answer:To prove that if every vertex of a convex polyhedron has at least four edges meeting at it, then at least one of its faces must be a triangle, we will proceed by contradiction. Let's assume the opposite: that there exists a convex polyhedron satisfying the given condition but having no triangular faces. 1. **Identify Polyhedron's Properties**: - Each vertex has at least four edges converging at it. - None of the faces are triangles, implying every face is a polygon with at least four edges (quadrilateral or higher). 2. **Calculate Average Angle for Each Face**: - Each face has at least four vertices. - The sum of interior angles (S_F) of a polygon with (n) sides is given by: [ S_F = 180^circ times (n-2) ] - The average angle (alpha) for a face with (n) sides is: [ alpha = frac{180^circ times (n-2)}{n} ] For (n geq 4): [ alpha = 180^circ left(1 - frac{2}{n}right) ] Since (frac{2}{n} leq frac{2}{4} = 0.5) for (n geq 4), thus: [ alpha geq 180^circ left(1 - 0.5right) = 90^circ ] Hence, the average angle of each face is at least (90^circ). 3. **Recalculate Average Angle by Grouping by Vertices**: - Consider summing the internal angles around each vertex. Since each vertex connects at least four edges, there are at least four angles converging at each vertex. - The sum of the internal angles around a vertex for a polyhedron is less than (360^circ) (since the polyhedron is convex). 4. **Analyze the Angles at Vertices**: - Let these angles around a particular vertex be (beta_1, beta_2, beta_3, beta_4, ldots) where (beta_i geq 0). - The sum at each vertex: [ sum beta_i < 360^circ ] - Since there are at least four such angles, [ k times beta_{text{avg}} < 360^circ ] - The average angle (beta_{text{avg}}) around a vertex must be: [ beta_{text{avg}} < frac{360^circ}{k} leq frac{360^circ}{4} = 90^circ ] Hence, the average angle of all the angles at vertices is less than (90^circ). 5. **Conclusion**: - We have derived the contradictory results: from considering the faces, the average angle is at least (90^circ), while from considering the vertices, it is strictly less than (90^circ). - This contradiction implies the initial assumption (that there is no triangular face) is false. Thus, there must be at least one triangular face in the convex polyhedron. [ boxed{} ]

answer:Given the vertex form of the parabola is y = a(x - h)^2 + k, knowing the vertex (3, -9), we substitute h = 3 and k = -9: [ y = a(x - 3)^2 - 9 ] We substitute the point (6, 27) into the equation to solve for a: [ 27 = a(6 - 3)^2 - 9 36 = 9a a = 4 ] Therefore, the equation of the parabola is: [ y = 4(x - 3)^2 - 9 ] To find the zeros, set y = 0: [ 0 = 4(x - 3)^2 - 9 4(x - 3)^2 = 9 (x - 3)^2 = frac{9}{4} x - 3 = pm frac{3}{2} x = 3 pm frac{3}{2} ] Thus, x = 4.5 or x = 1.5. The axis of symmetry is x = 3. The difference between the zeros is |4.5 - 1.5| = 3. Conclusion: The axis of symmetry is x = 3 and m - n = boxed{3}.

answer:Given f(x)=cos (omega x-frac{π}{6}) where omega > 0, and if f(x)leqslant f(frac{π}{4}) holds for any real number x, we aim to find the minimum value of omega. The condition f(x)leqslant f(frac{π}{4}) implies that the maximum value of f(x) occurs at x=frac{π}{4}. This is because the cosine function achieves its maximum value when its argument is closest to 2kπ, where k is an integer. Given the function f(x)=cos (omega x-frac{π}{6}), we substitute x=frac{π}{4} to find the condition for the maximum value of f(x): [ omega•frac{π}{4}-frac{π}{6}=2kπ, quad kin mathbb{Z}. ] Solving this equation for omega gives: [ omega = frac{2kπ + frac{π}{6}}{frac{π}{4}} = 8k + frac{2}{3}, quad kin mathbb{Z}. ] Given that omega > 0, we need to find the smallest positive value of omega. This occurs when k=0: [ omega = 8(0) + frac{2}{3} = frac{2}{3}. ] Therefore, the minimum value of omega that satisfies the given conditions is boxed{frac{2}{3}}.

answer:1. **Analyze the structure of the equation:** The given equation is: [ sqrt{2 x^{2}-2024 x+1023131} + sqrt{3 x^{2}-2025 x+1023132} + sqrt{4 x^{2}-2026 x+1023133} = sqrt{x^{2}-x+1} + sqrt{2 x^{2}-2 x+2} + sqrt{3 x^{2}-3 x+3} ] 2. **Observations on the expressions under the square roots:** We notice that the expressions (sqrt{x^2 - x + 1}), (sqrt{2x^2-2x+2}), and (sqrt{3x^2 -3x + 3}) on the right-hand side are all non-negative, as they are sums of squares and non-negative terms. 3. **Relate left-hand side terms with right-hand side terms:** Investigate how the terms are derived. Notice that for the left-hand side: [ 2 x^{2} - 2024 x + 1023131, quad 3 x^{2} - 2025 x + 1023132, quad 4 x^{2} - 2026 x + 1023133 ] can be viewed by shifting right-hand side expressions (x^2 - x + 1), (2 x^{2} - 2 x + 2), (3 x^{2} - 3 x + 3). 4. **Simplify the transformation:** Let's add (x^2 - 2023x + 1023130) to the corresponding right-hand terms: [ x^2 - x + 1 + (x^2 - 2023x + 1023130) = 2x^2 - 2024x + 1023131 ] The same holds for the other terms: [ 2x^2 - 2x + 2 + (x^2 - 2023x + 1023130) = 3x^2 - 2025x + 1023132 ] [ 3x^2 - 3x + 3 + (x^2 - 2023x + 1023130) = 4x^2 - 2026x + 1023133 ] 5. **Equate terms by their values:** Hence, we need the left-hand side and right-hand side to match, meaning we equate the shifts: The shifts ((x^2 - 2023x + 1023130)) turn out to be factorized: [ x^2 - 2023 x + 1023130 = (x - 1010)(x - 1013) ] 6. **Analyze critical value behavior:** Since all terms under square roots are positive (as previously indicated), we check for equality by analyzing within this interval. It's established that: [ text{LHS} < text{RHS when } 1010 < x < 1013 quad text{and } text{LHS} > text{RHS} quad text{otherwise}. ] 7. **Identify solutions:** The equation holds strictly equality only at: [ x = 1010 quad text{or} quad x = 1013 ] Therefore, the roots of the equation are (x = 1010) and (x = 1013). 8. **Sum of the roots:** Sum of roots ( 1010 + 1013 = 2023 ) # Conclusion: The sum of all roots of the equation is: [ boxed{2023} ]