answer:# Problem: Given an acute triangle ( triangle ABC ) with the circumcenter ( O ), draw the altitude from ( A ) to ( BC ) and denote the foot of the perpendicular by ( P ). Given that ( angle BCA geq angle ABC + 30^circ ), prove that ( angle CAB + angle COP < 90^circ ). 1. **Define Angles**: Let: [ angle CAB = alpha, quad angle ABC = beta, quad angle BCA = gamma ] Let ( angle COP = delta ). 2. **Mirror Points**: Suppose ( K ) and ( Q ) are the reflections of points ( A ) and ( P ) over the perpendicular bisector of ( BC ). 3. **Radius and Quadrilateral**: Let ( R ) be the circumradius of ( triangle ABC ). Therefore: [ OA = OB = OC = OK = R ] 4. **Rectangle Property**: Since ( KQPQ ) forms a rectangle, we have: [ QP = KA ] 5. **Angle Calculation**: In ( triangle AOK ), since ( OA = OK = R ), the angle is: [ angle AOK = angle AOB - angle KOB ] Noticing symmetry: [ angle AOK = angle AOB - angle AOC = 2gamma - 2beta geq 60^circ ] 6. **PQ Length**: Due to (angle BCA geq angle ABC + 30^circ), then ( gamma geq beta + 30^circ ) leading to: [ angle AOK geq 60^circ ] Therefore from triangle properties: [ KA geq R implies QP geq R ] 7. **Triangle Inequality**: Using the triangle inequality: [ OP + R = OQ + OC > QC = QP + PC geq R + PC implies OP > PC ] 8. **Triangle Relations**: Therefore, given all relations ( OP > PC ), angles satisfy: [ angle PCO > angle COP = delta ] 9. **Final Calculation**: From triangle properties: [ alpha = frac{1}{2} angle BOC = frac{1}{2}(180^circ - 2 angle PCO ) = 90^circ - angle PCO ] Conclusion: Combining the above results, we have: [ alpha + delta < 90^circ ] Hence, we have proved: [ boxed{angle CAB + angle COP < 90^circ} ]

answer:Given the problem: The parabola (C_{m}: y=x^{2}-mx+m+1 ) intersects the line segment joining points (A(0, 4)) and (B(4, 0)) at exactly two points. We need to determine the range of ( m ). 1. **Find the equation of the line segment (AB)**: The line segment (AB) can be expressed by finding the line passing through points (A(0, 4)) and (B(4, 0)). The slope ((m_s)) of this line is: [ m_s = frac{0 - 4}{4 - 0} = -1 ] Using the point-slope form of the line equation, with point (A(0, 4)): [ y - 4 = -1 (x - 0) y = -x + 4 ] 2. **Set the equations equal to find intersection points**: Substitute ( y = -x + 4 ) into the parabolic equation ( y = x^2 - mx + m + 1 ): [ -x + 4 = x^2 - mx + m + 1 ] Simplify this equation: [ x^2 - (m-1)x + (m-3) = 0 ] 3. **Ensure the quadratic has two distinct real roots**: The quadratic equation ( x^2 - (m-1)x + (m-3) = 0 ) must have two distinct real roots ( x_1 ) and ( x_2 ), such that ( 0 leq x_1 < x_2 leq 4 ). For this to happen, the discriminant must be positive: [ Delta = (m-1)^2 - 4(m-3) > 0 ] Simplify the discriminant: [ (m-1)^2 - 4(m-3) > 0 m^2 - 2m + 1 - 4m + 12 > 0 m^2 - 6m + 13 > 0 ] Solving the inequality: [ m^2 - 6m + 13 > 0 text{ always true since the parabola never touches the axis} ] 4. **Verify constraints for (x) within the interval [0, 4]**: * The vertex form of the quadratic to be within [0, 4] is: [ 0 leq frac{m-1}{2} leq 4 0 leq m-1 leq 8 1 leq m leq 9 ] * Evaluate ( f(0) geq 0 ): [ f(0) = m - 3 geq 0 m geq 3 ] * Evaluate ( f(4) geq 0 ): [ f(4) = 16 - 4(m-1) + m - 3 geq 0 16 - 4m + 4 + m - 3 geq 0 17 - 3m geq 0 m leq frac{17}{3} ] Combining these constraints, we find: [ 3 leq m leq frac{17}{3} ] # Conclusion: The value of ( m ) is in the range ( left[3, frac{17}{3}right] ). [ boxed{left[3, frac{17}{3}right]} ]

answer:If the average of the marks of 10 students in a class is 40, the total marks of all the students combined is: Average = Total Marks / Number of Students Total Marks = Average * Number of Students Total Marks = 40 * 10 Total Marks = 400 If the marks of each student are doubled, the new total marks will be: New Total Marks = Total Marks * 2 New Total Marks = 400 * 2 New Total Marks = 800 The new average will be: New Average = New Total Marks / Number of Students New Average = 800 / 10 New Average = 80 So, the new average after doubling the marks of each student will be boxed{80} .

answer:Since the given hyperbola is frac{x^{2}}{9} - frac{y^{2}}{16} = 1, We have a=3, b=4, and c = sqrt{16+9} = 5. By the definition of a hyperbola, we have |PF_1| - |PF_2| = pm 2a = pm 6. Squaring both sides, we get |PF_1|^2 + |PF_2|^2 - 2|PF_1||PF_2| = 36. Thus, |PF_1|^2 + |PF_2|^2 = 36 + 2|PF_1||PF_2|. Given that angle F_1 P F_2 = 90^{circ}, We have |PF_1|^2 + |PF_2|^2 = 100. Equating the two expressions for |PF_1|^2 + |PF_2|^2, we get: 36 + 2|PF_1||PF_2| = 100, 2|PF_1||PF_2| = 64, |PF_1||PF_2| = 32. Therefore, the area of triangle F_1 P F_2 is S_{triangle F_1 P F_2} = frac{1}{2}|PF_1||PF_2| = frac{1}{2} times 32 = boxed{16}.