answer：Alright, so I have this problem about an ellipse centered at the origin. The minor axis length is given as (2sqrt{2}), and there's a focus at (F(c, 0)) where (c > 0). The directrix corresponding to this focus intersects the x-axis at point (A). It's also given that the distance from the center (O) to the focus (F) is twice the distance from (F) to (A). Then, there's a line passing through (A) intersecting the ellipse at points (P) and (Q). Part (1) asks for the equation of the ellipse and its eccentricity. Let me start by recalling some basic properties of ellipses. The standard equation of an ellipse centered at the origin is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. The eccentricity (e) is given by (e = frac{c}{a}), where (c) is the distance from the center to a focus.Given that the minor axis is (2sqrt{2}), the semi-minor axis (b) is (sqrt{2}). So, (b^2 = 2). Now, the relationship between (a), (b), and (c) in an ellipse is (c^2 = a^2 - b^2). So, if I can find (a), I can find (c) and hence the eccentricity.The problem states that (|OF| = 2|FA|). Since (O) is the origin and (F) is at ((c, 0)), the distance (|OF|) is just (c). The directrix corresponding to the focus (F(c, 0)) is a vertical line at (x = frac{a^2}{c}). This directrix intersects the x-axis at point (A), so the coordinates of (A) are (left(frac{a^2}{c}, 0right)).The distance (|FA|) is the distance between (F(c, 0)) and (Aleft(frac{a^2}{c}, 0right)). Since both points are on the x-axis, this distance is just the absolute difference of their x-coordinates:[|FA| = left|frac{a^2}{c} - cright| = left|frac{a^2 - c^2}{c}right| = frac{a^2 - c^2}{c}]Because (a > c) in an ellipse, this simplifies to (frac{a^2 - c^2}{c}).Given that (|OF| = 2|FA|), substituting the expressions we have:[c = 2 times frac{a^2 - c^2}{c}]Multiplying both sides by (c):[c^2 = 2(a^2 - c^2)]Expanding:[c^2 = 2a^2 - 2c^2]Bringing all terms to one side:[c^2 + 2c^2 - 2a^2 = 0 implies 3c^2 - 2a^2 = 0]So,[3c^2 = 2a^2 implies frac{c^2}{a^2} = frac{2}{3}]But (frac{c^2}{a^2} = e^2), so:[e^2 = frac{2}{3} implies e = sqrt{frac{2}{3}} = frac{sqrt{6}}{3}]So, the eccentricity is (frac{sqrt{6}}{3}).Now, using the relationship (c^2 = a^2 - b^2), and knowing that (b^2 = 2), we can write:[c^2 = a^2 - 2]But from earlier, we have (3c^2 = 2a^2). Let's substitute (c^2) from the first equation into the second:[3(a^2 - 2) = 2a^2]Expanding:[3a^2 - 6 = 2a^2]Subtracting (2a^2) from both sides:[a^2 - 6 = 0 implies a^2 = 6]So, (a = sqrt{6}). Therefore, the equation of the ellipse is:[frac{x^2}{6} + frac{y^2}{2} = 1]Alright, that takes care of part (1). Now, moving on to part (2): If (overrightarrow{OP} cdot overrightarrow{OQ} = 0), find the equation of the line (PQ).First, let's interpret what (overrightarrow{OP} cdot overrightarrow{OQ} = 0) means. The dot product of vectors (overrightarrow{OP}) and (overrightarrow{OQ}) being zero implies that these vectors are perpendicular to each other. So, points (P) and (Q) lie on the ellipse such that the lines from the origin to (P) and (Q) are perpendicular.Given that the line passes through point (A), which we found earlier as (left(frac{a^2}{c}, 0right)). From part (1), (a^2 = 6) and (c^2 = frac{2}{3}a^2 = 4), so (c = 2). Therefore, point (A) is at (left(frac{6}{2}, 0right) = (3, 0)).So, the line (PQ) passes through (A(3, 0)). Let's denote the equation of line (PQ) as (y = k(x - 3)), where (k) is the slope.We need to find the points (P) and (Q) where this line intersects the ellipse. Substituting (y = k(x - 3)) into the ellipse equation:[frac{x^2}{6} + frac{[k(x - 3)]^2}{2} = 1]Expanding this:[frac{x^2}{6} + frac{k^2(x^2 - 6x + 9)}{2} = 1]Multiply through by 6 to eliminate denominators:[x^2 + 3k^2(x^2 - 6x + 9) = 6]Expanding further:[x^2 + 3k^2x^2 - 18k^2x + 27k^2 = 6]Combine like terms:[(1 + 3k^2)x^2 - 18k^2x + (27k^2 - 6) = 0]This is a quadratic equation in (x). Let's denote this as:[Ax^2 + Bx + C = 0]where:- (A = 1 + 3k^2)- (B = -18k^2)- (C = 27k^2 - 6)Let the roots of this quadratic be (x_1) and (x_2), corresponding to the x-coordinates of points (P) and (Q). The corresponding y-coordinates will be (y_1 = k(x_1 - 3)) and (y_2 = k(x_2 - 3)).Given that (overrightarrow{OP} cdot overrightarrow{OQ} = 0), we have:[x_1x_2 + y_1y_2 = 0]Substituting (y_1) and (y_2):[x_1x_2 + [k(x_1 - 3)][k(x_2 - 3)] = 0]Simplify:[x_1x_2 + k^2(x_1 - 3)(x_2 - 3) = 0]Expanding the second term:[x_1x_2 + k^2(x_1x_2 - 3x_1 - 3x_2 + 9) = 0]Factor out (x_1x_2):[x_1x_2(1 + k^2) - 3k^2(x_1 + x_2) + 9k^2 = 0]Now, using Vieta's formulas for the quadratic equation (Ax^2 + Bx + C = 0):- (x_1 + x_2 = -frac{B}{A} = frac{18k^2}{1 + 3k^2})- (x_1x_2 = frac{C}{A} = frac{27k^2 - 6}{1 + 3k^2})Substitute these into the equation:[left(frac{27k^2 - 6}{1 + 3k^2}right)(1 + k^2) - 3k^2left(frac{18k^2}{1 + 3k^2}right) + 9k^2 = 0]Let me compute each term step by step.First term:[left(frac{27k^2 - 6}{1 + 3k^2}right)(1 + k^2) = frac{(27k^2 - 6)(1 + k^2)}{1 + 3k^2}]Second term:[- 3k^2left(frac{18k^2}{1 + 3k^2}right) = frac{-54k^4}{1 + 3k^2}]Third term:[9k^2]So, combining all terms:[frac{(27k^2 - 6)(1 + k^2) - 54k^4}{1 + 3k^2} + 9k^2 = 0]Multiply through by (1 + 3k^2) to eliminate the denominator:[(27k^2 - 6)(1 + k^2) - 54k^4 + 9k^2(1 + 3k^2) = 0]Let me expand each part:First, expand ((27k^2 - 6)(1 + k^2)):[27k^2 times 1 + 27k^2 times k^2 - 6 times 1 - 6 times k^2 = 27k^2 + 27k^4 - 6 - 6k^2 = 21k^2 + 27k^4 - 6]Second, subtract (54k^4):[21k^2 + 27k^4 - 6 - 54k^4 = 21k^2 - 27k^4 - 6]Third, expand (9k^2(1 + 3k^2)):[9k^2 + 27k^4]Now, combine all terms:[21k^2 - 27k^4 - 6 + 9k^2 + 27k^4 = 0]Simplify:- The ( -27k^4) and (+27k^4) cancel out.- Combine (21k^2 + 9k^2 = 30k^2)- So, (30k^2 - 6 = 0)Solving for (k^2):[30k^2 = 6 implies k^2 = frac{6}{30} = frac{1}{5} implies k = pm frac{1}{sqrt{5}} = pm frac{sqrt{5}}{5}]So, the slopes (k) are (frac{sqrt{5}}{5}) and (-frac{sqrt{5}}{5}).Therefore, the equations of the lines (PQ) are:[y = frac{sqrt{5}}{5}(x - 3) quad text{and} quad y = -frac{sqrt{5}}{5}(x - 3)]To write these in standard form, multiply both sides by 5:1. For (k = frac{sqrt{5}}{5}): [ 5y = sqrt{5}(x - 3) implies sqrt{5}x - 5y - 3sqrt{5} = 0 ] Dividing through by (sqrt{5}) to simplify: [ x - sqrt{5}y - 3 = 0 ]2. For (k = -frac{sqrt{5}}{5}): [ 5y = -sqrt{5}(x - 3) implies sqrt{5}x + 5y - 3sqrt{5} = 0 ] Dividing through by (sqrt{5}): [ x + sqrt{5}y - 3 = 0 ]So, the equations of the line (PQ) are (x - sqrt{5}y - 3 = 0) and (x + sqrt{5}y - 3 = 0).Let me just double-check my steps to ensure I didn't make any mistakes. Starting from the condition (overrightarrow{OP} cdot overrightarrow{OQ} = 0), I substituted the coordinates of (P) and (Q) in terms of (x_1, x_2, y_1, y_2), used Vieta's formulas to express the sum and product of roots, and then solved for (k). The algebra seems correct, and the final equations of the lines make sense given the slopes and the point (A(3, 0)). I think that's solid. So, the ellipse equation is (frac{x^2}{6} + frac{y^2}{2} = 1) with eccentricity (frac{sqrt{6}}{3}), and the lines are (x - sqrt{5}y - 3 = 0) and (x + sqrt{5}y - 3 = 0).**Final Answer**(1) The equation of the ellipse is (boxed{dfrac{x^2}{6} + dfrac{y^2}{2} = 1}) and its eccentricity is (boxed{dfrac{sqrt{6}}{3}}).(2) The equations of the line (PQ) are (boxed{x - sqrt{5}y - 3 = 0}) and (boxed{x + sqrt{5}y - 3 = 0}).

answer：Okay, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively, and it's given that a = 7. Point D is the midpoint of side AC, and the dot product of vectors BD and AC is 25/2. I need to find the value of sin C when angle C is at its maximum.Alright, let's start by understanding the given information. We have triangle ABC with sides a, b, c opposite angles A, B, C. So, side a is opposite angle A, which is 7 units long. Point D is the midpoint of AC, so AD = DC. The dot product of vectors BD and AC is 25/2. Hmm, vectors. I remember that the dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. So, maybe I can use that property here. But first, let me try to express vectors BD and AC in terms of the triangle's sides or other vectors.Since D is the midpoint of AC, vector BD can be expressed in terms of vectors BA and BC. I think there's a formula for the midpoint in vectors. Let me recall: if D is the midpoint of AC, then vector BD is equal to (vector BA + vector BC)/2. Yeah, that sounds right.So, vector BD = (vector BA + vector BC)/2. Vector AC is vector C - vector A, but in terms of vectors BA and BC, vector AC is vector BC - vector BA. Wait, let me make sure. If we consider point A as the origin, then vector BA would be from A to B, and vector BC would be from B to C. So, vector AC is from A to C, which is vector BC - vector BA. Hmm, actually, no. If A is the origin, then vector AC is just vector C, and vector BA is vector A - vector B, but this might be getting confusing.Maybe it's better to express everything in terms of position vectors. Let me assign coordinates to the triangle to make it easier. Let's place point A at the origin (0,0), and point C at (c,0), so AC is along the x-axis. Then point B will be somewhere in the plane, say at coordinates (d,e). Then vector AC is (c,0), and vector BD is the vector from B to D. Since D is the midpoint of AC, its coordinates are (c/2, 0). So vector BD is (c/2 - d, 0 - e).Now, the dot product of BD and AC is given as 25/2. So, vector BD is (c/2 - d, -e), and vector AC is (c,0). The dot product is (c/2 - d)*c + (-e)*0 = c*(c/2 - d). This equals 25/2. So, c*(c/2 - d) = 25/2.Simplifying, that's (c²)/2 - c*d = 25/2. So, (c² - 2c*d)/2 = 25/2. Multiply both sides by 2: c² - 2c*d = 25.Hmm, okay. So, that's one equation. Now, I need to relate this to the sides of the triangle. Since a = 7, which is the side opposite angle A, so side BC is length a = 7. So, in my coordinate system, point B is at (d,e), so the distance from B to C is 7. The coordinates of C are (c,0), so distance BC is sqrt[(c - d)² + (0 - e)²] = 7. So, (c - d)² + e² = 49.Also, side AB is length c, since side opposite angle C is AB, which is from A(0,0) to B(d,e), so length AB is sqrt(d² + e²) = c. So, d² + e² = c².So, now I have three equations:1. c² - 2c*d = 252. (c - d)² + e² = 493. d² + e² = c²Let me see if I can solve these equations. From equation 3, d² + e² = c². From equation 2, (c - d)² + e² = 49. Let's expand equation 2: (c² - 2c*d + d²) + e² = 49. But from equation 3, d² + e² = c², so substitute that in: c² - 2c*d + c² = 49. So, 2c² - 2c*d = 49.But from equation 1, c² - 2c*d = 25. So, 2c² - 2c*d = 50. But equation 2 gives 2c² - 2c*d = 49. Wait, that's a contradiction. 50 = 49? That can't be right. Did I make a mistake somewhere?Let me check. Equation 1: c² - 2c*d = 25. Equation 2: (c - d)² + e² = 49. Equation 3: d² + e² = c².Expanding equation 2: c² - 2c*d + d² + e² = 49. But equation 3 says d² + e² = c², so substitute that in: c² - 2c*d + c² = 49 => 2c² - 2c*d = 49.But equation 1 is c² - 2c*d = 25, so 2c² - 2c*d = 50. But equation 2 gives 2c² - 2c*d = 49. So, 50 = 49? That's not possible. Hmm, maybe I messed up the vector BD.Wait, let's go back to the vector BD. I placed A at (0,0), C at (c,0), and B at (d,e). Then D is the midpoint of AC, so D is at (c/2, 0). So, vector BD is D - B, which is (c/2 - d, 0 - e). Vector AC is C - A, which is (c, 0). So, the dot product is (c/2 - d)*c + (-e)*0 = c*(c/2 - d). So, that's correct.So, equation 1 is correct: c² - 2c*d = 25.Equation 2 is distance BC: sqrt[(c - d)^2 + e^2] = 7, so (c - d)^2 + e^2 = 49.Equation 3 is distance AB: sqrt[d^2 + e^2] = c, so d^2 + e^2 = c^2.So, plugging equation 3 into equation 2: (c - d)^2 + (c^2 - d^2) = 49.Wait, hold on. If d² + e² = c², then e² = c² - d². So, equation 2 becomes (c - d)^2 + (c² - d²) = 49.Let me expand (c - d)^2: c² - 2c*d + d². So, equation 2 becomes c² - 2c*d + d² + c² - d² = 49. Simplify: 2c² - 2c*d = 49.But equation 1 is c² - 2c*d = 25. So, 2c² - 2c*d = 50. But equation 2 gives 2c² - 2c*d = 49. So, 50 = 49? That's impossible. Hmm, so something's wrong here.Wait, maybe my coordinate system is causing confusion. Let me try a different approach without coordinates.Given that D is the midpoint of AC, so AD = DC. The vector BD can be expressed as (BA + BC)/2, as I thought earlier. So, vector BD = (vector BA + vector BC)/2.Vector AC is vector C - vector A, which is vector AC. So, the dot product BD · AC = [(BA + BC)/2] · AC.Let me express BA and BC in terms of vectors. Vector BA is from A to B, which is vector B - vector A. Vector BC is from B to C, which is vector C - vector B.So, vector BD = (vector B - vector A + vector C - vector B)/2 = (vector C - vector A)/2. Wait, that simplifies to vector AC / 2. So, vector BD = vector AC / 2.But then the dot product BD · AC would be (vector AC / 2) · vector AC = (|AC|²)/2. So, |AC|² / 2 = 25/2. Therefore, |AC|² = 25, so |AC| = 5.But |AC| is side AB? Wait, no. In triangle ABC, side opposite angle A is BC, which is a = 7. Side opposite angle B is AC, which is b. Side opposite angle C is AB, which is c.Wait, so |AC| is side b. So, |AC| = b = 5. So, b = 5.Wait, that's interesting. So, from this, we get that side AC, which is opposite angle B, is 5 units long.But earlier, I thought that side AB is c, which is opposite angle C. So, side AB is c, AC is b = 5, and BC is a = 7.So, now, with sides a = 7, b = 5, and c unknown. We can use the Law of Cosines to relate the sides and angles.But the problem is asking for sin C when angle C is at its maximum. So, perhaps I need to find the maximum possible value of angle C given the constraints.Wait, but in a triangle, the angles are determined by the sides. So, if sides a, b, c are fixed, the angles are fixed. But in this case, we have sides a = 7, b = 5, and c is variable? Or is c fixed?Wait, no. Wait, from the earlier calculation, we found that |AC| = b = 5. So, sides a = 7, b = 5, and c is the side opposite angle C, which is AB.But in a triangle, the sides must satisfy the triangle inequality. So, a + b > c, a + c > b, b + c > a.So, 7 + 5 > c => c < 127 + c > 5 => c > -2 (which is always true)5 + c > 7 => c > 2So, c must be between 2 and 12.But angle C is opposite side c. So, as c increases, angle C increases, up to a point. But in a triangle, the largest angle is opposite the largest side. So, if c is the largest side, angle C is the largest angle.But in our case, sides a = 7, b = 5, c is variable between 2 and 12. So, when c is greater than 7, angle C becomes larger than angle A. So, to maximize angle C, we need to maximize c, but c can't exceed 12. However, when c approaches 12, angle C approaches 180 degrees, but in reality, c must be less than 12.But wait, actually, in a triangle, the maximum angle is less than 180 degrees, and the sides must satisfy the triangle inequality. So, the maximum value of angle C occurs when c is as large as possible, but still satisfying the triangle inequality. So, c approaches 12, but can't reach it.But in our case, is c fixed? Wait, earlier, from the vector dot product, we found that |AC| = b = 5. So, sides a = 7, b = 5, and c is variable. So, c can vary, but in this triangle, sides a, b, c must satisfy the triangle inequality.But wait, in the problem statement, it's given that a = 7, and D is the midpoint of AC. The dot product BD · AC = 25/2. From that, we found that |AC| = 5, so b = 5.So, sides a = 7, b = 5, and c is variable. So, c can vary between 2 and 12, as above.But then, how does angle C vary? Since angle C is opposite side c, as c increases, angle C increases. So, to maximize angle C, we need to maximize c. But c can't exceed 12, but c must satisfy the triangle inequality.Wait, but in reality, c can't be 12 because then the triangle would be degenerate. So, the maximum angle C occurs when c is as large as possible, approaching 12, but not reaching it. However, in reality, c can be up to just less than 12.But wait, in our case, is c fixed? Because from the vector calculation, we found that |AC| = 5, so side b = 5. So, sides a = 7, b = 5, and c is variable. So, c can vary, but in this triangle, sides a, b, c must satisfy the triangle inequality.Wait, but in the problem, is c given? Or is it variable? Hmm, the problem says "It is known that BD · AC = 25/2." From that, we found that |AC| = 5, so side b = 5. So, sides a = 7, b = 5, and c is variable.So, in this triangle, sides a = 7, b = 5, c is variable, and angle C is opposite side c. So, angle C can vary depending on the value of c.But we need to find sin C when angle C is at its maximum. So, when is angle C maximized? As I thought earlier, when c is maximized. But c can't exceed 12, but in reality, c can be up to just less than 12.But wait, is there another constraint? Because in the problem, we have BD · AC = 25/2, which gave us |AC| = 5. So, is c fixed? Or is it variable?Wait, no. Wait, in the problem, sides a, b, c are opposite angles A, B, C respectively, with a = 7. So, side a is fixed at 7, side b is AC, which we found to be 5, and side c is AB, which is variable.So, sides a = 7, b = 5, and c is variable, with c between 2 and 12. So, angle C is opposite side c, so as c increases, angle C increases.Therefore, to maximize angle C, we need to maximize c. So, c approaches 12, but can't reach it. However, in reality, we can find the maximum value of angle C when c is as large as possible.But wait, in our earlier calculation, we found that |AC| = 5, so side b = 5. So, sides a = 7, b = 5, c is variable. So, using the Law of Cosines, we can express angle C in terms of c.Law of Cosines: c² = a² + b² - 2ab cos C.So, cos C = (a² + b² - c²)/(2ab).So, as c increases, cos C decreases, so angle C increases.Therefore, to maximize angle C, we need to minimize cos C, which occurs when c is maximized.But c is constrained by the triangle inequality: c < a + b = 12.So, as c approaches 12, cos C approaches (49 + 25 - 144)/(2*7*5) = (-70)/70 = -1. So, angle C approaches 180 degrees, but that's not possible in a triangle.Wait, but in reality, c can't be 12 because the triangle would be degenerate. So, the maximum angle C is less than 180 degrees.But perhaps, instead of c approaching 12, there's another constraint from the given dot product.Wait, no. The dot product gave us |AC| = 5, so side b = 5. So, sides a = 7, b = 5, c variable.Wait, but in the problem, we are to find sin C when angle C is at its maximum. So, perhaps angle C is maximized when c is as large as possible, which is approaching 12, but we need to find the exact value.But maybe there's another way. Let's think about the Law of Sines.Law of Sines: a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius.So, if we can express sin C in terms of c, and then find when angle C is maximized.But since angle C is maximized when c is maximized, but c is limited by the triangle inequality.Wait, but perhaps there's a relation between sides a, b, c and the position of point D.Wait, earlier, we found that |AC| = 5, so b = 5. So, sides a = 7, b = 5, c is variable.But in the problem, point D is the midpoint of AC, so AD = DC = 5/2.Wait, but in our coordinate system earlier, we had AC = 5, so coordinates of A(0,0), C(5,0), D(2.5, 0). Point B is somewhere in the plane, say (d,e). Then, vector BD is (2.5 - d, -e), and vector AC is (5,0). Their dot product is 5*(2.5 - d) + 0*(-e) = 12.5 - 5d = 25/2 = 12.5. So, 12.5 - 5d = 12.5 => -5d = 0 => d = 0.Wait, so d = 0. So, point B is at (0, e). So, coordinates of B are (0, e). So, point B is somewhere along the y-axis.So, in this coordinate system, point A is (0,0), point C is (5,0), point B is (0,e). So, triangle ABC has vertices at (0,0), (5,0), and (0,e).So, side AB is from (0,0) to (0,e), so length AB is |e| = c.Side BC is from (0,e) to (5,0), so length BC is sqrt[(5)^2 + (e)^2] = sqrt(25 + e²) = a = 7.So, sqrt(25 + e²) = 7 => 25 + e² = 49 => e² = 24 => e = sqrt(24) = 2*sqrt(6). So, e = 2√6.So, point B is at (0, 2√6). So, coordinates are A(0,0), B(0, 2√6), C(5,0).So, side AB is c = 2√6, side BC is 7, side AC is 5.Therefore, sides a = 7, b = 5, c = 2√6.So, now, angle C is the angle at point C, which is at (5,0). So, angle C is the angle between sides AC and BC.So, to find sin C, we can use the Law of Sines or Law of Cosines.Law of Cosines: c² = a² + b² - 2ab cos C.So, (2√6)² = 7² + 5² - 2*7*5 cos C.Calculating:(2√6)² = 4*6 = 247² + 5² = 49 + 25 = 74So, 24 = 74 - 70 cos CSo, 70 cos C = 74 - 24 = 50So, cos C = 50 / 70 = 5/7Therefore, sin C = sqrt(1 - (5/7)^2) = sqrt(1 - 25/49) = sqrt(24/49) = (2√6)/7So, sin C = 2√6 / 7.But wait, the problem says "when angle C reaches its maximum value." But in this case, angle C is fixed because sides a, b, c are fixed. So, how can angle C vary?Wait, earlier, I thought that c was variable, but from the given dot product, we found that |AC| = 5, so side b = 5, and side a = 7, so sides a and b are fixed. Therefore, side c is fixed as well because in the coordinate system, we found c = 2√6.Wait, so in this case, triangle ABC is fixed with sides a = 7, b = 5, c = 2√6. So, angle C is fixed, and sin C is 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, maybe I misunderstood something.Wait, perhaps the initial assumption that vector BD = (vector BA + vector BC)/2 is incorrect. Let me double-check.Yes, in general, the midpoint D of AC can be expressed as (A + C)/2. So, vector BD = D - B = (A + C)/2 - B.But in terms of vectors BA and BC, let's see. Vector BA = A - B, vector BC = C - B.So, (vector BA + vector BC)/2 = (A - B + C - B)/2 = (A + C - 2B)/2 = (A + C)/2 - B = D - B = vector BD.So, that's correct. So, vector BD = (vector BA + vector BC)/2.Therefore, BD · AC = [(vector BA + vector BC)/2] · (vector C - vector A).But in our coordinate system, vector AC is (5,0), vector BD is (2.5 - d, -e). So, their dot product is 5*(2.5 - d) + 0*(-e) = 12.5 - 5d = 25/2 = 12.5. So, 12.5 - 5d = 12.5 => d = 0.So, point B must be at (0,e). So, in this case, triangle ABC is fixed with sides a = 7, b = 5, c = 2√6. So, angle C is fixed, and sin C is 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, perhaps my initial assumption that c is variable is incorrect, and in reality, c is fixed due to the given dot product.Wait, but in the problem statement, it's given that a = 7, and D is the midpoint of AC, and BD · AC = 25/2. From that, we found that |AC| = 5, so side b = 5. So, sides a = 7, b = 5, and c is determined by the triangle.So, in this case, triangle ABC is fixed with sides a = 7, b = 5, c = 2√6, so angle C is fixed, and sin C is 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, perhaps I'm missing something. Maybe the triangle isn't fixed, and c can vary, but with the given dot product constraint.Wait, let's think again. The dot product BD · AC = 25/2. From that, we found that |AC| = 5, so side b = 5. So, sides a = 7, b = 5, and c is variable. But in the coordinate system, we found that point B must be at (0,e), so c = e, and side BC = 7, so e = 2√6, so c = 2√6.So, in this case, c is fixed, so angle C is fixed. Therefore, sin C is fixed at 2√6 / 7.Wait, but the problem says "when angle C reaches its maximum value." So, perhaps the initial assumption that c is fixed is incorrect, and c can vary, but with the given dot product constraint.Wait, let's consider that. Suppose that point D is the midpoint of AC, and BD · AC = 25/2. So, regardless of the position of B, this condition must hold. So, perhaps c can vary, but with the condition that BD · AC = 25/2.Wait, but in our earlier coordinate system, we found that BD · AC = 25/2 implies that |AC| = 5, so side b = 5. So, side AC is fixed at 5, so sides a = 7, b = 5, and c is variable.But in that case, point B must lie somewhere such that BD · AC = 25/2, which in our coordinate system forced point B to be at (0,e). So, in that case, c is fixed at 2√6.Wait, maybe I need to consider that point B can move such that BD · AC remains 25/2, but AC is fixed at 5. So, perhaps c can vary, but with the condition that BD · AC = 25/2.Wait, but in our coordinate system, BD · AC = 25/2 forced point B to be at (0,e), so c is fixed. So, perhaps c is fixed, and angle C is fixed.But the problem says "when angle C reaches its maximum value." So, maybe I need to consider that point B can move such that BD · AC = 25/2, but AC is not fixed.Wait, but earlier, from BD · AC = 25/2, we found that |AC| = 5, so AC is fixed at 5. So, sides a = 7, b = 5, c is fixed at 2√6.Therefore, angle C is fixed, and sin C is 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, perhaps the problem is not in a fixed coordinate system, and AC is not fixed at 5, but rather, AC can vary, but BD · AC = 25/2.Wait, but in our earlier calculation, BD · AC = 25/2 led us to |AC| = 5, so AC is fixed at 5. So, sides a = 7, b = 5, c is fixed at 2√6.Therefore, angle C is fixed, and sin C is 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, perhaps the problem is not as I interpreted it.Wait, maybe I made a mistake in the vector calculation. Let me try again.Given that D is the midpoint of AC, so vector BD = (vector BA + vector BC)/2.Then, BD · AC = [(vector BA + vector BC)/2] · (vector AC).Expressed in terms of vectors, this is (vector BA + vector BC)/2 · (vector C - vector A).But vector BA = vector A - vector B, vector BC = vector C - vector B.So, vector BA + vector BC = (vector A - vector B) + (vector C - vector B) = vector A + vector C - 2 vector B.So, BD = (vector A + vector C - 2 vector B)/2.Then, BD · AC = [(vector A + vector C - 2 vector B)/2] · (vector C - vector A).Let me compute this dot product.Let me denote vector A as a, vector B as b, vector C as c.So, BD = (a + c - 2b)/2.AC = c - a.So, BD · AC = [(a + c - 2b)/2] · (c - a) = [ (a + c - 2b) · (c - a) ] / 2.Expanding the numerator:(a + c - 2b) · (c - a) = a · (c - a) + c · (c - a) - 2b · (c - a)= a · c - |a|² + |c|² - c · a - 2b · c + 2b · aSimplify:a · c - |a|² + |c|² - a · c - 2b · c + 2b · a= (-|a|² + |c|²) + (-2b · c + 2b · a)= (|c|² - |a|²) + 2b · (a - c)So, BD · AC = [ (|c|² - |a|²) + 2b · (a - c) ] / 2But in our case, BD · AC = 25/2.So,(|c|² - |a|²)/2 + b · (a - c) = 25/2But I'm not sure if this helps. Maybe it's better to use coordinate geometry.Let me place point A at (0,0), point C at (c,0), so AC is along the x-axis. Point D is the midpoint, so D is at (c/2, 0). Point B is at (d,e).Vector BD is D - B = (c/2 - d, -e).Vector AC is C - A = (c,0).Dot product BD · AC = (c/2 - d)*c + (-e)*0 = c*(c/2 - d) = 25/2.So, c*(c/2 - d) = 25/2 => c²/2 - c*d = 25/2 => c² - 2c*d = 25.Also, side BC is length a = 7, so distance from B(d,e) to C(c,0) is 7:sqrt( (c - d)^2 + (0 - e)^2 ) = 7 => (c - d)^2 + e² = 49.Side AB is length c, so distance from A(0,0) to B(d,e) is c:sqrt(d² + e²) = c => d² + e² = c².So, we have three equations:1. c² - 2c*d = 252. (c - d)^2 + e² = 493. d² + e² = c²From equation 3, e² = c² - d².Substitute into equation 2:(c - d)^2 + (c² - d²) = 49Expand (c - d)^2: c² - 2c*d + d²So, equation 2 becomes:c² - 2c*d + d² + c² - d² = 49Simplify:2c² - 2c*d = 49But from equation 1: c² - 2c*d = 25Multiply equation 1 by 2: 2c² - 4c*d = 50Subtract equation 2 from this:(2c² - 4c*d) - (2c² - 2c*d) = 50 - 49-2c*d = 1 => c*d = -1/2From equation 1: c² - 2c*d = 25But c*d = -1/2, so 2c*d = -1Thus, equation 1 becomes:c² - (-1) = 25 => c² + 1 = 25 => c² = 24 => c = sqrt(24) = 2*sqrt(6)So, c = 2√6.Then, from c*d = -1/2, d = (-1/2)/c = (-1/2)/(2√6) = -1/(4√6) = -√6/24Wait, that seems small. Let me check:c = 2√6, so c ≈ 4.899d = (-1/2)/c ≈ (-0.5)/4.899 ≈ -0.102So, d ≈ -0.102Then, from equation 3: d² + e² = c²So, e² = c² - d² ≈ (24) - (0.0104) ≈ 23.9896So, e ≈ sqrt(23.9896) ≈ 4.898So, point B is at (d,e) ≈ (-0.102, 4.898)So, in this case, sides are:a = 7, b = AC = c = 2√6 ≈ 4.899, c = AB = sqrt(d² + e²) = sqrt( ( (-√6/24)^2 + (sqrt(23.9896))^2 )) ≈ sqrt( (0.0104 + 23.9896) ) ≈ sqrt(24) = 2√6.Wait, that's interesting. So, side AB is also 2√6. So, sides a = 7, b = 2√6, c = 2√6.Wait, so triangle ABC has sides a = 7, b = 2√6, c = 2√6. So, it's an isoceles triangle with sides b = c = 2√6, and base a = 7.So, in this case, angles B and C are equal because sides b and c are equal.Wait, but in our coordinate system, point B is at (-√6/24, sqrt(23.9896)) ≈ (-0.102, 4.898). So, it's very close to the y-axis but slightly to the left.But in this case, angle C is at point C(2√6, 0). Wait, no, in our coordinate system, point C is at (c,0) = (2√6, 0). Wait, no, earlier, we set point C at (c,0), but in our calculation, c = 2√6, so point C is at (2√6, 0). But wait, in our earlier setup, we had AC = 5, but now c = 2√6 ≈ 4.899, which is approximately 5. So, that's consistent.Wait, but earlier, we found that |AC| = 5, so c = 5. But now, we have c = 2√6 ≈ 4.899, which is approximately 5. So, that's consistent because 2√6 ≈ 4.899 ≈ 5.Wait, but 2√6 is exactly sqrt(24), which is approximately 4.899, which is less than 5. So, |AC| = 2√6, not 5. Wait, but earlier, we thought |AC| = 5 because BD · AC = 25/2. But now, with c = 2√6, |AC| = 2√6, and BD · AC = 25/2.Wait, let me recast this.In our coordinate system, AC is from (0,0) to (c,0), so |AC| = c.From BD · AC = 25/2, we found that c*(c/2 - d) = 25/2.But from solving the equations, we found c = 2√6, d = -√6/24.So, |AC| = c = 2√6, not 5.Wait, so earlier, I thought |AC| = 5, but that was a mistake. Because in the coordinate system, AC is from (0,0) to (c,0), so |AC| = c.From BD · AC = 25/2, we have c*(c/2 - d) = 25/2.But from solving, we found c = 2√6, so |AC| = 2√6.Therefore, sides are a = 7, b = AC = 2√6, c = AB = 2√6.So, triangle ABC is isoceles with sides AB = AC = 2√6, and base BC = 7.Wait, but in this case, angle C is at point C, which is opposite side AB = 2√6.Wait, no, in triangle ABC, side opposite angle A is BC = a = 7, side opposite angle B is AC = b = 2√6, and side opposite angle C is AB = c = 2√6.So, since sides b and c are equal, angles B and C are equal.Therefore, angle C is equal to angle B.But the problem says "when angle C reaches its maximum value." So, in this case, angle C is fixed because the triangle is fixed with sides a = 7, b = c = 2√6.Therefore, angle C is fixed, and sin C is fixed at 2√6 / 7.But the problem says "when angle C reaches its maximum value." So, perhaps the initial assumption that c is fixed is incorrect, and c can vary, but with the given dot product constraint.Wait, but in our earlier calculation, we found that c is fixed at 2√6 due to the given dot product. So, perhaps angle C is fixed, and the maximum value is just sin C = 2√6 / 7.Alternatively, maybe I need to consider that point B can move such that BD · AC = 25/2, but AC is not fixed. So, AC can vary, and c can vary, but with the condition BD · AC = 25/2.But in that case, we need to express BD · AC in terms of variables and find the maximum angle C.Wait, let's try that approach.Let me denote AC = x, so |AC| = x.Since D is the midpoint, AD = DC = x/2.Vector BD = (vector BA + vector BC)/2.Vector AC = vector C - vector A.So, BD · AC = [(vector BA + vector BC)/2] · (vector C - vector A)Expressed in terms of vectors, this is (vector BA + vector BC)/2 · vector AC.But vector BA = vector A - vector B, vector BC = vector C - vector B.So, vector BA + vector BC = vector A + vector C - 2 vector B.Thus, BD = (vector A + vector C - 2 vector B)/2.So, BD · AC = [(vector A + vector C - 2 vector B)/2] · (vector C - vector A)Let me compute this:= [ (vector A + vector C - 2 vector B) · (vector C - vector A) ] / 2= [ vector A · (vector C - vector A) + vector C · (vector C - vector A) - 2 vector B · (vector C - vector A) ] / 2= [ (vector A · vector C - |A|²) + (|C|² - vector C · vector A) - 2 vector B · vector C + 2 vector B · vector A ] / 2Simplify:= [ (vector A · vector C - |A|² + |C|² - vector A · vector C) - 2 vector B · vector C + 2 vector B · vector A ] / 2= [ (-|A|² + |C|²) - 2 vector B · vector C + 2 vector B · vector A ] / 2= [ (|C|² - |A|²) + 2 vector B · (vector A - vector C) ] / 2But in our case, BD · AC = 25/2.So,(|C|² - |A|²)/2 + vector B · (vector A - vector C) = 25/2But I'm not sure how to proceed from here without coordinate geometry.Alternatively, let's use coordinate geometry again, but this time, let AC be variable.Let me place point A at (0,0), point C at (x,0), so AC = x. Point D is the midpoint, so D is at (x/2, 0). Point B is at (d,e).Vector BD = D - B = (x/2 - d, -e).Vector AC = C - A = (x,0).Dot product BD · AC = (x/2 - d)*x + (-e)*0 = x*(x/2 - d) = 25/2.So, x*(x/2 - d) = 25/2 => x²/2 - x*d = 25/2 => x² - 2x*d = 25.Also, side BC is length a = 7, so distance from B(d,e) to C(x,0) is 7:sqrt( (x - d)^2 + (0 - e)^2 ) = 7 => (x - d)^2 + e² = 49.Side AB is length c, so distance from A(0,0) to B(d,e) is c:sqrt(d² + e²) = c => d² + e² = c².So, we have three equations:1. x² - 2x*d = 252. (x - d)^2 + e² = 493. d² + e² = c²From equation 3, e² = c² - d².Substitute into equation 2:(x - d)^2 + (c² - d²) = 49Expand (x - d)^2: x² - 2x*d + d²So, equation 2 becomes:x² - 2x*d + d² + c² - d² = 49Simplify:x² - 2x*d + c² = 49But from equation 1: x² - 2x*d = 25So, substitute into equation 2:25 + c² = 49 => c² = 24 => c = 2√6So, c is fixed at 2√6.Therefore, side AB is fixed at 2√6, side BC is fixed at 7, and side AC is variable, but from equation 1: x² - 2x*d = 25.But from equation 3: d² + e² = c² = 24.So, we have:x² - 2x*d = 25andd² + e² = 24But from equation 2: (x - d)^2 + e² = 49Which we used to find c² = 24.So, in this case, x is variable, but c is fixed at 2√6.Wait, but if c is fixed, then angle C is fixed as well, because in triangle ABC, sides a = 7, b = x, c = 2√6.So, angle C is opposite side c = 2√6, so angle C can vary depending on x.Wait, but in our earlier calculation, we found that c is fixed at 2√6, so angle C is fixed.Wait, no, because side AC = x is variable, so sides a = 7, b = x, c = 2√6.So, using the Law of Cosines, angle C is opposite side c = 2√6, so:cos C = (a² + b² - c²)/(2ab) = (49 + x² - 24)/(2*7*x) = (x² + 25)/(14x)So, cos C = (x² + 25)/(14x)To find the maximum angle C, we need to minimize cos C.Because as cos C decreases, angle C increases.So, to minimize cos C, we need to minimize (x² + 25)/(14x).Let me denote f(x) = (x² + 25)/(14x). We need to find the minimum of f(x).To find the minimum, take derivative of f(x) with respect to x and set to zero.f(x) = (x² + 25)/(14x) = (x²)/(14x) + 25/(14x) = x/14 + 25/(14x)So, f(x) = (x)/14 + 25/(14x)Compute derivative f’(x):f’(x) = 1/14 - 25/(14x²)Set f’(x) = 0:1/14 - 25/(14x²) = 0 => 1/14 = 25/(14x²) => 1 = 25/x² => x² = 25 => x = 5 (since x > 0)So, the minimum of f(x) occurs at x = 5.Therefore, the minimum value of cos C is f(5) = (25 + 25)/(14*5) = 50/70 = 5/7.So, cos C = 5/7, which is the minimum value, so angle C is maximized when cos C is minimized, which is 5/7.Therefore, sin C = sqrt(1 - (5/7)^2) = sqrt(1 - 25/49) = sqrt(24/49) = 2√6 / 7.So, sin C = 2√6 / 7 when angle C is maximized.Therefore, the answer is 2√6 / 7.

answer：First, I need to determine the face value of the bill by adding the present worth and the true discount. Next, I'll calculate the banker's discount using the formula that involves the face value and the true discount. Finally, I'll perform the division to find the exact amount of the banker's discount.

answer：First, I'll define the variables involved in the problem. Let ( S_A ) represent the number of sprockets machine A produces per hour, and ( T_G ) represent the time it takes for machine G to produce 2000 sprockets.From the problem statement:- Machine A takes 10 hours longer than machine G to produce 2000 sprockets, so its time is ( T_G + 10 ).- Machine B takes 5 hours less than machine G to produce 2000 sprockets, so its time is ( T_G - 5 ).- Machine G produces 10% more sprockets per hour than machine A, so its rate is ( 1.10S_A ).- Machine B produces 15% more sprockets per hour than machine A, so its rate is ( 1.15S_A ).Next, I'll set up the equations based on the total production of 2000 sprockets for each machine:- For machine A: ( S_A times (T_G + 10) = 2000 )- For machine G: ( 1.10S_A times T_G = 2000 )- For machine B: ( 1.15S_A times (T_G - 5) = 2000 )I'll solve for ( T_G ) from the equation for machine G:[ T_G = frac{2000}{1.10S_A} ]Substituting ( T_G ) into the equation for machine A:[ S_A times left(frac{2000}{1.10S_A} + 10right) = 2000 ][ frac{2000}{1.10} + 10S_A = 2000 ][ 10S_A = 2000 - frac{2000}{1.10} ][ 10S_A = frac{2000}{11} ][ S_A = frac{200}{11} ][ S_A approx 18.18 ]Therefore, machine A produces approximately 18.18 sprockets per hour.

answer：First, I need to determine the total number of homework problems Isabel has, which is 72.0.Next, I'll consider the total number of sub tasks she needs to solve, which is 200.To find out how many sub tasks are in each problem, I'll divide the total number of sub tasks by the total number of problems.So, 200 sub tasks divided by 72.0 problems equals approximately 2.78 sub tasks per problem.Since it's not practical to have a fraction of a sub task, I'll round this to the nearest whole number, which is 3 sub tasks per problem.

answer：Okay, so I have this problem with two equations and I need to figure out which of the given options are true. Let me take it step by step.First, let's look at Equation I: x + y + z = 48, where x, y, and z are positive integers. The options are about whether this equation can be solved with consecutive integers, consecutive odd integers, etc. Similarly, Equation II is x + y + z + w = 52, and we need to check similar things for that.Starting with Equation I and Option A: Can it be solved with consecutive integers?Alright, if x, y, z are consecutive integers, that means they follow one after the other without gaps. So, let's denote them as y - 1, y, y + 1. Then, adding them up:(y - 1) + y + (y + 1) = 48Simplify that:y - 1 + y + y + 1 = 48The -1 and +1 cancel out, so we have:3y = 48Divide both sides by 3:y = 16So, y is 16, which means x = 15 and z = 17. Let me check if these add up to 48:15 + 16 + 17 = 48Yes, that works. So, Option A is correct.Moving on to Option B: Can Equation I be solved with consecutive odd integers?Consecutive odd integers would be numbers like 1, 3, 5, etc., each two apart. So, let me denote them as y - 2, y, y + 2. Then, adding them:(y - 2) + y + (y + 2) = 48Simplify:y - 2 + y + y + 2 = 48Again, the -2 and +2 cancel out:3y = 48So, y = 16Wait, but y is supposed to be an odd integer. 16 is even, so that doesn't work. Therefore, Equation I cannot be solved with consecutive odd integers. So, Option B is incorrect.Now, moving on to Equation II: x + y + z + w = 52, with x, y, z, w as positive integers. Let's check the options for this equation.Option C: Can Equation II be solved with consecutive integers?Let's denote them as y - 3, y - 1, y + 1, y + 3. Wait, no, that's for four consecutive integers. Actually, consecutive integers would be y, y + 1, y + 2, y + 3. Let me set them up that way.So, x = y, y = y + 1, z = y + 2, w = y + 3. Adding them up:y + (y + 1) + (y + 2) + (y + 3) = 52Simplify:4y + 6 = 52Subtract 6 from both sides:4y = 46Divide by 4:y = 11.5Hmm, y is 11.5, which is not an integer. Since we need positive integers, this doesn't work. So, Option C is incorrect.Option D: Can Equation II be solved with consecutive odd integers?Consecutive odd integers would be numbers like 1, 3, 5, 7, each two apart. Let me denote them as y, y + 2, y + 4, y + 6. Adding them:y + (y + 2) + (y + 4) + (y + 6) = 52Simplify:4y + 12 = 52Subtract 12:4y = 40Divide by 4:y = 10But y is supposed to be an odd integer, and 10 is even. So, this doesn't work. Therefore, Option D is incorrect.Option E: Can Equation II be solved with consecutive even integers?Consecutive even integers are numbers like 2, 4, 6, 8, each two apart. Let's denote them as y, y + 2, y + 4, y + 6. Adding them:y + (y + 2) + (y + 4) + (y + 6) = 52Simplify:4y + 12 = 52Subtract 12:4y = 40Divide by 4:y = 10Okay, y is 10, which is even, so that works. Let me check the numbers:10, 12, 14, 16Adding them up:10 + 12 + 14 + 16 = 52Yes, that adds up correctly. So, Option E is correct.So, summarizing:- Option A: Correct- Option B: Incorrect- Option C: Incorrect- Option D: Incorrect- Option E: CorrectTherefore, the correct answers are A and E.**Final Answer**The correct options are boxed{A} and boxed{E}.