answer:1. **Consider Triangle (MOC)**: Let’s examine triangle (MOC) where ( M) is the point where the angle bisector of (angle C) intersects the circumcircle of (triangle ABC) again (other than (C)), ( O) is the circumcenter, and (C) is the vertex of the triangle. 2. **Apply Stewart's Theorem**: Stewart's theorem for the triangle (MOC) with cevian (IC) (where (I) is the incenter) states that: [ CI^2 cdot MO + IO^2 cdot CM = CO^2 cdot IM + MO cdot CM cdot CO ] In this context, we simplify it to show the relationship between the segments on (CM) created by point (I). Suppose (CI = d) (distance from vertex to incenter), and denote the segments (x = IM), and (y = IC - IM). 3. **Use for (IC)**: For the sake of convenient calculation later, name the segments that (I) splits (CM) into (CI = a) and (IM = b), therefore, segment (CM) can be written in the form (CM = a + b). 4. **Noting the Lengths**: By the triangle property, we know that: [ CI cdot CM = CI cdot (IM + IC) = IO cdot IM ] 5. **Euler's Theorem Application**: Euler's theorem states the relation between the inradius (r) and circumradius (R) of a triangle, which is written as: [ d^2 = R(R - 2r) ] where (d) is the distance between the circumcenter (O) and the incenter (I). 6. **Connect the Important Identities**: Given that Euler's result (d^2 = R^2 - 2Rr) can be written in terms of the bisected segments: [ IO^2 = R^2 - 2Rr ] 7. **Final Calculation**: Given (c = CI), (m=CM), and letting (I) be the point dividing (CM) at lengths (a) and (b), with the known property (a cdot b = 2Rr), keeping (CI = a + b). Therefore: [ CM = a + b ] Under the angle bisector theorem and Stewart's theorem appropriately, (I) divides (CM) such that: [ a cdot b = 2Rr ] 8. **Conclusion**: The incenter (I) bisects the segment (CM) in such a way that the product of the segments ( a) and (b) formed is equal to the double product of the radius of the inscribed circle and the circumscribed circle. [ boxed{a cdot b = 2Rr} ]

answer:From the given information, we have the inequality f(log_4 x) > 2, which can be written as f(log_4 x) > f(frac{1}{2}). Since the even function f(x) is decreasing on (-infty, 0]: 1. f(x) is increasing on [0, +infty). 2. log_4 x > frac{1}{2} = log_4 2 or log_4 x < -frac{1}{2} = log_4 frac{1}{sqrt{2}}. Applying the properties of logarithmic functions, we find that either 0 < x < frac{1}{2} or x > 2. Therefore, the solution set for the inequality f(log_4 x) > 2 is boxed{(0, frac{1}{2}) cup (2, +infty)}.

answer:To solve the problem, let's analyze the transformation step by step: 1. The original function is y=3sin 2x. This function has a sine wave pattern with an amplitude of 3 and a period of pi (since the period of sin x is 2pi, and multiplying x by 2 halves the period to pi). 2. The goal is to transform this function into y=3sin left(2x-dfrac{pi }{5}right). This transformation involves a phase shift, which is determined by the argument of the sine function. 3. The phase shift can be calculated by setting the inside of the sine function equal to zero: 2x - dfrac{pi}{5} = 0. Solving for x gives x = dfrac{pi}{10}. This means the graph of y=3sin 2x needs to be shifted to the right by dfrac{pi}{10} units to match the graph of y=3sin left(2x-dfrac{pi }{5}right). 4. Therefore, the transformation from y=3sin 2x to y=3sin left(2x-dfrac{pi }{5}right) is achieved by shifting the graph to the right by dfrac{pi}{10} units. This is equivalent to saying y=3sin left[2left(x-dfrac{pi}{10}right)right]=3sin left(2x-dfrac{pi}{5}right). Thus, the correct answer is boxed{C}.

answer:Given x_1+x_2+…+x_n=-5, we can assume there are a numbers of -2, so there are 2a-5 numbers of 1. Since x_1^2+x_2^2+…+x_n^2=19, we have (-2)^2a+(2a-5)×1^2=6a-5=19. Solving this, we get a=4, and 2a-5=3. Therefore, x_1^5+x_2^5+…+x_n^5=4×(-2)^5+3×1^5=4×(-32)+3=-128+3=-125. Hence, the answer is boxed{-125}.