answer:1. **Define the points and transformations**: Let A' be the point corresponding to A obtained by a parallel translation by the vector overrightarrow{MN}. This translation means that A' is directly across the river from A and shifted by the length of the bridge MN. 2. **Express distances**: Since the bridge (MN) is perpendicular to the river's banks and the banks are considered parallel straight lines, the length of the path AMNB can be expressed as: [ AMNB = AM + MN + NB ] Given the transformation, we have: [ AM = A'N ] Therefore, the above length can be re-expressed as: [ AMNB = A'N + NB + MN ] 3. **Minimize the relevant sum**: Notice that the length of the bridge MN is constant, and hence does not affect the optimization. Therefore, to minimize the entire path A M N B, we just need to minimize: [ A'N + NB ] 4. **Determine the optimal position of N**: The sum A'N + NB is minimized when N lies directly on the line segment joining A' and B. This is because the shortest distance between any two points is a straight line. Hence, N should be the point where this line intersects the river bank closest to B. **Conclusion**: To ensure the path AMNB is the shortest, place the bridge MN such that point N is the intersection of the river bank and the straight line segment A'B. [ boxed{{N text{ is at the intersection point of the bank and the segment } A'B}} ]

answer:Let a = b + c, a + b = b + c + b = 2b + c, quad a - b = b + c - b = c. Now, substitute a + b and a - b into the expression: frac{a+b}{a-b} + frac{a-b}{a+b} = frac{2b + c}{c} + frac{c}{2b + c}. Let y = frac{2b+c}{c}. Then, frac{c}{2b+c} = frac{1}{y}, so the expression becomes y + frac{1}{y}. This simplifies to frac{y^2 + 1}{y}, which we recognize as a transformation to minimize. Setting frac{y^2 + 1}{y} = k, we get: y^2 - ky + 1 = 0. To find the minimum k, examine the discriminant: k^2 - 4 = (k-2)(k+2). The quadratic has no real solutions when k^2 - 4 < 0, which is when -2 < k < 2. The smallest positive k at which the quadratic becomes feasible is when k = 2. Verify this is achievable: if b = 0 and c = 1, then a = 1 and b = 0, and so: frac{a+b}{a-b} + frac{a-b}{a+b} = frac{1+0}{1-0} + frac{1-0}{1+0} = 1 + 1 = 2. Thus, the smallest possible positive value of the expression is boxed{2}.

answer:The difference in speed when including stoppages is 50 km/hr - 40 km/hr = 10 km/hr. This means that for every hour of travel, the bus is stopped for a distance equivalent to 10 km at its average moving speed. To find out how long it takes the bus to cover 10 km at its average moving speed of 50 km/hr, we can use the formula: Time = Distance / Speed Time = 10 km / 50 km/hr Converting 50 km/hr to km/min (since we want the answer in minutes), we know that there are 60 minutes in an hour, so: 50 km/hr = 50 km / 60 min = 5/6 km/min Now we can calculate the time: Time = 10 km / (5/6 km/min) Time = 10 km * (6/5 min/km) Time = (10 * 6) / 5 min Time = 60 / 5 min Time = 12 min Therefore, the bus stops for boxed{12} minutes per hour.

answer:1. **Define the three given expressions and suppose they are consecutive natural numbers:** Let the three given expressions be: [ x_1 = 2023 + a - b, quad x_2 = 2023 + b - c, quad x_3 = 2023 + c - a ] Assume that ( x_1, x_2, x_3 ) are consecutive natural numbers. 2. **Calculate the arithmetic mean of these expressions:** Add these three expressions and divide by 3. [ frac{(2023 + a - b) + (2023 + b - c) + (2023 + c - a)}{3} ] Simplify the expression inside: [ = frac{2023 + a - b + 2023 + b - c + 2023 + c - a}{3} ] Combine like terms: [ = frac{2023 + 2023 + 2023 + (a - a) + (b - b) + (c - c)}{3} ] Further simplify: [ = frac{3 times 2023}{3} = 2023 ] 3. **Conclude the nature of the arithmetic mean:** The arithmetic mean of the three expressions is a natural number, ( 2023 ). Note that for these three numbers to be consecutive, their arithmetic mean needs to be one of these numbers. Specifically, if they take the values ( n-1, n, n+1 ), the arithmetic mean will be ( n ). 4. **Assume one of the expressions equals ( 2023 ):** At least one of ( x_1, x_2, x_3 ) must be equal to ( 2023 ). Assume: [ 2023 + a - b = 2023 ] This simplifies to: [ a - b = 0 implies a = b ] 5. **Identify the contradiction:** This contradicts the condition that ( a, b, c ) are pairwise distinct natural numbers. Thus, at least one of the given conditions ( a neq b ), ( b neq c ), or ( c neq a ) must fail if this assumption were true. 6. **Conclusion:** Therefore, the assumption that all three numbers ( 2023 + a - b ), ( 2023 + b - c ), and ( 2023 + c - a ) can be consecutive natural numbers leads to a contradiction. [ boxed{text{Thus, the numbers cannot be three consecutive natural numbers.}} ]