answer:1. **To prove that the Reuleaux shape (R) with a constant width of (a+2r) cannot be rotated inside a regular hexagon frame with side length (frac{a+2r}{sqrt{3}}):** Consider three supporting strips of the Reuleaux shape (R) that are rotated (120^circ) relative to each other. The common area of these strips generally does not match the frame, which is a regular hexagon. To illustrate this, if we choose three strips where each strip touches the boundary arcs of (R) with radii (r) and (a+r) at the midpoints of opposite pairs of these arcs, two different side lengths will arise. - Let (A') and (C'') be the midpoints of the arcs with radii (r) and (a+r) centered at points (A) and (C) respectively. - The intersection of tangents at these points will form a point (H). Exhibiting these strips, the lengths of the common section's sides will alternately be (2A'H) and (2C''H), differing from each other due to symmetry considerations. Symmetrically, the lines (AA') and (CC'') pass through the common center (O) of triangle (ABC): [OHA'] and [OHC''] are right triangles with a common hypotenuse (OH), but distinct legs (A'O) and (C''O): [ begin{aligned} A'O &= A'A + AO = r + frac{a}{sqrt{3}}, C''O &= C''C - CO = a + r - frac{a}{sqrt{3}} = r + left(frac{sqrt{3}-1}{sqrt{3}}right)a. end{aligned} ] Clearly, (A'O > C''O), thus by the Pythagorean theorem, (A'H < C''H). This confirms our claim. 2. **Proving the Reuleaux shape cannot thus be rotated inside a regular triangle-shaped frame:** Similarly, consider points (B'), (C'), (A''), and (B'') as midpoints of the arcs on the boundary. The tangents at these points create support regular triangles whose side lengths differ. Both sets of tangents will coincide with point (O) forming inscribed radii: Given (O A' > O C''), it implies inherent inequality in our tangent lengths. 3. **Conclusion on the impossibility based on symmetry axes:** If rotation were possible, there would exist a position where the symmetry axes of (R) and the frame are pairwise parallel and would coincide, placing point (O) of frame over (R)'s midpoint. But, given in (R) (with (a>0)) (O) does not bisect sections, implying impossibility. 4. **Impossibility based on the contact points argument:** Reuleaux shape (R) is insertable notably in alignment with joining points of arcs with radii (r) and (a+r): Upon rotating, if contact occurs in an inner point of arcs, the intersection of supporting strips yields a nonsymmetric common area. Given (120^circ) symmetry, tangential lengths vary, hence differing from the frame's sides. 5. **Verification by specific computations for enclosing within an (a+2 r) width polygon:** Bollobás Béla provides proof indicating that (R) is non-rotatable within such a framework but possible in a rhombus frame. A Reuleaux shape may fit by aligning with an (a+2 r) hexagon's inscribable circle radius, but this is insufficient since tangents and intersecting lines offer discrepancy points. [boxed{}]

answer:The area of the square is (ZW^2). By applying the Pythagorean theorem to triangle XYZ, we have (XZ^2 = 25 + 4 = 29) square units. By applying the Pythagorean theorem to triangle XZW, we have (ZW^2 = 29 + 49 = 78) square units. Thus, the area of the fourth square is (boxed{78}) square units.

answer:**Analysis** This question tests the definition of an arithmetic sequence, which is determined by whether the difference between the subsequent term and the previous term is a constant. **Solution** Solution: (1) According to the definition of an arithmetic sequence, a_{n+1}+1-(a_n+1)=d, so {a_n+1} is an arithmetic sequence; (2) a_{n+1}^2-a_n^2=d(a_{n+1}+a_n), which is not a constant, hence it is not an arithmetic sequence; (3) a_{n+1}-a_n=d, so {a_{n+1}-a_n} is a constant sequence, thus it is an arithmetic sequence; (4) 2a_{n+1}-2a_n=2d, so {2a_n} is an arithmetic sequence; (5) (a_{n+1}+n+1)-(a_n+n)=d+1 is a constant, hence {a_n+n} is an arithmetic sequence. Therefore, the answer is boxed{D}.

answer:1. Given the conditions: ( n geq 2 ), ( x_1, x_2, cdots, x_n ) are real numbers satisfying: [ x_j > -1 quad text{for all} quad j=1,2,cdots,n ] and [ x_1 + x_2 + cdots + x_n = n ] 2. We are to prove: [ sum_{j=1}^{n} frac{1}{1+x_j} geq sum_{j=1}^{n} frac{x_j}{1+x_j^2} ] 3. Consider the difference between the two sums: [ sum_{j=1}^{n} left( frac{1}{1+x_j} - frac{x_j}{1+x_j^2} right) ] 4. Simplify each term in the sum: [ frac{1}{1+x_j} - frac{x_j}{1+x_j^2} = frac{(1+x_j^2) - x_j(1+x_j)}{(1+x_j)(1+x_j^2)} ] [ = frac{(1 + x_j^2) - x_j - x_j^2}{(1 + x_j)(1 + x_j^2)} ] [ = frac{1 - x_j}{(1 + x_j)(1 + x_j^2)} ] 5. Hence, we have: [ sum_{j=1}^{n} left( frac{1}{1+x_j} - frac{x_j}{1+x_j^2} right) = sum_{j=1}^{n} frac{1 - x_j}{(1 + x_j)(1 + x_j^2)} ] 6. To show (sum_{j=1}^{n} frac{1-x_j}{(1 + x_j)(1 + x_j^2)} geq 0), we need to analyze the function ( f(x) ): [ f(x) = frac{1 - x}{(1 + x)(1 + x^2)} quad text{for} quad x > -1 ] 7. At ( x = 1 ), the function is tangential to the line: [ y = frac{1 - x}{4} ] 8. Proving ( f(x) geq frac{1 - x}{4} ) for ( x > -1 ): - When ( x geq 1 ): [ (1 + x)(1 + x^2) geq 4 ] - When ( -1 < x < 1 ): [ (1 + x)(1 + x^2) leq 4 ] 9. Combine both scenarios to conclude that: [ frac{1 - x}{(1 + x)(1 + x^2)} geq frac{1 - x}{4} ] 10. Therefore: [ sum_{j=1}^{n} frac{1 - x_j}{(1 + x_j)(1 + x_j^2)} geq sum_{j=1}^{n} frac{1 - x_j}{4} ] 11. Since (sum_{j=1}^{n} x_j = n), we have: [ sum_{j=1}^{n} (1 - x_j) = n - sum_{j=1}^{n} x_j = n - n = 0 ] [ sum_{j=1}^{n} frac{1 - x_j}{4} = frac{1}{4} sum_{j=1}^{n} (1 - x_j) = frac{1}{4} cdot 0 = 0 ] 12. Thus: [ sum_{j=1}^{n} frac{1-x_j}{(1 + x_j)(1 + x_j^2)} geq 0 ] Conclusion: [ boxed{sum_{j=1}^{n} frac{1}{1+x_j} geq sum_{j=1}^{n} frac{x_j}{1+x_j^2}} ]