answer:1. **Understanding the structure of the icosahedron**: An icosahedron is a regular polyhedron with twenty faces. Each face is an equilateral triangle. 2. **Determining rotation symmetry**: We are interested in finding the smallest positive angle (theta) for a rotation around an axis that passes through two opposite vertices, such that the icosahedron occupies the same region of space. 3. **Calculating vertex angles and face arrangements**: - Each vertex in the icosahedron meets five equilateral triangles. - The measure of each angle of an equilateral triangle is (60^circ). - Since five triangles meet at each vertex, the total angle around each vertex is (5 times 60^circ = 300^circ). 4. **Symmetry and rotational order**: - Considering the vertices through which the axis passes, the rotational symmetry of the icosahedron involves rotating it around the axis so that it looks the same. - For such a rotation, the minimum angle of rotation (theta) must correspond to going through one subgroup of vertices cyclically. - Given each vertex has five connecting points, rotating by ( frac{360^circ}{5}) will map the icosahedron onto itself exactly. 5. **Calculating the smallest positive value of (theta)**: [ theta = frac{360^circ}{5} = 72^circ. ] # Conclusion [ boxed{72^circ} ]

answer:1. **Determine the inradius of the equilateral triangle:** Let the side length of the equilateral triangle be ( s ). The area ( A ) of the equilateral triangle is given by: [ A = frac{s^2 sqrt{3}}{4} ] The semiperimeter ( frac{3s}{2} ) and the inradius ( r ) are related by: [ A = frac{1}{2} times text{semiperimeter} times text{inradius} Rightarrow frac{s^2 sqrt{3}}{4} = frac{3s}{2} times r ] Solving for ( r ): [ r = frac{s sqrt{3}}{6} ] 2. **Determine the radius of the smaller inscribed circles:** The radius of the next smaller circle, which touches the previous circle and has two edges of the triangle as tangents, is scaled down by a factor of 3. Therefore, the radius of the smaller circle is: [ r_1 = frac{r}{3} = frac{s sqrt{3}}{18} ] 3. **Calculate the ratio of the areas of the circles:** The area of a circle is proportional to the square of its radius. Therefore, the ratio of the areas of the largest circle to the next smaller circle is: [ left( frac{r}{r_1} right)^2 = left( frac{frac{s sqrt{3}}{6}}{frac{s sqrt{3}}{18}} right)^2 = 3^2 = 9 ] 4. **Sum the areas of all the smaller circles:** The sequence of smaller circles extends infinitely in 3 different directions. Let ( A ) be the area of the largest circle and ( X ) be the combined area of all the smaller circles. The area of the largest circle is: [ A = pi r^2 = pi left( frac{s sqrt{3}}{6} right)^2 = frac{pi s^2}{12} ] The area of the first smaller circle is: [ A_1 = pi r_1^2 = pi left( frac{s sqrt{3}}{18} right)^2 = frac{pi s^2}{108} ] The total area of the smaller circles in one direction is a geometric series: [ sum_{k=1}^{infty} left( frac{1}{9} right)^k cdot A_1 = A_1 sum_{k=1}^{infty} left( frac{1}{9} right)^k = frac{A_1}{8} = frac{frac{pi s^2}{108}}{8} = frac{pi s^2}{864} ] Since there are 3 such sequences: [ X = 3 cdot frac{pi s^2}{864} = frac{pi s^2}{288} ] 5. **Calculate the ratio of the area of the largest circle to the combined area of all the smaller circles:** [ frac{A}{X} = frac{frac{pi s^2}{12}}{frac{pi s^2}{288}} = frac{288}{12} = 24 ] However, this result seems incorrect. Let's re-evaluate the geometric series sum correctly: [ X = 3 cdot sum_{k=1}^{infty} frac{1}{12} pi left( frac{1}{9} right)^k = 3 cdot frac{1}{12} pi cdot frac{frac{1}{9}}{1 - frac{1}{9}} = 3 cdot frac{1}{12} pi cdot frac{1}{8} = frac{1}{32} pi ] The area of the largest circle is: [ A = frac{1}{12} pi ] Therefore, the ratio is: [ frac{A}{X} = frac{frac{1}{12} pi}{frac{1}{32} pi} = frac{32}{12} = frac{8}{3} ] The final answer is (boxed{frac{8}{3}})

answer:**Answer:** First, from the function s(t) = 2t^3 - 5t^2 + 2, we find the derivative s'(t) = 6t^2 - 10t, and then calculate s'(2). Given the function s(t) = 2t^3 - 5t^2 + 2, s'(t) = 6t^2 - 10t, Therefore, s'(2) = 6 times 2^2 - 10 times 2 = 4. Hence, the correct option is boxed{A}.

answer:(1) Since ain (frac{pi }{2},pi ) and sin alpha =frac{1}{3}, we can find the cosine of alpha using the Pythagorean identity: cos alpha =-sqrt{1-{{sin }^{2}}alpha }=-frac{2sqrt{2}}{3}. Now, we can find sin 2alpha using the double angle formula: sin 2alpha =2sin alpha cos alpha =2cdot frac{1}{3}cdot (-frac{2sqrt{2}}{3})=boxed{-frac{4sqrt{2}}{9}}. (2) Given that sin (alpha +beta )=-frac{3}{5}, beta in (0,frac{pi }{2}), and ain (frac{pi }{2},pi ), we know that alpha +beta in (frac{pi }{2},frac{3pi }{2}). So, cos (alpha +beta )=-sqrt{1-{{sin }^{2}}(alpha +beta )}=-frac{4}{5}. Now, we can find sin beta using the sine difference formula: sin beta =sin [(alpha +beta )-alpha ]=sin (alpha +beta )cos alpha -cos (alpha +beta )sin alpha =-frac{3}{5}cdot (-frac{2sqrt{2}}{3})-(-frac{4}{5})cdot frac{1}{3}=boxed{frac{6sqrt{2}+4}{15}}.