answer:To find the length of the train, we can use the formula: Distance = Speed × Time First, we need to convert the speed from km/hr to m/s because the time is given in seconds. We know that 1 km = 1000 meters and 1 hour = 3600 seconds. So, the speed in m/s is: Speed = 120 km/hr × (1000 m / 1 km) × (1 hr / 3600 s) Speed = 120 × 1000 / 3600 Speed = 120000 / 3600 Speed = 33.33 m/s Now, we can calculate the distance, which is the length of the train, using the time it takes to cross the pole: Distance = Speed × Time Distance = 33.33 m/s × 15 s Distance = 499.95 m Therefore, the length of the train is approximately boxed{500} meters.

answer:Given the functional equation: [ f(f(m)^2 + 2f(n)^2) = m^2 + 2n^2 quad forall m, n in mathbb{N}^* ] We will demonstrate that the only function ( f: mathbb{N}^* rightarrow mathbb{N}^* ) that satisfies this condition is ( f(n) = n ). 1. **Injection of ( f )**: - Assume ( f(a) = f(b) ) for some ( a, b in mathbb{N}^* ). - Then, substituting ( m = a ) and ( n = 1 ) into the functional equation, we have: [ f(f(a)^2 + 2f(1)^2) = a^2 + 2 ] - Similarly, substituting ( m = b ) and ( n = 1 ): [ f(f(b)^2 + 2f(1)^2) = b^2 + 2 ] - Since ( f(a) = f(b) ), the left-hand sides of both equations are equal, so: [ a^2 + 2 = b^2 + 2 Rightarrow a^2 = b^2 Rightarrow a = b ] - Hence, ( f ) is injective. 2. **Inductive Hypothesis**: - We will use strong induction on ( n ) to prove that ( f(n) = n ). - **Base Case**: For ( n = 1 ): [ f(f(1)^2 + 2f(1)^2) = 1^2 + 2(1^2) = 3 Rightarrow f(3f(1)^2) = 3 ] - Since ( f(3) = 3 ) and ( f ) is a permutation, the only possible solution is ( f(1) = 1 ). 3. **Induction Step**: - Suppose ( f(k) = k ) for all ( k leq n ). - Consider the equation derived from the observation: [ (n+4)^2 + 2(n+1)^2 = n^2 + 2(n+3)^2 ] - Applying ( f ) to both sides: [ f((n+4)^2 + 2(n+1)^2) = f(n^2 + 2(n+3)^2) ] - By our hypothesis and injectiveness of ( f ), this simplifies to: [ f((n+4)^2) + 2 f((n+1)^2) = f(n)^2 + 2 f((n+3)^2) ] - Given the form of our induction hypothesis, it follows that: [ (n+4)^2 + 2(n+1)^2 = n^2 + 2(n+3)^2 Rightarrow n+4, n+1, n, n+3 ] - Thus, since ( f ) is a permutation and consistent with the recurrence, [ f(n+4) = n+4 ] - Therefore, by induction, ( f(n) = n ) for all ( n in mathbb{N}^* ). **Conclusion**: [ boxed{f(n) = n} ]

answer:Let's denote the rate at which Pipe A fills the tank as ( A ) and the rate at which the leak empties the tank as ( L ). Pipe A fills the tank in 4 hours, so its rate is ( frac{1}{4} ) of the tank per hour. The leak empties the tank in 12 hours, so its rate is ( frac{1}{12} ) of the tank per hour. When both Pipe A and the leak are at work, their combined rate is the rate of filling the tank by Pipe A minus the rate of emptying by the leak. So the combined rate ( C ) is: [ C = A - L ] [ C = frac{1}{4} - frac{1}{12} ] To find a common denominator, we can multiply the terms by 3/3 and 1/1 respectively: [ C = frac{3}{12} - frac{1}{12} ] [ C = frac{2}{12} ] [ C = frac{1}{6} ] So the combined rate of filling the tank with the leak present is ( frac{1}{6} ) of the tank per hour. To find out how long it takes to fill the tank at this rate, we take the reciprocal of the combined rate: [ Time = frac{1}{C} ] [ Time = frac{1}{frac{1}{6}} ] [ Time = 6 ] Therefore, it takes boxed{6} hours for Pipe A to fill the tank with the leak present.

answer:1. **Identify point properties:** Given a triangle (ABC) inscribed in a circle, with a point (P) on the arc (BC) that does not contain (A). From point (P), perpendiculars (PX), (PY), and (PZ) are dropped to sides (BC), (CA), and (AB), respectively. 2. **Collinearity and areas:** Points (X), (Y), and (Z) lie on a straight line (per problem (5.85(a))). Hence, we have the relation: [ S_{triangle PYZ} = S_{triangle PXZ} + S_{triangle PXY}. ] 3. **Area expressions:** The area of (triangle PYZ) is given by: [ S_{triangle PYZ} = frac{1}{2} times PY times PZ times sin alpha, ] where (alpha) is the angle between (PY) and (PZ). 4. **Decompose areas:** Similarly, we express the areas of (triangle PXZ) and (triangle PXY): [ S_{triangle PXZ} = frac{1}{2} times PX times PZ times sin beta, ] [ S_{triangle PXY} = frac{1}{2} times PX times PY times sin gamma. ] 5. **Equate and relate areas:** Substituting the areas into the earlier collinear relation, we get: [ frac{1}{2} times PY times PZ times sin alpha = frac{1}{2} times PX times PZ times sin beta + frac{1}{2} times PX times PY times sin gamma. ] 6. **Simplify the equations:** Cancelling the common factor of (frac{1}{2} PY times PZ) on both sides, we obtain: [ sin alpha = frac{PX times PZ times sin beta}{PY times PZ} + frac{PX times PY times sin gamma}{PX times PZ}. ] [ sin alpha = frac{PX times sin beta}{PY} + frac{PX times sin gamma}{PZ}. ] 7. **Rearrange and finalize:** Dividing both sides by (PX), we get: [ frac{sin alpha}{PX} = frac{sin beta}{PY} + frac{sin gamma}{PZ}. ] 8. **Conclude with side-sine ratio:** Notice that the sine values ( sin alpha, sin beta, sin gamma ) are proportional to the sides of the triangle, specifically: [ sin alpha : sin beta : sin gamma = BC : AC : AB. ] 9. **Final relation:** Thus, we substitute these into our previous result to get: [ frac{BC}{PX} = frac{AC}{PY} + frac{AB}{PZ}. ] # Conclusion: [ boxed{frac{BC}{PX} = frac{AC}{PY} + frac{AB}{PZ}} ]