answer：Okay, so I have this problem where I need to find the triangle ABC with the maximum area, given that BC is 2 times the fourth root of 3, and angle BAC is π/3, which is 60 degrees. Hmm, let me think about how to approach this.First, let me visualize the triangle. Triangle ABC, with points A, B, and C. The side BC is fixed at 2√[4]{3}. The angle at A, which is angle BAC, is 60 degrees. I need to maximize the area of this triangle.I remember that the area of a triangle can be calculated in a few ways. One common formula is (1/2)*base*height. But since I don't know the height here, maybe another formula would be better. Oh, right! The formula using two sides and the included angle: Area = (1/2)*ab*sinθ, where a and b are two sides, and θ is the included angle.In this case, angle BAC is 60 degrees, so θ is 60°, and sides AB and AC are the two sides forming this angle. So, if I can express the area in terms of AB and AC, maybe I can find a way to maximize it.Let me denote AB as c and AC as b. Then, the area would be (1/2)*b*c*sin(60°). Since sin(60°) is √3/2, the area becomes (√3/4)*b*c. So, to maximize the area, I need to maximize the product of b and c.But wait, I have a constraint here: side BC is fixed at 2√[4]{3}. So, I can't just let b and c be as large as possible. I need to relate b and c with BC.I remember the Law of Cosines, which relates the sides of a triangle with one of its angles. For triangle ABC, the Law of Cosines states that BC² = AB² + AC² - 2*AB*AC*cos(angle BAC). Plugging in the known values, that would be:(2√[4]{3})² = c² + b² - 2*b*c*cos(60°)Let me compute (2√[4]{3})² first. The fourth root of 3 is 3^(1/4), so squaring that gives 3^(1/2), which is √3. Then, multiplying by 4 (since 2 squared is 4), we get 4√3. So, the left side is 4√3.Now, cos(60°) is 0.5, so the equation becomes:4√3 = c² + b² - 2*b*c*(0.5)Simplify the right side: 2*b*c*(0.5) is just b*c. So, the equation is:4√3 = c² + b² - b*cSo, we have the equation c² + b² - b*c = 4√3.And we need to maximize the area, which is (√3/4)*b*c. So, to maximize the area, we need to maximize b*c, given that c² + b² - b*c = 4√3.Hmm, okay, so we have a constraint c² + b² - b*c = 4√3, and we need to maximize b*c.This seems like an optimization problem with a constraint. Maybe I can use calculus, like Lagrange multipliers, but since it's a two-variable problem, perhaps substitution would work.Let me denote x = b and y = c. Then, the constraint is x² + y² - x*y = 4√3, and we need to maximize P = x*y.So, how can I express one variable in terms of the other? Maybe solve the constraint for one variable.Let me rearrange the constraint:x² + y² - x*y = 4√3This is a quadratic in terms of x and y. Maybe I can treat this as a quadratic equation in x, treating y as a constant.So, x² - y*x + (y² - 4√3) = 0Using the quadratic formula, x = [y ± √(y² - 4*(1)*(y² - 4√3))]/2Simplify the discriminant:√(y² - 4*(y² - 4√3)) = √(y² - 4y² + 16√3) = √(-3y² + 16√3)So, x = [y ± √(-3y² + 16√3)] / 2Hmm, since x must be a real number, the discriminant must be non-negative:-3y² + 16√3 ≥ 0So, 3y² ≤ 16√3Therefore, y² ≤ (16√3)/3So, y ≤ sqrt(16√3 / 3) = sqrt((16/3)*√3) = sqrt(16/3 * 3^(1/2)) = sqrt(16 * 3^(-1/2)) = 4 * 3^(-1/4)Wait, let me compute that again:sqrt(16√3 / 3) = sqrt(16/3 * √3) = sqrt(16 / 3^{1/2}) = sqrt(16 * 3^{-1/2}) = 4 * 3^{-1/4}Which is 4 divided by 3^{1/4}, which is 4 / √[4]{3}So, y ≤ 4 / √[4]{3}Similarly, since the equation is symmetric in x and y, x must also be ≤ 4 / √[4]{3}But I'm not sure if this helps me directly. Maybe another approach.Alternatively, since the problem is symmetric in b and c, perhaps the maximum occurs when b = c? Let me test that.Assume b = c. Then, the constraint becomes:b² + b² - b*b = 4√3So, 2b² - b² = 4√3 => b² = 4√3 => b = sqrt(4√3) = 2*(3)^{1/4}So, if b = c = 2√[4]{3}, then the product b*c = (2√[4]{3})² = 4√3.So, the area would be (√3 / 4) * 4√3 = (√3 / 4) * 4√3 = (√3 * √3) = 3.But is this the maximum? Maybe not. Let me see.Wait, if I let b and c vary, perhaps I can get a larger product. Let me see.Let me consider the constraint equation: x² + y² - x*y = 4√3.Let me denote S = x + y and P = x*y. Maybe express the constraint in terms of S and P.We know that x² + y² = (x + y)^2 - 2xy = S² - 2P.So, substituting into the constraint:S² - 2P - P = 4√3 => S² - 3P = 4√3So, S² = 3P + 4√3But we need to maximize P, so perhaps express S² in terms of P.But I don't see a direct way to relate S and P without more information.Alternatively, maybe use the method of Lagrange multipliers.Define function f(x, y) = x*y, which we want to maximize, subject to the constraint g(x, y) = x² + y² - x*y - 4√3 = 0.The method of Lagrange multipliers says that at the maximum, the gradient of f is proportional to the gradient of g.So, ∇f = λ∇gCompute gradients:∇f = (y, x)∇g = (2x - y, 2y - x)So, setting up equations:y = λ(2x - y)x = λ(2y - x)So, from the first equation: y = λ(2x - y)From the second equation: x = λ(2y - x)We can write these as:y = 2λx - λy => y + λy = 2λx => y(1 + λ) = 2λx => y = (2λ / (1 + λ)) xSimilarly, from the second equation:x = 2λ y - λx => x + λx = 2λ y => x(1 + λ) = 2λ y => x = (2λ / (1 + λ)) yWait, so from the first equation, y = (2λ / (1 + λ)) x, and from the second, x = (2λ / (1 + λ)) y.Substituting the first into the second:x = (2λ / (1 + λ)) * (2λ / (1 + λ)) xSo, x = (4λ² / (1 + λ)^2) xAssuming x ≠ 0, we can divide both sides by x:1 = (4λ²) / (1 + λ)^2Multiply both sides by (1 + λ)^2:(1 + λ)^2 = 4λ²Expand left side:1 + 2λ + λ² = 4λ²Bring all terms to one side:1 + 2λ + λ² - 4λ² = 0 => 1 + 2λ - 3λ² = 0Multiply both sides by -1:3λ² - 2λ - 1 = 0Solve quadratic equation:λ = [2 ± sqrt(4 + 12)] / 6 = [2 ± sqrt(16)] / 6 = [2 ± 4]/6So, λ = (2 + 4)/6 = 6/6 = 1, or λ = (2 - 4)/6 = (-2)/6 = -1/3So, two possibilities: λ = 1 or λ = -1/3Case 1: λ = 1From the first equation: y = (2*1)/(1 + 1) x = (2/2)x = xSo, y = x. So, b = c.Which is the case we considered earlier, where b = c = 2√[4]{3}, leading to area 3.Case 2: λ = -1/3From the first equation: y = (2*(-1/3))/(1 + (-1/3)) x = (-2/3)/(2/3) x = (-2/3)*(3/2) x = -1*xSo, y = -xBut since x and y are lengths, they must be positive. So, y = -x would imply x = y = 0, which is not possible for a triangle. So, this case is invalid.Therefore, the only valid critical point is when b = c, so the triangle is isoceles with AB = AC.Thus, the maximum area occurs when AB = AC = 2√[4]{3}, and the area is 3.Wait, but let me double-check. Maybe I made a mistake in assuming that the maximum occurs when b = c. Let me see.Alternatively, perhaps using another method. Let me consider expressing one variable in terms of the other.From the constraint: x² + y² - x*y = 4√3Let me solve for y in terms of x.x² + y² - x*y = 4√3This is a quadratic in y: y² - x*y + (x² - 4√3) = 0Using quadratic formula:y = [x ± sqrt(x² - 4*(x² - 4√3))]/2Simplify discriminant:sqrt(x² - 4x² + 16√3) = sqrt(-3x² + 16√3)So, y = [x ± sqrt(-3x² + 16√3)] / 2Since y must be positive, we take the positive root:y = [x + sqrt(-3x² + 16√3)] / 2But this expression is complicated. Maybe instead, express P = x*y, and try to express P in terms of x.From the constraint, x² + y² - x*y = 4√3, and P = x*y.Let me express y² from the constraint: y² = 4√3 + x*y - x²So, y² = 4√3 + P - x²But y² must be positive, so 4√3 + P - x² > 0.But I'm not sure if that helps.Alternatively, perhaps express P in terms of x.From y = [x + sqrt(-3x² + 16√3)] / 2Then, P = x*y = x*[x + sqrt(-3x² + 16√3)] / 2So, P = [x² + x*sqrt(-3x² + 16√3)] / 2To find the maximum of P, take derivative with respect to x and set to zero.But this seems messy. Maybe another approach.Wait, perhaps using trigonometric substitution.Since angle BAC is 60°, maybe we can use coordinates.Let me place point A at the origin (0,0). Let me let AB lie along the x-axis, so point B is at (c, 0), where c is the length AB. Point C is somewhere in the plane, making angle BAC = 60°, so the coordinates of C can be expressed as (b*cos(60°), b*sin(60°)) = (b*(1/2), b*(√3/2)).Then, the coordinates are:A: (0,0)B: (c, 0)C: (b/2, (b√3)/2)Now, the distance BC is given as 2√[4]{3}.Compute the distance between B (c, 0) and C (b/2, (b√3)/2):Distance BC = sqrt[(c - b/2)^2 + (0 - (b√3)/2)^2] = sqrt[(c - b/2)^2 + ( (b√3)/2 )^2]Simplify:= sqrt[ (c² - c*b + b²/4) + (3b²/4) ]= sqrt[ c² - c*b + b²/4 + 3b²/4 ]= sqrt[ c² - c*b + b² ]So, we have sqrt(c² - c*b + b²) = 2√[4]{3}Square both sides:c² - c*b + b² = (2√[4]{3})² = 4√3Which is the same constraint as before: c² + b² - c*b = 4√3So, same equation.Now, the area of the triangle is (1/2)*base*height. If I take AB as the base, which is c, then the height is the y-coordinate of point C, which is (b√3)/2.So, area = (1/2)*c*(b√3)/2 = (√3/4)*b*c, which is the same as before.So, again, we need to maximize b*c given c² + b² - c*b = 4√3.Alternatively, maybe parametrize b and c using an angle.Let me set b = k and c = m, but I don't see an immediate way.Wait, perhaps using the method of substitution.Let me set t = b/c, so b = t*c.Then, substitute into the constraint:c² + (t*c)^2 - c*(t*c) = 4√3Simplify:c² + t² c² - t c² = 4√3Factor c²:c² (1 + t² - t) = 4√3So, c² = 4√3 / (1 + t² - t)Then, the product P = b*c = t*c² = t*(4√3 / (1 + t² - t)) = (4√3 t) / (1 + t² - t)So, now P is a function of t: P(t) = (4√3 t) / (1 + t² - t)To find the maximum of P(t), take derivative with respect to t and set to zero.Compute derivative P’(t):P’(t) = [4√3*(1 + t² - t) - 4√3 t*(2t - 1)] / (1 + t² - t)^2Set numerator equal to zero:4√3*(1 + t² - t) - 4√3 t*(2t - 1) = 0Factor out 4√3:4√3 [ (1 + t² - t) - t*(2t - 1) ] = 0Since 4√3 ≠ 0, the expression in brackets must be zero:(1 + t² - t) - t*(2t - 1) = 0Simplify:1 + t² - t - 2t² + t = 0Combine like terms:1 + (t² - 2t²) + (-t + t) = 0 => 1 - t² = 0So, 1 - t² = 0 => t² = 1 => t = ±1But since t = b/c and lengths are positive, t must be positive. So, t = 1.Thus, t = 1, so b = c.Therefore, the maximum occurs when b = c, which is consistent with the earlier result.So, when t = 1, b = c.Then, from the constraint:c² + c² - c*c = 4√3 => 2c² - c² = 4√3 => c² = 4√3 => c = sqrt(4√3) = 2*(3)^{1/4} = 2√[4]{3}So, b = c = 2√[4]{3}Thus, the product P = b*c = (2√[4]{3})² = 4√3Therefore, the maximum area is (√3 / 4)*4√3 = (√3 * √3) = 3So, the maximum area is 3.Wait, let me confirm once more.If I set b = c, then the triangle is isoceles with AB = AC = 2√[4]{3}, angle at A is 60°, so actually, this would make triangle ABC equilateral?Wait, no, because BC is 2√[4]{3}, and AB = AC = 2√[4]{3}, but in an equilateral triangle, all sides are equal, so BC would also be 2√[4]{3}. But in our case, BC is fixed at 2√[4]{3}, so if AB = AC = 2√[4]{3}, then indeed, triangle ABC is equilateral.Wait, but in an equilateral triangle, all angles are 60°, so angle BAC is 60°, which fits. So, this makes sense.Therefore, the triangle with maximum area is equilateral, with all sides equal to 2√[4]{3}, and area 3.Hence, the maximum area is 3.**Final Answer**The maximum area of triangle (ABC) is boxed{3}.

answer：To find the greatest whole number less than 100 with an odd number of positive factors, I need to understand the characteristics of such numbers.First, I recall that the number of positive factors of a number is determined by its prime factorization. If a number has a prime factorization of the form ( n = p_1^{a_1} times p_2^{a_2} times dots times p_k^{a_k} ), then the total number of positive factors is ( (a_1 + 1) times (a_2 + 1) times dots times (a_k + 1) ).For the number of factors to be odd, each ( (a_i + 1) ) must be odd. This implies that each exponent ( a_i ) in the prime factorization must be even. Therefore, the number must be a perfect square.Next, I list the perfect squares less than 100:- ( 1^2 = 1 )- ( 2^2 = 4 )- ( 3^2 = 9 )- ( 4^2 = 16 )- ( 5^2 = 25 )- ( 6^2 = 36 )- ( 7^2 = 49 )- ( 8^2 = 64 )- ( 9^2 = 81 )- ( 10^2 = 100 ) (exceeds 100)Among these, the largest perfect square less than 100 is 81. Therefore, the greatest whole number less than 100 with an odd number of positive factors is 81.

answer：First, I need to determine the percentage of students who scored 100 points. Since the total percentage must add up to 100%, I subtract the given percentages: 100% - 10% - 35% - 30% = 25%.Next, I'll calculate the mean score by multiplying each score by its corresponding percentage and summing the results:Mean = (0.10 × 70) + (0.35 × 80) + (0.30 × 90) + (0.25 × 100) = 7 + 28 + 27 + 25 = 87.To find the median, I'll consider the cumulative percentages. The median falls within the group where the cumulative percentage reaches 50%. The first two groups account for 45%, so the median must be in the third group, which scored 90 points.Finally, I'll find the difference between the mean and the median:Difference = Mean - Median = 87 - 90 = -3. Since we're looking for the absolute difference, the answer is 3.

answer：To find the least common multiple of 8 and 15, I'll start by determining their prime factorizations. 8 can be factored into 2 multiplied by itself three times, so 2³. 15 is the product of 3 and 5, which are both prime numbers, so 3¹ × 5¹. Next, I'll identify the highest power of each prime number present in these factorizations. For 2, the highest power is 2³, and for both 3 and 5, the highest power is 1. Finally, I'll multiply these highest powers together: 2³ × 3¹ × 5¹, which equals 120. Therefore, the least common multiple of 8 and 15 is 120.

answer：First, I need to determine the sampling ratio used for school C. Since 50 papers were drawn from 900, the ratio is 50/900, which simplifies to 1/18.Next, I'll apply this same ratio to the number of papers from schools A and B. For school A, with 1260 papers, the number of papers drawn would be 1260 multiplied by 1/18, resulting in 70 papers. For school B, with 720 papers, the number of papers drawn would be 720 multiplied by 1/18, resulting in 40 papers.Finally, I'll add up the papers drawn from all three schools: 70 from school A, 40 from school B, and 50 from school C. The total number of papers drawn for the research is 160.

answer：To find the smallest number that satisfies the given conditions, I will analyze each condition step by step.First, the number leaves a remainder of 1 when divided by 3. This means the number can be expressed as 3k + 1, where k is an integer. The possible numbers are 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, and so on.Next, the number leaves a remainder of 3 when divided by 5. This implies the number can be written as 5m + 3, where m is an integer. The possible numbers are 3, 8, 13, 18, 23, 28, 33, and so forth.Finally, the number leaves a remainder of 4 when divided by 6. This means the number can be represented as 6n + 4, where n is an integer. The possible numbers are 4, 10, 16, 22, 28, 34, 40, etc.To find the smallest number that satisfies all three conditions, I will look for the smallest common number in all three lists. From the lists above, the smallest common number is 28.Therefore, the smallest number that meets all the given conditions is 28.