answer:First, compute the transpose of mathbf{N}: [mathbf{N}^T = begin{pmatrix} x & 2x & 3x 2y & 3y & y 2z & z & 2z end{pmatrix}.] Next, compute the product mathbf{N}^T mathbf{N}: [mathbf{N}^T mathbf{N} = begin{pmatrix} x & 2x & 3x 2y & 3y & y 2z & z & 2z end{pmatrix} begin{pmatrix} x & 2y & 2z 2x & 3y & z 3x & y & 2z end{pmatrix} = begin{pmatrix} 14x^2 & 10xy & 10xz 10xy & 14y^2 & 10yz 10xz & 10yz & 9z^2 end{pmatrix}.] We want mathbf{N}^T mathbf{N} = 3mathbf{I}, therefore: [14x^2 = 14y^2 = 9z^2 = 3,] [x^2 = y^2 = frac{3}{14}, quad z^2 = frac{1}{3}.] Finally, compute x^2 + y^2 + z^2: [x^2 + y^2 + z^2 = frac{3}{14} + frac{3}{14} + frac{1}{3} = frac{6}{14} + frac{14}{42} = frac{20}{42} = frac{10}{21}.] Thus, we find that (x^2 + y^2 + z^2 = boxed{frac{10}{21}}).

answer:1. **Understanding the Problem:** We are given a unit cube (a cube with edge length 1) with eight equal spheres inside it. Each sphere is inscribed in one of the cube's trihedral angles (corners) and touches three adjacent spheres corresponding to neighboring corners of the cube. We need to find the radius of these spheres. 2. **Cube's Edge as Sum of Diameters:** Given that the sum of the diameters of two neighboring spheres is equal to the edge length of the cube (1 unit), let ( r ) be the radius of each sphere. Therefore, the diameter of each sphere is ( 2r ). 3. **Establish the Relationship:** Since the spheres are equal and inscribed in the cube's corners, each sphere touches three other spheres at the adjacent corners. Consequently, for two neighboring spheres: [ 2r + 2r = 1 ] 4. **Solve for the Radius:** Simplify the above equation to find ( r ): [ 4r = 1 r = frac{1}{4} ] 5. **Conclusion:** The radius of each sphere inside the unit cube is ( frac{1}{4} ). [ boxed{frac{1}{4}} ]

answer:1. Let the 100 non-negative real numbers be denoted as ( x_1, x_2, dots, x_{100} ), and their sum is given by: [ x_1 + x_2 + dots + x_{100} = 1. ] 2. We will place these numbers on the circumference of a circle in different permutations. The total number of distinct permutations of these 100 numbers is ( 99! ), because the circle is rotationally invariant. 3. Consider the sums of products of each pair of adjacent numbers when arranged on this circle. Denote these sums for different permutations as ( S_1, S_2, dots, S_{99!} ). 4. Let's compute the total sum of all ( S_i )'s: [ M = S_1 + S_2 + cdots + S_{99!}. ] 5. For a specific permutation, the sum ( S_i ) can be expressed as: [ S_i = x_{1} x_{2} + x_{2} x_{3} + cdots + x_{99} x_{100} + x_{100} x_1. ] 6. Each product ( x_i x_j ) where ( 1 leq i < j leq 100 ) appears exactly ( 2 cdot 98! ) times in the total sum ( M ). This is because for each such pair in one arrangement, there are exactly ( 98! ) permutations of the remaining numbers: [ sum_{i=1}^{100} x_i x_{i+1} text{ (considering } x_{101} text{ as } x_1). ] 7. Therefore: [ M = 2 cdot 98! sum_{1 leqslant i < j leqslant 100} x_i x_j. ] 8. Using the equation for the sum of all pairs ( sum_{1 leqslant i < j leqslant 100} x_i x_j ): [ sum_{1 leqslant i < j leqslant 100} x_i x_j = frac{1}{2} left[ left( sum_{i=1}^{100} x_i right)^2 - sum_{i=1}^{100} x_i^2 right]. ] 9. Substitute back into ( M ): [ M = 2 cdot 98! cdot frac{1}{2} left[ left( sum_{i=1}^{100} x_i right)^2 - sum_{i=1}^{100} x_i^2 right] = 98! left[ left( sum_{i=1}^{100} x_i right)^2 - sum_{i=1}^{100} x_i^2 right]. ] 10. Since ( sum_{i=1}^{100} x_i = 1 ): [ M = 98! left[ 1 - sum_{i=1}^{100} x_i^2 right]. ] 11. Using the inequality ( sum_{i=1}^{100} x_i^2 geq frac{left( sum_{i=1}^{100} x_i right)^2}{100} ) which follows from the Cauchy-Schwarz inequality: [ sum_{i=1}^{100} x_i^2 geq frac{1^2}{100} = frac{1}{100}. ] 12. Substitute this into our equation for ( M ): [ M leq 98! left[ 1 - frac{1}{100} right] = 98! cdot frac{99}{100} = frac{99!}{100}. ] 13. By the pigeonhole principle, there exists at least one ( S_i ) such that: [ S_i leq frac{M}{99!} = frac{1}{100} = 0.01. ] # Conclusion [ boxed{S_i leq 0.01} ]

answer:Since a_n, S_n, a_n^2 form an arithmetic sequence, and a_n > 0, we have 2S_n = a_n + a_n^2. When n=1, we get 2a_1 = a_1 + a_1^2, thus a_1 = 1. When n=2, 2(a_2 + 1) = a_2 + a_2^2, thus a_2 = 2. Therefore, the common difference d = a_2 - a_1 = 1. By the formula of the general term of an arithmetic sequence, we have a_{2009} = a_1 + 2008d = 1 + 2008 = 2009. Hence, the correct choice is: boxed{D}.