answer:Let's adopt the point of view and notations provided in the problem: 1. **Define the Key Points and Segment**: Let the endpoints of the given segment be M and N, with the midpoint of this segment labeled as O. 2. **Identify Constraints and Relations**: When the segment MN slides such that its endpoints move along the adjacent sides of the right angle angle ABC, point O will trace a path due to the geometric constraints. 3. **Analyze the Circular Path**: Considering the properties of a circle, since M and N glide along the perpendicular sides, O is always equidistant from the right-angled vertex B. Essentially, O lies on a circular path centered at B. 4. **Determine the Radius of the Circular Path**: The important observation is that since O is the midpoint of segment MN, it will always be at a distance of half the length of MN from vertex B. Consequently, the radius R of the circular path which O follows is: [ OB = frac{MN}{2} ] 5. **Conclude the Trajectory of O**: Thus, the midpoint O traces an arc of a circle with radius R = frac{MN}{2} centered at B. This arc lies within the angles formed by the perpendicular sides at B and within the confines of the right angle angle ABC. 6. **Final Formulation**: The trajectory of the midpoint (O) is part of a circle with radius (frac{MN}{2}), centered at point (B), and constrained within the right angle (angle ABC). [ boxed{text{The trajectory of the midpoint (O) is an arc of a circle with radius (frac{MN}{2}) centered at (B), lying inside (angle ABC).}} ]

answer:(1) Let the sides opposite angles A, B, C in triangle ABC be a, b, c, respectively. Since the area of triangle ABC is 3, we have bcsintheta = 3, and since 0 leq cdot leq 6, we also have 0 leq bccostheta leq 6, from which we can deduce that tantheta geq 1, thus theta in boxed{text{range}}. (2) f(theta) = 2sin^2theta - cos 2theta = 1 - cos^2theta - cos 2theta = 1 + sin 2theta - cos 2theta = 1 + 2sinthetacostheta Since theta in boxed{text{range}}, then 2theta - pi in boxed{text{range}}, thus, when theta = boxed{text{value}}, i.e., 2theta - pi = boxed{text{value}}, f(theta) reaches its maximum value of boxed{3}; when theta = boxed{text{value}}, i.e., 2theta - pi = boxed{text{value}}, f(theta) reaches its minimum value of boxed{2}.

answer:To determine the correct formula relating x and y, we review each option with the points provided: - **Point (0,80) Testing**: - **Option A:** y = 80 - 10x implies y = 80 - 10(0) = 80 - **Option B:** y = 80 - 5x^3 implies y = 80 - 5(0)^3 = 80 - **Option C:** y = 80 - 5x - 5x^2 implies y = 80 - 5(0) - 5(0)^2 = 80 - **Option D:** y = 80 - x - x^2 implies y = 80 - 0 - 0^2 = 80 - **Point (4,40) Testing**: - **Option A:** y = 80 - 10x implies y = 80 - 10(4) = 40 - **Option B:** y = 80 - 5x^3 implies y = 80 - 5(4)^3 = 80 - 320 = -240 - **Option C:** y = 80 - 5x - 5x^2 implies y = 80 - 5(4) - 5(4)^2 = 80 - 20 - 80 = -20 - **Option D:** y = 80 - x - x^2 implies y = 80 - 4 - 4^2 = 80 - 4 - 16 = 60 - **Point (1,70) Testing**: - **Option A:** y = 80 - 10x implies y = 80 - 10(1) = 70 Conclusion: From testing and verifying the given points, it's clear that: [textbf{(A) y=80-10x}] The final answer is boxed{textbf{(A)}}

answer:1. **Definition of Radical and ( A_l ) Sets**: - For a given integer ( n geq k ), we define the radical of ( n ), denoted by ( r(n) ), as the product of the prime numbers less than or equal to ( k ) that divide ( n ). This radical ( r(n) ) is a squarefree number (a number that is not divisible by any square number except 1). - For any squarefree number ( l ) whose prime factors are all less than or equal to ( k ), we define ( A_l ) as the set of all integers greater than or equal to ( k ) whose radical is ( l ). 2. **Initial Understanding of ( A_l )**: - Fix a squarefree ( l ) and let its prime factorization be ( l = p_1 cdots p_u ), where ( p_i ) are primes. - Let ( j ) be the minimum element in the set ( A_l ). If ( j ) has a prime factor strictly greater than ( k ), then ( j geq k(p_1 cdots p_u) ). 3. **Contradiction Argument**: - Define ( m ) as the largest integer such that ( (p_1 cdots p_u)^m < k ). - Since ( j > (p_1 cdots p_u)^m geq k ), this contradicts the minimality of ( j ). Thus, the minimum of ( A_l ) must have prime factors all less than or equal to ( k ). 4. **Recurrence Hypothesis**: - We use strong induction on ( n geq k ) to show that ( n ) is alicien if and only if its radical ( r(n) ) is alicien. 5. **Base Case**: - The base case for ( n = k ) holds since ( n = k ) and its radical is also ( k ). 6. **Induction Step**: - Assume the hypothesis is true for all integers less than ( m ), where ( m > k ). - Let ( r ) be the radical of ( m ). 7. **Case 1: ( r ) is Alicien**: - If ( r ) is alicien, it leads to a losing position ( x ), meaning ( x ) is the minimum element of ( A_{r(x)} ). Since ( m ) shares the same prime factors (less than or equal to ( k )), ( m ) also leads to ( x ), making ( m ) alicien. 8. **Case 2: ( m ) is Alicien**: - If ( m ) is alicien, it leads to a losing position ( y ). Let ( y ) be the minimum of ( A_{r(y)} ). If ( r > y ), then ( r ) is alicien because ( y ) is losing. - Since ( r = y ) can't hold, the only remaining case is ( y > r ). Thus, ( y ) leads to ( r ), making ( r ) alicien. 9. **Conclusion**: - A number ( n ) is alicien if and only if its radical is alicien. - Therefore, if ( l ) and ( l' ) have the same prime factors less than or equal to ( k ), their radical values are the same, implying either both are alicien or both are bobesque. [ boxed{} ]