💡Okay, so I have this function ( f(x) = frac{a x}{2 x + 3} ), and I need to find the value of ( a ) such that when I apply the function twice, I get back to ( x ). In other words, ( f(f(x)) = x ) for all ( x ). Hmm, that sounds interesting. I guess I need to compute ( f(f(x)) ) and set it equal to ( x ), then solve for ( a ). Let me try to do that step by step.First, let me write down the function again:( f(x) = frac{a x}{2 x + 3} )Now, I need to find ( f(f(x)) ). That means I substitute ( f(x) ) into itself. So, let me compute ( f(f(x)) ):( f(f(x)) = fleft( frac{a x}{2 x + 3} right) )To compute this, I'll replace every ( x ) in the original function with ( frac{a x}{2 x + 3} ). So, substituting into ( f(x) ):( fleft( frac{a x}{2 x + 3} right) = frac{a cdot left( frac{a x}{2 x + 3} right)}{2 cdot left( frac{a x}{2 x + 3} right) + 3} )Okay, let me simplify this expression step by step. First, the numerator:Numerator: ( a cdot left( frac{a x}{2 x + 3} right) = frac{a^2 x}{2 x + 3} )Denominator: ( 2 cdot left( frac{a x}{2 x + 3} right) + 3 )Let me compute the denominator:( 2 cdot left( frac{a x}{2 x + 3} right) = frac{2 a x}{2 x + 3} )So, adding 3 to that:( frac{2 a x}{2 x + 3} + 3 )To add these, I need a common denominator. The 3 can be written as ( frac{3(2 x + 3)}{2 x + 3} ):( frac{2 a x}{2 x + 3} + frac{3(2 x + 3)}{2 x + 3} = frac{2 a x + 6 x + 9}{2 x + 3} )So, the denominator simplifies to ( frac{2 a x + 6 x + 9}{2 x + 3} ).Now, putting the numerator and denominator together:( f(f(x)) = frac{frac{a^2 x}{2 x + 3}}{frac{2 a x + 6 x + 9}{2 x + 3}} )Since both the numerator and denominator have the same denominator ( 2 x + 3 ), they cancel out when we divide them:( f(f(x)) = frac{a^2 x}{2 a x + 6 x + 9} )So, now we have ( f(f(x)) = frac{a^2 x}{(2 a + 6) x + 9} ). According to the problem, this should be equal to ( x ) for all ( x ). So, we can set up the equation:( frac{a^2 x}{(2 a + 6) x + 9} = x )To solve for ( a ), let's cross-multiply to eliminate the denominator:( a^2 x = x cdot [(2 a + 6) x + 9] )Expanding the right-hand side:( a^2 x = (2 a + 6) x^2 + 9 x )Now, let's bring all terms to one side to set the equation to zero:( (2 a + 6) x^2 + 9 x - a^2 x = 0 )Factor out ( x ) from the terms:( (2 a + 6) x^2 + (9 - a^2) x = 0 )Since this equation must hold true for all ( x ), the coefficients of each power of ( x ) must be zero. That gives us two equations:1. Coefficient of ( x^2 ): ( 2 a + 6 = 0 )2. Coefficient of ( x ): ( 9 - a^2 = 0 )Let's solve the first equation:( 2 a + 6 = 0 )Subtract 6 from both sides:( 2 a = -6 )Divide both sides by 2:( a = -3 )Now, let's check the second equation with this value of ( a ):( 9 - a^2 = 0 )Substitute ( a = -3 ):( 9 - (-3)^2 = 9 - 9 = 0 )Which is true. So, both equations are satisfied when ( a = -3 ).Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from ( f(f(x)) ), substituting, simplifying, setting equal to ( x ), cross-multiplying, expanding, and equating coefficients. It all seems to check out. So, I think ( a = -3 ) is the correct answer.Just to be thorough, let me plug ( a = -3 ) back into the original function and compute ( f(f(x)) ) to see if it indeed equals ( x ).Original function with ( a = -3 ):( f(x) = frac{-3 x}{2 x + 3} )Compute ( f(f(x)) ):First, compute ( f(x) = frac{-3 x}{2 x + 3} )Then, ( f(f(x)) = fleft( frac{-3 x}{2 x + 3} right) )Substitute into ( f ):( fleft( frac{-3 x}{2 x + 3} right) = frac{-3 cdot left( frac{-3 x}{2 x + 3} right)}{2 cdot left( frac{-3 x}{2 x + 3} right) + 3} )Simplify numerator:( -3 cdot left( frac{-3 x}{2 x + 3} right) = frac{9 x}{2 x + 3} )Simplify denominator:( 2 cdot left( frac{-3 x}{2 x + 3} right) + 3 = frac{-6 x}{2 x + 3} + 3 )Convert 3 to have the same denominator:( frac{-6 x}{2 x + 3} + frac{3(2 x + 3)}{2 x + 3} = frac{-6 x + 6 x + 9}{2 x + 3} = frac{9}{2 x + 3} )So, denominator is ( frac{9}{2 x + 3} )Now, ( f(f(x)) = frac{frac{9 x}{2 x + 3}}{frac{9}{2 x + 3}} = frac{9 x}{2 x + 3} cdot frac{2 x + 3}{9} = x )Yes, it simplifies to ( x ). So, ( f(f(x)) = x ) when ( a = -3 ). That confirms my solution.I think I've covered all the steps and checked my work. I don't see any mistakes, so I'm confident that ( a = -3 ) is the correct answer.**Final Answer**The value of the real number ( a ) is boxed{-3}.

💡Okay, so I have this inequality to solve: 1/x is greater than 1/(x - 1). Hmm, that seems a bit tricky, but let me try to work through it step by step. First, I know that when dealing with inequalities involving fractions, I have to be careful about the denominators because they can't be zero, and the sign of the denominator affects the inequality. So, I should probably start by figuring out the domain of this inequality. That means I need to find the values of x where both denominators are defined. So, for 1/x to be defined, x can't be zero. And for 1/(x - 1) to be defined, x can't be 1. So, the domain is all real numbers except x = 0 and x = 1. That means I have to consider intervals around these points: (-∞, 0), (0, 1), and (1, ∞). Now, I want to solve the inequality 1/x > 1/(x - 1). Maybe I can subtract 1/(x - 1) from both sides to get everything on one side. Let me try that:1/x - 1/(x - 1) > 0.Okay, now I have a single rational expression. To combine these two fractions, I need a common denominator. The denominators are x and (x - 1), so the common denominator would be x(x - 1). Let me rewrite each fraction with this common denominator:( (x - 1) - x ) / [x(x - 1)] > 0.Simplifying the numerator: (x - 1) - x = x - 1 - x = -1. So now the inequality becomes:-1 / [x(x - 1)] > 0.Hmm, so I have -1 divided by the product of x and (x - 1). I need to find where this expression is greater than zero. Since the numerator is -1, which is negative, the sign of the entire expression depends on the denominator. Let me think about the denominator: x(x - 1). This product will be positive or negative depending on the values of x. I can analyze the sign of the denominator in each interval determined by the critical points x = 0 and x = 1.1. When x < 0: Let's pick a test point, say x = -1. Then x is negative, and (x - 1) is also negative (since -1 - 1 = -2). Multiplying two negatives gives a positive. So, the denominator is positive. The numerator is -1, so the entire expression is negative. Since we're looking for where the expression is greater than zero, this interval doesn't work.2. When 0 < x < 1: Let's pick x = 1/2. Then x is positive, and (x - 1) is negative (since 1/2 - 1 = -1/2). Multiplying a positive and a negative gives a negative. So, the denominator is negative. The numerator is -1, so the entire expression is positive (because a negative divided by a negative is positive). This interval satisfies the inequality.3. When x > 1: Let's pick x = 2. Then x is positive, and (x - 1) is also positive (since 2 - 1 = 1). Multiplying two positives gives a positive. So, the denominator is positive. The numerator is -1, so the entire expression is negative. This interval doesn't satisfy the inequality.So, from this analysis, the only interval where the expression -1/[x(x - 1)] is greater than zero is when 0 < x < 1. Wait, but I should double-check if there are any other considerations. For example, what happens at the critical points x = 0 and x = 1? Well, at x = 0, the original expression 1/x is undefined, so it's not included. Similarly, at x = 1, 1/(x - 1) is undefined, so that's also excluded. Also, I should make sure that I didn't make any mistakes in simplifying the inequality. Let me go through that again:Starting with 1/x > 1/(x - 1).Subtract 1/(x - 1) from both sides: 1/x - 1/(x - 1) > 0.Combine the fractions: [ (x - 1) - x ] / [x(x - 1)] > 0.Simplify numerator: x - 1 - x = -1.So, -1 / [x(x - 1)] > 0.Yes, that seems correct. So, the expression simplifies to -1 over the product of x and (x - 1). Another way to think about this is to consider the original inequality: 1/x > 1/(x - 1). Maybe I can cross-multiply, but I have to be careful because the direction of the inequality depends on the sign of the denominators. Wait, cross-multiplying might be tricky here because x and (x - 1) can be positive or negative depending on the interval. So, maybe it's safer to stick with the method I used earlier where I combined the fractions and analyzed the sign.But just to explore, if I try cross-multiplying, I have to consider two cases: when both denominators are positive or both are negative.Case 1: Both x and (x - 1) are positive. That would mean x > 1. In this case, multiplying both sides by x(x - 1) (which is positive) doesn't change the inequality direction:1*(x - 1) > 1*x.Simplify: x - 1 > x.Subtract x from both sides: -1 > 0.That's not true, so no solution in this case.Case 2: Both x and (x - 1) are negative. That would mean x < 0. Multiplying both sides by x(x - 1) (which is positive because negative times negative is positive) doesn't change the inequality direction:1*(x - 1) > 1*x.Simplify: x - 1 > x.Subtract x from both sides: -1 > 0.Again, not true, so no solution in this case.Case 3: One denominator is positive and the other is negative. That would happen when 0 < x < 1. In this interval, x is positive and (x - 1) is negative. So, multiplying both sides by x(x - 1) (which is negative) would reverse the inequality direction:1*(x - 1) < 1*x.Simplify: x - 1 < x.Subtract x from both sides: -1 < 0.That's always true, so in this interval, the inequality holds.So, this method also leads me to the same conclusion: the solution is 0 < x < 1.Therefore, the solution set is (0, 1), which corresponds to option C.

💡To determine the percentage of famous Europeans born in September, I first need to identify the number of people born in that month from the bar graph. Looking at the graph, the bar corresponding to September has a height that aligns with the number 12 on the vertical axis. This indicates that 12 people were born in September.Next, I know that the total number of famous Europeans surveyed is 150. To find the percentage of people born in September, I use the formula for percentage:[text{Percentage} = left( frac{text{Number of people born in September}}{text{Total number of people}} right) times 100]Plugging in the numbers:[text{Percentage} = left( frac{12}{150} right) times 100 = frac{12 times 100}{150} = frac{1200}{150} = 8]Therefore, the percentage of famous Europeans born in September is 8%.

💡First, I need to find the width of the canal at the top using the area formula for a trapezium.The area of a trapezium is given by:Area = 1/2 * (sum of parallel sides) * heightGiven:- Area = 10290 sq m- Height (depth) = 257.25 m- Bottom width = 4 m- Top width = x (unknown)Plugging the known values into the formula:10290 = 1/2 * (x + 4) * 257.25To eliminate the fraction, multiply both sides by 2:20580 = (x + 4) * 257.25Next, subtract 1029 from both sides to isolate the term with x:19551 = 257.25xFinally, divide both sides by 257.25 to solve for x:x ≈ 76So, the width of the canal at the top is approximately 76 meters.

💡First, determine the number of intervals along the path by dividing the total length by the spacing between trees: 80 meters ÷ 4 meters = 20 intervals.Since trees are planted at both ends, add 1 to the number of intervals to find the number of trees on one side: 20 + 1 = 21 trees.Finally, multiply by 2 to account for both sides of the path: 21 × 2 = 42 trees.

💡Okay, so I have this problem where there are 20 large uniform disks suspended from a thin string attached to the center O of the disk, and they're maintaining a horizontal position. Weights p₁, p₂, p₃ are placed at three different points A, B, C on the edge of the disk without disrupting the balance. I need to calculate the angles ∠AOB, ∠BOC, and ∠COA.Hmm, first, I need to visualize this setup. There's a disk hanging from a string at its center O, and it's horizontal. So, the disk is like a flat circle, and weights are placed on the edge at points A, B, and C. The disk remains balanced, so these weights must be arranged in such a way that the torques they create around the center O cancel each other out.Since the disk is uniform, its own center of mass is at O, and the weights are placed on the edge, which is a distance r from the center, where r is the radius of the disk. But the problem doesn't specify the radius, so maybe it will cancel out in the equations.I remember that for an object to be in rotational equilibrium, the sum of the torques about any point must be zero. Torque is given by τ = r × F, where r is the distance from the pivot point and F is the force applied. In this case, the pivot point is O, and the forces are the weights p₁, p₂, p₃.So, the torque due to each weight will be the weight multiplied by the distance from O, which is the radius r, and also multiplied by the sine of the angle between the weight's line of action and the radius. Since the weights are placed on the edge, their lines of action are vertical, and the radii OA, OB, OC are horizontal. So, the angle between the weight and the radius is 90 degrees, and sin(90°) = 1. Therefore, the torque for each weight is just p * r.But wait, actually, the torque is r * F * sin(theta), where theta is the angle between the radius and the force. Since the force is vertical and the radius is horizontal, the angle between them is 90 degrees, so sin(theta) is 1. So, the torque is indeed p * r for each weight.However, the direction of the torque depends on the position of the weight. If the weight is placed at a certain point, it will create a torque that tends to rotate the disk in a particular direction. For the disk to remain balanced, the sum of these torques must be zero.But how do the angles ∠AOB, ∠BOC, and ∠COA come into play? These angles are the central angles between the points where the weights are placed. So, if I can find these angles, I can determine how the weights are distributed around the disk.Let me think about the equilibrium condition. The torques created by the weights must balance each other out. Since the disk is uniform and the weights are on the edge, the only thing that matters is the distribution of the weights around the center.If I consider the torques, they must sum to zero vectorially. That means that the vector sum of the torques due to p₁, p₂, and p₃ must be zero.Each torque can be represented as a vector in the plane of the disk. The magnitude of each torque is p * r, and the direction is perpendicular to the radius at the point where the weight is applied. So, each torque vector will be tangent to the circle at points A, B, and C.Therefore, the problem reduces to finding three vectors (the torques) of magnitudes p₁r, p₂r, p₃r, each tangent to the circle at points A, B, and C, such that their vector sum is zero.This is similar to finding three vectors that form a closed triangle when added tip-to-tail. So, the vectors must form a triangle, meaning that the angles between them must satisfy certain conditions.But how do I relate this to the central angles ∠AOB, ∠BOC, and ∠COA?Well, the direction of each torque vector is tangent to the circle at the respective points. The angle between two torque vectors would be equal to the angle between their respective tangent directions, which is related to the central angles between the points.Specifically, the angle between the torque vectors at A and B would be equal to the central angle ∠AOB. Similarly, the angle between the torque vectors at B and C would be ∠BOC, and between C and A would be ∠COA.Wait, no, actually, the torque vectors are tangents, so the angle between two torque vectors is equal to the angle between their respective tangent lines. The tangent at a point on a circle is perpendicular to the radius at that point. So, the angle between two torque vectors would be equal to the angle between their respective radii.Wait, let me clarify. If I have two points A and B on the circle, the radius OA and OB make an angle ∠AOB. The tangent at A is perpendicular to OA, and the tangent at B is perpendicular to OB. Therefore, the angle between the two tangent vectors (which are the torque directions) would be equal to the angle between OA and OB, because each tangent is rotated 90 degrees from the radius.So, if ∠AOB is the central angle between A and B, then the angle between the torque vectors at A and B is also ∠AOB.Therefore, the three torque vectors form a triangle where the angles between them are equal to the central angles ∠AOB, ∠BOC, and ∠COA.Since the torque vectors must sum to zero, they form a closed triangle. Therefore, the triangle formed by the torque vectors must satisfy the law of sines and cosines.Let me denote the magnitudes of the torque vectors as τ₁ = p₁r, τ₂ = p₂r, τ₃ = p₃r.Since the torque vectors form a closed triangle, we can apply the law of sines to relate the angles and the magnitudes.The law of sines states that in any triangle, the ratio of the length of a side to the sine of the opposite angle is constant. So, in this case:τ₁ / sin(∠BOC) = τ₂ / sin(∠COA) = τ₃ / sin(∠AOB)Substituting τ₁ = p₁r, τ₂ = p₂r, τ₃ = p₃r, we get:p₁r / sin(∠BOC) = p₂r / sin(∠COA) = p₃r / sin(∠AOB)Since r is common in all terms, it cancels out:p₁ / sin(∠BOC) = p₂ / sin(∠COA) = p₃ / sin(∠AOB)So, we have the relationship:p₁ / sin(∠BOC) = p₂ / sin(∠COA) = p₃ / sin(∠AOB)This is a key equation that relates the weights to the central angles.Additionally, since the sum of the central angles around point O must be 360 degrees, we have:∠AOB + ∠BOC + ∠COA = 360°So, we have two equations:1. p₁ / sin(∠BOC) = p₂ / sin(∠COA) = p₃ / sin(∠AOB)2. ∠AOB + ∠BOC + ∠COA = 360°But we have three unknown angles, so we need another equation. However, the first equation actually provides two relationships because it equates three ratios. So, we can set up two equations from the first relationship.Let me denote:p₁ / sin(∠BOC) = p₂ / sin(∠COA) = p₃ / sin(∠AOB) = kWhere k is a constant.Therefore:p₁ = k sin(∠BOC)p₂ = k sin(∠COA)p₃ = k sin(∠AOB)So, we can express each weight in terms of the sine of the opposite angle and a constant k.Now, we can use the second equation, the sum of angles, to relate them.But we have three angles and three equations, so we can solve for the angles.Let me denote:Let’s call ∠AOB = α, ∠BOC = β, and ∠COA = γ.So, we have:α + β + γ = 360°And from the first relationship:p₁ = k sin βp₂ = k sin γp₃ = k sin αSo, we can write:sin α = p₃ / ksin β = p₁ / ksin γ = p₂ / kBut since α + β + γ = 360°, we can use the sine addition formula or other trigonometric identities to relate these angles.However, this might get complicated because we have three angles adding up to 360°, and their sines are proportional to the weights.Alternatively, we can use the law of cosines on the triangle formed by the torque vectors.Wait, the torque vectors form a triangle, so we can apply the law of cosines to each side.Let me consider the triangle formed by the torque vectors τ₁, τ₂, τ₃.In this triangle, the sides are τ₁, τ₂, τ₃, and the angles opposite these sides are ∠BOC, ∠COA, ∠AOB respectively.So, applying the law of cosines to each side:τ₁² = τ₂² + τ₃² - 2 τ₂ τ₃ cos(∠AOB)τ₂² = τ₁² + τ₃² - 2 τ₁ τ₃ cos(∠BOC)τ₃² = τ₁² + τ₂² - 2 τ₁ τ₂ cos(∠COA)Substituting τ₁ = p₁r, τ₂ = p₂r, τ₃ = p₃r, we get:(p₁r)² = (p₂r)² + (p₃r)² - 2 (p₂r)(p₃r) cos(∠AOB)Similarly for the other equations.Simplifying, since r² cancels out:p₁² = p₂² + p₃² - 2 p₂ p₃ cos(∠AOB)p₂² = p₁² + p₃² - 2 p₁ p₃ cos(∠BOC)p₃² = p₁² + p₂² - 2 p₁ p₂ cos(∠COA)So, we have three equations:1. p₁² = p₂² + p₃² - 2 p₂ p₃ cos(∠AOB)2. p₂² = p₁² + p₃² - 2 p₁ p₃ cos(∠BOC)3. p₃² = p₁² + p₂² - 2 p₁ p₂ cos(∠COA)These are the equations we need to solve for the angles ∠AOB, ∠BOC, and ∠COA.But solving these equations simultaneously might be a bit involved. Let me see if I can express each angle in terms of the others.From equation 1:cos(∠AOB) = (p₂² + p₃² - p₁²) / (2 p₂ p₃)Similarly, from equation 2:cos(∠BOC) = (p₁² + p₃² - p₂²) / (2 p₁ p₃)And from equation 3:cos(∠COA) = (p₁² + p₂² - p₃²) / (2 p₁ p₂)So, we can express each cosine of the angles in terms of the weights.But since the angles are related by α + β + γ = 360°, we can use these expressions to find the angles.However, we need to ensure that the angles we find satisfy this sum.Alternatively, since we have expressions for cos(α), cos(β), and cos(γ), we can use the fact that α + β + γ = 360° and try to find a relationship between them.But this might be complex because the sum of angles is 360°, and we have cosines of each angle.Alternatively, perhaps we can use the law of sines and the law of cosines together.From the law of sines, we have:p₁ / sin β = p₂ / sin γ = p₃ / sin α = kSo, sin α = p₃ / k, sin β = p₁ / k, sin γ = p₂ / kAnd from the angle sum:α + β + γ = 360°So, we can write:α = 360° - β - γThen, sin α = sin(360° - β - γ) = -sin(β + γ)But sin α = p₃ / k, so:-p₃ / k = sin(β + γ)But sin(β + γ) = sin β cos γ + cos β sin γSo,-p₃ / k = sin β cos γ + cos β sin γBut sin β = p₁ / k, sin γ = p₂ / kSo,-p₃ / k = (p₁ / k) cos γ + (p₂ / k) cos βMultiplying both sides by k:-p₃ = p₁ cos γ + p₂ cos βBut from the law of cosines, we have expressions for cos β and cos γ.From equation 2:cos β = (p₁² + p₃² - p₂²) / (2 p₁ p₃)From equation 3:cos γ = (p₁² + p₂² - p₃²) / (2 p₁ p₂)Substituting these into the equation:-p₃ = p₁ * [(p₁² + p₂² - p₃²) / (2 p₁ p₂)] + p₂ * [(p₁² + p₃² - p₂²) / (2 p₁ p₃)]Simplify:-p₃ = [ (p₁² + p₂² - p₃²) / (2 p₂) ] + [ (p₁² + p₃² - p₂²) / (2 p₁) ]Multiply both sides by 2 p₁ p₂ to eliminate denominators:-2 p₁ p₂ p₃ = p₁ (p₁² + p₂² - p₃²) + p₂ (p₁² + p₃² - p₂²)Expanding the right-hand side:= p₁³ + p₁ p₂² - p₁ p₃² + p₂ p₁² + p₂ p₃² - p₂³Combine like terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₁ p₃² + p₂ p₃² - p₂³Factor terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₂³ - p₁ p₃² + p₂ p₃²Hmm, this seems complicated. Maybe there's a better approach.Alternatively, perhaps we can use the fact that the sum of the angles is 360°, and use the cosine of the sum.But that might not be straightforward.Wait, another thought: since the torque vectors form a triangle, the angles between them are the central angles. So, perhaps we can use vector addition.Let me represent each torque vector as a vector in the plane.Let’s place point O at the origin. Let’s assume point A is at (r, 0). Then, the torque vector at A is tangential, so it points upward, in the positive y-direction. Similarly, the torque at B is at an angle ∠AOB from A, so its direction is ∠AOB + 90° from the x-axis. Similarly for C.Wait, no. The torque vector is tangent to the circle at the point where the weight is applied. So, if point A is at (r, 0), the torque vector at A is in the positive y-direction. If point B is at an angle ∠AOB from A, then its torque vector is at an angle ∠AOB + 90° from the x-axis.Similarly, point C is at an angle ∠BOC from B, so its torque vector is at an angle ∠AOB + ∠BOC + 90° from the x-axis.But since the sum of the torque vectors must be zero, their vector sum is zero.So, we can write the vector sum as:τ₁ (0, 1) + τ₂ (cos(∠AOB + 90°), sin(∠AOB + 90°)) + τ₃ (cos(∠AOB + ∠BOC + 90°), sin(∠AOB + ∠BOC + 90°)) = (0, 0)This gives us two equations (one for the x-component and one for the y-component):1. τ₂ cos(∠AOB + 90°) + τ₃ cos(∠AOB + ∠BOC + 90°) = 02. τ₁ + τ₂ sin(∠AOB + 90°) + τ₃ sin(∠AOB + ∠BOC + 90°) = 0But this seems complicated because we have angles inside the cosine and sine functions.Alternatively, perhaps we can use complex numbers to represent the torque vectors.Let’s denote each torque vector as a complex number. The torque at A is τ₁ e^{i(90°)} = τ₁ i. The torque at B is τ₂ e^{i(∠AOB + 90°)}. The torque at C is τ₃ e^{i(∠AOB + ∠BOC + 90°)}.The sum of these complex numbers must be zero:τ₁ i + τ₂ e^{i(∠AOB + 90°)} + τ₃ e^{i(∠AOB + ∠BOC + 90°)} = 0This is a complex equation, which can be separated into real and imaginary parts.But this might not be the most straightforward approach either.Wait, going back to the earlier equations from the law of cosines:cos(∠AOB) = (p₂² + p₃² - p₁²) / (2 p₂ p₃)cos(∠BOC) = (p₁² + p₃² - p₂²) / (2 p₁ p₃)cos(∠COA) = (p₁² + p₂² - p₃²) / (2 p₁ p₂)These are the expressions for the cosines of the angles. So, if we can compute these cosines, we can find the angles themselves using the arccosine function.But we need to ensure that the sum of the angles is 360°, so we need to check if the angles obtained from these cosines satisfy that condition.Alternatively, perhaps we can use these expressions directly to find the angles.Let me summarize:Given weights p₁, p₂, p₃, the central angles can be found using:∠AOB = arccos[(p₂² + p₃² - p₁²) / (2 p₂ p₃)]∠BOC = arccos[(p₁² + p₃² - p₂²) / (2 p₁ p₃)]∠COA = arccos[(p₁² + p₂² - p₃²) / (2 p₁ p₂)]But we need to ensure that the sum of these angles is 360°, so we might need to adjust for any discrepancies due to the arccosine function only giving values between 0° and 180°.Wait, actually, the central angles in a circle can be more than 180°, but the arccosine function only returns values between 0° and 180°. So, if the actual angle is greater than 180°, the arccosine will return 360° minus that angle.Therefore, we need to be careful with the interpretation of the angles.Alternatively, perhaps we can use the fact that the sum of the angles is 360°, and use the expressions for the cosines to find the angles.But this might require solving a system of equations, which could be complex.Alternatively, perhaps we can use the law of sines and the fact that the sum of the angles is 360° to find the angles.From the law of sines:p₁ / sin β = p₂ / sin γ = p₃ / sin α = kSo, sin α = p₃ / k, sin β = p₁ / k, sin γ = p₂ / kAnd α + β + γ = 360°So, we can write:α = 360° - β - γThen, sin α = sin(360° - β - γ) = -sin(β + γ)But sin α = p₃ / k, so:-p₃ / k = sin(β + γ)But sin(β + γ) = sin β cos γ + cos β sin γSo,-p₃ / k = (p₁ / k) cos γ + (p₂ / k) cos βMultiplying both sides by k:-p₃ = p₁ cos γ + p₂ cos βBut from the law of cosines, we have expressions for cos β and cos γ.From earlier:cos β = (p₁² + p₃² - p₂²) / (2 p₁ p₃)cos γ = (p₁² + p₂² - p₃²) / (2 p₁ p₂)Substituting these into the equation:-p₃ = p₁ * [(p₁² + p₂² - p₃²) / (2 p₁ p₂)] + p₂ * [(p₁² + p₃² - p₂²) / (2 p₁ p₃)]Simplify:-p₃ = [ (p₁² + p₂² - p₃²) / (2 p₂) ] + [ (p₁² + p₃² - p₂²) / (2 p₁) ]Multiply both sides by 2 p₁ p₂ to eliminate denominators:-2 p₁ p₂ p₃ = p₁ (p₁² + p₂² - p₃²) + p₂ (p₁² + p₃² - p₂²)Expanding the right-hand side:= p₁³ + p₁ p₂² - p₁ p₃² + p₂ p₁² + p₂ p₃² - p₂³Combine like terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₁ p₃² + p₂ p₃² - p₂³Factor terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₂³ - p₁ p₃² + p₂ p₃²Hmm, this seems complicated. Maybe there's a better approach.Alternatively, perhaps we can use the fact that the sum of the angles is 360°, and use the cosine of the sum.But that might not be straightforward.Wait, another thought: since the torque vectors form a triangle, the angles between them are the central angles. So, perhaps we can use vector addition.Let me represent each torque vector as a vector in the plane.Let’s place point O at the origin. Let’s assume point A is at (r, 0). Then, the torque vector at A is tangential, so it points upward, in the positive y-direction. Similarly, the torque at B is at an angle ∠AOB from A, so its direction is ∠AOB + 90° from the x-axis. Similarly for C.Wait, no. The torque vector is tangent to the circle at the point where the weight is applied. So, if point A is at (r, 0), the torque vector at A is in the positive y-direction. If point B is at an angle ∠AOB from A, then its torque vector is at an angle ∠AOB + 90° from the x-axis.Similarly, point C is at an angle ∠BOC from B, so its torque vector is at an angle ∠AOB + ∠BOC + 90° from the x-axis.But since the sum of the torque vectors must be zero, their vector sum is zero.So, we can write the vector sum as:τ₁ (0, 1) + τ₂ (cos(∠AOB + 90°), sin(∠AOB + 90°)) + τ₃ (cos(∠AOB + ∠BOC + 90°), sin(∠AOB + ∠BOC + 90°)) = (0, 0)This gives us two equations (one for the x-component and one for the y-component):1. τ₂ cos(∠AOB + 90°) + τ₃ cos(∠AOB + ∠BOC + 90°) = 02. τ₁ + τ₂ sin(∠AOB + 90°) + τ₃ sin(∠AOB + ∠BOC + 90°) = 0But this seems complicated because we have angles inside the cosine and sine functions.Alternatively, perhaps we can use complex numbers to represent the torque vectors.Let’s denote each torque vector as a complex number. The torque at A is τ₁ e^{i(90°)} = τ₁ i. The torque at B is τ₂ e^{i(∠AOB + 90°)}. The torque at C is τ₃ e^{i(∠AOB + ∠BOC + 90°)}.The sum of these complex numbers must be zero:τ₁ i + τ₂ e^{i(∠AOB + 90°)} + τ₃ e^{i(∠AOB + ∠BOC + 90°)} = 0This is a complex equation, which can be separated into real and imaginary parts.But this might not be the most straightforward approach either.Wait, going back to the earlier equations from the law of cosines:cos(∠AOB) = (p₂² + p₃² - p₁²) / (2 p₂ p₃)cos(∠BOC) = (p₁² + p₃² - p₂²) / (2 p₁ p₃)cos(∠COA) = (p₁² + p₂² - p₃²) / (2 p₁ p₂)These are the expressions for the cosines of the angles. So, if we can compute these cosines, we can find the angles themselves using the arccosine function.But we need to ensure that the sum of the angles is 360°, so we need to check if the angles obtained from these cosines satisfy that condition.Alternatively, perhaps we can use these expressions directly to find the angles.Let me summarize:Given weights p₁, p₂, p₃, the central angles can be found using:∠AOB = arccos[(p₂² + p₃² - p₁²) / (2 p₂ p₃)]∠BOC = arccos[(p₁² + p₃² - p₂²) / (2 p₁ p₃)]∠COA = arccos[(p₁² + p₂² - p₃²) / (2 p₁ p₂)]But we need to ensure that the sum of these angles is 360°, so we might need to adjust for any discrepancies due to the arccosine function only giving values between 0° and 180°.Wait, actually, the central angles in a circle can be more than 180°, but the arccosine function only returns values between 0° and 180°. So, if the actual angle is greater than 180°, the arccosine will return 360° minus that angle.Therefore, we need to be careful with the interpretation of the angles.Alternatively, perhaps we can use the fact that the sum of the angles is 360°, and use the expressions for the cosines to find the angles.But this might require solving a system of equations, which could be complex.Alternatively, perhaps we can use the law of sines and the fact that the sum of the angles is 360° to find the angles.From the law of sines:p₁ / sin β = p₂ / sin γ = p₃ / sin α = kSo, sin α = p₃ / k, sin β = p₁ / k, sin γ = p₂ / kAnd α + β + γ = 360°So, we can write:α = 360° - β - γThen, sin α = sin(360° - β - γ) = -sin(β + γ)But sin α = p₃ / k, so:-p₃ / k = sin(β + γ)But sin(β + γ) = sin β cos γ + cos β sin γSo,-p₃ / k = (p₁ / k) cos γ + (p₂ / k) cos βMultiplying both sides by k:-p₃ = p₁ cos γ + p₂ cos βBut from the law of cosines, we have expressions for cos β and cos γ.From earlier:cos β = (p₁² + p₃² - p₂²) / (2 p₁ p₃)cos γ = (p₁² + p₂² - p₃²) / (2 p₁ p₂)Substituting these into the equation:-p₃ = p₁ * [(p₁² + p₂² - p₃²) / (2 p₁ p₂)] + p₂ * [(p₁² + p₃² - p₂²) / (2 p₁ p₃)]Simplify:-p₃ = [ (p₁² + p₂² - p₃²) / (2 p₂) ] + [ (p₁² + p₃² - p₂²) / (2 p₁) ]Multiply both sides by 2 p₁ p₂ to eliminate denominators:-2 p₁ p₂ p₃ = p₁ (p₁² + p₂² - p₃²) + p₂ (p₁² + p₃² - p₂²)Expanding the right-hand side:= p₁³ + p₁ p₂² - p₁ p₃² + p₂ p₁² + p₂ p₃² - p₂³Combine like terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₁ p₃² + p₂ p₃² - p₂³Factor terms:= p₁³ + p₂ p₁² + p₁ p₂² - p₂³ - p₁ p₃² + p₂ p₃²This seems quite involved. Maybe there's a simplification I'm missing.Alternatively, perhaps we can consider specific cases or assume certain relationships between the weights to simplify the problem.For example, if all weights are equal, p₁ = p₂ = p₃, then the angles would each be 120°, since the torques would balance each other symmetrically.But in the general case, where the weights are different, the angles will vary.Given that, perhaps the best approach is to use the expressions from the law of cosines and compute each angle individually, then check if their sum is 360°, adjusting for any discrepancies due to the arccosine function.So, to summarize, the angles can be calculated as:∠AOB = arccos[(p₂² + p₃² - p₁²) / (2 p₂ p₃)]∠BOC = arccos[(p₁² + p₃² - p₂²) / (2 p₁ p₃)]∠COA = arccos[(p₁² + p₂² - p₃²) / (2 p₁ p₂)]And we need to ensure that the sum of these angles is 360°, taking into account that arccosine returns angles between 0° and 180°, so if the actual angle is greater than 180°, we need to subtract it from 360°.Therefore, the final answer is:∠AOB = arccos[(p₂² + p₃² - p₁²) / (2 p₂ p₃)]∠BOC = arccos[(p₁² + p₃² - p₂²) / (2 p₁ p₃)]∠COA = arccos[(p₁² + p₂² - p₃²) / (2 p₁ p₂)]But to express this in a boxed format as requested, I think the user expects the expressions for the cosines of the angles, not the angles themselves, since the angles depend on the specific values of p₁, p₂, p₃.Therefore, the final answer is:cos(∠AOB) = (p₂² + p₃² - p₁²) / (2 p₂ p₃)cos(∠BOC) = (p₁² + p₃² - p₂²) / (2 p₁ p₃)cos(∠COA) = (p₁² + p₂² - p₃²) / (2 p₁ p₂)So, boxed:cos(∠AOB) = boxed{dfrac{p_2^2 + p_3^2 - p_1^2}{2 p_2 p_3}}cos(∠BOC) = boxed{dfrac{p_1^2 + p_3^2 - p_2^2}{2 p_1 p_3}}cos(∠COA) = boxed{dfrac{p_1^2 + p_2^2 - p_3^2}{2 p_1 p_2}}