💡Alright, so I've got these two trigonometric identities to prove, and I'm a bit nervous because they look pretty complex. Let me take them one at a time and see if I can figure them out.Starting with part (a):[ cos frac{2 pi}{7} cos frac{4 pi}{7} cos frac{8 pi}{7} = frac{1}{8} ]Hmm, okay. So I need to show that the product of these three cosines equals 1/8. I remember that sometimes products of cosines can be simplified using certain trigonometric identities, but I'm not exactly sure which one applies here. Maybe I can use the identity for the product of cosines in terms of sine? I think there's something like that.Let me recall: I think the formula is something like:[ cos A cos 2A cos 4A dots cos 2^{n-1}A = frac{sin 2^n A}{2^n sin A} ]Is that right? I think so. So in this case, my angles are 2π/7, 4π/7, and 8π/7. Let me see if these fit into a pattern where each angle is double the previous one.Starting with 2π/7, doubling it gives 4π/7, and doubling that gives 8π/7. Perfect! So this seems to fit the formula where n = 3, because there are three terms, and each term is double the previous angle.So applying the formula:[ cos frac{2 pi}{7} cos frac{4 pi}{7} cos frac{8 pi}{7} = frac{sin (2^3 cdot frac{2 pi}{7})}{2^3 sin frac{2 pi}{7}} ]Calculating the exponent:2^3 = 8, so:[ frac{sin (8 cdot frac{2 pi}{7})}{8 sin frac{2 pi}{7}} ]Simplify the argument of the sine:8 * (2π/7) = 16π/7Now, 16π/7 is more than 2π, so I can subtract 2π to make it within the standard range:16π/7 - 2π = 16π/7 - 14π/7 = 2π/7So sin(16π/7) = sin(2π/7)Therefore, the expression becomes:[ frac{sin frac{2 pi}{7}}{8 sin frac{2 pi}{7}} ]The sin(2π/7) terms cancel out, leaving:1/8So that proves part (a). Okay, that wasn't too bad once I remembered the formula.Now, moving on to part (b):[ cos frac{2 pi}{9} cos frac{4 pi}{9} cos frac{8 pi}{9} = -frac{1}{8} ]Similar structure here: the angles are 2π/9, 4π/9, and 8π/9. Again, each angle is double the previous one. So maybe I can use the same formula as before.Applying the formula:[ cos frac{2 pi}{9} cos frac{4 pi}{9} cos frac{8 pi}{9} = frac{sin (2^3 cdot frac{2 pi}{9})}{2^3 sin frac{2 pi}{9}} ]Calculating the exponent:2^3 = 8, so:[ frac{sin (8 cdot frac{2 pi}{9})}{8 sin frac{2 pi}{9}} ]Simplify the argument of the sine:8 * (2π/9) = 16π/9Again, 16π/9 is more than 2π, so subtract 2π:16π/9 - 2π = 16π/9 - 18π/9 = -2π/9But sine is an odd function, so sin(-x) = -sin(x). Therefore:sin(-2π/9) = -sin(2π/9)So the expression becomes:[ frac{-sin frac{2 pi}{9}}{8 sin frac{2 pi}{9}} ]The sin(2π/9) terms cancel out, leaving:-1/8So that proves part (b). Wait a second, in part (a), the angle after subtracting 2π was positive, but in part (b), it became negative. That's why one result is positive and the other is negative. That makes sense because the sine function is positive in the first and second quadrants and negative in the third and fourth. So depending on where the angle falls after subtraction, the sine can be positive or negative.I think I've got it. Both parts use the same formula, but the difference in the angle after subtracting 2π leads to a positive or negative result. That's pretty cool how the formula works out.I should probably double-check my steps to make sure I didn't make any mistakes. For part (a), starting with the formula, plugging in the angles, simplifying the sine argument, and then canceling out. It all seems to flow logically. For part (b), similar steps, but with a negative sine due to the angle being negative after subtraction. Yeah, that seems right.I wonder if there's another way to approach this without using that formula. Maybe by using multiple-angle identities or something else. But honestly, this method seems straightforward once you remember the product formula. I guess it's one of those things where knowing the right identity can really simplify the problem.Another thought: these angles are all related to the seventh and ninth roots of unity. Maybe there's a connection there, but I'm not sure. It might be going into more complex analysis, which is probably beyond what I need for this problem.In any case, I think I've got the proofs down using the product-to-sine formula. It's a neat trick, and I'm glad I remembered it. I should probably review that formula again to make sure I don't forget it for future problems.Overall, I feel confident about these solutions now. The key was recognizing the pattern in the angles and applying the appropriate identity. It's a good reminder of how important it is to remember these trigonometric product formulas—they can really make or break a problem like this.

💡Okay, so I have this inequality to prove, and it's given that ( x_i ) and ( y_i ) are real numbers with ( y_i > 0 ) for each ( i ) from 1 to ( n ). The inequality is:[left(y_{1}+y_{2}+cdots+y_{n}right)left(frac{x_{1}^{2}}{y_{1}}+frac{x_{2}^{2}}{y_{2}}+cdots+frac{x_{n}^{2}}{y_{n}}right) geqslant left(x_{1}+x_{2}+cdots+x_{n}right)^{2}]Hmm, this looks a bit intimidating at first, but maybe I can break it down. I remember there are some inequalities like Cauchy-Schwarz or AM-GM that might be useful here. Let me think about Cauchy-Schwarz because it often deals with sums and products.The Cauchy-Schwarz inequality states that for any real numbers ( a_i ) and ( b_i ):[left( sum_{i=1}^{n} a_i b_i right)^2 leq left( sum_{i=1}^{n} a_i^2 right) left( sum_{i=1}^{n} b_i^2 right)]Wait, but in our case, the inequality is in the opposite direction. Let me see. Maybe I can manipulate the given expression to fit into the Cauchy-Schwarz framework.Looking at the left-hand side of the inequality we need to prove, it's the product of two sums: one is the sum of ( y_i )s, and the other is the sum of ( x_i^2 / y_i ). The right-hand side is the square of the sum of ( x_i )s.Let me try to set up two sequences ( a_i ) and ( b_i ) such that when I apply Cauchy-Schwarz, I get the desired inequality.If I let ( a_i = sqrt{y_i} ) and ( b_i = frac{x_i}{sqrt{y_i}} ), then:- ( a_i^2 = y_i )- ( b_i^2 = frac{x_i^2}{y_i} )- ( a_i b_i = sqrt{y_i} cdot frac{x_i}{sqrt{y_i}} = x_i )So, applying Cauchy-Schwarz to these sequences:[left( sum_{i=1}^{n} a_i b_i right)^2 leq left( sum_{i=1}^{n} a_i^2 right) left( sum_{i=1}^{n} b_i^2 right)]Substituting the values:[left( sum_{i=1}^{n} x_i right)^2 leq left( sum_{i=1}^{n} y_i right) left( sum_{i=1}^{n} frac{x_i^2}{y_i} right)]Which is exactly the inequality we needed to prove, but in reverse. So, this shows that:[left( sum_{i=1}^{n} y_i right) left( sum_{i=1}^{n} frac{x_i^2}{y_i} right) geq left( sum_{i=1}^{n} x_i right)^2]Wait, so by choosing ( a_i = sqrt{y_i} ) and ( b_i = frac{x_i}{sqrt{y_i}} ), applying Cauchy-Schwarz gives us the desired inequality. That seems to work!But just to make sure I didn't make any mistakes, let me test this with a simple case where ( n = 2 ). Let’s say ( y_1 = 1 ), ( y_2 = 1 ), ( x_1 = 1 ), ( x_2 = 1 ).Left-hand side:[(1 + 1)left( frac{1^2}{1} + frac{1^2}{1} right) = 2 times (1 + 1) = 4]Right-hand side:[(1 + 1)^2 = 4]So, equality holds here, which makes sense because in Cauchy-Schwarz, equality holds when ( a_i ) and ( b_i ) are proportional. In this case, ( a_i = sqrt{y_i} = 1 ) and ( b_i = frac{x_i}{sqrt{y_i}} = 1 ), so they are indeed proportional.Let me try another example where ( y_1 = 2 ), ( y_2 = 1 ), ( x_1 = 2 ), ( x_2 = 1 ).Left-hand side:[(2 + 1)left( frac{2^2}{2} + frac{1^2}{1} right) = 3 times (2 + 1) = 9]Right-hand side:[(2 + 1)^2 = 9]Again, equality holds. Hmm, maybe I need a case where the inequality is strict. Let's try ( y_1 = 1 ), ( y_2 = 2 ), ( x_1 = 1 ), ( x_2 = 1 ).Left-hand side:[(1 + 2)left( frac{1^2}{1} + frac{1^2}{2} right) = 3 times left(1 + 0.5right) = 3 times 1.5 = 4.5]Right-hand side:[(1 + 1)^2 = 4]So, 4.5 ≥ 4, which holds true. That’s good.Wait, but in the previous examples, sometimes equality holds, sometimes it's strict. So, the inequality is always true, and equality holds when ( frac{x_i}{sqrt{y_i}} ) is proportional to ( sqrt{y_i} ), meaning ( x_i ) is proportional to ( y_i ).Let me think about another approach. Maybe using the AM-GM inequality? But I'm not sure how to apply AM-GM here because we have products and sums in a more complex way.Alternatively, maybe expanding both sides and comparing term by term. Let me try that.Left-hand side:[left( sum_{i=1}^{n} y_i right) left( sum_{j=1}^{n} frac{x_j^2}{y_j} right) = sum_{i=1}^{n} sum_{j=1}^{n} y_i cdot frac{x_j^2}{y_j}]Right-hand side:[left( sum_{i=1}^{n} x_i right)^2 = sum_{i=1}^{n} x_i^2 + 2 sum_{1 leq i < j leq n} x_i x_j]So, the left-hand side is a double sum, and the right-hand side is the square of the sum. To compare them, maybe I can subtract the right-hand side from the left-hand side and show that it's non-negative.Let’s compute:[left( sum_{i=1}^{n} y_i right) left( sum_{j=1}^{n} frac{x_j^2}{y_j} right) - left( sum_{i=1}^{n} x_i right)^2 geq 0]Expanding the left-hand side:[sum_{i=1}^{n} sum_{j=1}^{n} y_i cdot frac{x_j^2}{y_j} - sum_{i=1}^{n} x_i^2 - 2 sum_{1 leq i < j leq n} x_i x_j]Simplify the first term:[sum_{i=1}^{n} sum_{j=1}^{n} frac{y_i}{y_j} x_j^2 - sum_{i=1}^{n} x_i^2 - 2 sum_{1 leq i < j leq n} x_i x_j]Notice that when ( i = j ), the term is ( x_j^2 ), so:[sum_{i=1}^{n} x_i^2 cdot frac{y_i}{y_i} + sum_{i neq j} frac{y_i}{y_j} x_j^2 - sum_{i=1}^{n} x_i^2 - 2 sum_{1 leq i < j leq n} x_i x_j]Simplify:[sum_{i=1}^{n} x_i^2 + sum_{i neq j} frac{y_i}{y_j} x_j^2 - sum_{i=1}^{n} x_i^2 - 2 sum_{1 leq i < j leq n} x_i x_j]The ( sum x_i^2 ) terms cancel out:[sum_{i neq j} frac{y_i}{y_j} x_j^2 - 2 sum_{1 leq i < j leq n} x_i x_j]Now, let's write this as:[sum_{i=1}^{n} sum_{j=1, j neq i}^{n} frac{y_i}{y_j} x_j^2 - 2 sum_{1 leq i < j leq n} x_i x_j]Hmm, this seems a bit messy. Maybe I can rearrange the terms or factor them differently. Alternatively, perhaps completing the square or using another inequality.Wait, another thought: maybe using the concept of weighted averages or something related to the Cauchy-Schwarz in another form.Alternatively, consider the expression as a quadratic form. Let me think of vectors ( mathbf{x} = (x_1, x_2, ldots, x_n) ) and ( mathbf{y} = (y_1, y_2, ldots, y_n) ). Then, the left-hand side is ( mathbf{y} cdot left( frac{mathbf{x}^2}{mathbf{y}} right) ), which is similar to ( mathbf{y} cdot mathbf{z} ) where ( z_i = frac{x_i^2}{y_i} ). But I'm not sure if that helps directly.Wait, going back to the Cauchy-Schwarz approach, I think that was the right path. Let me just recap:By setting ( a_i = sqrt{y_i} ) and ( b_i = frac{x_i}{sqrt{y_i}} ), we have:[left( sum a_i b_i right)^2 leq left( sum a_i^2 right) left( sum b_i^2 right)]Which translates to:[left( sum x_i right)^2 leq left( sum y_i right) left( sum frac{x_i^2}{y_i} right)]Which is exactly the inequality we needed to prove. So, that seems solid.But just to make sure, let me think about another perspective. Maybe using Lagrange multipliers or optimization, but that might be overcomplicating.Alternatively, think about the inequality as a form of the Cauchy-Schwarz inequality in the context of inner products. If I define an inner product with weights ( y_i ), then maybe this inequality is a specific case.But I think the Cauchy-Schwarz approach is sufficient and straightforward. It directly applies and gives the result without much hassle.So, in summary, by choosing appropriate sequences and applying the Cauchy-Schwarz inequality, we can prove the given inequality. The key was to recognize the structure that matches Cauchy-Schwarz and set up the sequences accordingly.I think I've got it. The proof using Cauchy-Schwarz seems correct, and testing with specific examples supports the inequality. I don't see any flaws in this reasoning, so I'm confident this is the right approach.

💡Okay, so I have two problems here about finding the standard equation of a hyperbola. Let me try to tackle them one by one. Starting with problem (1): It says the hyperbola passes through points P(3, 15/4) and Q(-16/3, 5). Hmm, okay. I remember that the standard form of a hyperbola can be either vertical or horizontal. Since the problem doesn't specify, I might need to figure out which one it is based on the points given.Wait, the initial solution used an equation of the form mx² + ny² = 1, which is a general conic section equation. Since it's a hyperbola, mn should be negative, which makes sense because hyperbolas have one positive and one negative term. So, if I let the equation be mx² + ny² = 1, then plugging in the points P and Q should give me two equations to solve for m and n.Let me write that down:For point P(3, 15/4):m*(3)² + n*(15/4)² = 1So, 9m + (225/16)n = 1For point Q(-16/3, 5):m*(-16/3)² + n*(5)² = 1So, (256/9)m + 25n = 1Now I have a system of two equations:1) 9m + (225/16)n = 12) (256/9)m + 25n = 1I need to solve for m and n. Let me see. Maybe I can use substitution or elimination. Let's try elimination.First, let me make the coefficients of m or n the same. Let's try to eliminate m. To do that, I can find the least common multiple of 9 and 256/9. Hmm, that might be messy. Alternatively, maybe I can express m from one equation and substitute into the other.From equation 1:9m = 1 - (225/16)nSo, m = [1 - (225/16)n]/9Simplify:m = (1/9) - (225/144)nSimplify 225/144: divide numerator and denominator by 9, get 25/16So, m = (1/9) - (25/16)nNow plug this into equation 2:(256/9)*[(1/9) - (25/16)n] + 25n = 1Let me compute each term step by step.First, (256/9)*(1/9) = 256/81Second, (256/9)*(-25/16)n = (256*(-25))/(9*16) n = (-6400)/144 n = (-400)/9 nSo, equation becomes:256/81 - (400/9)n + 25n = 1Combine the n terms:- (400/9)n + 25n = (-400/9 + 225/9)n = (-175/9)nSo, equation is:256/81 - (175/9)n = 1Subtract 256/81 from both sides:- (175/9)n = 1 - 256/81Compute 1 - 256/81:1 is 81/81, so 81/81 - 256/81 = -175/81So:- (175/9)n = -175/81Divide both sides by -175/9:n = (-175/81) / (-175/9) = (175/81) * (9/175) = 9/81 = 1/9So, n = 1/9Now plug back into m = (1/9) - (25/16)n:m = (1/9) - (25/16)*(1/9) = (1/9) - (25/144)Convert to common denominator:1/9 = 16/144So, 16/144 - 25/144 = -9/144 = -1/16So, m = -1/16Therefore, the equation is:(-1/16)x² + (1/9)y² = 1But to write it in standard form, we can rearrange:y²/9 - x²/16 = 1Which is a vertical hyperbola centered at the origin, with a² = 9 and b² = 16.Okay, that seems to make sense. Let me just verify by plugging in the points.For P(3, 15/4):y²/9 = (225/16)/9 = 225/(16*9) = 25/16x²/16 = 9/16So, 25/16 - 9/16 = 16/16 = 1. Correct.For Q(-16/3, 5):y²/9 = 25/9x²/16 = (256/9)/16 = 256/(9*16) = 16/9So, 25/9 - 16/9 = 9/9 = 1. Correct.Alright, so problem (1) seems solved.Moving on to problem (2): The distance between the foci is c = sqrt(6), it passes through (-5, 2), and the foci are on the x-axis.Okay, so since the foci are on the x-axis, it's a horizontal hyperbola. The standard form is x²/a² - y²/b² = 1, with c² = a² + b².Given c = sqrt(6), so c² = 6.So, 6 = a² + b².We need another equation because we have two variables, a² and b². The hyperbola passes through (-5, 2), so plugging that into the equation:(-5)²/a² - (2)²/b² = 125/a² - 4/b² = 1So, now we have:1) 25/a² - 4/b² = 12) a² + b² = 6Hmm, so we have two equations with two variables. Let me denote A = a² and B = b² to make it easier.So, equations become:1) 25/A - 4/B = 12) A + B = 6From equation 2, B = 6 - APlug into equation 1:25/A - 4/(6 - A) = 1Let me solve this equation for A.First, write it as:25/A - 4/(6 - A) = 1Multiply both sides by A(6 - A) to eliminate denominators:25*(6 - A) - 4*A = A(6 - A)Compute each term:25*(6 - A) = 150 - 25A-4*A = -4ASo, left side: 150 - 25A - 4A = 150 - 29ARight side: A(6 - A) = 6A - A²So, equation becomes:150 - 29A = 6A - A²Bring all terms to one side:150 - 29A - 6A + A² = 0Simplify:A² - 35A + 150 = 0Now, solve quadratic equation:A² - 35A + 150 = 0Use quadratic formula:A = [35 ± sqrt(35² - 4*1*150)] / 2Compute discriminant:35² = 12254*1*150 = 600So, sqrt(1225 - 600) = sqrt(625) = 25Thus, A = [35 ±25]/2So, two solutions:A = (35 +25)/2 = 60/2 = 30A = (35 -25)/2 = 10/2 = 5So, A = 30 or A = 5But from equation 2, A + B = 6, so if A = 30, then B = 6 -30 = -24, which is impossible because B = b² must be positive.Therefore, A = 5, so B = 6 -5 =1Therefore, a² =5 and b²=1Thus, the equation is:x²/5 - y²/1 =1Simplify:x²/5 - y² =1Let me verify by plugging in (-5,2):(-5)²/5 - (2)² =25/5 -4=5 -4=1. Correct.Also, check c² = a² + b²=5 +1=6, which matches c= sqrt(6). Correct.So, problem (2) is solved.Wait, but in the initial solution, for problem (2), the equation was written as x²/5 - y²=1, which is the same as what I got. So, that seems correct.I think I've got both problems solved. Let me just recap:Problem (1): The hyperbola passes through two points, so I set up the general conic equation, solved for m and n, found they were negative reciprocals, leading to the standard form y²/9 -x²/16=1.Problem (2): Given c, a point, and foci on x-axis, set up the standard horizontal hyperbola, used the point to form an equation, solved the system, found a²=5 and b²=1, leading to x²/5 - y²=1.Everything seems consistent. I don't see any mistakes in my reasoning, and the checks with the given points worked out. So, I think I'm confident with these solutions.**Final Answer**(1) The standard equation is boxed{dfrac{y^2}{9} - dfrac{x^2}{16} = 1}.(2) The standard equation is boxed{dfrac{x^2}{5} - y^2 = 1}.

💡To compute ( frac{50!}{48!} ) without using a calculator, I start by understanding the factorial notation. The factorial of a number ( n ), denoted as ( n! ), is the product of all positive integers from 1 up to ( n ).So, ( 50! ) means ( 50 times 49 times 48 times cdots times 1 ), and ( 48! ) means ( 48 times 47 times cdots times 1 ).When I divide ( 50! ) by ( 48! ), the expression becomes:[frac{50!}{48!} = frac{50 times 49 times 48 times cdots times 1}{48 times 47 times cdots times 1}]I notice that the terms from ( 48 ) downwards appear in both the numerator and the denominator. These common terms cancel each other out, simplifying the expression to:[50 times 49]Next, I calculate the product of 50 and 49:[50 times 49 = 2450]Therefore, the value of ( frac{50!}{48!} ) is 2450.

💡Alright, so I have this cube problem to solve. Let me try to visualize it first. The cube is labeled ABCD A₁B₁C₁D₁, and each edge has length a. Points M and N are on segments BD and CC₁ respectively. The line MN forms an angle of π/4 with the plane ABCD and an angle of π/6 with the plane BB₁C₁C. I need to find the length of segment MN and the radius of a sphere centered on MN that's tangent to both planes ABCD and BB₁C₁C.Okay, let's start by assigning coordinates to the cube. Since it's a cube, all edges are equal, so I can place it in a 3D coordinate system. Let me assign the coordinates as follows:- A = (0, 0, 0)- B = (a, 0, 0)- C = (a, a, 0)- D = (0, a, 0)- A₁ = (0, 0, a)- B₁ = (a, 0, a)- C₁ = (a, a, a)- D₁ = (0, a, a)So, point M is on BD, which goes from B (a, 0, 0) to D (0, a, 0). Let's parameterize point M. If I let M divide BD in some ratio, say, k:1-k, then the coordinates of M can be written as:M = (a - ka, ka, 0) = (a(1 - k), a k, 0)Similarly, point N is on CC₁, which goes from C (a, a, 0) to C₁ (a, a, a). Let's parameterize N as well. If I let N divide CC₁ in some ratio, say, m:1-m, then the coordinates of N are:N = (a, a, m a)So, the coordinates are:M = (a(1 - k), a k, 0)N = (a, a, m a)Now, the vector MN can be found by subtracting the coordinates of M from N:MN = (a - a(1 - k), a - a k, m a - 0) = (a k, a(1 - k), m a)So, the direction vector of MN is (a k, a(1 - k), m a). Let's denote this as (x, y, z) for simplicity:x = a ky = a(1 - k)z = m aNow, the problem states that MN forms an angle of π/4 with the plane ABCD and an angle of π/6 with the plane BB₁C₁C.First, let's recall that the angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane. So, if θ is the angle between the line and the plane, then the angle between the line and the normal to the plane is 90° - θ.But, in terms of vectors, the angle between the line and the plane can be found using the sine of the angle between the line and the normal to the plane. Alternatively, sometimes it's easier to use the direction ratios and the normal vector of the plane.Let me think about the plane ABCD. Since ABCD is the base of the cube, it's the xy-plane. The normal vector to plane ABCD is along the z-axis, which is (0, 0, 1).Similarly, the plane BB₁C₁C is the right face of the cube. This plane is the yz-plane shifted along the x-axis. The normal vector to this plane would be along the x-axis, which is (1, 0, 0).So, the angle between MN and plane ABCD is π/4, which is 45 degrees. The angle between MN and plane BB₁C₁C is π/6, which is 30 degrees.Let me recall that the angle φ between a line with direction vector (x, y, z) and a plane with normal vector (n₁, n₂, n₃) is given by:sin φ = |(x n₁ + y n₂ + z n₃)| / (|MN| |n|)But wait, actually, the angle between the line and the plane is equal to the angle between the line and its projection onto the plane. Alternatively, it's equal to the complement of the angle between the line and the normal to the plane.So, if θ is the angle between the line and the normal, then the angle between the line and the plane is 90° - θ.Therefore, sin(φ) = cos(θ), where φ is the angle between the line and the plane, and θ is the angle between the line and the normal.So, for the plane ABCD, the normal is (0, 0, 1). The direction vector of MN is (a k, a(1 - k), m a). So, the angle between MN and the normal is θ, and the angle between MN and the plane is φ = 90° - θ.Given that φ = π/4, so θ = π/2 - π/4 = π/4.Therefore, the angle between MN and the normal to ABCD is π/4.Similarly, for the plane BB₁C₁C, the normal is (1, 0, 0). The angle between MN and this normal is θ', and the angle between MN and the plane is φ' = 90° - θ' = π/6. Therefore, θ' = π/2 - π/6 = π/3.So, now, we can write the expressions for the angles.First, for the angle with plane ABCD:cos(θ) = |(MN · n)| / |MN| |n|Where n is the normal vector (0, 0, 1).So, MN · n = z-component of MN, which is m a.Therefore,cos(π/4) = |m a| / |MN|Similarly, for the angle with plane BB₁C₁C:cos(θ') = |(MN · n')| / |MN| |n'|Where n' is the normal vector (1, 0, 0).So, MN · n' = x-component of MN, which is a k.Therefore,cos(π/3) = |a k| / |MN|So, now, let's write these equations.First, cos(π/4) = |m a| / |MN|cos(π/4) is √2 / 2, so:√2 / 2 = |m a| / |MN| ...(1)Second, cos(π/3) = |a k| / |MN|cos(π/3) is 1/2, so:1/2 = |a k| / |MN| ...(2)Also, |MN| is the magnitude of the vector MN:|MN| = sqrt( (a k)^2 + (a(1 - k))^2 + (m a)^2 )Let me compute |MN|:|MN| = a sqrt( k² + (1 - k)² + m² )So, |MN| = a sqrt( k² + (1 - 2k + k²) + m² ) = a sqrt( 2k² - 2k + 1 + m² )So, |MN| = a sqrt(2k² - 2k + 1 + m² )Now, from equation (2):1/2 = |a k| / |MN|So,1/2 = (a k) / (a sqrt(2k² - 2k + 1 + m² )) )Simplify:1/2 = k / sqrt(2k² - 2k + 1 + m² )Similarly, from equation (1):√2 / 2 = |m a| / |MN| = (m a) / (a sqrt(2k² - 2k + 1 + m² )) = m / sqrt(2k² - 2k + 1 + m² )So,√2 / 2 = m / sqrt(2k² - 2k + 1 + m² )So, now, we have two equations:1) 1/2 = k / sqrt(2k² - 2k + 1 + m² )2) √2 / 2 = m / sqrt(2k² - 2k + 1 + m² )Let me denote sqrt(2k² - 2k + 1 + m² ) as S for simplicity.Then, equations become:1) 1/2 = k / S => S = 2k2) √2 / 2 = m / S => S = m / (√2 / 2) = 2m / √2 = √2 mSo, from equation 1: S = 2kFrom equation 2: S = √2 mTherefore, 2k = √2 m => m = (2k) / √2 = √2 kSo, m = √2 kNow, since S = 2k, and S = sqrt(2k² - 2k + 1 + m² ), let's substitute m = √2 k into S:S = sqrt(2k² - 2k + 1 + (√2 k)^2 ) = sqrt(2k² - 2k + 1 + 2k² ) = sqrt(4k² - 2k + 1 )But S is also equal to 2k, so:2k = sqrt(4k² - 2k + 1 )Let me square both sides:(2k)^2 = 4k² - 2k + 14k² = 4k² - 2k + 1Subtract 4k² from both sides:0 = -2k + 1So,-2k + 1 = 0 => 2k = 1 => k = 1/2So, k = 1/2Then, m = √2 k = √2 * (1/2) = √2 / 2So, now, we have k = 1/2 and m = √2 / 2Therefore, the coordinates of M and N are:M = (a(1 - 1/2), a*(1/2), 0) = (a/2, a/2, 0)N = (a, a, (√2 / 2) a )So, vector MN is:N - M = (a - a/2, a - a/2, (√2 / 2) a - 0 ) = (a/2, a/2, (√2 / 2) a )Therefore, the direction vector is (a/2, a/2, (√2 / 2) a )So, the length of MN is the magnitude of this vector:|MN| = sqrt( (a/2)^2 + (a/2)^2 + ( (√2 / 2) a )^2 )Compute each term:(a/2)^2 = a² / 4Similarly, the second term is a² / 4Third term: ( (√2 / 2) a )^2 = (2 / 4) a² = a² / 2So, |MN| = sqrt( a² / 4 + a² / 4 + a² / 2 ) = sqrt( (a² / 4 + a² / 4) + a² / 2 ) = sqrt( a² / 2 + a² / 2 ) = sqrt( a² ) = aSo, the length of MN is a.Wait, that's interesting. So, despite the angles, the length of MN is equal to the edge length of the cube.But let me double-check my calculations.First, the direction vector MN is (a/2, a/2, (√2 / 2) a )So, squared terms:(a/2)^2 = a² / 4(a/2)^2 = a² / 4(√2 / 2 a)^2 = (2 / 4) a² = a² / 2Adding them up: a² / 4 + a² / 4 + a² / 2 = (a² / 4 + a² / 4) + a² / 2 = a² / 2 + a² / 2 = a²So, sqrt(a²) = a. Correct.So, the length of MN is a.Alright, that's part (a). Now, part (b): the radius of a sphere with its center on segment MN that is tangent to planes ABCD and BB₁C₁C.So, the sphere is tangent to both planes ABCD and BB₁C₁C, and its center lies on MN.Let me recall that the distance from the center of the sphere to each plane must be equal to the radius R.So, if the center is on MN, then its coordinates can be parameterized along MN.Let me parameterize the center O of the sphere as a point on MN.Since MN is from M (a/2, a/2, 0) to N (a, a, (√2 / 2) a ), we can write the parametric equations for MN.Let me define a parameter t, where t ranges from 0 to 1.So, when t = 0, we are at M, and when t = 1, we are at N.Thus, the coordinates of O can be written as:O = M + t (N - M ) = (a/2, a/2, 0) + t (a/2, a/2, (√2 / 2) a )So,O = ( a/2 + t a/2, a/2 + t a/2, 0 + t (√2 / 2) a )Simplify:O = ( a/2 (1 + t), a/2 (1 + t), (√2 / 2) a t )So, O = ( a/2 (1 + t), a/2 (1 + t), (√2 / 2) a t )Now, the distance from O to plane ABCD is the z-coordinate of O, since ABCD is the xy-plane (z=0). So, distance from O to ABCD is:d1 = |z| = (√2 / 2) a tSimilarly, the distance from O to plane BB₁C₁C. Plane BB₁C₁C is the plane x = a, right? Because BB₁C₁C is the right face of the cube, which is at x = a.So, the distance from O to plane BB₁C₁C is |x - a|, since the plane is x = a.So, distance d2 = |x - a| = | a/2 (1 + t) - a | = | a/2 (1 + t - 2) | = | a/2 (t - 1) | = (a/2) |t - 1|Since t ranges from 0 to 1, (t - 1) is negative or zero, so |t - 1| = 1 - tThus, d2 = (a/2)(1 - t)Now, since the sphere is tangent to both planes, the distances d1 and d2 must both equal the radius R.Therefore,(√2 / 2) a t = Rand(a/2)(1 - t) = RSo, we have two equations:1) (√2 / 2) a t = R2) (a / 2)(1 - t) = RSet them equal to each other:(√2 / 2) a t = (a / 2)(1 - t)We can cancel a / 2 from both sides:√2 t = 1 - tBring terms with t to one side:√2 t + t = 1Factor t:t (√2 + 1) = 1Thus,t = 1 / (√2 + 1 )Rationalize the denominator:t = [1 / (√2 + 1)] * [ (√2 - 1) / (√2 - 1) ] = (√2 - 1) / ( (√2)^2 - 1^2 ) = (√2 - 1) / (2 - 1) = √2 - 1So, t = √2 - 1Now, plug t back into equation for R:R = (√2 / 2) a t = (√2 / 2) a (√2 - 1 )Simplify:√2 * (√2 - 1 ) = 2 - √2So,R = (2 - √2 ) / 2 * a = (2 - √2 ) a / 2So, R = [ (2 - √2 ) / 2 ] aAlternatively, R = a (2 - √2 ) / 2So, that's the radius.Let me just verify the calculations.From the two distance equations:(√2 / 2) a t = R(a / 2)(1 - t) = RSet equal:(√2 / 2) a t = (a / 2)(1 - t)Cancel a/2:√2 t = 1 - t√2 t + t = 1t (√2 + 1 ) = 1t = 1 / (√2 + 1 ) = √2 - 1Then, R = (√2 / 2) a (√2 - 1 ) = (2 - √2 ) a / 2Yes, that seems correct.So, the radius is (2 - √2 ) a / 2**Final Answer**a) The length of segment ( MN ) is (boxed{a}).b) The radius of the sphere is (boxed{dfrac{a(2 - sqrt{2})}{2}}).

💡Okay, so I have this problem about two lathes, A and B, that are processing parts with a diameter of 100 cm. To check the quality, they've taken samples from each lathe—six pieces of data for each. For lathe A, the diameters are 99, 100, 98, 100, 100, and 103. For lathe B, they are 99, 100, 102, 99, 100, and 100. The first part of the problem asks me to calculate the mean and variance of both sets of data. The second part wants me to determine which lathe processes parts with more stable quality based on these calculations.Alright, starting with the mean. The mean is just the average of the numbers. So for lathe A, I need to add up all the diameters and divide by the number of samples, which is 6. Let me do that step by step.For lathe A: 99 + 100 + 98 + 100 + 100 + 103. Let me add these up. 99 plus 100 is 199, plus 98 is 297, plus 100 is 397, plus another 100 is 497, and finally plus 103 is 600. So the total is 600. Now, dividing by 6 gives me 600 / 6 = 100. So the mean for lathe A is 100 cm.Now for lathe B: 99 + 100 + 102 + 99 + 100 + 100. Let's add these up. 99 plus 100 is 199, plus 102 is 301, plus 99 is 400, plus 100 is 500, and plus another 100 is 600. So the total is also 600. Dividing by 6 gives me 600 / 6 = 100. So the mean for lathe B is also 100 cm.Okay, so both lathes have the same mean diameter of 100 cm. That's interesting. Now, moving on to variance. Variance measures how spread out the numbers are from the mean. The formula for variance is the average of the squared differences from the mean.So for lathe A, I need to calculate each diameter's deviation from the mean, square it, and then take the average. Let's do that.For lathe A:- 99: (99 - 100) = -1, squared is 1- 100: (100 - 100) = 0, squared is 0- 98: (98 - 100) = -2, squared is 4- 100: (100 - 100) = 0, squared is 0- 100: (100 - 100) = 0, squared is 0- 103: (103 - 100) = 3, squared is 9Now, adding these squared differences: 1 + 0 + 4 + 0 + 0 + 9 = 14. Then, variance is 14 divided by 6, which is approximately 2.333.For lathe B:- 99: (99 - 100) = -1, squared is 1- 100: (100 - 100) = 0, squared is 0- 102: (102 - 100) = 2, squared is 4- 99: (99 - 100) = -1, squared is 1- 100: (100 - 100) = 0, squared is 0- 100: (100 - 100) = 0, squared is 0Adding these squared differences: 1 + 0 + 4 + 1 + 0 + 0 = 6. Then, variance is 6 divided by 6, which is 1.So, the variance for lathe A is approximately 2.333, and for lathe B, it's 1.Now, the second part of the problem asks which lathe has more stable quality. Since variance measures the spread, a lower variance indicates more stability because the data points are closer to the mean. Therefore, lathe B, with a variance of 1, has more stable quality compared to lathe A, which has a variance of approximately 2.333.I think that's it. Both lathes have the same average diameter, but lathe B's parts are less varied, meaning they are more consistent in size, which is better for quality control.