answer：Okay, so I have this problem where points A and B are moving on two lines, y equals sqrt(2)/2 x and y equals -sqrt(2)/2 x. The distance between A and B is 2 sqrt(2). Point P is defined such that the vector OP is the sum of vectors OA and OB. I need to find the trajectory of point P, which is called C1. Then, in part II, I have to find the equation of a hyperbola C2 that has one focus at the upper vertex of C1 and has an asymptote x + 2y = 0.Alright, starting with part I. Let me visualize this. Points A and B are moving on two lines that pass through the origin with slopes sqrt(2)/2 and -sqrt(2)/2. These lines are symmetric with respect to the x-axis. The distance between A and B is fixed at 2 sqrt(2). Point P is the vector sum of OA and OB, so it's like the diagonal of the parallelogram formed by OA and OB.Let me denote the coordinates of A as (x1, y1) and B as (x2, y2). Since A is on the line y = sqrt(2)/2 x, y1 = sqrt(2)/2 x1. Similarly, since B is on y = -sqrt(2)/2 x, y2 = -sqrt(2)/2 x2.So, the coordinates of A are (x1, sqrt(2)/2 x1) and B are (x2, -sqrt(2)/2 x2).The distance between A and B is given as 2 sqrt(2). So, using the distance formula:sqrt[(x1 - x2)^2 + (y1 - y2)^2] = 2 sqrt(2)Substituting y1 and y2:sqrt[(x1 - x2)^2 + (sqrt(2)/2 x1 + sqrt(2)/2 x2)^2] = 2 sqrt(2)Let me square both sides to eliminate the square root:(x1 - x2)^2 + (sqrt(2)/2 (x1 + x2))^2 = (2 sqrt(2))^2Simplify each term:First term: (x1 - x2)^2 = x1^2 - 2x1x2 + x2^2Second term: (sqrt(2)/2 (x1 + x2))^2 = (2/4)(x1 + x2)^2 = (1/2)(x1^2 + 2x1x2 + x2^2)Right side: (2 sqrt(2))^2 = 8So, putting it all together:x1^2 - 2x1x2 + x2^2 + (1/2)(x1^2 + 2x1x2 + x2^2) = 8Let me expand the second term:(1/2)x1^2 + x1x2 + (1/2)x2^2Now, adding the first term:x1^2 - 2x1x2 + x2^2 + (1/2)x1^2 + x1x2 + (1/2)x2^2Combine like terms:x1^2 + (1/2)x1^2 = (3/2)x1^2-2x1x2 + x1x2 = (-x1x2)x2^2 + (1/2)x2^2 = (3/2)x2^2So, overall:(3/2)x1^2 - x1x2 + (3/2)x2^2 = 8Hmm, that seems a bit complicated. Maybe I can factor it differently or find a substitution.Alternatively, since point P is defined as OP = OA + OB, so P has coordinates (x1 + x2, y1 + y2). Let me denote P as (x, y). So:x = x1 + x2y = y1 + y2 = (sqrt(2)/2 x1) + (-sqrt(2)/2 x2) = sqrt(2)/2 (x1 - x2)So, from these, I can express x1 and x2 in terms of x and y.Let me solve for x1 and x2.From x = x1 + x2, we have x1 = x - x2.Substitute into y:y = sqrt(2)/2 (x1 - x2) = sqrt(2)/2 ((x - x2) - x2) = sqrt(2)/2 (x - 2x2)So, y = sqrt(2)/2 x - sqrt(2) x2Let me solve for x2:sqrt(2) x2 = sqrt(2)/2 x - yDivide both sides by sqrt(2):x2 = (1/2)x - y / sqrt(2)Similarly, from x = x1 + x2, x1 = x - x2 = x - [(1/2)x - y / sqrt(2)] = (1/2)x + y / sqrt(2)So, x1 = (1/2)x + y / sqrt(2)x2 = (1/2)x - y / sqrt(2)Now, let me substitute x1 and x2 into the earlier equation we had:(3/2)x1^2 - x1x2 + (3/2)x2^2 = 8But this might be messy. Alternatively, maybe I can use the expressions for x1 and x2 in terms of x and y and substitute back into the distance equation.Wait, let's recall that we had:(x1 - x2)^2 + (sqrt(2)/2 (x1 + x2))^2 = 8But from x = x1 + x2 and y = sqrt(2)/2 (x1 - x2), we can express (x1 - x2) as y * 2 / sqrt(2) = sqrt(2) y.Similarly, (x1 + x2) is x.So, substituting back into the distance equation:(x1 - x2)^2 + (sqrt(2)/2 (x1 + x2))^2 = (sqrt(2) y)^2 + (sqrt(2)/2 x)^2 = 2 y^2 + (1/2) x^2 = 8So, 2 y^2 + (1/2) x^2 = 8Multiply both sides by 2 to eliminate the fraction:4 y^2 + x^2 = 16Divide both sides by 16 to write in standard form:x^2 / 16 + y^2 / 4 = 1So, that's an ellipse with semi-major axis 4 along the x-axis and semi-minor axis 2 along the y-axis.Therefore, the trajectory equation C1 is x²/16 + y²/4 = 1.Okay, that seems solid. Let me just recap:- Expressed A and B in terms of x1 and x2.- Used the distance formula to get an equation involving x1 and x2.- Expressed P's coordinates in terms of x1 and x2.- Expressed x1 and x2 in terms of P's coordinates.- Substituted back into the distance equation to eliminate x1 and x2, resulting in the ellipse equation.Good, that seems to make sense.Now, moving on to part II. We have to find the equation of a hyperbola C2. It has one focus at the upper vertex of C1, and one asymptote is x + 2y = 0.First, let's recall that C1 is an ellipse with equation x²/16 + y²/4 = 1. The upper vertex of this ellipse is at (0, 2). So, the hyperbola C2 has a focus at (0, 2).Also, one of its asymptotes is x + 2y = 0, which can be rewritten as y = (-1/2)x.So, the hyperbola has a vertical transverse axis because the focus is on the y-axis, and the asymptotes are lines with slopes ±m, where m is related to the hyperbola's parameters.Wait, but the asymptote given is y = (-1/2)x, which is a negative slope. Since hyperbolas with vertical transverse axes have asymptotes of the form y = ±(a/b)x, but in this case, the slope is negative, but since it's just one asymptote, the other would be y = (1/2)x? Wait, no, because hyperbola asymptotes are symmetric. If one asymptote is y = (-1/2)x, the other should be y = (1/2)x, but actually, no, because the hyperbola can be oriented either way.Wait, hold on. Let me think.If the hyperbola has a vertical transverse axis, its asymptotes are of the form y = ±(a/b)x. If it has a horizontal transverse axis, the asymptotes are of the form y = ±(b/a)x.But in our case, the focus is at (0, 2), which is on the y-axis, so the hyperbola likely has a vertical transverse axis.Given that, the asymptotes would be y = ±(a/b)x. But one of the asymptotes is given as x + 2y = 0, which is y = (-1/2)x. So, that would correspond to one of the asymptotes with slope -1/2.Therefore, the slopes of the asymptotes are ±1/2.Wait, but if it's a vertical hyperbola, the slopes are ±(a/b). So, (a/b) = 1/2.Therefore, a/b = 1/2, so b = 2a.Also, for hyperbolas, the relationship between a, b, and c is c² = a² + b², where c is the distance from the center to each focus.Given that the hyperbola has a focus at (0, 2), and assuming the center is at (0, k), but wait, the problem says "one focus at the upper vertex of C1", which is (0, 2). So, the hyperbola has a focus at (0, 2). But is the center of the hyperbola at (0, 0)? Or somewhere else?Wait, the problem doesn't specify the center, so we need to figure that out.But since the hyperbola has a vertical transverse axis, the standard form is:(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1But since one asymptote is x + 2y = 0, which passes through the origin, and if the hyperbola's asymptotes pass through the center, then the center must be at the intersection of the asymptotes.But the asymptotes are x + 2y = 0 and, presumably, another one. Wait, the given asymptote is x + 2y = 0, but for a hyperbola, there are two asymptotes. Since the hyperbola is symmetric, the other asymptote should be x - 2y = 0? Wait, no, because the slopes are ±1/2.Wait, no, if one asymptote is y = (-1/2)x, then the other should be y = (1/2)x. So, their equations are y = (1/2)x and y = (-1/2)x.But the given asymptote is x + 2y = 0, which is y = (-1/2)x. So, the other asymptote is y = (1/2)x, which can be written as x - 2y = 0.Therefore, the asymptotes intersect at the origin (0,0). So, the center of the hyperbola is at (0,0).But wait, the hyperbola has a focus at (0, 2). So, the center is at (0,0), and one focus is at (0, 2). So, the distance from the center to the focus is c = 2.For a hyperbola with vertical transverse axis, the standard form is:y² / a² - x² / b² = 1And the relationship is c² = a² + b².We also know that the slopes of the asymptotes are ±(a/b). From the asymptotes, we have slopes of ±1/2, so a/b = 1/2, which implies b = 2a.So, we have c = 2, and c² = a² + b².Substituting b = 2a:c² = a² + (2a)² = a² + 4a² = 5a²But c = 2, so c² = 4 = 5a² => a² = 4/5 => a = 2/sqrt(5)Then, b = 2a = 4/sqrt(5), so b² = 16/5.Therefore, the equation of the hyperbola is:y² / (4/5) - x² / (16/5) = 1To make it look nicer, multiply numerator and denominator by 5:(5y²)/4 - (5x²)/16 = 1So, the equation is (5y²)/4 - (5x²)/16 = 1.Let me just verify:- Center at (0,0), correct.- Asymptotes: y = ±(a/b)x = ±(sqrt(4/5)/sqrt(16/5))x = ±(2/sqrt(5))/(4/sqrt(5))x = ±(1/2)x, which matches the given asymptote x + 2y = 0 (which is y = -1/2 x) and the other asymptote y = 1/2 x.- Focus at (0, c) where c = sqrt(a² + b²) = sqrt(4/5 + 16/5) = sqrt(20/5) = sqrt(4) = 2, which is correct.So, everything checks out.Therefore, the equation of the hyperbola C2 is (5y²)/4 - (5x²)/16 = 1.**Final Answer**(I) The trajectory equation of point ( P ) is (boxed{dfrac{x^2}{16} + dfrac{y^2}{4} = 1}).(II) The equation of hyperbola ( C_2 ) is (boxed{dfrac{5y^2}{4} - dfrac{5x^2}{16} = 1}).

answer：First, I will list the number of cookies eaten by each family member and the dog based on the information provided.Charlie ate 15 cookies, his father ate 10 cookies, and his mother ate 5 cookies. The grandmother took 12 cookies but only ate half of them, which is 6 cookies. The dog stole 3 cookies and ate 75% of them, which amounts to 2.25 cookies.Next, I will sum up all the cookies eaten by each family member and the dog to find the total number of cookie pieces consumed.Adding these together:15 (Charlie) + 10 (Father) + 5 (Mother) + 6 (Grandmother) + 2.25 (Dog) = 38.25 cookies.Therefore, the total number of cookie pieces eaten by the family and the dog is 38.25.

answer：Okay, so I have this problem about an acute triangle ABC. The sides opposite angles A, B, and C are a, b, c respectively. They gave me an equation involving cosines and sines of angles: cos C + (cos B - sqrt(3) sin B) cos A = 0. Also, they told me that a = 2 sqrt(3). Part (1) asks me to find the area of triangle ABC when b = 2 sqrt(2). Hmm, okay. Let me try to unpack this step by step.First, the equation given is cos C + (cos B - sqrt(3) sin B) cos A = 0. That looks a bit complicated, but maybe I can simplify it using some trigonometric identities. I remember that in a triangle, the sum of angles is pi, so C = pi - A - B. Therefore, cos C = cos(pi - A - B) = -cos(A + B). That might help.So, substituting cos C with -cos(A + B), the equation becomes:- cos(A + B) + (cos B - sqrt(3) sin B) cos A = 0.Now, let's expand cos(A + B). I know that cos(A + B) = cos A cos B - sin A sin B. So, substituting that in:- (cos A cos B - sin A sin B) + (cos B - sqrt(3) sin B) cos A = 0.Let me distribute the negative sign:- cos A cos B + sin A sin B + cos B cos A - sqrt(3) sin B cos A = 0.Hmm, interesting. The -cos A cos B and +cos A cos B cancel each other out. So, we're left with:sin A sin B - sqrt(3) sin B cos A = 0.I can factor out sin B:sin B (sin A - sqrt(3) cos A) = 0.Since the triangle is acute, all angles are less than pi/2, so sin B is not zero. Therefore, the other factor must be zero:sin A - sqrt(3) cos A = 0.Let me solve for A. Dividing both sides by cos A (since cos A is positive in an acute triangle):tan A = sqrt(3).So, A = pi/3 or 60 degrees. Got it. So angle A is 60 degrees.Now, knowing that angle A is 60 degrees, and sides a = 2 sqrt(3), b = 2 sqrt(2), I can use the Law of Cosines to find side c. The Law of Cosines says:a^2 = b^2 + c^2 - 2bc cos A.Plugging in the known values:(2 sqrt(3))^2 = (2 sqrt(2))^2 + c^2 - 2 * 2 sqrt(2) * c * cos(60°).Calculating each term:(4 * 3) = (4 * 2) + c^2 - 2 * 2 sqrt(2) * c * (1/2).Simplify:12 = 8 + c^2 - 2 sqrt(2) * c.Subtract 8 from both sides:4 = c^2 - 2 sqrt(2) c.Rearranging:c^2 - 2 sqrt(2) c - 4 = 0.This is a quadratic equation in c. Let me use the quadratic formula:c = [2 sqrt(2) ± sqrt( (2 sqrt(2))^2 + 16 ) ] / 2.Calculating discriminant:(2 sqrt(2))^2 = 8, so sqrt(8 + 16) = sqrt(24) = 2 sqrt(6).So, c = [2 sqrt(2) ± 2 sqrt(6)] / 2 = sqrt(2) ± sqrt(6).Since c must be positive, we take the positive root:c = sqrt(2) + sqrt(6).Okay, so now I have sides a, b, c. To find the area, I can use the formula:Area = (1/2) * b * c * sin A.Since A is 60 degrees, sin A = sqrt(3)/2.Plugging in the values:Area = (1/2) * 2 sqrt(2) * (sqrt(2) + sqrt(6)) * (sqrt(3)/2).Simplify step by step:First, (1/2) * 2 sqrt(2) = sqrt(2).Then, sqrt(2) * (sqrt(2) + sqrt(6)) = sqrt(2)*sqrt(2) + sqrt(2)*sqrt(6) = 2 + sqrt(12) = 2 + 2 sqrt(3).Then, multiply by sqrt(3)/2:(2 + 2 sqrt(3)) * sqrt(3)/2 = (2 sqrt(3) + 6)/2 = sqrt(3) + 3.So, the area is 3 + sqrt(3). That seems reasonable.Moving on to part (2): Find the range of values for 2b + c.Hmm, okay. So, 2b + c. I need to express this in terms of angles or something else to find its range.Given that a = 2 sqrt(3), and angle A is pi/3, maybe I can use the Law of Sines. The Law of Sines says:a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius.So, let's compute 2R:2R = a / sin A = (2 sqrt(3)) / (sqrt(3)/2) = (2 sqrt(3)) * (2 / sqrt(3)) ) = 4.So, 2R = 4, which means R = 2.Therefore, b = 2R sin B = 4 sin B, and c = 2R sin C = 4 sin C.But since in triangle ABC, angles A + B + C = pi, and A is pi/3, so C = pi - pi/3 - B = 2pi/3 - B.Therefore, c = 4 sin(2pi/3 - B).So, 2b + c = 2*(4 sin B) + 4 sin(2pi/3 - B) = 8 sin B + 4 sin(2pi/3 - B).Let me simplify sin(2pi/3 - B). Using the sine subtraction formula:sin(2pi/3 - B) = sin(2pi/3) cos B - cos(2pi/3) sin B.We know that sin(2pi/3) = sqrt(3)/2 and cos(2pi/3) = -1/2.So, sin(2pi/3 - B) = (sqrt(3)/2) cos B - (-1/2) sin B = (sqrt(3)/2) cos B + (1/2) sin B.Therefore, 2b + c = 8 sin B + 4*(sqrt(3)/2 cos B + 1/2 sin B) = 8 sin B + 2 sqrt(3) cos B + 2 sin B.Combine like terms:8 sin B + 2 sin B = 10 sin B, so 2b + c = 10 sin B + 2 sqrt(3) cos B.Hmm, so 2b + c is expressed as 10 sin B + 2 sqrt(3) cos B. That looks like a linear combination of sine and cosine, which can be written as a single sine function with a phase shift.Recall that A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and tan phi = B / A.Wait, actually, it's A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and tan phi = B / A. Wait, no, actually, it's A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and tan phi = B / A. Wait, let me double-check.Actually, the formula is A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and tan phi = B / A. Wait, no, that's not exactly right. Let me recall:A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and phi = arctan(B / A). Wait, no, actually, it's:A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2) and phi is such that cos phi = A / C and sin phi = B / C.Wait, maybe it's better to write it as A sin x + B cos x = C sin(x + phi), where C = sqrt(A^2 + B^2), and phi = arctan(B / A). Hmm, not exactly. Let me think.Alternatively, A sin x + B cos x = sqrt(A^2 + B^2) sin(x + phi), where phi = arctan(B / A). Wait, actually, no. The correct identity is:A sin x + B cos x = sqrt(A^2 + B^2) sin(x + phi), where phi = arctan(B / A). Wait, is that correct?Wait, let me derive it. Let me write:A sin x + B cos x = C sin(x + phi).Expanding the right side:C sin x cos phi + C cos x sin phi.Comparing coefficients:A = C cos phi,B = C sin phi.Therefore, C = sqrt(A^2 + B^2),and tan phi = B / A.So, yes, that's correct. So, in our case, A = 10, B = 2 sqrt(3).Therefore, C = sqrt(10^2 + (2 sqrt(3))^2) = sqrt(100 + 12) = sqrt(112) = 4 sqrt(7).And tan phi = (2 sqrt(3)) / 10 = sqrt(3)/5.So, phi = arctan(sqrt(3)/5). Let me compute sin phi and cos phi.Since tan phi = sqrt(3)/5, we can imagine a right triangle with opposite side sqrt(3), adjacent side 5, hypotenuse sqrt( (sqrt(3))^2 + 5^2 ) = sqrt(3 + 25) = sqrt(28) = 2 sqrt(7).Therefore, sin phi = sqrt(3) / (2 sqrt(7)) = sqrt(21)/14,and cos phi = 5 / (2 sqrt(7)) = 5 sqrt(7)/14.So, 2b + c = 4 sqrt(7) sin(x + phi), where x is angle B.Wait, no, in our case, x is angle B, so 2b + c = 4 sqrt(7) sin(B + phi).But since angle B is part of an acute triangle, we need to find the range of B.Given that triangle ABC is acute, all angles are less than pi/2. So, angle A is pi/3, which is 60 degrees, so angles B and C must each be less than pi/2.Since A + B + C = pi, and A = pi/3, so B + C = 2 pi / 3.Given that both B and C are less than pi/2, so B < pi/2, and C = 2 pi / 3 - B < pi/2.So, 2 pi / 3 - B < pi/2 => B > 2 pi / 3 - pi / 2 = pi / 6.Therefore, B is between pi / 6 and pi / 2.So, pi / 6 < B < pi / 2.Therefore, B + phi is between pi / 6 + phi and pi / 2 + phi.So, sin(B + phi) will have its maximum when B + phi is pi / 2, because sine reaches maximum at pi / 2.But wait, since B is less than pi / 2, and phi is positive (since tan phi = sqrt(3)/5 > 0), so B + phi is less than pi / 2 + phi.But the maximum value of sin(B + phi) is 1, which occurs when B + phi = pi / 2.Similarly, the minimum value occurs when B + phi is as small as possible, which is pi / 6 + phi.So, the range of sin(B + phi) is from sin(pi / 6 + phi) to 1.Thus, the range of 2b + c is from 4 sqrt(7) sin(pi / 6 + phi) to 4 sqrt(7).So, let me compute sin(pi / 6 + phi).Using the sine addition formula:sin(pi / 6 + phi) = sin(pi / 6) cos phi + cos(pi / 6) sin phi.We know sin(pi / 6) = 1/2, cos(pi / 6) = sqrt(3)/2.We already have sin phi = sqrt(21)/14 and cos phi = 5 sqrt(7)/14.So, sin(pi / 6 + phi) = (1/2)(5 sqrt(7)/14) + (sqrt(3)/2)(sqrt(21)/14).Simplify each term:First term: (1/2)(5 sqrt(7)/14) = 5 sqrt(7)/28.Second term: (sqrt(3)/2)(sqrt(21)/14) = (sqrt(63))/28 = (3 sqrt(7))/28.So, adding them together:5 sqrt(7)/28 + 3 sqrt(7)/28 = 8 sqrt(7)/28 = 2 sqrt(7)/7.Therefore, sin(pi / 6 + phi) = 2 sqrt(7)/7.So, the minimum value of 2b + c is 4 sqrt(7) * (2 sqrt(7)/7) = 4 sqrt(7) * 2 sqrt(7) / 7.Compute that:4 * 2 = 8,sqrt(7) * sqrt(7) = 7,So, 8 * 7 / 7 = 8.And the maximum value is 4 sqrt(7) * 1 = 4 sqrt(7).Therefore, the range of 2b + c is (8, 4 sqrt(7)].Wait, but hold on. Since B is strictly greater than pi / 6, and strictly less than pi / 2, does that mean sin(B + phi) is strictly greater than sin(pi / 6 + phi) and strictly less than sin(pi / 2 + phi)?Wait, pi / 2 + phi is greater than pi / 2, but sine of angles beyond pi / 2 starts decreasing. So, sin(pi / 2 + phi) is actually equal to cos(phi), which is 5 sqrt(7)/14, which is approximately 0.59.But wait, sin(pi / 2 + phi) = cos(phi) = 5 sqrt(7)/14 ≈ 0.59, which is less than 1.Wait, but earlier I thought that the maximum of sin(B + phi) is 1, but actually, since B + phi can go up to pi / 2 + phi, which is less than pi, but sin(pi / 2 + phi) = cos(phi) is less than 1.Wait, hold on, maybe I made a mistake here.Wait, when B approaches pi / 2, B + phi approaches pi / 2 + phi.But sin(pi / 2 + phi) = cos(phi) = 5 sqrt(7)/14 ≈ 0.59.But wait, if B is approaching pi / 2, then B + phi is approaching pi / 2 + phi, but sin(pi / 2 + phi) is cos(phi), which is less than 1.But earlier, I thought that the maximum of sin(B + phi) is 1, but actually, since B + phi cannot exceed pi / 2 + phi, which is less than pi, but sin(pi / 2 + phi) is less than 1.Wait, but that contradicts the earlier conclusion.Wait, maybe I need to think differently.Wait, no, actually, the maximum of sin(theta) is 1, which occurs at theta = pi / 2.So, if B + phi can reach pi / 2, then sin(B + phi) can reach 1.But in our case, B is less than pi / 2, and phi is positive, so B + phi can be equal to pi / 2 if B = pi / 2 - phi.But is pi / 2 - phi within the range of B?Given that B must be greater than pi / 6, so we need to check if pi / 2 - phi > pi / 6.Compute pi / 2 - phi > pi / 6.Which implies phi < pi / 2 - pi / 6 = pi / 3.Is phi < pi / 3?We have tan phi = sqrt(3)/5 ≈ 0.3464.Compute arctan(0.3464). Since tan(pi/6) ≈ 0.577, tan(pi/12) ≈ 0.2679.So, arctan(0.3464) is between pi/12 and pi/6, approximately 0.33 radians or about 19 degrees.Which is less than pi / 3 (≈1.047 radians or 60 degrees). So, yes, phi < pi / 3.Therefore, pi / 2 - phi > pi / 6, so there exists a value of B = pi / 2 - phi, which is within the range pi / 6 < B < pi / 2.Therefore, sin(B + phi) can reach 1 when B = pi / 2 - phi.Therefore, the maximum value of 2b + c is 4 sqrt(7).And the minimum value is when B is pi / 6, so sin(pi / 6 + phi) = 2 sqrt(7)/7, so 4 sqrt(7) * 2 sqrt(7)/7 = 8.But wait, when B approaches pi / 6, sin(B + phi) approaches sin(pi / 6 + phi) = 2 sqrt(7)/7, so 2b + c approaches 8.But since B must be greater than pi / 6, 2b + c must be greater than 8.Similarly, when B approaches pi / 2, sin(B + phi) approaches sin(pi / 2 + phi) = cos(phi) = 5 sqrt(7)/14 ≈ 0.59, so 2b + c approaches 4 sqrt(7) * 5 sqrt(7)/14 = 4 * 5 * 7 /14 = 20 / 2 = 10.Wait, hold on, that contradicts my earlier conclusion.Wait, no, 4 sqrt(7) * (5 sqrt(7)/14) = (4 * 5 * 7) /14 = 20 / 2 = 10.But earlier, I thought the maximum was 4 sqrt(7), but now it seems that when B approaches pi / 2, 2b + c approaches 10, which is less than 4 sqrt(7) ≈ 10.583.Wait, so perhaps I made a mistake in thinking that sin(B + phi) can reach 1.Wait, let's think again.We have 2b + c = 4 sqrt(7) sin(B + phi).We need to find the range of sin(B + phi) as B varies from pi / 6 to pi / 2.So, when B = pi / 6, sin(B + phi) = sin(pi / 6 + phi) = 2 sqrt(7)/7 ≈ 0.7559.When B = pi / 2, sin(B + phi) = sin(pi / 2 + phi) = cos(phi) = 5 sqrt(7)/14 ≈ 0.5916.Wait, but earlier, I thought that when B = pi / 2 - phi, sin(B + phi) = sin(pi / 2) = 1.But if B = pi / 2 - phi, then B + phi = pi / 2.But is B = pi / 2 - phi within the allowed range of B?Given that B must be greater than pi / 6, so pi / 2 - phi > pi / 6.Which is equivalent to phi < pi / 2 - pi / 6 = pi / 3.As we saw earlier, phi ≈ 0.33 radians < pi / 3 ≈ 1.047 radians.Therefore, yes, B = pi / 2 - phi is within the allowed range.Therefore, sin(B + phi) can reach 1, making 2b + c = 4 sqrt(7).But wait, when B = pi / 2 - phi, which is approximately pi / 2 - 0.33 ≈ 1.308 radians or about 75 degrees, which is still less than pi / 2 (90 degrees). So, that is allowed.Therefore, sin(B + phi) can reach 1, so 2b + c can reach 4 sqrt(7).But when B approaches pi / 2, sin(B + phi) approaches sin(pi / 2 + phi) = cos(phi) ≈ 0.5916, so 2b + c approaches 4 sqrt(7) * 0.5916 ≈ 4 sqrt(7) * (5 sqrt(7)/14) = 10.Wait, but 4 sqrt(7) * (5 sqrt(7)/14) = (4 * 5 * 7) /14 = 20 / 2 = 10.So, as B approaches pi / 2, 2b + c approaches 10.But earlier, when B = pi / 2 - phi, 2b + c = 4 sqrt(7) ≈ 10.583.So, the maximum value is 4 sqrt(7), and the minimum value is 8.But wait, when B approaches pi / 6, sin(B + phi) approaches sin(pi / 6 + phi) = 2 sqrt(7)/7 ≈ 0.7559, so 2b + c approaches 4 sqrt(7) * 0.7559 ≈ 4 sqrt(7) * (2 sqrt(7)/7) = 8.Therefore, the range of 2b + c is from 8 to 4 sqrt(7).But wait, when B approaches pi / 2, 2b + c approaches 10, which is less than 4 sqrt(7) ≈ 10.583.So, the maximum is 4 sqrt(7), and the minimum is 8.But wait, when B approaches pi / 2, 2b + c approaches 10, which is less than the maximum of 4 sqrt(7). So, the range is from 8 to 4 sqrt(7).But wait, 4 sqrt(7) is approximately 10.583, which is greater than 10, so 2b + c can go up to 10.583 and down to 8.But when B is pi / 2, 2b + c is 10, which is within the range.So, the range is 8 < 2b + c ≤ 4 sqrt(7).Therefore, the range is (8, 4 sqrt(7)].I think that's the answer.**Final Answer**(1) The area of triangle ABC is boxed{3 + sqrt{3}}.(2) The range of values for 2b + c is boxed{(8, 4sqrt{7}]}.

answer：First, I need to determine the relative speed between the train and the man. Since both are moving in the same direction, the relative speed is the difference between the train's speed and the man's speed.Next, I'll convert the train's speed from kilometers per hour to meters per second to match the units of the train's length and the time taken to cross the man.Once I have the train's speed in meters per second, I'll use the formula for distance, which is speed multiplied by time, to find the relative speed. This will help me establish an equation involving the train's speed, the man's speed, and the known time.After setting up the equation, I'll solve for the man's speed by isolating it on one side of the equation. This will give me the man's speed in meters per second.Finally, I'll convert the man's speed back to kilometers per hour if needed, to present the answer in the desired unit.

answer：Okay, so I need to prove that (a equiv b pmod{m}) if and only if (a - b) is divisible by (m). Hmm, let me try to understand what this means.First, I remember that (a equiv b pmod{m}) means that when you divide both (a) and (b) by (m), they leave the same remainder. So, like, if (a = 14) and (b = 7) with (m = 7), then both leave a remainder of 0 when divided by 7, so (14 equiv 7 pmod{7}). That makes sense.Now, the statement says that this congruence is true if and only if (a - b) is divisible by (m). So, I need to show two things: 1. If (a equiv b pmod{m}), then (a - b) is divisible by (m).2. If (a - b) is divisible by (m), then (a equiv b pmod{m}).Let me start with the first part. Suppose (a equiv b pmod{m}). By definition, this means that (m) divides (a - b), right? Wait, is that the definition? Or is it that they have the same remainder?I think it's both. If they have the same remainder when divided by (m), then their difference should be a multiple of (m). Let me write that down.If (a) and (b) have the same remainder when divided by (m), then we can write (a = mq + r) and (b = mp + r), where (q) and (p) are integers, and (r) is the common remainder, which is less than (m).Subtracting these two equations, we get (a - b = mq + r - (mp + r) = m(q - p)). So, (a - b) is equal to (m) times some integer ((q - p)), which means (m) divides (a - b). Okay, that seems to work. So, if (a equiv b pmod{m}), then (a - b) is divisible by (m).Now, for the other direction. Suppose (a - b) is divisible by (m). That means there exists some integer (k) such that (a - b = mk). I need to show that (a equiv b pmod{m}), which means they have the same remainder when divided by (m).Let me think about dividing (a) and (b) by (m). Let’s say (a = mq + r) and (b = mp + s), where (r) and (s) are the remainders when (a) and (b) are divided by (m), respectively, and (0 leq r, s < m).Then, (a - b = mq + r - (mp + s) = m(q - p) + (r - s)). But we know that (a - b = mk), so:(m(q - p) + (r - s) = mk)This simplifies to:(r - s = mk - m(q - p) = m(k - q + p))So, (r - s) is a multiple of (m). But since both (r) and (s) are between 0 and (m - 1), the difference (r - s) must be less than (m) in absolute value. The only multiple of (m) that is less than (m) in absolute value is 0. Therefore, (r - s = 0), which means (r = s).So, (a) and (b) have the same remainder when divided by (m), which means (a equiv b pmod{m}).Wait, does that cover all cases? What if (r - s) was negative? Well, since (r) and (s) are both non-negative and less than (m), their difference can't be less than (-m) or greater than (m). So, the only multiple of (m) in that range is 0. That makes sense.So, both directions have been proven. If (a equiv b pmod{m}), then (a - b) is divisible by (m), and if (a - b) is divisible by (m), then (a equiv b pmod{m}). Therefore, the statement is true.I think I've covered all the steps. Maybe I should double-check if I missed anything. Let me see:1. Defined congruence in terms of remainders.2. Showed that congruence implies the difference is a multiple of (m).3. Assumed the difference is a multiple of (m) and showed that the remainders must be equal.4. Considered the range of remainders to conclude that the difference must be zero.Yeah, that seems solid. I don't see any gaps in the logic. I guess this proves the statement.

answer：Okay, so I have this problem here: I need to find the absolute value of (2 + 2√3i) raised to the sixth power. Hmm, absolute value of a complex number... I remember that the absolute value, or modulus, of a complex number a + bi is calculated as the square root of (a squared plus b squared). So, for a complex number z = a + bi, |z| = √(a² + b²). Alright, so first, maybe I should find the modulus of the complex number 2 + 2√3i before I do anything else. Let me write that down. The modulus is √(2² + (2√3)²). Let me compute that step by step. First, 2 squared is 4. Then, (2√3) squared is (2)² times (√3)², which is 4 times 3, so that's 12. So, adding those together, I get 4 + 12, which is 16. Then, the square root of 16 is 4. So, the modulus of 2 + 2√3i is 4. Now, the problem asks for the modulus of this complex number raised to the sixth power. I think there's a property of moduli that says |z^n| = |z|^n. So, if I have a complex number z and I raise it to the power of n, the modulus of that is just the modulus of z raised to the power of n. So, applying that property here, |(2 + 2√3i)^6| should be equal to |2 + 2√3i|^6. Since I already found that |2 + 2√3i| is 4, this simplifies to 4^6. Now, I need to compute 4 raised to the sixth power. Let me calculate that. 4 squared is 16, 4 cubed is 64, 4 to the fourth is 256, 4 to the fifth is 1024, and 4 to the sixth is 4096. So, 4^6 equals 4096. Therefore, the modulus of (2 + 2√3i)^6 is 4096. Wait, let me double-check to make sure I didn't make any mistakes. First, the modulus calculation: 2 squared is 4, (2√3) squared is 12, 4 + 12 is 16, square root of 16 is 4. That seems correct. Then, raising 4 to the sixth power: 4^2 is 16, 4^3 is 64, 4^4 is 256, 4^5 is 1024, 4^6 is 4096. Yep, that all checks out. I think I might also recall that when you raise a complex number to a power, you can use De Moivre's Theorem, which involves converting the complex number to polar form. Maybe I should try that method as well to confirm my answer. So, let's see. The complex number 2 + 2√3i can be expressed in polar form as r*(cosθ + i sinθ), where r is the modulus and θ is the argument. We already found that r is 4. Now, to find θ, which is the angle made with the positive real axis. To find θ, we can use tanθ = (imaginary part)/(real part) = (2√3)/2 = √3. So, tanθ = √3. I know that tan(60°) is √3, so θ is 60 degrees, or π/3 radians. So, in polar form, the complex number is 4*(cos(π/3) + i sin(π/3)). Now, using De Moivre's Theorem, (r*(cosθ + i sinθ))^n = r^n*(cos(nθ) + i sin(nθ)). Applying this, (4*(cos(π/3) + i sin(π/3)))^6 = 4^6*(cos(6*(π/3)) + i sin(6*(π/3))). Simplifying the angle, 6*(π/3) is 2π. So, this becomes 4096*(cos(2π) + i sin(2π)). Now, cos(2π) is 1, and sin(2π) is 0. Therefore, the result is 4096*(1 + 0i) = 4096. So, the modulus of this result is just 4096, which matches what I found earlier. Hmm, so both methods give me the same answer, which is reassuring. I think I can confidently say that the modulus of (2 + 2√3i)^6 is 4096. Just to make sure I didn't overlook anything, let me think about the properties of complex numbers again. The modulus of a product is the product of the moduli, and the modulus of a power is the power of the modulus. So, |z^n| = |z|^n. Since I applied that correctly, and my calculations for the modulus and the exponentiation seem right, I don't see any issues here. I guess another way to think about it is that when you raise a complex number to a power, you're essentially scaling its modulus by that exponent and rotating it by multiples of its argument. But since we're only asked for the modulus, the rotation part doesn't affect the final answer, only the scaling does. So, focusing on the modulus and raising it to the sixth power is sufficient. Alright, I think I've covered all the bases here. Calculated the modulus, applied the exponent, double-checked using polar form and De Moivre's Theorem, and everything lines up. I don't think I made any calculation errors, and the logic seems solid. So, I'm pretty sure the answer is 4096.