answer:Part (a) To find the length of (OC), we start by determining the length of segment (overline{AB}). 1. **Calculate the length of ( overline{AB} )**: [ |AB| = sqrt{(40-0)^2 + (0-30)^2} = sqrt{40^2 + 30^2} = sqrt{1600 + 900} = sqrt{2500} = 50 ] Since (overline{OC}) is perpendicular to (overline{AB}), the area of triangle (triangle AOB) can also be calculated using (OC). 2. **Calculate the area of (triangle AOB)** using both the base-height form and coordinate geometry form: ( text{Area} = frac{1}{2} times AB times OC = frac{1}{2} times 50 times OC text{Area from coordinates} = frac{1}{2} times 40 times 30 = 600 ) Set the two expressions for area equal to each other, then solve for (OC): [ frac{1}{2} times 50 times OC = 600 implies 25 times OC = 600 implies OC = frac{600}{25} = 24 ] **Conclusion part (a):** The length of (OC) is: [ boxed{24} ] Part (b) Since (OC) is perpendicular to (AB), triangle (triangle ACO) is similar to triangle (triangle AOB). 1. **Determine the ratio** of the lengths (frac{AC}{AO} = frac{AO}{AB}): [ frac{AC}{30} = frac{30}{50} implies AC = frac{30 times 30}{50} = 18 ] 2. **Determine the coordinates of point (C)** along line (AB): The coordinates of (C) are found by dividing the segment (AB) in the proportion (frac{18}{50}): [ (x_C, y_C) = left(0 + 40 times frac{18}{50}, 30 + (0-30) times frac{18}{50}right) = left(frac{720}{50}, frac{1200}{50}right) = left(frac{72}{5}, frac{96}{5}right) ] **Conclusion part (b):** The coordinates of point (C) are: [ boxed{left(frac{72}{5}, frac{96}{5}right)} ] Part (c) The circle passing through (O), (A), and (B) has (overline{AB}) as its diameter. The center (M) of this circle is the midpoint of (overline{AB}). 1. **Calculate the midpoint (M) of (overline{AB})**: [ M = left(frac{0+40}{2}, frac{30+0}{2}right) = (20, 15) ] 2. **Calculate the distance ( CM )** between (C) and (M): Using the coordinates of (C) found in part (b), compute: [ CM = sqrt{left(20 - frac{72}{5}right)^2 + left(15 - frac{96}{5}right)^2} ] 3. **Simplify the components** of the distance: [ 20 - frac{72}{5} = frac{100}{5} - frac{72}{5} = frac{28}{5} 15 - frac{96}{5} = frac{75}{5} - frac{96}{5} = -frac{21}{5} ] Substitute these back and simplify: [ CM = sqrt{left(frac{28}{5}right)^2 + left(-frac{21}{5}right)^2} = sqrt{frac{784}{25} + frac{441}{25}} = sqrt{frac{1225}{25}} = sqrt{49} = 7 ] **Conclusion part (c):** The length of (CM) is: [ boxed{7} ]

answer:First, let's find the focus F of the given parabola x^2=4y. The general form of a parabola with its vertex at the origin is x^2=4py, where p is the distance from the vertex to the focus. Comparing this to the given equation, we have: 4p = 4 Rightarrow p = 1. Thus, the focus F is at (0, p) = (0, 1). For any point A(x_1, y_1) on x^2=4y, we have y_1 = frac{x_1^2}{4}. Since F lies on line l, and the line l passes through A and B which lie on the parabola, the midpoint M of chord AB lies on this line as well. The midpoint M has coordinates (frac{x_1+x_2}{2}, frac{y_1+y_2}{2}). However, since M lies on the line that passes through the focus, the y-coordinate of M must be exactly halfway between y_1 and the y-coordinate of the focus (1). That leads us to: frac{y_1+y_2}{2} = frac{y_1 + 1}{2}. Given that A and B are on the parabola, we have: y_1 = frac{x_1^2}{4}, quad y_2 = frac{x_2^2}{4}. Substitute y_1 and y_2 into the midpoint M formula: frac{y_1+y_2}{2} = frac{frac{x_1^2}{4} + frac{x_2^2}{4}}{2} = frac{x_1^2+x_2^2}{8}. Now, substitute the given expression for y_1: frac{frac{x_1^2}{4} + 1}{2} = frac{x_1^2+x_2^2}{8} Rightarrow frac{x_1^2 + 4}{8} = frac{x_1^2+x_2^2}{8}. Solving for x_2^2, we get: x_2^2 = 4. The x-coordinate of M is: frac{x_1+x_2}{2} = frac{x_1+sqrt{4}}{2} = frac{x_1+2}{2}. Because we are looking for a relation between x and y and not x_1 or x_2, we need to express x_1 in terms of y. Remember that x_1=2sqrt{y_1}. We have: x = frac{2sqrt{y_1}+2}{2} = sqrt{y_1}+1. We square both sides to eliminate the square root: x^2 = (sqrt{y_1}+1)^2 = y_1 + 2sqrt{y_1} + 1. Substitute back frac{x_1^2}{4} for y_1: x^2 = frac{x_1^2}{4} + 2left(frac{x_1}{2}right) + 1. Knowing that x_1 = xM at this point and y_1 = yM, where (x_M, y_M) are the coordinates of the midpoint M: x_M^2 = y_M + x_M + 1. So, the equation of the trajectory of the midpoint M is: x^2 = y + x + 1. Or, when simplified: x^2 - x - y - 1 = 0. Therefore, the equation of the trajectory of the midpoint M of chord AB is (boxed{x^2 - x - y - 1 = 0}).

answer:# Problem: Determine all real solutions to the equation ( f_n(x) = 2x ) where: - ( f_1(x) = sqrt{x^2 + 48} ) - ( f_n(x) = sqrt{x^2 + 6f_{n-1}(x)} ) for ( n geq 2 ) 1. **Assumption and Initial Consideration**: - We claim that ( x = 4 ) is a solution. We'll verify this first: [ f_1(4) = sqrt{4^2 + 48} = sqrt{16 + 48} = sqrt{64} = 8 ] Since ( 2 times 4 = 8 ), for ( n = 1 ), ( x = 4 ) satisfies ( f_1(x) = 2x ). 2. **Non-negativity Condition**: - Since ( f_n(x) geq 0 ) for all ( x ), any solution must be non-negative. Therefore, we restrict our attention to ( x geq 0 ). 3. **Induction Hypothesis**: - Suppose for some ( x ), ( x < 4 ). We will show by induction that ( f_n(x) > 2x ): - **Base Case (n = 1)**: [ f_1(x) = sqrt{x^2 + 48} ] [ text{We need to show that } sqrt{x^2 + 48} > 2x ] Squaring both sides: [ x^2 + 48 > 4x^2 ] [ 48 > 3x^2 quad Rightarrow quad x^2 < 16 ] Since ( x < 4 ), ( x^2 < 16 ) holds true. - **Inductive Step**: - Assume ( f_n(x) > 2x ) [ f_{n+1}(x) = sqrt{x^2 + 6f_n(x)} ] We need to show that: [ sqrt{x^2 + 6f_n(x)} > 2x ] Squaring both sides: [ x^2 + 6f_n(x) > 4x^2 ] By the inductive hypothesis, ( f_n(x) > 2x ): [ 6f_n(x) > 12x quad Rightarrow quad x^2 + 12x < x^2 + 6f_n(x) ] Since ( x < 4 ): [ 3x^2 < 12x quad Rightarrow quad 4x^2 < x^2 + 12x ] Hence, [ sqrt{x^2 + 6f_n(x)} > 2x quad Rightarrow quad f_{n+1}(x) > 2x ] By induction, for all ( n ), ( f_n(x) > 2x ) if ( x < 4 ). 4. **Argument for ( x > 4 )**: - A similar argument shows that if ( x > 4 ), then ( f_n(x) < 2x ). 5. **Conclusion**: - Given that ( f_n(x) = 2x ) can neither hold for ( x < 4 ) (since ( f_n(x) > 2x )) nor for ( x > 4 ) (since ( f_n(x) < 2x )), the only possible solution is ( x = 4 ). Thus, the only real solution to ( f_n(x) = 2x ) is: [ boxed{4} ]

answer:Let's denote the first term of the geometric progression as ( a ) and the common ratio as ( r ). Given that the common ratio ( r = 3 ). The sum of the first ( n ) terms of a geometric progression is given by the formula: [ S_n = frac{a(1 - r^n)}{1 - r} ] For the first 6 terms, the sum ( S_6 ) is: [ S_6 = frac{a(1 - r^6)}{1 - r} ] For the first ( m ) terms, the sum ( S_m ) is: [ S_m = frac{a(1 - r^m)}{1 - r} ] According to the problem, the ratio of ( S_6 ) to ( S_m ) is 28: [ frac{S_6}{S_m} = 28 ] Substituting the expressions for ( S_6 ) and ( S_m ) we get: [ frac{frac{a(1 - r^6)}{1 - r}}{frac{a(1 - r^m)}{1 - r}} = 28 ] Simplifying, we get: [ frac{1 - r^6}{1 - r^m} = 28 ] Since ( r = 3 ), we can substitute it into the equation: [ frac{1 - 3^6}{1 - 3^m} = 28 ] [ frac{1 - 729}{1 - 3^m} = 28 ] [ frac{-728}{1 - 3^m} = 28 ] Now, we solve for ( 3^m ): [ 1 - 3^m = frac{-728}{28} ] [ 1 - 3^m = -26 ] [ 3^m = 1 + 26 ] [ 3^m = 27 ] Since ( 3^3 = 27 ), we have: [ 3^m = 3^3 ] Therefore, ( m = 3 ). So, the second sum consists of the first boxed{3} terms of the geometric progression.