.

(9)[x] is rounded, that is, x? +y？ The area of this circle is 4π. So choose a.