answer:1. Start with the tangent-segment theorem which states that PA=AB=PA'=A'B'=3. 2. Drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let the center of the smaller circle be S and the center of the larger circle be L. If the radius of the smaller circle is y, then the radius of the larger circle is 2y based on the problem statement. 3. The segment from P to the outer circle's tangent point and back to P' will be 3 + 3 = 6. The line segment connecting the centers S and L will be the sum of the radii, y + 2y = 3y. 4. The distance from P to the line segment SL is the radius of the larger circle, which is 2y. Thus, using the Pythagorean theorem on the triangle formed between P, S, and the tangent points, we have y^2 + 3^2 = (2y)^2 leading to y^2 + 9 = 4y^2. 5. Solving for y, we find 4y^2 - y^2 = 9 Rightarrow 3y^2 = 9 Rightarrow y^2 = 3 Rightarrow y = sqrt{3}. 6. Therefore, the area of the smaller circle is pi y^2 = pi (sqrt{3})^2 = 3pi. boxed{3pi}

answer:(1) **Analysis** Using the formula for the area of a triangle, we get ab=4. Then, applying the cosine rule, we get 2^2=a^2+b^2−2abcos C. **Solution** By the given, the area of ΔABC, S= frac{1}{2}absin C= frac{sqrt{3}}{4}ab= sqrt{3}, we get ab=4. Hence, 4=a^2+b^2−ab=(a+b)^2−3ab, so (a+b)^2=4+3ab=16, thus a+b=4. Solving the system of equations begin{cases}ab=4 a+b=4end{cases}, we find the values of a and b. Using the cosine rule, we get 2^2=a^2+b^2−2abcos C, which means 4=a^2+b^2−ab=(a+b)^2−3ab, so (a+b)^2=4+3ab=16, hence a+b=4, solving the system of equations begin{cases}ab=4 a+b=4end{cases}, we find a=2,b=2, thus, the perimeter of ΔABC is 6. Therefore, the answer is boxed{6}. (2) **Analysis** Based on the line l: frac{x}{a} + frac{y}{b} = 1 (a > 0, b > 0) passing through the point (3, 2), we derive frac{3}{a} + frac{2}{b} = 1, then apply the AM-GM inequality. **Solution** Since the line l: frac{x}{a} + frac{y}{b} = 1 (a > 0, b > 0) passes through the point (3, 2), we have frac{3}{a} + frac{2}{b} = 1, thus a+b=(a+b)(frac{3}{a} + frac{2}{b})=3+ frac{3b}{a}+ frac{2a}{b}+2geq5+2sqrt{6}, equality holds when frac{3b}{a}= frac{2a}{b}, i.e., sqrt{2}a= sqrt{3}b=2sqrt{3}+3sqrt{2}, therefore, the minimum value of a+b is 5+2sqrt{6}. Hence, the answer is boxed{5+2sqrt{6}}. (3) **Analysis** According to the problem, the area of the region where the distance to all three vertices is less than 2 is exactly the area of a semicircle with radius 2, which is 2pi. Thus, the area of the region where the distance to all three vertices is greater than 2 is 30-2pi. The probability can be found using the formula for geometric probability. **Solution** From the problem, the area of the region where the distance to all three vertices is less than 2 is exactly the area of a semicircle with radius 2, which is 2pi, thus, the area of the region where the distance to all three vertices is greater than 2 is 30-2pi, using the formula for geometric probability, the probability that it lands in the region where the distance to all three vertices is greater than 2 is frac{30-2pi}{30}= frac{15-pi}{15}. Therefore, the answer is boxed{frac{15-pi}{15}}. (4) **Analysis** By analyzing the given conditions and the relationship between the sum of the first n terms of an arithmetic sequence and its general term, we can examine the truth of the given statements. **Solution** From the given, we have a_7=S_7-S_6 < 0, a_6=S_6-S_5 > 0, thus d=a_7-a_6 < 0, statement ① is correct; S_{11}= frac{a_1+a_{11}}{2}times11= frac{11}{2}times2a_6=11a_6 > 0, statement ② is correct; S_{12}= frac{a_1+a_{12}}{2}times11= frac{11}{2}times(a_6+a_7), and a_6+a_7=S_7-S_5 > 0, thus S_{12} > 0, statement ③ is incorrect; The sequence is increasing, and since a_6 > 0, a_7 < 0, it follows that the maximum term in the sequence {S_n} is S_6, statement ④ is incorrect. In conclusion, the correct statement numbers are ①②. Therefore, the answer is boxed{①②}.

answer:When a=0, the function f(x) = ax^3 + x + 1 = x + 1 is a monotonically increasing function without any extreme value, thus options B and D are eliminated. When a > 0, the function f(x) = ax^3 + x + 1 is a monotonically increasing function without any extreme value, thus option A is eliminated. Therefore, the correct answer is boxed{text{C}}.

answer:Solution: (1) Let the common difference of the arithmetic sequence {a_n} be d, since a_5=8, a_7=12, we have begin{cases} a_1+4d=8 a_1+6d=12 end{cases}, solving this gives a_1=0, d=2. Therefore, the general formula for the sequence {a_n} is a_n=2(n-1)=2n-2. (2) Let the common ratio of {b_n} be q(q > 0). Since a_n=2n-2, we have b_3=a_3=4, thus T_2= dfrac{b_3}{q^2}+ dfrac{b_3}{q}= dfrac{4}{q^2}+ dfrac{4}{q}=3, solving this gives q=2 or q=- dfrac{2}{3} (discard this solution), therefore b_1=1, T_n= dfrac{1-2^n}{1-2}=2^n-1. Thus, the answers are: (1) The general formula for the sequence {a_n} is boxed{a_n=2n-2}. (2) The sum of the first n terms of the geometric sequence is boxed{T_n=2^n-1}.