answer:Solution: (Ⅰ) Let the medians on sides AC and AB be CD and BE, respectively. Since BG = frac{2}{3}BE, CG = frac{2}{3}CD, we have BG + CG = frac{2}{3}(BE + CD) = 6 (a constant value). Therefore, the trajectory of G is an ellipse with foci at B and C, where 2a = 6 and c = sqrt{5}. Thus, a = 3, b = 2, and the equation of the ellipse is frac{x^2}{9} + frac{y^2}{4} = 1. Since when point G is on the x-axis, points A, B, and C are collinear, which cannot form triangle ABC, the y-coordinate of G cannot be 0. Therefore, the equation of the trajectory of the centroid G of triangle ABC is boxed{frac{x^2}{9} + frac{y^2}{4} = 1 (y neq 0)}. (Ⅱ) According to the problem, when P is at the vertex of the short axis of the ellipse, angle BPC is maximized, and cosangle BPC is minimized. Therefore, cosangle BPC = frac{3^2 + 3^2 - (2sqrt{5})^2}{2 times 3 times 3} = boxed{-frac{1}{9}}.

answer:1. **Proof that towers taller than 2 are pointless to construct**: - Assume there is a tower with height h > 2. - If we remove the topmost floor from this tower and create a new tower of height 1, the scene changes as described: - The shortened tower h becomes h-1 in height, and this new 1-floor tower emerges. - Originally, towers of height h would obscure towers of height h-1 completely from the inspector's view. - Now, since h > 2, there will be a change where the new tower can be seen from all towers including the shortened one. - This results in an increase in the visibility of the new 1-floor tower, thereby increasing the cumulative visibility significantly. - From this change, we conclude that having towers taller than 2 would always be less effective compared to maintaining them at height 1 or 2. 2. **Case Study: All towers are either of height 1 or 2**: - Let there be X towers of height 1 and Y towers of height 2. - According to the problem's constraint, the total height of all towers combined must sum to 30: [ X + 2Y = 30 ] - The total visibility score for the inspector can be expressed as the product of the number of height 1 towers and height 2 towers: [ XY = (30 - 2Y)Y ] - Simplify the expression: [ XY = (30 - 2Y)Y = 2Y(15 - Y) ] - This expression represents a symmetric parabola which reaches its maximum when (Y = 7.5). - For integer values of (Y), the values that maximize the expression are (Y = 7) and (Y = 8): * For (Y = 7): [ X = 30 - 2(7) = 30 - 14 = 16 ] [ XY = 16 times 7 = 112 ] * For (Y = 8): [ X = 30 - 2(8) = 30 - 16 = 14 ] [ XY = 14 times 8 = 112 ] 3. **Conclusion**: - The maximum visibility sum for the inspector is achieved either with (16) towers of height (1) and (7) towers of height (2), or with (14) towers of height (1) and (8) towers of height (2). [ boxed{text{Nужно построить либо 16 одноэтажных и 7 двухэтажных, либо 14 одноэтажных и 8 двухэтажных башен}} ]

answer:To solve for the value range of arg (z - b + b mathrm{i}), given that z satisfies the conditions arg (z + a + a mathrm{i}) = frac{pi}{4} and arg (z - a - a mathrm{i}) = frac{5pi}{4} where a in mathbf{R}^{+} and a > b > 0, we can proceed as follows: 1. Let's start by analyzing the condition arg (z + a + a mathrm{i}) = frac{pi}{4}. This means that the complex number z + a + a mathrm{i} lies on a line that makes an angle of frac{pi}{4} with the positive real axis. 2. Similarly, the condition arg (z - a - a mathrm{i}) = frac{5pi}{4} means that the complex number z - a - a mathrm{i} lies on a line that makes an angle of frac{5pi}{4} with the positive real axis. 3. Geometrically, these two conditions show that the point z must lie on the line segment AB where: - Point ( A ) is located at ( (a, a) ). This is derived from adding (a) and (a mathrm{i} ) components. - Point ( B ) is located at ( (-a, -a) ). This is derived from subtracting (a) and (a mathrm{i} ) components. 4. To find the argument arg (z - b + b mathrm{i}), we consider the line segment CP where: - Point ( C ) is located at ( (b, -b) ), which are the real and imaginary components given in the problem. - Point ( P ) is any point on line segment AB. 5. The slope of line segment AC is given by: [ k_{AC} = frac{a - (-b)}{a - b} = frac{a + b}{a - b} ] 6. The slope of line segment BC is given by: [ k_{BC} = frac{(-a) - (-b)}{(-a) - b} = frac{(-a + b)}{-a - b} = frac{a - b}{a + b} ] 7. The argument arg (z - b + b mathrm{i}) can now be viewed as the angle formed between the positive x-axis and the line segment CP. Consequently, the lower bound of the range is the arctangent of the slope (k_{AC}): [ arg (z - b + b mathrm{i}) bigg|_{text{lower}} = arctan left( frac{a + b}{a - b} right) ] 8. The upper bound of the range is the angle pi added to the arctangent of the slope (k_{BC}): [ arg (z - b + b mathrm{i}) bigg|_{text{upper}} = pi + arctan left( frac{a - b}{a + b} right) ] 9. Therefore, combining these bounds, the range of arg (z - b + b mathrm{i}) is: [ boxed{left[ arctan left( frac{a + b}{a - b} right), pi + arctan left( frac{a - b}{a + b} right) right]} ] This completes the solution.

answer:First, we need to find the modular inverse of 5 modulo 26. We look for an integer m such that (5 cdot m) equiv 1 pmod{26}. Testing several values, we find that 5 cdot 21 = 105, and 105 equiv 1 pmod{26}, so the inverse of 5 modulo 26 is 21. Now, we solve the congruence: [ 5x equiv 14 pmod{26} ] [ x equiv 14 cdot 21 pmod{26} ] [ x equiv 294 pmod{26} ] Simplifying 294 mod 26 gives 294 = 11 cdot 26 + 12, so 294 equiv 12 pmod{26}. Thus, the smallest positive integer that satisfies the given congruence is boxed{12}.