answer:1. **Define the problem and the parameters**: Consider a circle in which n sectors are colored. Each sector has an angular measure of less than frac{pi}{n^2 - n + 1}. 2. **Focus on the range of rotation angles**: We will analyze the possible rotation angles of the circle, which range from 0 to 2pi. 3. **Consider pairs of sectors**: Let us study what happens when we choose any two sectors. Suppose one sector has angular measure alpha and another has angular measure beta. 4. **Understand the effect of rotation on sector overlap**: If we rotate the circle, the image of the first sector will intersect with the second sector if the rotation angle falls within a segment of length alpha + beta on the unit circle. 5. **Use the given condition on the angle size**: Since each sector's angular measure is less than frac{pi}{n^2 - n + 1}, we have: [ alpha + beta leq frac{2pi}{n^2 - n + 1} ] 6. **Analyze all possible pairs of sectors**: There are binom{n}{2} ways to choose 2 sectors out of the n sectors. This gives us n(n - 1) pairs of sectors. 7. **Calculate total length of rotation intervals**: The sum of all intervals alpha + beta corresponding to these pairs would be: [ n(n - 1) cdot frac{2pi}{n^2 - n + 1} ] 8. **Simplify the expression**: [ n(n - 1) cdot frac{2pi}{n^2 - n + 1} = frac{2pi n(n - 1)}{n^2 - n + 1} ] 9. **Compare total intervals with full rotation**: Notice that: [ frac{2pi n(n - 1)}{n^2 - n + 1} < 2pi ] 10. **Conclude the existence of a valid rotation angle**: Since the total length of these rotation intervals is less than 2pi, there must exist at least one rotation angle for which none of the sectors overlap after the rotation. Thus, it is possible to rotate the circle such that all the colored sectors move into uncolored parts of the circle. blacksquare

answer:**Analysis** This problem mainly examines the properties of the sine function and the simplification and evaluation of trigonometric functions. **Solution** Since T=pi, then omega=2, so f(x)=sin (2x+varphi), thus g(x)=sin (2x+varphi)+1, hence the function g(x)=sin (2x+varphi)+1 < 1, implies sin (2x+varphi) < 0, thus 2kpi-pi < 2x+varphi < 2kpi, kin mathbb{Z}; Given xin(-frac{pi}{3},-frac{pi}{12}), thus 2xin(-frac{2pi}{3},-frac{pi}{6}), 2x+varphiin(-frac{2pi}{3}+varphi,-frac{pi}{6}+varphi), therefore begin{cases}-frac{2pi}{3}+varphigeqslant -pi -frac{pi}{6}+varphileqslant 0end{cases}, hence -frac{pi}{3}leqslant varphileqslant frac{pi}{6}, Given varphiin[-frac{pi}{2},0], thus varphiin[-frac{pi}{3},0], therefore, when varphi takes the minimum value -frac{pi}{3}, the value of g(frac{pi}{4}) is frac{3}{2}. Hence, the correct option is boxed{C}.

answer:First, calculate angle C using the fact that the sum of angles in a triangle is 180^circ: [ angle C = 180^circ - 45^circ - 60^circ = 75^circ. ] Then, applying the Law of Sines, we have: [ frac{BC}{sin A} = frac{AC}{sin B}. ] Substituting in the known values and solving for BC: [ BC = AC cdot frac{sin A}{sin B} = 6 cdot frac{sin 45^circ}{sin 60^circ}. ] We utilize known sine values: [ sin 45^circ = frac{sqrt{2}}{2}, quad sin 60^circ = frac{sqrt{3}}{2}. ] Thus, [ BC = 6 cdot frac{frac{sqrt{2}}{2}}{frac{sqrt{3}}{2}} = 6 cdot frac{sqrt{2}}{sqrt{3}} = 6 cdot frac{sqrt{6}}{3} = 2sqrt{6}. ] Thus, the length of BC is boxed{2sqrt{6}}.

answer:# Step-by-Step Solution Part 1: Cost Price of Jerseys Let's denote the cost price of one B jersey as x yuan. Therefore, the cost price of one A jersey becomes (x + 20) yuan. According to the given information, the number of A jerseys that can be purchased for 30,000 yuan is three times the number of B jerseys that can be purchased for 9,000 yuan. This can be mathematically represented as: [ frac{30000}{x+20} = 3 times frac{9000}{x} ] Solving this equation for x: [ frac{30000}{x+20} = frac{27000}{x} ] Cross-multiplying gives: [ 30000x = 27000(x + 20) ] Expanding and simplifying: [ 30000x = 27000x + 540000 ] [ 3000x = 540000 ] [ x = 180 ] Therefore, the cost price of one B jersey is 180 yuan, and the cost price of one A jersey is 180 + 20 = 200 yuan. Final Answer for Part 1: [ boxed{text{The cost price of one A jersey is 200 yuan, and the cost price of one B jersey is 180 yuan.}} ] Part 2: Function between Profit W and m Given the constraints, the number of A jerseys, denoted as m, must satisfy: [ left{begin{array}{c} m leq 2 times (210 - m) m geq 100 end{array}right. ] Solving the inequality: [ m leq 420 - 2m ] [ 3m leq 420 ] [ m leq 140 ] Therefore, 100 leq m leq 140. The profit W from selling these goods is given by: [ W = (320 - 200)m + (280 - 180)(210 - m) ] Simplifying: [ W = 120m + 100(210 - m) ] [ W = 120m + 21000 - 100m ] [ W = 20m + 21000 ] Final Answer for Part 2: [ boxed{W = 20m + 21000 text{ for } 100 leq m leq 140} ] Part 3: Maximum Profit after Supporting Children Let Q represent the profit after supporting children in impoverished mountainous areas: [ Q = (20 - a)m + 21000 ] When 0 < a < 20, Q increases with m, thus at m = 140, Q is maximized: [ Q_{text{max}} = (20 - a)140 + 21000 = 23800 - 140a ] When a = 20, Q is constant regardless of m, thus: [ Q = 21000 ] When a > 20, Q decreases with m, thus at m = 100, Q is maximized: [ Q_{text{max}} = (20 - a)100 + 21000 = 23000 - 100a ] Final Answer for Part 3: [ boxed{begin{array}{l} text{When } 0 < a < 20, text{ the maximum profit is } (23800 - 140a) text{ yuan.} text{When } a = 20, text{ the maximum profit is } 21000 text{ yuan.} text{When } a > 20, text{ the maximum profit is } (23000 - 100a) text{ yuan.} end{array}} ]