answer:# Problem: Given n points M_1, M_2, ldots, M_n: (a) In the plane; (b) In space; Determine under what conditions these points can be positioned such that no three points form a triangle with an obtuse angle. (a) In the Plane: 1. **Convex Polygon Consideration**: - Consider a convex polygon T with vertices chosen from our points and containing the rest of the points inside it. This can be done similar to the beginning of the solution to Problem 25. 2. **Case Analysis on Number of Vertices T**: - **More than Four Vertices**: Since a convex polygon with more than four vertices must have at least one obtuse angle, it follows that some three points among these vertices form an obtuse-angled triangle. ( text{Therefore, } n leq 4. ) - **Exactly Four Vertices**: A convex polygon with exactly four vertices is a quadrilateral. Only if the quadrilateral is a rectangle will all angles be non-obtuse. If there is a point inside such a rectangle, it would form an obtuse angle with at least two vertices of the rectangle and this point inside. ( text{Therefore, } n leq 4. ) - **Three Vertices**: If the number of vertices is exactly three, forming a triangle ABC, then no points lie inside this triangle. Thus, it is possible for n = 3 where these three points form an acute-angled or right-angled triangle. Therefore, the possible values for ( n ) are 3 (vertices of an acute or right triangle) or 4 (vertices of a rectangle). [ boxed{ text{n} = 3 text{ or } text{n} = 4} ] (b) In Space: 1. **Convex Polyhedron Consideration**: - Consider the convex hull T, which is the smallest convex polyhedron containing all the points M_1, M_2, ldots, M_n. This is done similarly to the construction on the plane. 2. **Exact Layout**: - It is known that in space, a system of up to eight points can be arranged to form a rectangular parallelepiped (cube), where no triangle formed by these points has an obtuse angle. 3. **Maximum Number**: - Let’s show that n leq 8: Assume each point M_i belongs to a convex polyhedron T. If each face of T can contain at most four points thus same logic as the plane case a face containing more points would mean it had an obtuse angle somewhere. We represent n such points in space such that they form a rectangular cuboid. - Eight points can be the vertices of a rectangular parallelepiped. This implies: - No triangle with vertices among these points has an obtuse angle. - The solution would imply no face or edge cross or exceeded beyond convexity thus maintaining acute or 90-degree vertices. If any more than eight points are attempted, placing another point would fail as vertices would fail convex condition or exceed the physical cube form. Therefore, the solution valid remains: [ boxed{ text{n} = 8} ] # Conclusion: - For the plane, the possible values for n are 3 or 4 points. - For space, the possible value for n is 8 points.

answer:1. Given a triangle ( triangle ABC ) with ( AB = 86 ) and ( AC = 97 ). A circle with center at point ( A ) and radius ( AB ) intersects side ( BC ) at points ( B ) and ( X ). 2. Let ( BX = x ) and ( CX = y ). We need to calculate the length of ( BC ), which is ( x + y ), given that ( BX ) and ( CX ) are integers. 3. Using the Power of a Point theorem for point ( C ), considering the tangents from ( C ) to the circle (through point ( X )), we have the equation for the power of point ( C ): [ y(y + x) = AC^2 - AB^2 ] 4. Substitute the given values: [ y(y + x) = 97^2 - 86^2 ] 5. Calculate ( 97^2 ) and ( 86^2 ): [ 97^2 = 9409 ] [ 86^2 = 7396 ] 6. Substitute back into the power of point equation: [ y(y + x) = 9409 - 7396 ] [ y(y + x) = 2013 ] 7. We now need to find integer pairs ((y, x)) that satisfy ( y(y + x) = 2013 ). First, factorize 2013: [ 2013 = 3 times 11 times 61 ] 8. Given that ( y ) and ( y + x ) are integers, we need to find ( y ) as a factor of 2013 such that : [ y(y + x) = 2013 ] We consider the factors of 2013: 1, 3, 11, 33, 61, 183, 671, 2013. 9. Considering the inequalities derived from the triangle inequality and checking each factor, we find that the only valid choice for ( y ) that makes ( x ) an integer and satisfies the conditions of the problem, is ( y = 33 ): [ 33(61) = 2013 quad text{(since} quad 61 = y + x, quad text{we have that} quad y + x = 61) ] 10. Thus, the length of ( BC ) is: [ BC = x + y = 61 ] # Conclusion: [ boxed{61} ]

answer:# Step-by-Step Solution Part (Ⅰ): Finding the Monthly Profit Function 1. **For 0 < x < 70 units:** The total cost consists of a fixed cost of 4 million yuan and a variable cost of frac{1}{2}x^2 + 40x million yuan. The revenue from selling x units at 100 million yuan each is 100x million yuan. Therefore, the profit y can be calculated as: [ y = text{Revenue} - text{Total Cost} ] [ y = 100x - left(frac{1}{2}x^2 + 40x + 4right) ] [ y = -frac{1}{2}x^2 + 60x - 4 ] 2. **For x geqslant 70 units:** The total cost changes to 101x + frac{6400}{x} - 2060 million yuan, including the fixed cost. The revenue remains the same, 100x million yuan. Thus, the profit y is: [ y = 100x - left(101x + frac{6400}{x} - 2060 + 4right) ] [ y = 1660 - left(x + frac{6400}{x}right) ] Hence, the function relationship of monthly profit y with monthly output x is: [ y = left{ begin{array}{ll} -frac{1}{2}x^2 + 60x - 4, & 0 < x < 70 text{ and } x in mathbb{N} 1660 - left(x + frac{6400}{x}right), & x geqslant 70 text{ and } x in mathbb{N} end{array} right. ] Part (Ⅱ): Maximizing Monthly Profit 1. **For 0 < x < 70:** The profit function can be rewritten as: [ y = -frac{1}{2}(x - 60)^2 + 1400 ] This is a parabola opening downwards, with its vertex at (60, 1400). Therefore, at x = 60, the profit reaches its maximum of 1400 million yuan. 2. **For x geqslant 70:** The profit function is: [ y = 1660 - left(x + frac{6400}{x}right) ] Using the AM-GM inequality, we have: [ x + frac{6400}{x} geq 2sqrt{x cdot frac{6400}{x}} = 2sqrt{6400} = 160 ] Thus, the maximum profit y is: [ y leq 1660 - 160 = 1500 ] This maximum occurs when x = 80, as per the equality condition of AM-GM. Conclusion To achieve the maximum monthly profit, the company should produce 80 units per month, resulting in a maximum profit of approximately boxed{1500} million yuan.

answer:1. We start with the given relation ( c^4 = a^4 + b^4 ). 2. From this relation, it follows that ( c > a ) and ( c > b ). This is because the fourth power of a positive real number preserves the order, hence if ( c^4 ) is greater than ( a^4 + b^4 ), ( c ) must be greater than both ( a ) and ( b ). 3. Next, let's consider the inequality based on summing squares: [ c^4 = a^4 + b^4 < a^4 + b^4 + 2a^2b^2 = (a^2 + b^2)^2 ] Here, we have used the elementary algebraic fact that: [ a^4 + b^4 < a^4 + b^4 + 2a^2b^2 ] so, [ c^4 < (a^2 + b^2)^2. ] 4. Taking the square root of both sides to reduce the fourth power to a second power, we have: [ c^2 < a^2 + b^2 ] This inequality ( c^2 < a^2 + b^2 ) is crucial because it implies that the square of the largest side of the triangle ((c) being the largest side as previously established) is less than the sum of the squares of the other two sides. 5. According to the Euclidean geometry, if in a triangle, the square of one side is less than the sum of the squares of the other two sides, the triangle is acute-angled (all angles less than 90 degrees). 6. Therefore, the given triangle with sides ( a, b, ) and ( c ) must be an acute-angled triangle. [ boxed{text{The triangle is acute-angled.}} ]