answer:1. **Establish Cone Dimensions:** Define key dimensions and parameters of the cone. Let the radius of the base of the cone be ( R ) and the height of the cone be ( m ). Define (alpha) as the angle between the slant height of the cone and the radius of the base of the cone. 2. **Relate ( R ) and ( m ) to ( r ):** The coordinates of the sphere in the cone dictate that: [ R = frac{r}{tan frac{alpha}{2}} ] [ m = R tan alpha = frac{r tan alpha}{tan frac{alpha}{2}} ] 3. **Cone Volume Formula:** The volume ( V ) of the cone is given by: [ V = frac{1}{3} pi R^2 m ] 4. **Substitute ( R ) and ( m ):** Substitute the expressions for ( R ) and ( m ) into the volume formula: [ V = frac{1}{3} pi left( frac{r}{tan frac{alpha}{2}} right)^2 left( frac{r tan alpha}{tan frac{alpha}{2}} right) ] 5. **Simplify Volume Expression:** Simplify the volume expression: [ V = frac{1}{3} pi left( frac{r^2}{tan^2 frac{alpha}{2}} right) left( frac{r tan alpha}{tan frac{alpha}{2}} right) ] [ V = frac{1}{3} pi frac{r^3 tan alpha}{tan^3 frac{alpha}{2}} ] 6. **Reformulate as a Function of ( beta ):** Let ( beta = frac{alpha}{2} ), then ( alpha = 2 beta ): [ V = frac{r^3 pi}{3} cdot frac{tan 2 beta}{tan^3 beta} ] Recall that: [ tan 2 beta = frac{2 tan beta}{1 - tan^2 beta} ] Substitute this into the volume expression: [ V = frac{r^3 pi}{3} cdot frac{2 tan beta}{tan^3 beta (1 - tan^2 beta)} ] [ V = frac{2 r^3 pi}{3} cdot frac{1}{tan^2 beta (1 - tan^2 beta)} ] 7. **Introduce New Variable for Simplification:** Let ( x = tan^2 beta ): [ V = frac{2 r^3 pi}{3} cdot frac{1}{x (1 - x)} ] 8. **Optimize ( x ):** To minimize ( V ), find ( x ): [ text{Define } y = frac{1}{x (1-x)} ] The function to minimize: [ y = x (1 - x) Rightarrow text{Solve} y' = 0: ] [ y' = 1 - 2x Rightarrow 1 - 2x = 0 Rightarrow x = frac{1}{2} ] 9. **Minimum Volume Calculation:** Plug ( x = frac{1}{2} ) into volume equation: [ V = frac{2 r^3 pi}{3} cdot frac{1}{frac{1}{2} left(1-frac{1}{2}right)} ] [ V = frac{2 r^3 pi}{3} cdot frac{1}{frac{1}{2} cdot frac{1}{2}} = frac{2 r^3 pi}{3} cdot 4 = frac{8 r^3 pi}{3} ] 10. **Conclusion:** Therefore, the minimal volume of the cone is: [ boxed{frac{8 r^3 pi}{3}} ]

answer:Given a cube (ABCD - A_1B_1C_1D_1) with edge length 2. The cube is situated in a right-angle coordinate system (O-xyz), where the vertex (A) remains on the (z)-axis, and the vertex (B_1) stays on the (xOy) plane. We need to determine the minimum length of (OC). Let us consider the position of the vertices: 1. (A) is at ((0, 0, z)). 2. (A_1) will be directly below (A) at ((0, 0, z-2)) since the edge length of the cube is 2. 3. (B_1) lies in the plane (xOy) so its coordinates are ((x_1, y_1, 0)). 4. (B) directly above (B_1) is at ((x_1, y_1, z-2)). 5. (C) is diagonally across from (A) on the plane. Because (ABC_1) forms a right triangle (considering the cube's edges), we have: - (AB = 2) - (AC = 2sqrt{2}) To find (OC), consider the right triangle (AB_1C): 1. As triangle ( triangle AB_1C ) is a right triangle, and being a square's face diagonal, (AB_1C) will intersect diagonally through the face. Since the side length of the square face is 2, the diagonal (AC) will be: [ AC = 2sqrt{2} ] Therefore, the vertical distance (C) from point (A) to the line perpendicular from the plane (xOy): 2. (CM) (distance (C) to the plane parallel to (xOy)): [ CM = sqrt{(2 sin(45^circ))^2 + (2 cos(45^circ))^2} = sqrt{2 + 2} = sqrt{4} = 2 ] Since ( C ) is at the endpoint of the cube edge above origin (O), we need to calculate: [ OM = sqrt{Otext{A}^frac{2}{ } + Otext{B}_1 ^frac{2}{ }} = sqrt{2} ] Taking endpoints (C), (O), and (M) it corresponds to the diagonal distance to sides of additional cubes which remains perpendicular: Thus vertical height: (OC) minimum: [ boxed{sqrt{6} - sqrt{2}} ]

answer:To form an eight-digit palindrome, the number is represented as ABCCBA, where A, B, C, and D represent digits. The eight-digit format imposes that the number looks like ABCDDCBA where the digits D and C are mirrored in the middle and similarly A and B on the ends. - For A, we can choose 1, 2, or 3 (3 options) because it cannot be 0 (as we are forming an eight-digit number). - For B, C, and D, each can also be either 1, 2, or 3, allowing for 3 options each. Hence, the total number of eight-digit palindromes that can be formed using the digits 1, 2, or 3 for each place is 3 times 3 times 3 times 3 = 3^4 = 81. boxed{81}

answer:1. **Expression for g(n+2)**: [ g(n+2) = frac{3 + 2sqrt{3}}{6} left(frac{3+sqrt{3}}{6}right)^{n+2} + frac{3 - 2sqrt{3}}{6} left(frac{3-sqrt{3}}{6}right)^{n+2} ] Simplifying further using the identity left(frac{3+sqrt{3}}{6}right)^{n+2} = left(frac{3+sqrt{3}}{6}right)^n left(frac{3+sqrt{3}}{6}right)^2: [ g(n+2) = frac{3 + 2sqrt{3}}{6} left(frac{3+sqrt{3}}{6}right)^2 left(frac{3+sqrt{3}}{6}right)^n + frac{3 - 2sqrt{3}}{6} left(frac{3-sqrt{3}}{6}right)^2 left(frac{3-sqrt{3}}{6}right)^n ] Calculating the squares of the ratios: [ left(frac{3+sqrt{3}}{6}right)^2 = frac{3+2sqrt{3}}{4}, quad left(frac{3-sqrt{3}}{6}right)^2 = frac{3-2sqrt{3}}{4} ] Plugging these back into g(n+2): [ g(n+2) = frac{3 + 2sqrt{3}}{6} cdot frac{3+2sqrt{3}}{4} cdot g(n) + frac{3 - 2sqrt{3}}{6} cdot frac{3-2sqrt{3}}{4} cdot g(n) ] 2. **Computing g(n+2) - g(n)**: [ g(n+2) - g(n) = left(frac{(3+2sqrt{3})^2}{24} + frac{(3-2sqrt{3})^2}{24}right) g(n) - g(n) ] Simplifying the coefficients: [ frac{(3+2sqrt{3})^2 + (3-2sqrt{3})^2}{24} = frac{18 + 12}{24} = frac{30}{24} = frac{5}{4} ] [ g(n+2) - g(n) = left(frac{5}{4} - 1right) g(n) = frac{1}{4} g(n) ] 3. **Conclusion**: [ frac{1{4} g(n)} ] The final answer is boxed{A}