💡Okay, so I have this problem here: Given a family of subsets ( mathscr{A} ) of a set ( X ) where the size of ( mathscr{A} ) is ( t ) and ( t ) is at least 2, I need to prove that there are at least ( t ) distinct subsets of the form ( A triangle B ) where ( A ) and ( B ) are in ( mathscr{A} ). Alright, first, let me make sure I understand the problem correctly. We have a collection of subsets, each subset is from the set ( X ), and there are ( t ) such subsets. The task is to show that if I take all possible symmetric differences between pairs of these subsets, there will be at least ( t ) different results. So, symmetric difference, ( A triangle B ), is the set of elements that are in either ( A ) or ( B ) but not in both. It's like the XOR operation for sets. Okay, got that.Let me think about how to approach this. Maybe I can start by considering a specific example to get some intuition. Suppose ( t = 2 ). Then ( mathscr{A} ) has two subsets, say ( A_1 ) and ( A_2 ). The symmetric differences I can form are ( A_1 triangle A_1 ), ( A_1 triangle A_2 ), ( A_2 triangle A_1 ), and ( A_2 triangle A_2 ). But wait, ( A triangle A ) is the empty set for any set ( A ). So, in this case, the symmetric differences would be the empty set and ( A_1 triangle A_2 ). So, there are two distinct symmetric differences, which is equal to ( t ). So, for ( t = 2 ), the statement holds.Now, let me try ( t = 3 ). Suppose ( mathscr{A} = {A_1, A_2, A_3} ). The symmetric differences would be all possible ( A_i triangle A_j ) for ( i, j ) from 1 to 3. That gives us 9 symmetric differences, but many of them will be duplicates because ( A_i triangle A_j = A_j triangle A_i ) and ( A_i triangle A_i = emptyset ). But the problem is not asking for all possible symmetric differences, just that there are at least ( t ) distinct ones. So, in this case, we need to show there are at least 3 distinct symmetric differences. Let me list them out:1. ( A_1 triangle A_1 = emptyset )2. ( A_1 triangle A_2 )3. ( A_1 triangle A_3 )4. ( A_2 triangle A_1 = A_1 triangle A_2 )5. ( A_2 triangle A_2 = emptyset )6. ( A_2 triangle A_3 )7. ( A_3 triangle A_1 = A_1 triangle A_3 )8. ( A_3 triangle A_2 = A_2 triangle A_3 )9. ( A_3 triangle A_3 = emptyset )So, the distinct symmetric differences are ( emptyset ), ( A_1 triangle A_2 ), and ( A_1 triangle A_3 ), and ( A_2 triangle A_3 ). That's four distinct subsets, which is more than ( t = 3 ). So, the statement holds here as well.Hmm, so maybe for larger ( t ), the number of distinct symmetric differences increases, but the problem only requires at least ( t ). So, perhaps I can generalize this approach.Let me think about how to construct ( t ) distinct symmetric differences. Maybe I can fix one subset and take its symmetric difference with all the others. Let's say I fix ( A_1 ) and consider ( A_1 triangle A_i ) for each ( i ) from 1 to ( t ). So, that would give me ( t ) symmetric differences: ( A_1 triangle A_1, A_1 triangle A_2, ldots, A_1 triangle A_t ). Now, ( A_1 triangle A_1 ) is the empty set, and the others are ( A_1 triangle A_i ) for ( i geq 2 ). I need to show that these ( t ) subsets are all distinct. Suppose, for contradiction, that two of them are equal. That is, suppose ( A_1 triangle A_i = A_1 triangle A_j ) for some ( i neq j ). If I take the symmetric difference of both sides with ( A_1 ), I get ( (A_1 triangle A_i) triangle A_1 = (A_1 triangle A_j) triangle A_1 ). But symmetric difference is associative and commutative, so ( (A_1 triangle A_i) triangle A_1 = A_i ) and similarly ( (A_1 triangle A_j) triangle A_1 = A_j ). Therefore, ( A_i = A_j ). But this contradicts the fact that all subsets in ( mathscr{A} ) are distinct. Hence, our assumption that ( A_1 triangle A_i = A_1 triangle A_j ) must be false. Therefore, all ( t ) subsets ( A_1 triangle A_i ) are distinct. Wait, but in my earlier example with ( t = 3 ), I had more than ( t ) distinct symmetric differences. So, does this method always give exactly ( t ) distinct symmetric differences? No, because when I fix ( A_1 ) and take symmetric differences with all ( A_i ), I get ( t ) subsets, but some of them might be duplicates if the family ( mathscr{A} ) has certain properties. But in the general case, as I showed, they are all distinct because if two were equal, it would imply two subsets in ( mathscr{A} ) are equal, which they are not.So, this seems like a solid approach. By fixing one subset and taking symmetric differences with all others, we can generate ( t ) distinct subsets. But wait, in the problem statement, it says "at least ( t ) distinct subsets of the form ( A triangle B )". So, I think my approach shows that there are exactly ( t ) distinct subsets when considering ( A_1 triangle A_i ), but actually, the total number of symmetric differences is more because we can have ( A_i triangle A_j ) for any ( i, j ). But the problem only requires at least ( t ), so my approach suffices because I can find ( t ) distinct ones by fixing one subset. Let me recap:1. Let ( mathscr{A} = {A_1, A_2, ldots, A_t} ).2. Consider the family ( {A_1 triangle A_i mid 1 leq i leq t} ).3. Show that all these are distinct by contradiction: if ( A_1 triangle A_i = A_1 triangle A_j ), then ( A_i = A_j ), which is impossible since all ( A_i ) are distinct.4. Therefore, there are ( t ) distinct symmetric differences.This seems to work. But let me think about edge cases. What if ( t = 1 )? But the problem states ( t geq 2 ), so that's fine. Another thought: what if ( X ) is empty? Then all subsets are empty, but ( t geq 2 ) would mean we have multiple empty sets, which contradicts the distinctness. So, ( X ) must be non-empty if ( t geq 2 ). Wait, actually, if ( X ) is empty, the only subset is the empty set, so ( t ) cannot be greater than 1. Therefore, ( X ) must be non-empty when ( t geq 2 ). So, in the problem, since ( t geq 2 ), ( X ) is non-empty, and all subsets in ( mathscr{A} ) are distinct. Therefore, my approach holds. I think I've got it. By fixing one subset and taking symmetric differences with all others, I can generate ( t ) distinct subsets, which proves the statement.

💡Okay, so I have this determinant to evaluate:[begin{vmatrix} 1 & x & x^2 1 & x + 1 & (x+1)^2 1 & x & (x+1)^2 end{vmatrix}]Hmm, determinants. I remember that for a 3x3 matrix, there are a few methods to calculate the determinant. One common method is expansion by minors, which involves breaking down the determinant into smaller 2x2 determinants. Another method is using row operations to simplify the matrix before calculating. Maybe I can try both approaches and see which one is easier.Let me first recall the formula for the determinant of a 3x3 matrix. If I have a matrix:[begin{bmatrix}a & b & c d & e & f g & h & iend{bmatrix}]Then the determinant is:[a(ei - fh) - b(di - fg) + c(dh - eg)]So, applying this formula to the given matrix:First row: 1, x, x²Second row: 1, x+1, (x+1)²Third row: 1, x, (x+1)²So, plugging into the formula:1 * [(x+1)(x+1)² - (x+1)² * x] - x * [1*(x+1)² - 1*(x+1)²] + x² * [1*x - 1*(x+1)]Wait, let me write that step by step.First term: 1 * determinant of the submatrix:[begin{vmatrix} x+1 & (x+1)^2 x & (x+1)^2 end{vmatrix}]Which is (x+1)*(x+1)^2 - x*(x+1)^2. Let me compute that:(x+1)^3 - x*(x+1)^2. Hmm, factor out (x+1)^2:(x+1)^2 [ (x+1) - x ] = (x+1)^2 [1] = (x+1)^2.Okay, so the first term is (x+1)^2.Second term: -x * determinant of:[begin{vmatrix} 1 & (x+1)^2 1 & (x+1)^2 end{vmatrix}]Wait, both rows are the same here. So the determinant of a matrix with identical rows is zero. So the second term is -x * 0 = 0.Third term: x² * determinant of:[begin{vmatrix} 1 & x+1 1 & x end{vmatrix}]Which is 1*x - 1*(x+1) = x - (x+1) = x - x -1 = -1.So the third term is x²*(-1) = -x².Putting it all together:First term: (x+1)^2Second term: 0Third term: -x²So total determinant is (x+1)^2 - x².Let me expand (x+1)^2:(x+1)^2 = x² + 2x +1So subtract x²: x² + 2x +1 - x² = 2x +1.Wait, but in the initial calculation, I thought the first term was (x+1)^2, which is x² + 2x +1, and then subtract x², so that gives 2x +1.But in the initial thought process, I thought the answer was x+1. Hmm, that's conflicting.Wait, maybe I made a mistake in my calculation. Let me double-check.First term: 1 * determinant of:[begin{vmatrix} x+1 & (x+1)^2 x & (x+1)^2 end{vmatrix}]Which is (x+1)*(x+1)^2 - x*(x+1)^2 = (x+1)^3 - x*(x+1)^2.Factor out (x+1)^2: (x+1)^2*(x+1 - x) = (x+1)^2*(1) = (x+1)^2.Okay, that seems correct.Second term: -x * determinant of:[begin{vmatrix} 1 & (x+1)^2 1 & (x+1)^2 end{vmatrix}]Which is indeed zero because both rows are the same.Third term: x² * determinant of:[begin{vmatrix} 1 & x+1 1 & x end{vmatrix}]Which is 1*x - 1*(x+1) = x - x -1 = -1.So third term is x²*(-1) = -x².So total determinant is (x+1)^2 - x².Compute that:(x+1)^2 = x² + 2x +1Subtract x²: x² + 2x +1 - x² = 2x +1.Wait, so the determinant is 2x +1.But in the initial thought process, I thought it was x+1. So which one is correct?Wait, maybe I made a mistake in the initial thought process. Let me check again.Wait, in the initial thought process, I had:First determinant: (x+1)^2Third determinant: -1So total determinant: (x+1)^2 - x² = x+1.But wait, that would be if (x+1)^2 - x² = x+1, which is not true because (x+1)^2 - x² = 2x +1.So that suggests that the initial thought process had an error.Wait, perhaps I confused the expansion.Wait, let me go back.The determinant is:1*( (x+1)(x+1)^2 - x*(x+1)^2 ) - x*(1*(x+1)^2 - 1*(x+1)^2 ) + x²*(1*x - 1*(x+1))So first term: (x+1)^3 - x*(x+1)^2 = (x+1)^2*(x+1 -x) = (x+1)^2Second term: -x*(0) = 0Third term: x²*(x - x -1) = x²*(-1) = -x²So total determinant: (x+1)^2 - x² = (x² + 2x +1) - x² = 2x +1.Therefore, the correct answer should be 2x +1.But in the initial thought process, I thought it was x+1. So I must have made a mistake there.Alternatively, maybe I miscalculated the first term.Wait, let me compute the first determinant again:[begin{vmatrix} x+1 & (x+1)^2 x & (x+1)^2 end{vmatrix}]Which is (x+1)*(x+1)^2 - x*(x+1)^2.Compute (x+1)*(x+1)^2: that's (x+1)^3.Compute x*(x+1)^2: that's x*(x+1)^2.So the determinant is (x+1)^3 - x*(x+1)^2.Factor out (x+1)^2: (x+1)^2*(x+1 -x) = (x+1)^2*(1) = (x+1)^2.Yes, that's correct.So the first term is (x+1)^2.Third term is -x².So total determinant is (x+1)^2 - x² = 2x +1.Therefore, the correct answer is 2x +1.Wait, but in the initial thought process, I thought it was x+1. So I must have made a mistake in that initial process.Alternatively, maybe the initial thought process was incorrect.Wait, perhaps the initial thought process had a different approach.Wait, let me try another method to verify.Another method is to perform row operations to simplify the determinant.Given the matrix:Row1: 1, x, x²Row2: 1, x+1, (x+1)²Row3: 1, x, (x+1)²Let me subtract Row1 from Row2 and Row3.So, Row2' = Row2 - Row1:1-1=0, (x+1)-x=1, (x+1)² - x²= (x² + 2x +1) -x²=2x+1Similarly, Row3' = Row3 - Row1:1-1=0, x -x=0, (x+1)² -x²=2x +1So the new matrix is:Row1: 1, x, x²Row2': 0, 1, 2x+1Row3': 0, 0, 2x+1Now, the determinant of a triangular matrix is the product of the diagonal elements.So determinant = 1*1*(2x+1) = 2x +1.Yes, that confirms the determinant is 2x +1.Therefore, the initial thought process had an error, and the correct answer is 2x +1.So, in conclusion, the determinant evaluates to 2x +1.

💡Okay, so I have this problem: evaluate the product of cosines, specifically cos(π/7) multiplied by cos(2π/7) multiplied by cos(4π/7). Hmm, that seems a bit tricky because the angles are all different and they're not the standard angles I usually work with. I remember that sometimes products of cosines can be simplified using trigonometric identities, so maybe I can apply some formulas here.First, let me write down the expression again to make sure I have it correctly:cos(π/7) * cos(2π/7) * cos(4π/7)I wonder if there's a way to use the double-angle formula or maybe some product-to-sum identities. I recall that the product of cosines can sometimes be converted into a sum, but I'm not sure if that's the best approach here. Alternatively, maybe I can use the identity that relates the product of sines and cosines to something simpler.Wait, I think I remember something about multiple-angle formulas or perhaps using complex numbers. Maybe if I express each cosine in terms of exponentials, I can multiply them together and see if something cancels out. But that might get too complicated with the exponents.Another thought: I remember that for certain angles, especially those that are fractions of π, there are known products or sums that can be used. For example, I know that cos(π/3) is 1/2, but that's not directly helpful here. Maybe I can relate these angles to each other somehow.Let me consider the angles: π/7, 2π/7, and 4π/7. I notice that each subsequent angle is double the previous one. That seems like a pattern. Maybe I can use that to my advantage. If I let θ = π/7, then the angles become θ, 2θ, and 4θ. So the expression becomes cosθ * cos2θ * cos4θ.Hmm, that might be easier to handle. So, let's set θ = π/7, so the expression is cosθ * cos2θ * cos4θ. I wonder if there's a product formula for cosθ * cos2θ * cos4θ. Maybe I can find a general formula for such products.I recall that there's a formula for the product of multiple cosines when the angles are in geometric progression. Let me try to recall or derive it. If I have cosθ * cos2θ * cos4θ, perhaps I can use the double-angle identity multiple times.The double-angle identity is sin2θ = 2sinθcosθ, which can be rearranged to cosθ = sin2θ / (2sinθ). Maybe I can apply this identity to each cosine term.Let's try that. Starting with cosθ:cosθ = sin2θ / (2sinθ)Similarly, cos2θ = sin4θ / (2sin2θ)And cos4θ = sin8θ / (2sin4θ)So, substituting these into the product:cosθ * cos2θ * cos4θ = [sin2θ / (2sinθ)] * [sin4θ / (2sin2θ)] * [sin8θ / (2sin4θ)]Now, let's multiply these together:= [sin2θ * sin4θ * sin8θ] / [2sinθ * 2sin2θ * 2sin4θ]Simplifying the denominator: 2 * 2 * 2 = 8, and sinθ * sin2θ * sin4θ.So, the expression becomes:= [sin8θ] / [8sinθ]Because sin2θ cancels with sin2θ, sin4θ cancels with sin4θ, leaving just sin8θ in the numerator and 8sinθ in the denominator.So, cosθ * cos2θ * cos4θ = sin8θ / (8sinθ)Now, substituting back θ = π/7:sin8θ = sin(8π/7)And sinθ = sin(π/7)So, the expression becomes:sin(8π/7) / (8sin(π/7))Now, I need to evaluate sin(8π/7). Let's see, 8π/7 is more than π, so it's in the third quadrant where sine is negative. Also, 8π/7 can be written as π + π/7, so sin(8π/7) = sin(π + π/7) = -sin(π/7) because sine is negative in the third quadrant and sin(π + x) = -sinx.Therefore, sin(8π/7) = -sin(π/7)Substituting back into the expression:= [-sin(π/7)] / (8sin(π/7)) = -1/8So, the product cos(π/7) * cos(2π/7) * cos(4π/7) equals -1/8.Wait, let me double-check that. I used the identity for cosθ * cos2θ * cos4θ and transformed it into sin8θ / (8sinθ). Then, since 8θ was 8π/7, which is π + π/7, so sine of that is -sin(π/7). Dividing that by 8sin(π/7) gives -1/8. That seems correct.Alternatively, I could have used another method, like considering the roots of unity or using more advanced trigonometric identities, but this approach seems straightforward and gets me to the answer without too much complexity.I think that's solid. So, the final answer is -1/8.**Final Answer**The value of cos frac{pi}{7}cos frac{2pi}{7}cos frac{4pi}{7} is boxed{-dfrac{1}{8}}.

💡Alright, so I have this problem where I need to find all values of ( a ) such that the roots ( x_1, x_2, x_3 ) of the polynomial ( x^3 - 6x^2 + ax + a ) satisfy the condition ( (x_1 - 3)^3 + (x_2 - 3)^3 + (x_3 - 3)^3 = 0 ). Hmm, okay, let me try to break this down step by step.First, I remember that for polynomials, Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. Maybe that can help here. The polynomial is a cubic, so it has three roots, and I can express the sum of the roots, the sum of the products of the roots two at a time, and the product of all three roots in terms of the coefficients.Given the polynomial ( x^3 - 6x^2 + ax + a ), let me write down Vieta's formulas for this:1. ( x_1 + x_2 + x_3 = 6 ) (since the coefficient of ( x^2 ) is -6)2. ( x_1x_2 + x_1x_3 + x_2x_3 = a ) (since the coefficient of ( x ) is ( a ))3. ( x_1x_2x_3 = -a ) (since the constant term is ( a ))Okay, so I have these relationships. Now, the condition given is ( (x_1 - 3)^3 + (x_2 - 3)^3 + (x_3 - 3)^3 = 0 ). That looks a bit complicated, but maybe I can simplify it.Let me think about expanding ( (x_i - 3)^3 ). The expansion of ( (x - 3)^3 ) is ( x^3 - 9x^2 + 27x - 27 ). So, if I expand each term, I can write the sum as:( (x_1^3 - 9x_1^2 + 27x_1 - 27) + (x_2^3 - 9x_2^2 + 27x_2 - 27) + (x_3^3 - 9x_3^2 + 27x_3 - 27) = 0 )Let me combine like terms:( (x_1^3 + x_2^3 + x_3^3) - 9(x_1^2 + x_2^2 + x_3^2) + 27(x_1 + x_2 + x_3) - 81 = 0 )So, that's the expanded form. Now, I need to express each of these sums in terms of the known Vieta's formulas.First, I know ( x_1 + x_2 + x_3 = 6 ). That's straightforward.Next, I need ( x_1^2 + x_2^2 + x_3^2 ). I remember that ( x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_1x_3 + x_2x_3) ). Plugging in the known values:( (6)^2 - 2(a) = 36 - 2a )So, ( x_1^2 + x_2^2 + x_3^2 = 36 - 2a )Now, what about ( x_1^3 + x_2^3 + x_3^3 )? Hmm, I think there's a formula for that as well. Let me recall. I believe it's ( (x_1 + x_2 + x_3)^3 - 3(x_1 + x_2 + x_3)(x_1x_2 + x_1x_3 + x_2x_3) + 3x_1x_2x_3 ).Let me verify that. Yes, that seems right. So, plugging in the known values:( (6)^3 - 3(6)(a) + 3(-a) = 216 - 18a - 3a = 216 - 21a )So, ( x_1^3 + x_2^3 + x_3^3 = 216 - 21a )Now, let's plug all these back into the expanded condition:( (216 - 21a) - 9(36 - 2a) + 27(6) - 81 = 0 )Let me compute each term step by step.First term: ( 216 - 21a )Second term: ( -9(36 - 2a) = -324 + 18a )Third term: ( 27(6) = 162 )Fourth term: ( -81 )Now, combine all these:( (216 - 21a) + (-324 + 18a) + 162 - 81 = 0 )Let me compute the constants first:216 - 324 + 162 - 81216 - 324 is -108-108 + 162 is 5454 - 81 is -27Now, the terms with ( a ):-21a + 18a = -3aSo, altogether, we have:-27 - 3a = 0So, solving for ( a ):-3a = 27Divide both sides by -3:a = -9Wait, that can't be right. Let me check my calculations again because I feel like I might have made a mistake.Wait, let's go back step by step.First, expanding ( (x_i - 3)^3 ):Yes, that's ( x_i^3 - 9x_i^2 + 27x_i - 27 ). So, the sum is ( sum x_i^3 - 9sum x_i^2 + 27sum x_i - 81 = 0 ). That seems correct.Then, ( sum x_i = 6 ), correct.( sum x_i^2 = (6)^2 - 2a = 36 - 2a ), correct.( sum x_i^3 = (6)^3 - 3*6*a + 3*(-a) = 216 - 18a - 3a = 216 - 21a ), correct.Plugging into the equation:( (216 - 21a) - 9*(36 - 2a) + 27*6 - 81 = 0 )Compute each part:216 - 21a-9*(36 - 2a) = -324 + 18a27*6 = 162-81So, adding all together:216 - 21a - 324 + 18a + 162 - 81Compute constants:216 - 324 = -108-108 + 162 = 5454 - 81 = -27Variables:-21a + 18a = -3aSo, total: -27 - 3a = 0Thus, -3a = 27 => a = -9Wait, but let me think about this. If a = -9, does that make sense? Let me check if that's consistent with the original polynomial.The polynomial becomes ( x^3 - 6x^2 -9x -9 ). Let me see if this polynomial has roots such that the condition holds.Alternatively, maybe I made a mistake in the expansion or in the formula for ( sum x_i^3 ). Let me double-check that.The formula for ( x_1^3 + x_2^3 + x_3^3 ) is indeed ( (x_1 + x_2 + x_3)^3 - 3(x_1 + x_2 + x_3)(x_1x_2 + x_1x_3 + x_2x_3) + 3x_1x_2x_3 ). Plugging in the values:( 6^3 - 3*6*a + 3*(-a) = 216 - 18a - 3a = 216 - 21a ). That seems correct.Hmm, maybe I didn't make a mistake. So, a = -9 is the solution. But let me think again.Wait, another approach: Maybe instead of expanding everything, I can use substitution. Let me set ( y_i = x_i - 3 ). Then, the condition becomes ( y_1^3 + y_2^3 + y_3^3 = 0 ).If I can express the polynomial in terms of ( y ), then maybe I can use Vieta's formulas on the new polynomial.Let me try that.Let ( y = x - 3 ), so ( x = y + 3 ). Substitute into the original polynomial:( (y + 3)^3 - 6(y + 3)^2 + a(y + 3) + a = 0 )Let me expand this:First, ( (y + 3)^3 = y^3 + 9y^2 + 27y + 27 )Second, ( -6(y + 3)^2 = -6(y^2 + 6y + 9) = -6y^2 - 36y - 54 )Third, ( a(y + 3) = ay + 3a )Fourth, ( +a )Now, combine all terms:( y^3 + 9y^2 + 27y + 27 - 6y^2 - 36y - 54 + ay + 3a + a )Combine like terms:- ( y^3 ): 1y^3- ( y^2 ): 9y^2 - 6y^2 = 3y^2- ( y ): 27y - 36y + ay = (a - 9)y- Constants: 27 - 54 + 3a + a = (4a - 27)So, the transformed polynomial is ( y^3 + 3y^2 + (a - 9)y + (4a - 27) = 0 )Now, the roots of this polynomial are ( y_1, y_2, y_3 ), and we have the condition ( y_1^3 + y_2^3 + y_3^3 = 0 )Using Vieta's formulas for this new polynomial:1. ( y_1 + y_2 + y_3 = -3 ) (since the coefficient of ( y^2 ) is 3, and Vieta's formula is -coefficient)2. ( y_1y_2 + y_1y_3 + y_2y_3 = a - 9 )3. ( y_1y_2y_3 = -(4a - 27) )Now, we need ( y_1^3 + y_2^3 + y_3^3 = 0 ). There's a formula for the sum of cubes:( y_1^3 + y_2^3 + y_3^3 = (y_1 + y_2 + y_3)^3 - 3(y_1 + y_2 + y_3)(y_1y_2 + y_1y_3 + y_2y_3) + 3y_1y_2y_3 )Plugging in the known values:( (-3)^3 - 3*(-3)*(a - 9) + 3*(-(4a - 27)) )Compute each term:1. ( (-3)^3 = -27 )2. ( -3*(-3)*(a - 9) = 9(a - 9) = 9a - 81 )3. ( 3*(-(4a - 27)) = -12a + 81 )Now, sum them up:-27 + (9a - 81) + (-12a + 81)Combine like terms:-27 - 81 + 81 + 9a - 12aSimplify constants:-27 - 81 + 81 = -27Variables:9a - 12a = -3aSo, total: -27 - 3a = 0Thus, -3a = 27 => a = -9Wait, so both methods give me a = -9. But earlier, when I thought a = -9 might not make sense, I wasn't sure. Maybe I was overcomplicating it.Let me check if a = -9 satisfies the original condition.So, if a = -9, the polynomial becomes ( x^3 - 6x^2 -9x -9 ). Let me find its roots.Hmm, finding roots of a cubic can be tricky, but maybe I can try rational roots. The possible rational roots are factors of 9 over factors of 1, so ±1, ±3, ±9.Let me test x=3: ( 27 - 54 -27 -9 = -63 ≠ 0 )x=1: 1 -6 -9 -9 = -23 ≠ 0x=-1: -1 -6 +9 -9 = -7 ≠ 0x=9: 729 - 486 -81 -9 = 153 ≠ 0x=-3: -27 -54 +27 -9 = -63 ≠ 0Hmm, no rational roots. Maybe it's irreducible over rationals. So, perhaps the roots are irrational or complex.But regardless, the condition is about the sum of cubes of (x_i - 3). Since we transformed the polynomial and used Vieta's, the algebra seems consistent.Alternatively, maybe I can think about the sum ( y_1^3 + y_2^3 + y_3^3 = 0 ). If the product ( y_1y_2y_3 = 0 ), then at least one of the y_i is zero, meaning at least one root x_i is 3. But in our case, when a = -9, does x=3 satisfy the polynomial?Let me check: ( 3^3 - 6*3^2 + (-9)*3 + (-9) = 27 - 54 -27 -9 = -63 ≠ 0 ). So, x=3 is not a root. Therefore, ( y_i ) cannot be zero, so ( y_1y_2y_3 ≠ 0 ). Hmm, but from the transformed polynomial, ( y_1y_2y_3 = -(4a -27) ). When a = -9, this becomes -(4*(-9) -27) = -(-36 -27) = -(-63) = 63 ≠ 0. So, none of the y_i are zero, which is consistent.But then, how does ( y_1^3 + y_2^3 + y_3^3 = 0 ) hold? It must be due to some cancellation among the cubes. Maybe the roots are arranged such that their cubes sum to zero.Alternatively, perhaps I made a mistake in assuming that a = -9 is the only solution. Let me think again.Wait, in the transformed polynomial, we have ( y_1^3 + y_2^3 + y_3^3 = 0 ). Using the identity:( y_1^3 + y_2^3 + y_3^3 = (y_1 + y_2 + y_3)^3 - 3(y_1 + y_2 + y_3)(y_1y_2 + y_1y_3 + y_2y_3) + 3y_1y_2y_3 )We set this equal to zero:( (-3)^3 - 3*(-3)*(a - 9) + 3*(-(4a -27)) = 0 )Which simplifies to:-27 + 9(a - 9) -12a +81 = 0Wait, let me recompute that step:-27 + 9(a - 9) + 3*(-(4a -27)) = 0Which is:-27 + 9a -81 -12a +81 = 0Combine constants:-27 -81 +81 = -27Variables:9a -12a = -3aSo, -27 -3a = 0 => -3a = 27 => a = -9Yes, same result. So, seems consistent.Alternatively, maybe there's another approach. Let me think about the identity for sum of cubes. There's another identity that says if ( y_1 + y_2 + y_3 = S ), then ( y_1^3 + y_2^3 + y_3^3 = S^3 - 3S(y_1y_2 + y_1y_3 + y_2y_3) + 3y_1y_2y_3 ). Which is what I used.Alternatively, if ( y_1 + y_2 + y_3 = 0 ), then ( y_1^3 + y_2^3 + y_3^3 = 3y_1y_2y_3 ). But in our case, ( y_1 + y_2 + y_3 = -3 ), not zero. So, that identity doesn't directly apply.But wait, in our case, ( y_1 + y_2 + y_3 = -3 ), and we have ( y_1^3 + y_2^3 + y_3^3 = 0 ). So, using the identity:( 0 = (-3)^3 - 3*(-3)*(a - 9) + 3*(-(4a -27)) )Which is the same as before, leading to a = -9.So, seems like a = -9 is the only solution.Wait, but let me think again. Is there a possibility that the polynomial could have multiple roots or something? Or maybe a = -9 is the only solution.Alternatively, maybe I can think about the original condition ( (x_1 -3)^3 + (x_2 -3)^3 + (x_3 -3)^3 = 0 ). If I let ( y_i = x_i -3 ), then the condition is ( y_1^3 + y_2^3 + y_3^3 = 0 ). Now, if I consider the polynomial in terms of y, as I did before, and use Vieta's, it seems the only solution is a = -9.Alternatively, maybe I can think about specific cases. Suppose one of the roots is 3, then ( y_i = 0 ), and the other two roots would satisfy ( y_j^3 + y_k^3 = 0 ), meaning ( y_j = -y_k ). But in that case, the sum ( y_j + y_k = 0 ), but since ( y_1 + y_2 + y_3 = -3 ), if one y is zero, the other two must sum to -3. So, ( y_j + y_k = -3 ), and ( y_j^3 + y_k^3 = 0 ). Let me see if that's possible.Let me set ( y_j = -y_k ). Then, ( y_j + y_k = 0 ), but we need ( y_j + y_k = -3 ). Contradiction, unless y_j and y_k are complex. So, maybe that's possible, but in that case, the roots would be complex, but the polynomial has real coefficients, so complex roots come in pairs. So, if one root is 3, and the other two are complex conjugates, their cubes would also be complex conjugates, and their sum would be real. But in our case, the sum is zero, which is real.But wait, in our case, when a = -9, the polynomial doesn't have x=3 as a root, as we saw earlier. So, that approach might not help.Alternatively, maybe I can think about the equation ( y_1^3 + y_2^3 + y_3^3 = 0 ) and use the fact that ( y_1 + y_2 + y_3 = -3 ). Maybe I can express ( y_1^3 + y_2^3 + y_3^3 ) in terms of ( (y_1 + y_2 + y_3) ) and other symmetric sums.But I think I already did that using the identity.Alternatively, maybe I can consider that if ( y_1^3 + y_2^3 + y_3^3 = 0 ), then perhaps the roots satisfy some symmetry. For example, maybe two roots are negatives of each other, but given that their sum is -3, that might not be possible unless the third root is -3.Wait, if ( y_1 = -y_2 ), then ( y_1 + y_2 = 0 ), so ( y_3 = -3 ). Then, ( y_1^3 + y_2^3 + y_3^3 = y_1^3 + (-y_1)^3 + (-3)^3 = 0 + (-27) = -27 ≠ 0 ). So, that doesn't work.Alternatively, maybe all three roots are equal? If ( y_1 = y_2 = y_3 ), then ( 3y_1 = -3 ) => ( y_1 = -1 ). Then, ( y_1^3 + y_2^3 + y_3^3 = 3*(-1)^3 = -3 ≠ 0 ). So, that doesn't work either.Alternatively, maybe two roots are equal, and the third is different. Let me suppose ( y_1 = y_2 ). Then, ( 2y_1 + y_3 = -3 ), so ( y_3 = -3 - 2y_1 ). Then, ( y_1^3 + y_1^3 + y_3^3 = 2y_1^3 + (-3 - 2y_1)^3 = 0 ).Let me compute ( (-3 - 2y_1)^3 ):( (-3)^3 + 3*(-3)^2*(-2y_1) + 3*(-3)*(-2y_1)^2 + (-2y_1)^3 )= -27 + 3*9*(-2y_1) + 3*(-3)*(4y_1^2) + (-8y_1^3)= -27 -54y_1 -36y_1^2 -8y_1^3So, the equation becomes:2y_1^3 + (-27 -54y_1 -36y_1^2 -8y_1^3) = 0Combine like terms:2y_1^3 -8y_1^3 = -6y_1^3-36y_1^2-54y_1-27So, equation: -6y_1^3 -36y_1^2 -54y_1 -27 = 0Divide both sides by -3:2y_1^3 + 12y_1^2 + 18y_1 + 9 = 0Let me see if this cubic has any real roots. Maybe try rational roots: possible roots are ±1, ±3, ±9, ±1/2, etc.Try y = -1: 2*(-1)^3 + 12*(-1)^2 + 18*(-1) + 9 = -2 + 12 -18 + 9 = 1 ≠ 0y = -3: 2*(-27) + 12*9 + 18*(-3) + 9 = -54 + 108 -54 + 9 = 9 ≠ 0y = -1/2: 2*(-1/8) + 12*(1/4) + 18*(-1/2) + 9 = -1/4 + 3 -9 + 9 = 2.75 ≠ 0Hmm, no obvious rational roots. Maybe it's irreducible, meaning no real roots. So, this approach might not help.Alternatively, maybe I can think about the derivative of the polynomial to check for multiple roots, but that might be overcomplicating.Alternatively, maybe I can consider that the only solution is a = -9, as derived earlier, and accept that as the answer.Wait, but let me think again. When I transformed the polynomial, I got ( y^3 + 3y^2 + (a -9)y + (4a -27) = 0 ). If I set ( y_1^3 + y_2^3 + y_3^3 = 0 ), and using the identity, I found a = -9. So, that seems consistent.Alternatively, maybe I can plug a = -9 back into the transformed polynomial and see if the sum of cubes is zero.So, when a = -9, the transformed polynomial is:( y^3 + 3y^2 + (-9 -9)y + (4*(-9) -27) = y^3 + 3y^2 -18y -63 = 0 )Now, let me compute ( y_1^3 + y_2^3 + y_3^3 ) using Vieta's:From the polynomial, ( y_1 + y_2 + y_3 = -3 ), ( y_1y_2 + y_1y_3 + y_2y_3 = -18 ), ( y_1y_2y_3 = 63 )Using the identity:( y_1^3 + y_2^3 + y_3^3 = (-3)^3 - 3*(-3)*(-18) + 3*63 )Compute each term:-27 - 3*(-3)*(-18) + 189First term: -27Second term: -3*(-3)*(-18) = -3*54 = -162Third term: 3*63 = 189So, total: -27 -162 + 189 = (-189) + 189 = 0Yes, it works! So, when a = -9, the sum of cubes is indeed zero. Therefore, a = -9 is the correct solution.I think I've convinced myself that a = -9 is the correct answer. I initially doubted it, but after checking, it seems consistent.

💡Alright, so I have this problem here where I need to define a function ( f(x) ) on the real numbers that satisfies two conditions: ( f(x) = f(4 - x) ) and ( f(2 - x) + f(x - 2) = 0 ). Additionally, I know that ( f(2) = 1 ), and I need to find the value of ( f(2014) ). The options are A: -1, B: 0, C: 1, D: Cannot be determined.Okay, let me start by understanding what each condition means.First, ( f(x) = f(4 - x) ). This seems like a symmetry condition. If I plug in ( x ) into the function, it's the same as plugging in ( 4 - x ). So, this suggests that the function is symmetric around the line ( x = 2 ). Because if you take a point ( x ) and its mirror image across ( x = 2 ), which is ( 4 - x ), the function values are the same. So, the graph of ( f(x) ) is symmetric with respect to ( x = 2 ).Next, the second condition is ( f(2 - x) + f(x - 2) = 0 ). This looks like an odd function condition, but shifted. Let me see. If I let ( y = x - 2 ), then ( f(2 - x) = f(-y) ) and ( f(x - 2) = f(y) ). So, substituting, we get ( f(-y) + f(y) = 0 ), which implies ( f(-y) = -f(y) ). Therefore, ( f ) is an odd function with respect to the point ( x = 2 ). Wait, actually, it's an odd function around ( x = 2 ), meaning that if you shift the function so that ( x = 2 ) becomes the origin, it becomes an odd function.So, combining these two properties: ( f(x) ) is symmetric about ( x = 2 ) and it's odd about ( x = 2 ). Hmm, that seems a bit conflicting at first glance because symmetry usually suggests evenness, but here it's combined with oddness.Let me try to formalize this.From the first condition: ( f(x) = f(4 - x) ).From the second condition: ( f(2 - x) + f(x - 2) = 0 ), which as I saw earlier, implies ( f(-y) = -f(y) ) if ( y = x - 2 ). So, ( f(-y) = -f(y) ), meaning ( f ) is odd about the origin if we shift the function by 2 units.Wait, perhaps I should consider shifting the function to make the center at 0. Let me define a new function ( g(x) = f(x + 2) ). Then, ( g(-x) = f(-x + 2) ). Let's see what the conditions become in terms of ( g ).First condition: ( f(x) = f(4 - x) ). Substituting ( x = y - 2 ), we get ( f(y - 2) = f(4 - (y - 2)) = f(6 - y) ). So, ( g(y - 2) = g(6 - y) ). Hmm, not sure if that helps.Wait, maybe I should express ( f(4 - x) ) in terms of ( g ). Since ( f(x) = g(x - 2) ), then ( f(4 - x) = g((4 - x) - 2) = g(2 - x) ). So, the first condition becomes ( g(x - 2) = g(2 - x) ). Which is ( g(-x) = g(x) ). So, ( g ) is an even function.From the second condition: ( f(2 - x) + f(x - 2) = 0 ). In terms of ( g ), ( f(2 - x) = g(2 - x - 2) = g(-x) ) and ( f(x - 2) = g(x - 2 - 2) = g(x - 4) ). So, the condition becomes ( g(-x) + g(x - 4) = 0 ).But since ( g ) is even, ( g(-x) = g(x) ). So, substituting, we get ( g(x) + g(x - 4) = 0 ). Therefore, ( g(x - 4) = -g(x) ).So, ( g(x - 4) = -g(x) ). Let me shift the argument by 4: ( g(x) = -g(x + 4) ). Then, shifting again: ( g(x + 4) = -g(x + 8) ). So, combining these, ( g(x) = -g(x + 4) = -(-g(x + 8)) = g(x + 8) ). Therefore, ( g ) is periodic with period 8.So, ( g(x + 8) = g(x) ). Therefore, ( g ) is periodic with period 8.Since ( g(x) = f(x + 2) ), this implies that ( f(x + 10) = g(x + 8) = g(x) = f(x + 2) ). Wait, that seems a bit off. Let me check.Wait, ( g(x) = f(x + 2) ), so ( g(x + 8) = f(x + 10) ). But ( g(x + 8) = g(x) ), so ( f(x + 10) = f(x + 2) ). Therefore, ( f(x + 8) = f(x) ). So, the function ( f ) is periodic with period 8.Wait, hold on. Let me make sure.If ( g(x + 8) = g(x) ), then ( f(x + 10) = g(x + 8) = g(x) = f(x + 2) ). So, ( f(x + 10) = f(x + 2) ). Therefore, ( f(x + 8) = f(x) ). So, the period is 8.Therefore, ( f ) is periodic with period 8.Given that, ( f(2014) = f(2014 mod 8) ). Let's compute ( 2014 div 8 ).8 × 251 = 2008, so 2014 - 2008 = 6. Therefore, ( 2014 mod 8 = 6 ). So, ( f(2014) = f(6) ).Now, I need to find ( f(6) ).Given that ( f(x) = f(4 - x) ), let's plug in x = 6: ( f(6) = f(4 - 6) = f(-2) ).From the second condition, ( f(2 - x) + f(x - 2) = 0 ). Let me plug in x = 4: ( f(2 - 4) + f(4 - 2) = f(-2) + f(2) = 0 ). So, ( f(-2) + f(2) = 0 ). Since ( f(2) = 1 ), then ( f(-2) = -1 ).Therefore, ( f(6) = f(-2) = -1 ).So, ( f(2014) = -1 ).Therefore, the answer is A: -1.Wait, let me double-check my steps.1. Defined ( g(x) = f(x + 2) ) to shift the function so that the center is at 0.2. From ( f(x) = f(4 - x) ), we got ( g(-x) = g(x) ), so ( g ) is even.3. From ( f(2 - x) + f(x - 2) = 0 ), we got ( g(-x) + g(x - 4) = 0 ). Since ( g ) is even, ( g(-x) = g(x) ), so ( g(x) + g(x - 4) = 0 ), leading to ( g(x - 4) = -g(x) ).4. Then, shifting arguments, we found that ( g(x + 8) = g(x) ), so ( g ) is periodic with period 8.5. Therefore, ( f(x + 8) = f(x) ), so ( f ) is periodic with period 8.6. Calculated ( 2014 mod 8 = 6 ), so ( f(2014) = f(6) ).7. Using ( f(x) = f(4 - x) ), ( f(6) = f(-2) ).8. From ( f(2 - x) + f(x - 2) = 0 ), set x = 4: ( f(-2) + f(2) = 0 ), so ( f(-2) = -f(2) = -1 ).9. Therefore, ( f(6) = -1 ), so ( f(2014) = -1 ).Everything seems to check out. I don't see any errors in the reasoning.**Final Answer**boxed{A}

💡Okay, so I have this problem here in polar coordinates. It says that the line θ = π/6 intersects the circle ρ = 2 cos(θ - π/6), and I need to find the length of the chord obtained from this intersection. The options are A:1, B:2, C:3, D:4. Hmm, let me try to figure this out step by step.First, I remember that in polar coordinates, θ = π/6 is a straight line. It's like a ray starting from the origin and making an angle of π/6 with the positive x-axis. Since ρ can be any real number, this line extends infinitely in both directions.Next, the circle is given by ρ = 2 cos(θ - π/6). I think this is a polar equation of a circle, but I'm not entirely sure about its properties. Maybe I should convert both equations to Cartesian coordinates to better visualize them.Let me recall the conversion formulas from polar to Cartesian coordinates:- x = ρ cos θ- y = ρ sin θStarting with the line θ = π/6. In Cartesian coordinates, this would be a line with a slope of tan(π/6). Since tan(π/6) is 1/√3, the equation of the line is y = (1/√3)x. That seems straightforward.Now, for the circle ρ = 2 cos(θ - π/6). I think this is a circle that's shifted in some way. Maybe I can rewrite it using the cosine of a difference identity. The formula is cos(A - B) = cos A cos B + sin A sin B. So, cos(θ - π/6) = cos θ cos(π/6) + sin θ sin(π/6).Let me compute cos(π/6) and sin(π/6). I remember that cos(π/6) is √3/2 and sin(π/6) is 1/2. So substituting these values in, we get:ρ = 2 [cos θ (√3/2) + sin θ (1/2)] ρ = 2*(√3/2 cos θ + 1/2 sin θ) Simplifying, ρ = √3 cos θ + sin θHmm, now I need to convert this into Cartesian coordinates. Since ρ = √3 cos θ + sin θ, and knowing that ρ = √(x² + y²), cos θ = x/ρ, and sin θ = y/ρ, let me substitute these into the equation.So, ρ = √3 (x/ρ) + (y/ρ) Multiplying both sides by ρ to eliminate the denominator: ρ² = √3 x + yBut ρ² is x² + y², so substituting that in: x² + y² = √3 x + yNow, let me rearrange this equation to get it into the standard form of a circle. I'll move the linear terms to the left side:x² - √3 x + y² - y = 0To complete the square for both x and y terms, I need to add and subtract appropriate constants.For the x terms: x² - √3 x. The coefficient of x is -√3, so half of that is -√3/2, and squaring it gives (3/4). So, I add and subtract 3/4.For the y terms: y² - y. The coefficient of y is -1, half of that is -1/2, squaring it gives 1/4. So, I add and subtract 1/4.Putting it all together:(x² - √3 x + 3/4) - 3/4 + (y² - y + 1/4) - 1/4 = 0 Which simplifies to:(x - √3/2)² + (y - 1/2)² - (3/4 + 1/4) = 0 So,(x - √3/2)² + (y - 1/2)² = 1Ah, so this is a circle with center at (√3/2, 1/2) and radius 1. Wait, hold on, because I had to subtract 3/4 and 1/4, which added up to 1, so the radius squared is 1, so radius is 1.Wait, but in the equation, it's (x - √3/2)² + (y - 1/2)² = 1, so the center is at (√3/2, 1/2) and radius 1. Got it.Now, the line θ = π/6 in Cartesian coordinates is y = (1/√3)x, as I found earlier. So, I need to find where this line intersects the circle, and then find the length of the chord between those two intersection points.Alternatively, since the line passes through the origin, and the circle is centered at (√3/2, 1/2) with radius 1, maybe I can find the distance from the center of the circle to the line, and then use the formula for the length of a chord.Yes, that might be easier. The formula for the length of a chord is 2√(r² - d²), where r is the radius and d is the distance from the center to the line.So, first, let me find the distance from the center (√3/2, 1/2) to the line y = (1/√3)x.The general formula for the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / √(a² + b²).First, let me write the line y = (1/√3)x in standard form. Subtract y from both sides:(1/√3)x - y = 0So, a = 1/√3, b = -1, c = 0.Now, plugging in the center (√3/2, 1/2):Distance d = |(1/√3)(√3/2) + (-1)(1/2) + 0| / √((1/√3)² + (-1)²)Simplify numerator:(1/√3)(√3/2) = (1/√3)*(√3)/2 = 1/2(-1)(1/2) = -1/2So, numerator is |1/2 - 1/2| = |0| = 0Wait, the distance is zero? That means the line passes through the center of the circle.So, if the line passes through the center, then the chord length is just the diameter of the circle, which is 2r = 2*1 = 2.Therefore, the length of the chord is 2, which is option B.But wait, let me double-check my calculations because sometimes I might make a mistake.So, the line is y = (1/√3)x, which is the same as x - √3 y = 0. Wait, did I write the standard form correctly earlier?Wait, y = (1/√3)x can be rewritten as (1/√3)x - y = 0, which is correct. So, a = 1/√3, b = -1, c = 0.Then, plugging in the center (√3/2, 1/2):(1/√3)(√3/2) = (√3 / √3)/2 = 1/2(-1)(1/2) = -1/2So, 1/2 - 1/2 = 0, so the distance is indeed 0. So, the line passes through the center, so the chord is the diameter, which is 2.Therefore, the answer should be B:2.Alternatively, just to make sure, let me solve the equations together.We have the line y = (1/√3)x and the circle (x - √3/2)² + (y - 1/2)² = 1.Substitute y = (1/√3)x into the circle equation:(x - √3/2)² + ((1/√3)x - 1/2)² = 1Let me compute each term:First term: (x - √3/2)² = x² - √3 x + (3/4)Second term: ((1/√3)x - 1/2)² = (1/3)x² - (1/√3)x + 1/4Adding both terms:x² - √3 x + 3/4 + (1/3)x² - (1/√3)x + 1/4 = 1Combine like terms:(1 + 1/3)x² + (-√3 - 1/√3)x + (3/4 + 1/4) = 1Simplify:(4/3)x² + (-√3 - 1/√3)x + 1 = 1Subtract 1 from both sides:(4/3)x² + (-√3 - 1/√3)x = 0Factor out x:x [ (4/3)x - √3 - 1/√3 ] = 0So, either x = 0 or (4/3)x - √3 - 1/√3 = 0If x = 0, then y = (1/√3)*0 = 0. So, one point is (0,0).For the other solution:(4/3)x = √3 + 1/√3Multiply both sides by 3/4:x = (3/4)(√3 + 1/√3)Simplify:x = (3√3)/4 + 3/(4√3) = (3√3)/4 + (√3)/4 = (4√3)/4 = √3So, x = √3, then y = (1/√3)*√3 = 1So, the other point is (√3, 1)Therefore, the two points of intersection are (0,0) and (√3,1)Now, let's find the distance between these two points.Distance formula: √[(√3 - 0)² + (1 - 0)²] = √[3 + 1] = √4 = 2So, the length of the chord is 2, which confirms my earlier conclusion.Therefore, the correct answer is B:2.**Final Answer**boxed{B}