🧠:[<think>Okay, so I have this matrix A, which is a 2x2 matrix with entries 3, 3, 2, 2. The first part asks me to calculate A to the power of n, where n is a positive integer. The second part is to solve the equation X^2015 = A, where X is also a 2x2 matrix. Hmm, let's start with part a.First, I need to compute A^n. For matrices, one common way to compute powers is by diagonalization if the matrix is diagonalizable. So maybe I should check if A is diagonalizable. To do that, I need to find its eigenvalues and eigenvectors. Let me recall that the eigenvalues are the roots of the characteristic equation det(A - λI) = 0.So, let's compute the characteristic polynomial of A. The matrix A is:[3, 3][2, 2]So, A - λI is:[3 - λ, 3 ][2 , 2 - λ]The determinant of this matrix is (3 - λ)(2 - λ) - (3*2) = (3 - λ)(2 - λ) - 6.Let me compute that. First, expand (3 - λ)(2 - λ):3*2 - 3λ - 2λ + λ² = 6 - 5λ + λ²Then subtract 6: 6 -5λ + λ² -6 = λ² -5λ.So the characteristic equation is λ² -5λ = 0. Therefore, the eigenvalues are λ = 0 and λ = 5. Okay, so A has two distinct eigenvalues, 0 and 5. Since it has two distinct eigenvalues, it should be diagonalizable. That's good news because diagonal matrices are easy to raise to powers.Now, to diagonalize A, I need to find a matrix P such that P⁻¹AP = D, where D is the diagonal matrix with eigenvalues 0 and 5. Then A^n = P D^n P⁻¹.So let's find the eigenvectors corresponding to each eigenvalue.Starting with λ = 5. We solve (A - 5I)v = 0.A - 5I is:[3 -5, 3 ] = [-2, 3][2 , 2 -5] [2, -3]So the system is:-2v₁ + 3v₂ = 02v₁ - 3v₂ = 0Which are essentially the same equation. Let's take the first one: -2v₁ + 3v₂ = 0 => 2v₁ = 3v₂ => v₁ = (3/2)v₂. So an eigenvector can be [3/2, 1] or any scalar multiple. Let's take [3, 2] to make it integers.Now for λ = 0. Solve (A - 0I)v = 0 => Av = 0.So the system is:3v₁ + 3v₂ = 02v₁ + 2v₂ = 0Dividing the first equation by 3: v₁ + v₂ = 0Dividing the second by 2: v₁ + v₂ = 0So same equation. So eigenvectors satisfy v₁ = -v₂. So an eigenvector is [1, -1], for example.Therefore, the matrix P can be constructed with eigenvectors as columns. So P is:[3, 1][2, -1]Let me verify that. Let's compute P⁻¹AP and see if it's diagonal.But maybe first, let's compute P⁻¹. The inverse of a 2x2 matrix [a, b; c, d] is (1/(ad - bc))[d, -b; -c, a].So determinant of P is (3)(-1) - (1)(2) = -3 - 2 = -5. So P⁻¹ is (1/-5)[-1, -1; -2, 3] ?Wait:Wait, original matrix P is:[3, 1][2, -1]So ad - bc is 3*(-1) - 1*2 = -3 -2 = -5. So inverse is (1/-5)*[ -1, -1; -2, 3 ]? Wait:Wait, formula is [d, -b; -c, a] divided by determinant.So here, d is -1, b is 1, c is 2, a is 3. So:[ -1, -1 ][ -2, 3 ]Then divide by determinant -5. So:P⁻¹ = (1/-5)*[ -1, -1; -2, 3 ] = (1/5)[1, 1; 2, -3]Because multiplying numerator and denominator by -1.So P⁻¹ is:[1/5, 1/5][2/5, -3/5]Now let's check if P⁻¹AP is diagonal.Compute AP first. A is:[3, 3][2, 2]Multiply by P:First column: A*[3; 2] = [3*3 + 3*2; 2*3 + 2*2] = [9 +6; 6 +4] = [15; 10]Second column: A*[1; -1] = [3*1 + 3*(-1); 2*1 + 2*(-1)] = [3 -3; 2 -2] = [0; 0]So AP is:[15, 0][10, 0]Now multiply by P⁻¹: P⁻¹ is:[1/5, 1/5][2/5, -3/5]So P⁻¹ * AP:First column: [1/5*15 + 1/5*10, ...] Wait, actually, let's do it step by step.Wait, actually, AP is:First column: [15; 10], second column: [0; 0]So P⁻¹ times AP:First element: (1/5)(15) + (1/5)(10) = 3 + 2 = 5Second element: (2/5)(15) + (-3/5)(10) = 6 - 6 = 0Third element: (1/5)(0) + (1/5)(0) = 0Fourth element: (2/5)(0) + (-3/5)(0) = 0Wait, but AP's columns are [15;10] and [0;0], so when multiplying P⁻¹ * AP, each column is P⁻¹ multiplied by the corresponding column of AP.First column: P⁻¹ * [15;10] = [ (1/5)(15) + (1/5)(10) ; (2/5)(15) + (-3/5)(10) ] = [3 + 2; 6 -6] = [5; 0]Second column: P⁻¹ * [0;0] = [0;0]Therefore, P⁻¹AP is:[5, 0][0, 0]Which is the diagonal matrix D with eigenvalues 5 and 0. Perfect, so that works.Therefore, A = P D P⁻¹, so A^n = P D^n P⁻¹. Since D is diagonal, D^n is just diagonal entries raised to the nth power. So D^n is [5^n, 0; 0, 0^n]. But 0^n is 0 for n >=1. So D^n is [5^n, 0; 0, 0].Therefore, A^n = P [5^n, 0; 0, 0] P⁻¹.Let me compute that. Let's write it out:A^n = [3, 1; 2, -1] * [5^n, 0; 0, 0] * [1/5, 1/5; 2/5, -3/5]First, compute the product of the first two matrices:[3, 1; 2, -1] * [5^n, 0; 0, 0] = [3*5^n + 1*0, 3*0 + 1*0; 2*5^n + (-1)*0, 2*0 + (-1)*0] = [3*5^n, 0; 2*5^n, 0]Then multiply this result by P⁻¹:[3*5^n, 0; 2*5^n, 0] * [1/5, 1/5; 2/5, -3/5]Compute first row first column: (3*5^n)(1/5) + 0*(2/5) = (3*5^n)/5 = 3*5^{n-1}First row second column: (3*5^n)(1/5) + 0*(-3/5) = same as above, 3*5^{n-1}Second row first column: (2*5^n)(1/5) + 0*(2/5) = 2*5^{n-1}Second row second column: (2*5^n)(1/5) + 0*(-3/5) = 2*5^{n-1}Therefore, A^n is:[3*5^{n-1}, 3*5^{n-1}][2*5^{n-1}, 2*5^{n-1}]Wait, let's check again:Wait, the multiplication is:Row 1 of first matrix [3*5^n, 0] times column 1 of P⁻¹ [1/5, 2/5]^T? Wait, no, wait, when you multiply [3*5^n, 0; 2*5^n, 0] with [1/5, 1/5; 2/5, -3/5], each element is computed as:First matrix is:Row 1: [3*5^n, 0]Row 2: [2*5^n, 0]Second matrix is:Column 1: [1/5, 2/5]Column 2: [1/5, -3/5]So entry (1,1) is (3*5^n)(1/5) + 0*(2/5) = 3*5^{n}/5 = 3*5^{n-1}Entry (1,2) is (3*5^n)(1/5) + 0*(-3/5) = same 3*5^{n-1}Entry (2,1) is (2*5^n)(1/5) + 0*(2/5) = 2*5^{n-1}Entry (2,2) is (2*5^n)(1/5) + 0*(-3/5) = 2*5^{n-1}Therefore, the resulting matrix is:[3*5^{n-1}, 3*5^{n-1}][2*5^{n-1}, 2*5^{n-1}]So indeed, A^n is equal to 5^{n-1} times the original matrix A, since A is:[3, 3][2, 2]So A^n = 5^{n-1} * A.Let me verify this for n=1. A^1 = 5^{0} * A = 1 * A = A. Correct.For n=2, let's compute A^2:Original A:[3, 3][2, 2]A^2 = [3*3 + 3*2, 3*3 + 3*2; 2*3 + 2*2, 2*3 + 2*2] = [9 +6, 9 +6; 6 +4, 6 +4] = [15, 15; 10, 10]Which is 5^{1} * A = 5 * [3, 3; 2, 2] = [15, 15; 10, 10]. Correct. So that works.Similarly, for n=3, A^3 should be 5^{2} * A = 25 * A = [75,75;50,50]. Let's compute A^3 = A^2 * A = [15,15;10,10] * [3,3;2,2] = [15*3 +15*2, 15*3 +15*2; 10*3 +10*2,10*3 +10*2] = [45 +30, 45 +30; 30 +20, 30 +20] = [75,75;50,50]. Correct. So the formula seems to hold.Therefore, the general formula is A^n = 5^{n-1} A. So part a) is solved.Moving on to part b), solve X^{2015} = A, where X is a 2x2 matrix. So we need to find a matrix X such that when raised to the 2015th power, we get A.Given that A is diagonalizable, and we have A = P D P⁻¹ where D is diagonal with entries 5 and 0. So perhaps X can be expressed as P E P⁻¹, where E is some matrix such that E^{2015} = D. Because then X^{2015} = (P E P⁻¹)^{2015} = P E^{2015} P⁻¹ = P D P⁻¹ = A.Therefore, if we can find a matrix E such that E^{2015} = D, then X = P E P⁻¹ is a solution. Since D is diagonal, E can be a diagonal matrix whose entries are the 2015th roots of the diagonal entries of D. However, D has entries 5 and 0. So, the 2015th root of 5 is 5^{1/2015}, and the 2015th root of 0 is 0. Therefore, E can be diag(5^{1/2015}, 0). Then, E^{2015} = diag(5, 0) = D. Therefore, X = P E P⁻¹.Let me check this. Let’s compute X = P E P⁻¹, where E is diag(5^{1/2015}, 0). Then X^{2015} = P E^{2015} P⁻¹ = P D P⁻¹ = A. So that works.Therefore, the solution X is P diag(5^{1/2015}, 0) P⁻¹. Let's compute this matrix.First, let's note that E = [5^{1/2015}, 0; 0, 0]. Then P E is:[3, 1; 2, -1] * [5^{1/2015}, 0; 0, 0] = [3*5^{1/2015}, 0; 2*5^{1/2015}, 0]Then multiply by P⁻¹:[3*5^{1/2015}, 0; 2*5^{1/2015}, 0] * [1/5, 1/5; 2/5, -3/5]Compute the first row first column: 3*5^{1/2015}*(1/5) + 0*(2/5) = (3/5)*5^{1/2015} = 3*5^{1/2015 -1} = 3*5^{-(2014/2015)}.Wait, but 5^{1/2015} divided by 5 is 5^{(1/2015 -1)} = 5^{-2014/2015}.Alternatively, 5^{1/2015}/5 = 5^{(1/2015 - 1)} = 5^{-2014/2015}.But perhaps it's better to write it as (1/5) * 5^{1/2015} = 5^{-1} * 5^{1/2015} = 5^{( -1 + 1/2015)} = 5^{-2014/2015}.So entry (1,1): 3 * 5^{-2014/2015}Entry (1,2): 3*5^{1/2015}*(1/5) = same as (1,1), so 3*5^{-2014/2015}Entry (2,1): 2*5^{1/2015}*(1/5) = 2*5^{-2014/2015}Entry (2,2): 2*5^{1/2015}*(1/5) = same as (2,1), so 2*5^{-2014/2015}Therefore, X = [3*5^{-2014/2015}, 3*5^{-2014/2015}; 2*5^{-2014/2015}, 2*5^{-2014/2015}]Alternatively, factor out 5^{-2014/2015}:X = 5^{-2014/2015} * [3, 3; 2, 2] = 5^{-2014/2015} * A.But wait, 5^{-2014/2015} is equal to 5^{(1 - 2015)/2015} = 5^{1/2015 -1} = 5^{-2014/2015}. So indeed, X is 5^{-2014/2015} times A.But let's check if this is correct. Let's compute X^{2015}:If X = c * A, where c = 5^{-2014/2015}, then X^{2015} = c^{2015} * A^{2015}.But A^{2015} = 5^{2014} * A, as per part a). Wait, A^n = 5^{n-1} A. So A^{2015} = 5^{2014} A.Therefore, X^{2015} = (c)^{2015} * 5^{2014} A.But c = 5^{-2014/2015}, so c^{2015} = 5^{-2014}.Therefore, X^{2015} = 5^{-2014} * 5^{2014} A = A. Correct. So this works.Therefore, X = 5^{-2014/2015} * A.But let's write this out explicitly. Since 5^{-2014/2015} is 5^{-1 + 1/2015} = 5^{-1} * 5^{1/2015} = (1/5) * 5^{1/2015}.But maybe it's better to leave it as 5^{-2014/2015} unless simplification is needed.Therefore, the matrix X is:[3*5^{-2014/2015}, 3*5^{-2014/2015}][2*5^{-2014/2015}, 2*5^{-2014/2015}]Alternatively, written as 5^{-2014/2015} times the original matrix A. So X = 5^{-2014/2015} * A.But since A is rank 1 (since its determinant is 0), X is also rank 1. Therefore, this is the only solution? Wait, but matrix roots can sometimes have multiple solutions. However, since D has a zero eigenvalue, and we are taking the 2015th root, but 0's root is unique (only 0). For the eigenvalue 5, we take the real 2015th root, which is unique because 5 is positive and 2015 is odd. Therefore, in this case, the only real matrix solution is X = P E P⁻¹, which gives the matrix above.Alternatively, since A is a rank 1 matrix, and any power of a rank 1 matrix is also a scalar multiple of itself. So if X is a multiple of A, say X = kA, then X^{2015} = k^{2015} A^{2015} = k^{2015} * 5^{2014} A. Set this equal to A, so we have k^{2015} *5^{2014} A = A. Since A is non-zero, we require k^{2015} *5^{2014} =1. Therefore, k = (5^{-2014})^{1/2015} = 5^{-2014/2015}, which matches our previous result. Therefore, the only solution is X = 5^{-2014/2015} A.Therefore, part b) solution is X equals A multiplied by 5^{-2014/2015}.Alternatively, written as:X = (1/5^{2014})^{1/2015} * A = 5^{-2014/2015} * A.So, to write the matrix explicitly:X = left(begin{array}{ll}3 cdot 5^{-2014/2015} & 3 cdot 5^{-2014/2015} 2 cdot 5^{-2014/2015} & 2 cdot 5^{-2014/2015}end{array}right)Which can be factored as 5^{-2014/2015} times the original matrix A.Therefore, this should be the solution for part b.But let me verify with n=1. If we set 2015=1, then X^1 = A => X = A. According to the formula, 5^{-2014/2015} with 2015=1 is 5^{-2014/1} = 5^{-2014}, which would be incorrect. Wait, but 2015 is given as the exponent, so in the problem statement, n=2015 is fixed. So in this case, the formula is correct for n=2015.But if we think, for example, of n=2, and if the equation was X^2 = A, then similarly, X would be 5^{-1/2} A. Then X^2 = (5^{-1/2})^2 A^2 = 5^{-1} *5^{1} A = A. Correct. So the formula generalizes. Therefore, for exponent k, X = 5^{-(k-1)/k} A. Then X^k = 5^{-(k-1)} A^k = 5^{-(k-1)} *5^{k-1} A = A. Correct. So the formula works.Therefore, in general, X = 5^{-(2014/2015)} A is the solution.But we can write 5^{-2014/2015} as 5^{-1 + 1/2015} = 5^{-1} * 5^{1/2015} = (1/5) * 5^{1/2015}. But unless required, perhaps it's better to leave it as 5^{-2014/2015}.Alternatively, rational exponents are acceptable in the answer. Therefore, the solution is X = 5^{-2014/2015} * A.But the problem says "Solve the equation X^{2015}=A, X ∈ M_{2}(square)". The square probably denotes the real numbers. So since 5 is positive, 5^{-2014/2015} is a real number, so X is a real matrix. Therefore, this is the only real solution.Alternatively, could there be other solutions? For example, if we consider complex matrices, there could be more solutions, but since the problem specifies M_2(□), which might be M_2(R), the real matrices. So likely, this is the only solution.Therefore, the answers are:a) A^n = 5^{n-1} A, which is 5^{n-1} times the original matrix.b) X = 5^{-2014/2015} A, which can be written explicitly as the original matrix scaled by that factor.I should check if there are other possible solutions, but given that A is diagonalizable with eigenvalues 5 and 0, and we are looking for real matrices X such that X^2015 = A. Since 0 can only come from 0 in X, and 5 must come from (5)^{1/2015}, which is unique in real numbers. Therefore, the only real solution is the one constructed. So I think this is the unique solution.Final Answera) boxed{A^{n} = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}}b) boxed{X = 5^{-frac{2014}{2015}} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}}</think>To solve the given problem, we start with part (a) by diagonalizing the matrix ( A ). Part (a)1. Find Eigenvalues: The characteristic polynomial of ( A ) is: [ det(A - lambda I) = lambda^2 - 5lambda = 0 ] The eigenvalues are ( lambda = 0 ) and ( lambda = 5 ).2. Find Eigenvectors: - For ( lambda = 5 ), solving ( (A - 5I)v = 0 ) gives the eigenvector ( begin{pmatrix} 3 2 end{pmatrix} ). - For ( lambda = 0 ), solving ( Av = 0 ) gives the eigenvector ( begin{pmatrix} 1 -1 end{pmatrix} ).3. Diagonalization: - Matrix ( P ) with eigenvectors as columns: [ P = begin{pmatrix} 3 & 1 2 & -1 end{pmatrix} ] - Matrix ( P^{-1} ): [ P^{-1} = frac{1}{-5} begin{pmatrix} -1 & -1 -2 & 3 end{pmatrix} = frac{1}{5} begin{pmatrix} 1 & 1 2 & -3 end{pmatrix} ] - Diagonal matrix ( D ): [ D = begin{pmatrix} 5 & 0 0 & 0 end{pmatrix} ]4. Compute ( A^n ): Using ( A = PDP^{-1} ), we get: [ A^n = PD^nP^{-1} ] Since ( D^n = begin{pmatrix} 5^n & 0 0 & 0 end{pmatrix} ), we compute: [ A^n = frac{1}{5} begin{pmatrix} 3 cdot 5^n & 3 cdot 5^n 2 cdot 5^n & 2 cdot 5^n end{pmatrix} = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix} ] Part (b)1. Find ( X ): We need ( X^{2015} = A ). Using the diagonalization of ( A ), we find ( X ) such that ( X = P E P^{-1} ) where ( E^{2015} = D ). For ( D ) with eigenvalues 5 and 0, ( E ) must be: [ E = begin{pmatrix} 5^{1/2015} & 0 0 & 0 end{pmatrix} ] Therefore, ( X = P E P^{-1} ).2. Compute ( X ): Simplifying, we get: [ X = 5^{-2014/2015} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix} ] Final Answersa) (boxed{A^{n} = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}})b) (boxed{X = 5^{-frac{2014}{2015}} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}}),<think>Okay, let's tackle part (a) first. I need to calculate A raised to the power of n, where A is a 2x2 matrix. The matrix given is:A = [[3, 3], [2, 2]]Hmm, so I need to find a pattern or formula for A^n. Maybe diagonalizing the matrix would help? But first, let me check if A is diagonalizable. To do that, I need to find its eigenvalues and eigenvectors.The characteristic equation is det(A - λI) = 0. Let's compute the determinant:|A - λI| = (3 - λ)(2 - λ) - (3)(2) = (3 - λ)(2 - λ) - 6Expanding (3 - λ)(2 - λ):= 6 - 3λ - 2λ + λ² - 6Simplify:= λ² - 5λ + 6 - 6 = λ² - 5λSo the characteristic equation is λ² - 5λ = 0, which factors to λ(λ - 5) = 0. Therefore, the eigenvalues are λ₁ = 0 and λ₂ = 5.Okay, so the eigenvalues are 0 and 5. Since there are two distinct eigenvalues, the matrix is diagonalizable. Good, that should make things easier.Now, let's find the eigenvectors for each eigenvalue.Starting with λ₁ = 0:Solve (A - 0I)v = 0 => A v = 0The matrix A is:[3 3][2 2]So, the system is:3x + 3y = 02x + 2y = 0Divide the first equation by 3: x + y = 0Divide the second equation by 2: x + y = 0So, the eigenvectors for λ=0 are all vectors where x = -y. So, an eigenvector can be [1, -1]^T.Now for λ₂ = 5:Solve (A - 5I)v = 0A - 5I = [[3 - 5, 3], [2, 2 - 5]] = [[-2, 3], [2, -3]]The system is:-2x + 3y = 02x - 3y = 0These equations are scalar multiples of each other. Let's take the first one: -2x + 3y = 0 => 2x = 3y => x = (3/2)y. So, an eigenvector can be [3, 2]^T.So now, we have eigenvalues 0 and 5 with eigenvectors [1, -1]^T and [3, 2]^T respectively.Therefore, we can diagonalize A as:A = PDP^{-1}Where P is the matrix of eigenvectors, D is the diagonal matrix of eigenvalues.So, P = [[1, 3], [-1, 2]]Let me check the determinant of P to make sure it's invertible. det(P) = (1)(2) - (-1)(3) = 2 + 3 = 5 ≠ 0. Good.Then P^{-1} = (1/det(P)) * [[2, -3], [1, 1]]Wait, the inverse of a 2x2 matrix [[a, b], [c, d]] is (1/(ad - bc))[[d, -b], [-c, a]]So for P = [[1, 3], [-1, 2]], the inverse is (1/5)*[[2, -3], [1, 1]]Therefore, P^{-1} = [[2/5, -3/5], [1/5, 1/5]]Then, D = [[0, 0], [0, 5]]So, A^n = P D^n P^{-1}Since D is diagonal, D^n is [[0^n, 0], [0, 5^n]]But 0^n is 0 for n ≥ 1, which it is here since n ∈ ℕ*.Therefore, D^n = [[0, 0], [0, 5^n]]Therefore, compute A^n:A^n = P D^n P^{-1}Let's compute this multiplication step by step.First, compute P D^n:P D^n = [[1, 3], [-1, 2]] * [[0, 0], [0, 5^n]] = [[1*0 + 3*0, 1*0 + 3*5^n], [-1*0 + 2*0, -1*0 + 2*5^n]] = [[0, 3*5^n], [0, 2*5^n]]Then, multiply this by P^{-1}:A^n = [[0, 3*5^n], [0, 2*5^n]] * [[2/5, -3/5], [1/5, 1/5]]Let's compute each element:First row, first column: 0*(2/5) + 3*5^n*(1/5) = (3*5^n)/5 = 3*5^{n-1}First row, second column: 0*(-3/5) + 3*5^n*(1/5) = Same as above, 3*5^{n-1}Second row, first column: 0*(2/5) + 2*5^n*(1/5) = (2*5^n)/5 = 2*5^{n-1}Second row, second column: 0*(-3/5) + 2*5^n*(1/5) = Same as above, 2*5^{n-1}Wait, so putting this together:A^n = [[3*5^{n-1}, 3*5^{n-1}], [2*5^{n-1}, 2*5^{n-1}]]Hmm, so A^n is a scalar multiple of the original matrix A. Let me check with n=1:A^1 should be A. Plugging n=1 into the formula:3*5^{0} = 3*1 = 3, 2*5^{0} = 2*1 = 2. So the matrix is [[3,3],[2,2]], which is correct. Good.Check n=2:A^2 = A * A:Original A:3 32 2Multiplying:First row: 3*3 + 3*2 = 9 + 6 = 15; 3*3 + 3*2 = 15Second row: 2*3 + 2*2 = 6 + 4 = 10; 2*3 + 2*2 = 10So A^2 = [[15,15],[10,10]]Using the formula: 3*5^{1} = 15, 2*5^{1}=10. So same result. Perfect.Therefore, the general formula is A^n = 5^{n-1} * [[3,3],[2,2]], which simplifies to [[3*5^{n-1}, 3*5^{n-1}], [2*5^{n-1}, 2*5^{n-1}}].So that's part (a). I think that's solid.Now part (b): Solve the equation X^{2015} = A, where X is a 2x2 matrix.So we need to find a matrix X such that when raised to the 2015th power, we get A. Since we have A = [[3,3],[2,2]], and from part (a) we know that A = 5^{0} * [[3,3],[2,2]] (since n=1). And A^n is 5^{n-1} * A.Wait, so if we have X^{2015} = A, then perhaps X is related to the 2015th root of A.But how do we take roots of matrices? One approach is if A is diagonalizable, then we can take roots by diagonalizing.From part (a), A is diagonalizable as A = P D P^{-1}, where D is diagonal with entries 0 and 5. So, if we want X such that X^{2015} = A, then perhaps X can be written as P D^{1/2015} P^{-1}, but we have to be careful with the eigenvalues.But since D has entries 0 and 5, taking the 2015th root of 0 is still 0, and the 2015th root of 5 is 5^{1/2015}. So, if we set D' = [[0, 0], [0, 5^{1/2015}]], then (D')^{2015} = D = [[0, 0], [0, 5]]. Wait, but A = P D P^{-1} where D is [[0,0],[0,5]]. So then, if X = P D' P^{-1}, then X^{2015} = P (D')^{2015} P^{-1} = P D P^{-1} = A. That works.Therefore, X can be expressed as P D' P^{-1}, where D' is [[0,0],[0,5^{1/2015}]].But let's compute that.First, D' = [[0, 0], [0, 5^{1/2015}]]Then, X = P D' P^{-1}We already have P and P^{-1} from part (a):P = [[1, 3], [-1, 2]]P^{-1} = [[2/5, -3/5], [1/5, 1/5]]So let's compute P D':First, compute P D':[[1*0 + 3*0, 1*0 + 3*5^{1/2015}], [-1*0 + 2*0, -1*0 + 2*5^{1/2015}]] = [[0, 3*5^{1/2015}], [0, 2*5^{1/2015}]]Then multiply by P^{-1}:X = [[0, 3*5^{1/2015}], [0, 2*5^{1/2015}]] * [[2/5, -3/5], [1/5, 1/5]]Compute each entry:First row first column: 0*(2/5) + 3*5^{1/2015}*(1/5) = (3/5)*5^{1/2015} = 3*5^{(1/2015 - 1)} = 3*5^{-2014/2015}Wait, but let's compute it step by step:3*5^{1/2015}*(1/5) = 3/5 *5^{1/2015} = 3*5^{-1 + 1/2015} = 3*5^{(-2015 +1)/2015} = 3*5^{-2014/2015}Similarly for the other entries.First row, first column: 0 + 3*5^{1/2015}/5 = 3*5^{(1/2015 -1)} = 3*5^{-2014/2015}First row, second column: 0*(-3/5) + 3*5^{1/2015}*(1/5) = same as above: 3*5^{-2014/2015}Second row, first column: 0*(2/5) + 2*5^{1/2015}/5 = 2*5^{-2014/2015}Second row, second column: 0*(-3/5) + 2*5^{1/2015}*(1/5) = same as above: 2*5^{-2014/2015}Therefore, X = [[3*5^{-2014/2015}, 3*5^{-2014/2015}], [2*5^{-2014/2015}, 2*5^{-2014/2015}}]But 5^{-2014/2015} can be written as 5^{(1 - 2015)/2015} = 5^{1/2015 -1} = 5^{-1 + 1/2015} = (1/5) *5^{1/2015}Alternatively, we can write it as (5^{1/2015}) /5.But perhaps it's better to leave it as 5^{-2014/2015}, which is the same as (5^{1/2015})^{-2014} = 5^{-2014/2015}But perhaps we can write 5^{-2014/2015} = 5^{( -2014)/2015} = 5^{- (2014/2015)}.Alternatively, note that 5^{-2014/2015} = (5^{2015})^{-2014/2015} = 5^{-2014} ? Wait, no.Wait, 5^{-2014/2015} is just 1 / 5^{2014/2015}But maybe there's a simpler way to express it. Alternatively, since 2014 = 2015 -1, so:5^{-2014/2015} =5^{ - (2015 -1)/2015 } =5^{ -1 + 1/2015 }= 5^{-1} *5^{1/2015}= (1/5)*5^{1/2015}Ah, yes, so 5^{-2014/2015}= (5^{1/2015})/5Therefore, X can be written as:(5^{1/2015}/5) * [[3,3],[2,2]] = (5^{1/2015}/5) * ABut wait, A is [[3,3],[2,2]], so:X = (5^{1/2015}/5) * A = 5^{(1/2015 -1)} * A = 5^{-2014/2015} * AWhich matches our previous result.But since scalar multiplication commutes with matrix multiplication, we can write X as a scalar times A. Let's check if this works.Suppose X = c * A, where c is a scalar. Then X^{2015} = c^{2015} * A^{2015}But from part (a), A^{2015} = 5^{2014} * ATherefore, X^{2015} = c^{2015} *5^{2014} * AWe need this to equal A, so:c^{2015} *5^{2014} * A = ASince A is non-zero, we can equate the scalars:c^{2015} *5^{2014} =1 => c^{2015} =5^{-2014} => c=5^{-2014/2015}Which is exactly what we have. Therefore, X =5^{-2014/2015} * ATherefore, X can be written as:X= [[3*5^{-2014/2015}, 3*5^{-2014/2015}], [2*5^{-2014/2015}, 2*5^{-2014/2015}}]Alternatively, factor out the scalar:X=5^{-2014/2015} * [[3,3],[2,2]]Which is the same as (1/5^{2014/2015}) * AAlternatively, since 5^{-2014/2015} is the same as (5^{1/2015})^{-2014}, but perhaps leaving it as 5^{-2014/2015} is simplest.Alternatively, in radical form, 5^{-2014/2015} = 1 / (5^{2014})^{1/2015} = 1 / 5^{2014/2015}But 2014/2015 can't be simplified, so perhaps that's the way to write it.Therefore, the solution is X=5^{-2014/2015} * ABut let's verify this. Let's compute X^{2015}:If X= c * A where c=5^{-2014/2015}, then X^{2015}= c^{2015} * A^{2015}But A^{2015}=5^{2014}*A as per part (a).Therefore, X^{2015}= (5^{-2014/2015})^{2015} *5^{2014}*A=5^{-2014} *5^{2014}*A=1*A=APerfect, that works.But is this the only solution? Hmm. The equation X^{2015}=A. In scalar cases, roots can have multiple solutions, but with matrices, it's more complicated. However, since A is diagonalizable and we've found a diagonalizable solution via its eigenvalues, and since 0 is an eigenvalue, but raising to a power would keep 0 as 0, but 5 is positive real, so the real 2015th root is unique. Therefore, perhaps this is the only real solution. But we need to confirm if there are other possible matrices X that satisfy X^{2015}=A.However, given that A has eigenvalues 0 and 5, any matrix X that satisfies X^{2015}=A must have eigenvalues that are 0 and 5^{1/2015}, since eigenvalues of X would need to satisfy λ^{2015}=0 or 5. So λ=0 or λ=5^{1/2015}. Therefore, if X is diagonalizable, then it must be similar to the diagonal matrix with entries 0 and 5^{1/2015}. However, since A itself is diagonalizable, and we have constructed such an X via its diagonalization, then this should be the only solution if we require X to be diagonalizable. But if we allow non-diagonalizable solutions, maybe there are others? For example, if X is a Jordan block, but since A is diagonalizable, any X that is a root of A must also be diagonalizable, because if X had a Jordan block, then X^{2015} would also have a Jordan block, but A is diagonal, hence diagonalizable. Therefore, X must be diagonalizable. Hence, the solution we found is the only one.Therefore, the solution is X=5^{-2014/2015} * A.Alternatively, writing in terms of the matrix entries:X= [[3*5^{-2014/2015}, 3*5^{-2014/2015}], [2*5^{-2014/2015}, 2*5^{-2014/2015}}]But we can write 5^{-2014/2015} as 5^{(1 - 2015)/2015} = 5^{1/2015 -1} = 5^{-1 + 1/2015} which might not simplify further.Alternatively, expressing as (1/5) *5^{1/2015}, so:X= (1/5) *5^{1/2015} * [[3,3],[2,2]]But this is equivalent.Alternatively, we can write 5^{1/2015} as the 2015th root of 5, so:X= frac{1}{5} cdot sqrt[2015]{5} cdot ABut in terms of exact form, using exponents is probably preferable.Alternatively, note that 5^{-2014/2015} =5^{-1 +1/2015}=5^{-1} cdot5^{1/2015}= (5^{1/2015})/5, so:X= (5^{1/2015}/5) * ASo depending on how we want to present it, but all forms are equivalent.Therefore, the solution is X equals A multiplied by 5^{-2014/2015}, which can be written as above.So, summarizing:For part (a), A^n = 5^{n-1} * [[3,3],[2,2]]For part (b), X = 5^{-2014/2015} * [[3,3],[2,2]]But let's check dimensions and ensure everything is okay. Since X is a 2x2 matrix, and we're scaling A by a scalar, that's fine. Also, verifying with n=2015 in part (a), A^{2015}=5^{2014}*A, then X^{2015}= (5^{-2014/2015} * A)^{2015}=5^{-2014} * A^{2015}=5^{-2014}*5^{2014}*A= A. Correct.Therefore, the solutions are as above.Final Answera) (boxed{A^{n} = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}})b) (boxed{X = 5^{-frac{2014}{2015}} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}})</think>To solve part (a), we start by finding the eigenvalues and eigenvectors of the matrix ( A ). The characteristic polynomial is calculated as follows:[det(A - lambda I) = begin{vmatrix} 3 - lambda & 3 2 & 2 - lambda end{vmatrix} = (3 - lambda)(2 - lambda) - 6 = lambda^2 - 5lambda]The eigenvalues are (lambda_1 = 0) and (lambda_2 = 5). The corresponding eigenvectors are found to be ([1, -1]^T) for (lambda_1 = 0) and ([3, 2]^T) for (lambda_2 = 5). Using these eigenvectors, we form the matrix ( P ) and its inverse ( P^{-1} ):[P = begin{pmatrix} 1 & 3 -1 & 2 end{pmatrix}, quad P^{-1} = frac{1}{5} begin{pmatrix} 2 & -3 1 & 1 end{pmatrix}]The matrix ( A ) can be diagonalized as ( A = PDP^{-1} ), where ( D ) is the diagonal matrix of eigenvalues:[D = begin{pmatrix} 0 & 0 0 & 5 end{pmatrix}]Calculating ( A^n ):[A^n = PD^nP^{-1} = P begin{pmatrix} 0 & 0 0 & 5^n end{pmatrix} P^{-1}]After performing the matrix multiplications, we find:[A^n = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}]For part (b), we need to find a matrix ( X ) such that ( X^{2015} = A ). Using the diagonalization of ( A ), we consider the matrix ( D' ) with entries ( 0 ) and ( 5^{1/2015} ):[D' = begin{pmatrix} 0 & 0 0 & 5^{1/2015} end{pmatrix}]The matrix ( X ) is then given by ( X = P D' P^{-1} ). After performing the matrix multiplications and simplifying, we find:[X = 5^{-2014/2015} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}] Final Answera) (boxed{A^{n} = 5^{n-1} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}})b) (boxed{X = 5^{-frac{2014}{2015}} begin{pmatrix} 3 & 3 2 & 2 end{pmatrix}})]

🔑:## SUBJECT Ia) A^{2}=5 A(1 text{p})A^{n}=5^{n-1} cdot A(0.5 text{p})Completion by induction (0.5 text{p})b) operatorname{det} X=0(1 text{p})begin{aligned}& X^{2}=t cdot X, t in square, t=operatorname{Tr}(X)(1 text{p}) & X^{2015}=t^{2014} cdot X Rightarrow X=frac{1}{t^{2014}} cdot A(1 text{p}) & X^{2015}=frac{1}{t^{2014 cdot 2015}} cdot 5^{2014} cdot A Rightarrow t^{2014}=frac{1}{sqrt[2015]{5^{2014}}}(1 text{p}) & X=frac{1}{sqrt[2015]{5^{2014}}} cdot A(1 text{p})end{aligned}

🧠:[<think>Alright, so I have this problem here: Given the digits 1, 2, 3, and 4, find the probability that a two-digit number formed by selecting two different digits is divisible by 3. Hmm, okay. Let me think through this step by step.First, probability problems usually involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, I need to find how many two-digit numbers can be formed from those digits without repetition, and then determine how many of those are divisible by 3. Then, the probability would just be the ratio of those two numbers.Let me start with the total number of possible two-digit numbers. The digits available are 1, 2, 3, 4. Since we're forming two-digit numbers, the first digit can't be zero, but none of these digits are zero anyway, so that's not a concern here. For the first digit, we can choose any of the 4 digits. Once we've chosen the first digit, we can't use it again for the second digit, so there are 3 remaining digits. Therefore, the total number of two-digit numbers possible is 4 * 3 = 12. Let me list them out to verify:12, 13, 14,21, 23, 24,31, 32, 34,41, 42, 43.Yep, that's 12 numbers. So total outcomes = 12.Now, the next part is figuring out how many of these numbers are divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. So, for each two-digit number, I need to check if the sum of the two digits is a multiple of 3.Let me go through each number:12: 1 + 2 = 3 → divisible by 3.13: 1 + 3 = 4 → not divisible by 3.14: 1 + 4 = 5 → not divisible by 3.21: 2 + 1 = 3 → divisible by 3.23: 2 + 3 = 5 → not divisible by 3.24: 2 + 4 = 6 → divisible by 3.31: 3 + 1 = 4 → not divisible by 3.32: 3 + 2 = 5 → not divisible by 3.34: 3 + 4 = 7 → not divisible by 3.41: 4 + 1 = 5 → not divisible by 3.42: 4 + 2 = 6 → divisible by 3.43: 4 + 3 = 7 → not divisible by 3.So, the numbers divisible by 3 are: 12, 21, 24, 42. That's 4 numbers. So favorable outcomes = 4.Therefore, the probability is 4/12. Simplifying that, divide numerator and denominator by 4: 1/3.Wait, let me double-check. 4 divided by 12 is indeed 1/3. But just to make sure I didn't miss any numbers. Let me go through the list again:12 (3), 13 (4), 14 (5),21 (3), 23 (5), 24 (6),31 (4), 32 (5), 34 (7),41 (5), 42 (6), 43 (7).So, the sums are: 3,4,5,3,5,6,4,5,7,5,6,7. The sums divisible by 3 are 3,3,6,6. Therefore, four numbers. So yes, 4/12 = 1/3. Hmm, so probability is 1/3.But let me think if there's another way to approach this without listing all numbers, in case maybe I made a mistake in counting.Another approach: instead of listing all possible numbers, think about the possible pairs of digits. Since the order matters here (12 is different from 21), permutations are considered. So total permutations are 4P2 = 12, as we saw before.For divisibility by 3, the sum of the digits must be divisible by 3. So, possible sums when adding two different digits from 1,2,3,4:Possible pairs (regardless of order):1+2=3,1+3=4,1+4=5,2+3=5,2+4=6,3+4=7.So, the possible sums are 3,4,5,5,6,7. So, the sums that are divisible by 3 are 3 and 6. Each of these sums corresponds to certain pairs:Sum=3: (1,2)Sum=6: (2,4)Now, each pair can be arranged in two different ways (since we're forming two-digit numbers). For example, (1,2) can be 12 or 21, and (2,4) can be 24 or 42. So, for each of these pairs, there are 2 numbers. So, number of favorable outcomes would be 2 pairs * 2 arrangements = 4 numbers, which matches what I found before.Therefore, the probability is 4/12 = 1/3. So, the answer is 1/3.Alternatively, maybe using combinations. Wait, since the order matters here, but if we think in terms of combinations first, where order doesn't matter, then convert to permutations.Total number of combinations is 4C2 = 6. Each combination can be arranged in 2 ways. So, 6 combinations * 2 = 12 permutations.From the combinations, how many have sums divisible by 3? As above, the combinations with sum 3 and 6. So (1,2) and (2,4). So 2 combinations. Each gives 2 permutations, so 4 numbers. Therefore, same result.So, 2 out of 6 combinations, which is 1/3. But since each combination counts for two permutations, the probability is still 2/6 = 1/3. Hence, same answer.So, regardless of whether I count permutations directly or use combinations and adjust for order, the probability is 1/3.Therefore, I think the answer is 1/3.Final AnswerThe probability is boxed{dfrac{1}{3}}.</think>Given the digits 1, 2, 3, and 4, we need to find the probability that a two-digit number formed by selecting two different digits is divisible by 3.First, we calculate the total number of possible two-digit numbers. Since we have 4 digits and each digit can be used only once, the total number of permutations is:[4 times 3 = 12]Next, we determine how many of these numbers are divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. We list all possible pairs of digits and check their sums:- The pairs (1, 2) and (2, 4) have sums 3 and 6 respectively, which are divisible by 3.Each of these pairs can form two different two-digit numbers:- The pair (1, 2) forms 12 and 21.- The pair (2, 4) forms 24 and 42.Thus, there are 4 favorable numbers: 12, 21, 24, and 42.The probability is the ratio of favorable outcomes to the total number of outcomes:[frac{4}{12} = frac{1}{3}]Therefore, the probability that a two-digit number formed by selecting two different digits from 1, 2, 3, and 4 is divisible by 3 is boxed{dfrac{1}{3}}.,<think>Okay, so I need to find the probability that a two-digit number formed by selecting two different digits from 1, 2, 3, and 4 is divisible by 3. Hmm, let's start by understanding the problem.First, the digits available are 1, 2, 3, and 4. We have to form two-digit numbers using two different digits. Then, among all possible such numbers, we need to figure out how many are divisible by 3. The probability will be the number of favorable outcomes (those divisible by 3) divided by the total number of possible outcomes.Let me break this down step by step.Step 1: Total number of two-digit numbers possible.Since we are selecting two different digits from 4 digits, the first digit can be any of the 4 digits, and the second digit can be any of the remaining 3 digits. So, that would be 4 * 3 = 12 possible two-digit numbers.Wait, let me verify that. For example, if the first digit is 1, the second can be 2, 3, or 4. Similarly, if the first is 2, the second can be 1, 3, or 4. So yes, each of the 4 digits has 3 possibilities, leading to 12 numbers. So total outcomes are 12.Step 2: Numbers divisible by 3.Now, how do I check if a number is divisible by 3? The rule is that if the sum of the digits is divisible by 3, then the number itself is divisible by 3. So, for example, 12: 1 + 2 = 3, which is divisible by 3. Therefore, 12 is divisible by 3. Similarly, 21: 2 + 1 = 3, also divisible by 3. Okay, so the strategy is to look at all possible two-digit numbers and check which ones have digits that sum to a multiple of 3.Alternatively, maybe there's a smarter way. Instead of listing all numbers, perhaps find all pairs of digits where the sum is divisible by 3, and then count how many two-digit numbers those pairs can form.Let me try both methods to cross-check.Method 1: Listing all possible two-digit numbers and checking each.The digits are 1, 2, 3, 4. So possible two-digit numbers (with distinct digits) are:12, 13, 14,21, 23, 24,31, 32, 34,41, 42, 43.That's 12 numbers. Let me list them again to make sure I didn't miss any:Starting with 1: 12, 13, 14.Starting with 2: 21, 23, 24.Starting with 3: 31, 32, 34.Starting with 4: 41, 42, 43.Yes, 12 numbers total. Now let's check which are divisible by 3.12: 1 + 2 = 3 → Divisible by 3.13: 1 + 3 = 4 → Not divisible.14: 1 + 4 = 5 → Not divisible.21: 2 + 1 = 3 → Divisible.23: 2 + 3 = 5 → Not divisible.24: 2 + 4 = 6 → Divisible.31: 3 + 1 = 4 → Not divisible.32: 3 + 2 = 5 → Not divisible.34: 3 + 4 = 7 → Not divisible.41: 4 + 1 = 5 → Not divisible.42: 4 + 2 = 6 → Divisible.43: 4 + 3 = 7 → Not divisible.So the numbers divisible by 3 are: 12, 21, 24, 42. That's 4 numbers.Wait, let me count again. 12, 21, 24, 42. Yes, four numbers.So there are 4 favorable outcomes. Total outcomes are 12. Therefore, the probability is 4/12, which simplifies to 1/3. So the probability is 1/3.But let me try the other method to confirm.Method 2: Finding pairs of digits where the sum is divisible by 3.Possible pairs (since order matters in two-digit numbers, but the sum is the same regardless of order). Wait, actually, if we consider pairs where the sum is divisible by 3, each such pair can form two different numbers (except maybe if the digits are same, but here digits are different). But in our case, since the two digits are different, each pair corresponds to two numbers. Wait, but let's see.First, let's list all possible pairs (without considering order) and check their sums.The possible pairs from 1,2,3,4 are:1 and 2: sum 3 → divisible by 3.1 and 3: sum 4 → not divisible.1 and 4: sum 5 → not divisible.2 and 3: sum 5 → not divisible.2 and 4: sum 6 → divisible by 3.3 and 4: sum 7 → not divisible.So, the pairs with sum divisible by 3 are (1,2) and (2,4). Wait, that's two pairs. Each pair can form two numbers, right? For example, (1,2) can form 12 and 21. Similarly, (2,4) can form 24 and 42. So that's 2 pairs × 2 numbers each = 4 numbers. Which matches the previous method.Therefore, the number of favorable outcomes is 4. So probability is 4/12 = 1/3.Wait, but hold on. When I listed the pairs, I considered unordered pairs. But each unordered pair corresponds to two ordered pairs (two numbers). So, since there are two unordered pairs that sum to a multiple of 3, each gives two numbers. So 2×2=4 numbers. Therefore, 4 favorable numbers.Therefore, same as before. So probability 4/12=1/3.But wait, hold on. Let me check if there are more pairs. The possible pairs:1-2, 1-3, 1-4, 2-3, 2-4, 3-4. So 6 unordered pairs. But when I calculated the sums, only two of them sum to 3 or 6. 1+2=3, 2+4=6. So those two pairs. So yes, two unordered pairs, each leading to two numbers. So 4 numbers.Alternatively, if there was a pair where the sum is 9, which is also divisible by 3, but in our case, the maximum sum is 3+4=7, so no. So only sums 3 and 6. So two pairs.Therefore, the two methods agree.Therefore, the probability is 1/3.Wait, but just to be thorough, let's check the two-digit numbers again.12: Divisible by 3? 12 ÷ 3 = 4. Yes.21: 21 ÷ 3 = 7. Yes.24: 24 ÷ 3 = 8. Yes.42: 42 ÷ 3 = 14. Yes.So four numbers. So four out of twelve. 4/12 reduces to 1/3. So that's the probability.Alternatively, maybe another way: The total possible numbers are 12. For divisibility by 3, since each number is equally likely (assuming random selection of digits), the probability should be equal to the number of favorable pairs divided by total pairs. Since each pair (a,b) is equally likely, and there are 4 favorable numbers.Alternatively, if we think of it as forming a two-digit number by randomly selecting two different digits and arranging them, then the probability is 4/12=1/3.But let me check once again to make sure I didn't miss any numbers. Let's list all 12 numbers again:12,13,14,21,23,24,31,32,34,41,42,43.Check divisibility by 3:12: 1+2=3 ✔️13: 1+3=4 ❌14:1+4=5 ❌21:2+1=3 ✔️23:2+3=5 ❌24:2+4=6 ✔️31:3+1=4 ❌32:3+2=5 ❌34:3+4=7 ❌41:4+1=5 ❌42:4+2=6 ✔️43:4+3=7 ❌So exactly four numbers: 12,21,24,42. That's correct. Therefore, 4 out of 12, which simplifies to 1/3. So the probability is 1/3.Wait a second, but is there a way to calculate this without listing all numbers? For example, using combinations or permutations?Let me think. Since we have digits 1,2,3,4. We need two-digit numbers, so permutations of 2 digits out of 4. The number of permutations is 4P2 = 4*3=12, which we already know.Now, to find how many permutations result in a number divisible by 3. As before, the key is the sum of the digits. So for each permutation (a,b), check if a + b is divisible by 3.Alternatively, think of it as combinations where the sum is divisible by 3, then multiply by 2 (since each combination can be arranged in two ways). So if there are C combinations, then there are 2C permutations.From earlier, we saw that the combinations (unordered pairs) that sum to a multiple of 3 are (1,2) and (2,4). So C=2. Therefore, number of permutations is 2*2=4. Therefore, 4/12=1/3. So same result.Alternatively, another way: For a two-digit number, the probability that the sum of its digits is divisible by 3. How many pairs (a,b) with a ≠ b satisfy a + b ≡ 0 mod 3.The possible digits: 1,2,3,4.So each digit has a residue modulo 3:1 ≡ 1 mod32 ≡ 2 mod33 ≡ 0 mod34 ≡ 1 mod3So the residues are [1,2,0,1].We need two different digits with residues that add up to 0 mod3. So possible combinations:0 + 0 ≡0, but we don't have two 0s.1 + 2 ≡0, since 1+2=3≡0 mod3.2 +1≡0.1 +2≡0.Similarly, 0 +0≡0, but only one 0 (digit 3). So, pairs where residues are (1,2), (2,1), or (0,0). But since we can't have two 0s, only (1,2) or (2,1).But since residues are:Digits:1:1, 2:2, 3:0, 4:1.So possible pairs:Looking for pairs where one is 1 mod3 and the other is 2 mod3, or both 0 mod3.But since only digit 3 is 0 mod3, and we can't pair it with itself (digits must be different). So, pairs of residues (1,2) or (2,1). Each such pair will sum to 0 mod3.So how many such pairs?First, count the number of digits with residue 1: digits 1 and 4. So two digits.Number of digits with residue 2: digit 2. Only one digit.Number of digits with residue 0: digit 3. Only one digit.So pairs with residues 1 and 2: 2 (from residue 1) *1 (from residue 2) = 2 pairs. Each of these pairs can form two numbers (since order matters). So 2*2=4 numbers.Additionally, pairs with residue 0 and 0: but we only have one 0, so no such pairs.Alternatively, if we had two 0s, but we don't. So total favorable pairs (unordered) are 2. Each gives two numbers, so 4 numbers.Therefore, the same as before. So the probability is 4/12=1/3.Therefore, regardless of the method, the probability is 1/3.I think that's thorough. Let me check if there's any chance I missed a number. Wait, for example, 33 is not allowed because digits must be different. Similarly, 11,22, etc., are invalid. So in our list, all numbers are two different digits. So all 12 numbers are valid and accounted for.Yes, the four numbers that are divisible by 3 are 12,21,24,42. So 4 out of 12. Therefore, 1/3.So I think the answer is 1/3.Final AnswerThe probability is boxed{dfrac{1}{3}}.</think>Given the digits 1, 2, 3, and 4, we need to find the probability that a two-digit number formed by selecting two different digits is divisible by 3.Step 1: Calculate the total number of two-digit numbers possible.We can form two-digit numbers by selecting the first digit in 4 ways and the second digit in 3 ways (since the digits must be different). This gives:[4 times 3 = 12 text{ possible two-digit numbers}]Step 2: Determine the numbers divisible by 3.A number is divisible by 3 if the sum of its digits is divisible by 3. We list all possible two-digit numbers and check their divisibility:- 12: (1 + 2 = 3) (divisible by 3)- 21: (2 + 1 = 3) (divisible by 3)- 24: (2 + 4 = 6) (divisible by 3)- 42: (4 + 2 = 6) (divisible by 3)Other numbers (13, 14, 23, 31, 32, 34, 41, 43) do not have digit sums divisible by 3.Step 3: Count the favorable outcomes.There are 4 numbers (12, 21, 24, 42) that are divisible by 3.Step 4: Calculate the probability.The probability is the number of favorable outcomes divided by the total number of possible outcomes:[frac{4}{12} = frac{1}{3}]Thus, the probability that a two-digit number formed by selecting two different digits from 1, 2, 3, and 4 is divisible by 3 is (boxed{dfrac{1}{3}}).]

🔑:First, we list all possible two-digit numbers that can be formed by selecting two different digits from 1, 2, 3, and 4. There are a total of 12 such numbers: 12, 21, 13, 31, 14, 41, 23, 32, 24, 42, 34, and 43.Next, we determine which of these numbers are divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. Using this rule, we find that the following four numbers are divisible by 3: 12, 21, 24, and 42.Therefore, the probability that a two-digit number formed by selecting two different digits from 1, 2, 3, and 4 is divisible by 3 is given by:[ P = frac{text{Number of two-digit numbers divisible by 3}}{text{Total number of two-digit numbers}} = frac{4}{12} = boxed{frac{1}{3}} ]

🧠:[<think>Okay, let's see. I have this problem about a triangle ABC where A and B are two interior angles. There's a vector given as (overrightarrow{a} = sqrt{2}cosfrac{A+B}{2}overrightarrow{i} + sinfrac{A-B}{2}overrightarrow{j}), and its magnitude is (frac{sqrt{6}}{2}). The questions are about whether (tan A cdot tan B) is a constant and finding the maximum value of (tan C) along with the shape of the triangle then.First, for part (1), I need to check if (tan A cdot tan B) is constant. Since A, B, and C are angles of a triangle, we know that (A + B + C = pi). So, (C = pi - (A + B)). Maybe that relation will be useful here.Given the vector (overrightarrow{a}), its magnitude is given by the square root of the sum of the squares of its components. Let me compute (|overrightarrow{a}|^2) first because it's given as (frac{sqrt{6}}{2}), so squaring both sides will simplify things.Calculating (|overrightarrow{a}|^2):[|overrightarrow{a}|^2 = left( sqrt{2}cosfrac{A+B}{2} right)^2 + left( sinfrac{A-B}{2} right)^2][= 2cos^2left(frac{A+B}{2}right) + sin^2left(frac{A-B}{2}right)]This is equal to (left(frac{sqrt{6}}{2}right)^2 = frac{6}{4} = frac{3}{2}).So,[2cos^2left(frac{A+B}{2}right) + sin^2left(frac{A-B}{2}right) = frac{3}{2}]Hmm, let's denote (theta = frac{A+B}{2}) and (phi = frac{A-B}{2}). Then, the equation becomes:[2cos^2theta + sin^2phi = frac{3}{2}]But since (A + B = 2theta) and (A - B = 2phi), we can express (A = theta + phi) and (B = theta - phi). Also, since (A + B + C = pi), then (C = pi - 2theta).But maybe instead of substituting variables, I can use trigonometric identities to simplify the original equation.First, recall that (cos^2 x = frac{1 + cos 2x}{2}), so:[2 cdot frac{1 + cos(A + B)}{2} + sin^2left(frac{A - B}{2}right) = frac{3}{2}]Simplify:[1 + cos(A + B) + sin^2left(frac{A - B}{2}right) = frac{3}{2}]Subtract 1:[cos(A + B) + sin^2left(frac{A - B}{2}right) = frac{1}{2}]Since (A + B = pi - C), which is (pi - C), so (cos(A + B) = cos(pi - C) = -cos C). Therefore:[-cos C + sin^2left(frac{A - B}{2}right) = frac{1}{2}]So:[sin^2left(frac{A - B}{2}right) = frac{1}{2} + cos C]But maybe we can express this in terms of angles A and B. Let's see. Alternatively, perhaps express in terms of A and B directly. Since (A + B = pi - C), so maybe substituting that into another equation.Alternatively, we might need to relate (tan A cdot tan B). Let me recall that in a triangle, (tan A + tan B + tan C = tan A tan B tan C). But I don't know if that's helpful here.Wait, maybe we can express (tan A cdot tan B) in terms of angles. Let's see. Since (A + B = pi - C), so:[tan(A + B) = tan(pi - C) = -tan C]But (tan(A + B) = frac{tan A + tan B}{1 - tan A tan B}), so:[frac{tan A + tan B}{1 - tan A tan B} = -tan C]Therefore,[tan A + tan B = -tan C (1 - tan A tan B)]But since (A + B + C = pi), so all angles are positive and less than π, so tan C is positive (since C is less than π and greater than 0). Wait, but here we have (-tan C), which would be negative. Hmm, but (tan A) and (tan B) are positive as well because angles A and B are between 0 and π. So left side is positive, right side is negative? That seems conflicting. Wait, maybe I made a mistake here.Wait, (A + B = pi - C), so (A + B) is greater than 0 and less than π. Therefore, (tan(A + B)) is defined unless (A + B = pi/2), but in that case, it's infinite. Wait, but since (A + B = pi - C), then unless (C = pi/2), (A + B) is (pi/2), but in that case, C is right angle. Otherwise, tan(A + B) is defined. But anyway, proceeding:Since (tan(A + B) = -tan C), and if we let (T = tan A tan B), then:From the identity:[tan(A + B) = frac{tan A + tan B}{1 - T} = -tan C]But we need to find T. However, unless we can find another relation involving T, perhaps from the given condition of the vector's magnitude.Alternatively, perhaps express (sin^2left(frac{A - B}{2}right)) in terms of (cos(A - B)). Using the identity:[sin^2 x = frac{1 - cos 2x}{2}]So,[sin^2left(frac{A - B}{2}right) = frac{1 - cos(A - B)}{2}]Substituting back into the equation:[-cos C + frac{1 - cos(A - B)}{2} = frac{1}{2}]Multiply both sides by 2:[-2cos C + 1 - cos(A - B) = 1]Simplify:[-2cos C - cos(A - B) = 0]Therefore,[cos(A - B) = -2cos C]But (C = pi - (A + B)), so (cos C = cos(pi - (A + B)) = -cos(A + B)). Therefore,[cos(A - B) = -2(-cos(A + B)) = 2cos(A + B)]So,[cos(A - B) = 2cos(A + B)]Now, this is an equation involving angles A and B. Let's expand both sides using trigonometric identities.Recall that:[cos(A - B) = cos A cos B + sin A sin B][cos(A + B) = cos A cos B - sin A sin B]Therefore, substituting into the equation:[cos A cos B + sin A sin B = 2(cos A cos B - sin A sin B)]Expand the right-hand side:[cos A cos B + sin A sin B = 2cos A cos B - 2sin A sin B]Bring all terms to the left-hand side:[cos A cos B + sin A sin B - 2cos A cos B + 2sin A sin B = 0]Combine like terms:[(-cos A cos B) + 3sin A sin B = 0]Multiply both sides by -1:[cos A cos B - 3sin A sin B = 0]Which can be written as:[cos A cos B = 3sin A sin B]Divide both sides by (cos A cos B) (assuming (cos A cos B neq 0), which is valid because in a triangle, angles A and B are between 0 and π, so their cosines can't be zero unless the angle is π/2, but if either A or B is π/2, then C would be less than π/2, but let's check if that's possible later):[1 = 3tan A tan B]Therefore,[tan A tan B = frac{1}{3}]So, part (1): Yes, (tan A cdot tan B) is a constant value, which is (frac{1}{3}).Now, moving to part (2): Find the maximum value of (tan C) and determine the shape of the triangle at this time.First, since we already have (C = pi - (A + B)), and from part (1), we have (tan A tan B = 1/3). Let's express (tan C) in terms of A and B.From the earlier identity:[tan C = tan(pi - (A + B)) = -tan(A + B)]But also, from the identity:[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B}]But since (tan A tan B = 1/3), substitute that:[tan(A + B) = frac{tan A + tan B}{1 - 1/3} = frac{tan A + tan B}{2/3} = frac{3}{2}(tan A + tan B)]Therefore,[tan C = -tan(A + B) = -frac{3}{2}(tan A + tan B)]But since angles A, B, C are all positive and less than π, and C is positive, (tan C) is positive. However, the right-hand side has a negative sign, which seems contradictory. Wait, maybe I made a mistake here.Wait, (tan C = tan(pi - (A + B)) = -tan(A + B)), but since (A + B) is less than π (because C is positive), so (A + B) is between 0 and π. Therefore, (tan(A + B)) can be positive or negative?Wait, no. Since (A + B = pi - C), and (C) is between 0 and π, so (A + B) is between 0 and π as well. So (A + B) is between 0 and π. Therefore, if (A + B < pi/2), then (tan(A + B)) is positive, and if (A + B > pi/2), then (tan(A + B)) is negative because tangent is positive in first and third quadrants. Wait, but (A + B) is between 0 and π, so it's either in the first or second quadrant. So (tan(A + B)) is positive if (A + B) is in first quadrant (0 to π/2), and negative if in second quadrant (π/2 to π). But since (C = pi - (A + B)), when (A + B < pi/2), then (C > pi/2), which would make C an obtuse angle. However, in such a case, (tan C) would be negative, which can't be because C is an angle between 0 and π. Wait, no. Wait, if C is greater than π/2, then (tan C) is negative because tangent is positive in first and third quadrants. But angles in a triangle are between 0 and π, so C is between 0 and π, and if C is between π/2 and π, then (tan C) is negative. However, tangent of an angle in a triangle is usually considered as a positive quantity? Wait, maybe the problem didn't specify, but in reality, (tan C) can be negative if C is obtuse. However, in a triangle, angles are between 0 and π, so C is acute (0 to π/2) or obtuse (π/2 to π). If C is acute, (tan C) is positive; if C is obtuse, (tan C) is negative. But since we are asked for the maximum value of (tan C), perhaps we need to consider both possibilities. But maybe the maximum occurs when (tan C) is positive, so C is acute.Alternatively, perhaps we need to maximize the absolute value, but the problem just says "maximum value", so likely considering positive maximum. Let me check.But first, let's clarify the relation. We have:[tan C = -tan(A + B)]But if (A + B < pi/2), then (tan(A + B)) is positive, so (tan C) is negative. But since (C = pi - (A + B)), if (A + B < pi/2), then (C > pi/2), which is an obtuse angle, so (tan C) is indeed negative. Conversely, if (A + B > pi/2), then (C < pi/2), so (tan C) is positive.But in the equation:[tan C = -frac{3}{2}(tan A + tan B)]Therefore, depending on whether (A + B) is less than or greater than (pi/2), (tan C) is positive or negative.But since we are to find the maximum value of (tan C), we need to consider when (C) is acute (i.e., (C < pi/2)) so that (tan C) is positive, and we can maximize that positive value. Alternatively, if the problem allows considering negative values, but since it's asking for the maximum value, likely the positive maximum.So, assuming (C < pi/2), which implies (A + B > pi/2), and (tan C) is positive. Therefore, we can consider:[tan C = frac{3}{2}(tan A + tan B)]Wait, but according to the previous equation, it's (tan C = -frac{3}{2}(tan A + tan B)). However, if (C) is acute, then (A + B = pi - C > pi/2), so (tan(A + B)) is negative (since (A + B) is in the second quadrant), so (-tan(A + B)) is positive. Therefore, yes, (tan C) is positive.Therefore, (tan C = frac{3}{2}(tan A + tan B)). Wait, but how?Wait, step back:From above,[tan C = -tan(A + B) = -left( frac{tan A + tan B}{1 - tan A tan B} right)]But since (tan A tan B = frac{1}{3}), we substitute:[tan C = -left( frac{tan A + tan B}{1 - frac{1}{3}} right) = -left( frac{tan A + tan B}{frac{2}{3}} right) = -frac{3}{2}(tan A + tan B)]But since (C) is acute, (A + B > pi/2), so (tan(A + B)) is negative (as (A + B) is in the second quadrant), so (-tan(A + B)) is positive, which matches (tan C) being positive. Therefore, the correct expression is:[tan C = frac{3}{2}(tan A + tan B)]Wait, no. Wait, no, the expression is:[tan C = -frac{3}{2}(tan A + tan B)]But since (tan(A + B)) is negative (because (A + B) is in second quadrant), so:[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B} = text{negative}]Therefore, (tan A + tan B) must be negative because the denominator (1 - frac{1}{3} = frac{2}{3}) is positive. But (tan A) and (tan B) are both positive because angles A and B are between 0 and π/2 (since (A + B > pi/2) and both A and B are less than π). Wait, but if (A + B > pi/2), but individually, A or B could be greater than π/2? Wait, no, because in a triangle, the sum of any two angles must be less than π. Wait, A + B = π - C. Since C is positive, A + B < π. So angles A and B are each less than π. But if A + B > π/2, which is the case when C < π/2, then both A and B could be acute or one of them obtuse. Wait, but if A is obtuse (greater than π/2), then B must be less than π/2 because A + B < π. But if A is obtuse, then (tan A) is negative, which contradicts the earlier assumption. Wait, no: in a triangle, angles are between 0 and π, so an angle greater than π/2 (obtuse) would have a negative tangent because tangent is positive in first and third quadrants. Wait, no. Wait, tangent is positive in first (0 to π/2) and third (π to 3π/2) quadrants. But in a triangle, angles are between 0 and π, so angles are in first or second quadrants. In the second quadrant (π/2 to π), tangent is negative. Therefore, if angle A is obtuse, (tan A) is negative. But in our case, angles A and B must be such that (A + B > pi/2) if C is acute. However, if one of them is obtuse, its tangent would be negative, but the other angle would have to compensate.Wait, but this seems conflicting with the earlier result that (tan A tan B = 1/3). If one of them is negative, their product would be negative, but 1/3 is positive. Therefore, both (tan A) and (tan B) must be positive. Therefore, angles A and B must both be acute (i.e., less than π/2). Therefore, even though (A + B > pi/2), both A and B are acute angles, so (tan A) and (tan B) are positive. Therefore, (tan A + tan B) is positive, but since (A + B) is in the second quadrant, (tan(A + B)) is negative, so the expression:[tan C = -tan(A + B) = -(text{negative}) = positive]Therefore, (tan C = frac{3}{2}(tan A + tan B)), but with the negative sign included:Wait, no. Wait, according to the derivation:[tan C = -frac{3}{2}(tan A + tan B)]But since (tan C) is positive (because C is acute), then:[-frac{3}{2}(tan A + tan B) > 0 implies tan A + tan B < 0]But this contradicts the fact that (tan A) and (tan B) are positive. Therefore, there must be a mistake in the sign somewhere.Wait, going back to the original identity:We had:[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B}]But since (A + B = pi - C), so:[tan(pi - C) = -tan C = frac{tan A + tan B}{1 - tan A tan B}]Therefore,[-tan C = frac{tan A + tan B}{1 - tan A tan B}]Hence,[tan C = -frac{tan A + tan B}{1 - tan A tan B}]But since (1 - tan A tan B = 1 - frac{1}{3} = frac{2}{3}), so:[tan C = -frac{tan A + tan B}{frac{2}{3}} = -frac{3}{2}(tan A + tan B)]But as established, if C is acute, then (A + B) is in the second quadrant, so (tan(A + B)) is negative, so (-tan(A + B)) is positive. Therefore, (tan C) is positive, which matches with the left-hand side. However, according to the right-hand side, (tan C = -frac{3}{2}(tan A + tan B)). But since (tan A) and (tan B) are positive, the right-hand side is negative, which contradicts the left-hand side being positive. Therefore, there must be an error in the derivation.Wait, no. Wait, actually, if (A + B = pi - C), then (tan(A + B) = tan(pi - C) = -tan C). Therefore,[tan(A + B) = -tan C implies -tan C = frac{tan A + tan B}{1 - tan A tan B}]Therefore,[tan C = -frac{tan A + tan B}{1 - tan A tan B}]But since (1 - tan A tan B = frac{2}{3}), then:[tan C = -frac{3}{2}(tan A + tan B)]But in a triangle where C is acute, (tan C) is positive. Therefore, the negative sign indicates that (tan A + tan B) is negative. But (tan A) and (tan B) are positive (since angles A and B are acute), so their sum is positive. Hence, this leads to a contradiction. Therefore, there must be a mistake in the reasoning.Wait, let's check the original step. When I wrote:From (tan(A + B) = -tan C), so:[frac{tan A + tan B}{1 - tan A tan B} = -tan C]Therefore,[tan C = -frac{tan A + tan B}{1 - tan A tan B}]But since (1 - tan A tan B = frac{2}{3}), substituting gives:[tan C = -frac{3}{2}(tan A + tan B)]But since C is acute (as we are considering the case where (A + B > pi/2)), (tan C) is positive. Therefore, the right-hand side must also be positive, which implies that (tan A + tan B) is negative. But both A and B are acute angles (since their sum is greater than (pi/2) but each is less than (pi)), so (tan A) and (tan B) are positive. Therefore, their sum is positive, leading to a negative right-hand side. Contradiction. Therefore, there must be an error in my earlier steps.Wait, maybe I need to revisit the equation derived earlier: (cos(A - B) = 2cos(A + B)). Let's check the steps again.Original steps:Given the vector magnitude equation led to:[-cos C + sin^2left( frac{A - B}{2} right) = frac{1}{2}]Then using (sin^2 x = frac{1 - cos 2x}{2}):[-cos C + frac{1 - cos(A - B)}{2} = frac{1}{2}]Multiply by 2:[-2cos C + 1 - cos(A - B) = 1]Simplify:[-2cos C - cos(A - B) = 0 implies cos(A - B) = -2cos C]Then, since (C = pi - (A + B)), (cos C = -cos(A + B)), so:[cos(A - B) = -2(-cos(A + B)) = 2cos(A + B)]Thus, (cos(A - B) = 2cos(A + B)). This seems correct.Expanding both sides:Left side: (cos(A - B) = cos A cos B + sin A sin B)Right side: (2cos(A + B) = 2(cos A cos B - sin A sin B))Setting equal:[cos A cos B + sin A sin B = 2cos A cos B - 2sin A sin B]Bring all terms to left:[-cos A cos B + 3sin A sin B = 0 implies cos A cos B = 3sin A sin B]Divide both sides by (cos A cos B):[1 = 3tan A tan B implies tan A tan B = frac{1}{3}]This part seems correct, so part (1) is indeed a constant.Now, returning to the problem with (tan C). Maybe instead of using the identity involving (tan(A + B)), I can express (tan C) in terms of (tan A) and (tan B) using another approach.Given that (tan A tan B = frac{1}{3}), and (A + B = pi - C), so we can write (A + B = pi - C). Let me denote (A + B = pi - C), and we can use the formula for (tan(A + B)):[tan(A + B) = tan(pi - C) = -tan C]But also,[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B} = frac{tan A + tan B}{1 - frac{1}{3}} = frac{3}{2}(tan A + tan B)]Thus,[-tan C = frac{3}{2}(tan A + tan B)]Therefore,[tan C = -frac{3}{2}(tan A + tan B)]But (tan C) is positive (if C is acute) or negative (if C is obtuse). However, since angles A and B are acute (as established from (tan A tan B = 1/3) and their product is positive, so both positive), their sum (tan A + tan B) is positive. Therefore, (tan C = -frac{3}{2}(tan A + tan B)) is negative. But this contradicts the fact that if C is acute, (tan C) is positive. Therefore, this suggests that C must be obtuse, leading to a negative (tan C), but in that case, the problem asks for the maximum value of (tan C). Since if C is obtuse, (tan C) is negative, but we might need to consider the maximum in terms of absolute value or clarify whether the problem wants the maximum positive value. Alternatively, perhaps the maximum occurs when C is acute, and we can find a maximum positive value.But according to the equation (tan C = -frac{3}{2}(tan A + tan B)), which is negative if C is acute (since right-hand side is negative), but we saw that if C is acute, then (A + B > pi/2), which would make (A + B) in the second quadrant, hence (tan(A + B)) is negative, so (-tan(A + B)) is positive. Therefore, there's a contradiction here. So where is the mistake?Wait, maybe the formula (tan(A + B) = frac{tan A + tan B}{1 - tan A tan B}) is not valid when (A + B) is in the second quadrant. Wait, no. The formula is derived from the identity for tangent of sum, which is valid except when (A + B = pi/2 + kpi). So as long as (A + B neq pi/2), the formula is valid. Therefore, even if (A + B) is in the second quadrant, the formula holds, but (tan(A + B)) is negative. Therefore, we have:[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B} = text{negative}]Therefore, in the equation:[tan C = -tan(A + B) = -(text{negative}) = text{positive}]But according to the formula, (tan C = -frac{3}{2}(tan A + tan B)), which is negative because (tan A + tan B) is positive. This inconsistency suggests that there is a sign error.Wait, let me re-express this step:From:[tan(A + B) = -tan C]But (A + B = pi - C), so (tan(pi - C) = -tan C). Therefore:[tan(A + B) = tan(pi - C) = -tan C]So,[-tan C = frac{tan A + tan B}{1 - tan A tan B}]Therefore,[tan C = -frac{tan A + tan B}{1 - tan A tan B}]But since (1 - tan A tan B = frac{2}{3}), then:[tan C = -frac{3}{2}(tan A + tan B)]However, if (C) is acute, then (tan C) is positive, which implies that (-frac{3}{2}(tan A + tan B)) is positive, so (tan A + tan B) must be negative. But (tan A) and (tan B) are both positive (as angles A and B are acute), so their sum is positive. Contradiction. Therefore, the only way this can hold is if (C) is obtuse, leading to (tan C) negative, and (tan A + tan B) positive, which aligns with the equation.But the problem asks for the maximum value of (tan C). If (C) is obtuse, (tan C) is negative. So the maximum value (the least negative) would occur when (tan C) is closest to zero. Alternatively, if considering absolute value, the maximum |tan C| would be when C approaches π, but that's not possible in a triangle. Alternatively, perhaps there's a miscalculation here.Wait, perhaps we need to express (tan C) in terms of angle C only. Let's think differently.From part (1), we have (tan A tan B = 1/3). Also, since (A + B = pi - C), we can use the formula for (tan(A + B)):[tan(A + B) = tan(pi - C) = -tan C]But also,[tan(A + B) = frac{tan A + tan B}{1 - tan A tan B}]So,[-tan C = frac{tan A + tan B}{1 - 1/3} = frac{3}{2}(tan A + tan B)]Thus,[tan A + tan B = -frac{2}{3} tan C]But we need to relate this to angle C. Let me use the identity that in a triangle:[A + B + C = pi]So,[A + B = pi - C]Let me denote (A = alpha), (B = beta), so (alpha + beta = pi - C), and (tan alpha tan beta = 1/3). We need to express (tan C) in terms of (alpha) and (beta), then find its maximum.Alternatively, since (C = pi - (alpha + beta)), (tan C = tan(pi - (alpha + beta)) = -tan(alpha + beta)). But as before.Alternatively, express (tan C = -tan(alpha + beta) = -frac{tan alpha + tan beta}{1 - tan alpha tan beta} = -frac{tan alpha + tan beta}{1 - 1/3} = -frac{3}{2}(tan alpha + tan beta)). Therefore,[tan C = -frac{3}{2}(tan alpha + tan beta)]But since (alpha + beta = pi - C), and angles (alpha) and (beta) are positive and each less than π.But we need to express (tan C) in terms of C. Let me try.Let me denote (D = alpha - beta). Then, we have:[alpha = frac{(alpha + beta) + (alpha - beta)}{2} = frac{pi - C + D}{2}][beta = frac{(alpha + beta) - (alpha - beta)}{2} = frac{pi - C - D}{2}]But we also have from part (1) that (tan alpha tan beta = 1/3). Let me express (tan alpha tan beta) in terms of C and D.First, using the identity:[tan alpha tan beta = frac{sin alpha sin beta}{cos alpha cos beta} = frac{1}{3}]But from earlier steps, we had (cos alpha cos beta = 3 sin alpha sin beta). So,[frac{sin alpha sin beta}{cos alpha cos beta} = frac{1}{3} implies 3 sin alpha sin beta = cos alpha cos beta]Which is consistent with previous results.Alternatively, maybe we can use the formula for (tan alpha + tan beta):[tan alpha + tan beta = frac{sin alpha}{cos alpha} + frac{sin beta}{cos beta} = frac{sin(alpha + beta)}{cos alpha cos beta}]Because:[sin(alpha + beta) = sin alpha cos beta + cos alpha sin beta]But not sure if that helps.Wait, but we have (sin(alpha + beta) = sin(pi - C) = sin C).So,[tan alpha + tan beta = frac{sin(alpha + beta)}{cos alpha cos beta} = frac{sin C}{cos alpha cos beta}]But from the earlier equation, (cos alpha cos beta = 3 sin alpha sin beta). Let me express (sin alpha sin beta) in terms of something.Alternatively, using the identity:[cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta]But we also have from part (1):[cos(alpha - beta) = 2cos(alpha + beta) = 2cos(pi - C) = -2cos C]Therefore,[-2cos C = cos alpha cos beta + sin alpha sin beta]But we also have (cos alpha cos beta = 3 sin alpha sin beta) from earlier. Substitute into the equation:[-2cos C = 3 sin alpha sin beta + sin alpha sin beta = 4 sin alpha sin beta]Thus,[sin alpha sin beta = -frac{1}{2}cos C]But since (sin alpha sin beta) is a product of sines of two angles between 0 and π, which are positive, so left side is positive. Therefore, the right side must also be positive, which implies (-frac{1}{2}cos C) is positive. Therefore, (cos C) is negative. Hence, angle C is obtuse (since cosine is negative in the second quadrant, which for angles in a triangle, C is between π/2 and π). Therefore, C is obtuse, so (tan C) is negative.Therefore, returning to the expression:[tan C = -frac{3}{2}(tan alpha + tan beta)]But (tan alpha + tan beta = frac{sin C}{cos alpha cos beta}). From earlier, we have:[cos alpha cos beta = 3 sin alpha sin beta]And,[sin alpha sin beta = -frac{1}{2}cos C]Therefore,[cos alpha cos beta = 3 left(-frac{1}{2}cos C right) = -frac{3}{2}cos C]Therefore,[tan alpha + tan beta = frac{sin C}{cos alpha cos beta} = frac{sin C}{ -frac{3}{2}cos C } = -frac{2}{3} tan C]Substitute back into the expression for (tan C):[tan C = -frac{3}{2} left( -frac{2}{3} tan C right ) = -frac{3}{2} cdot left( -frac{2}{3} tan C right ) = tan C]Which simplifies to:[tan C = tan C]This tautology suggests that the equation is consistent but doesn't help in finding the maximum.Alternative approach: Express (tan C) in terms of angle C.Given that C is obtuse (as established), and (A + B = pi - C). From the earlier equation (cos(alpha - beta) = -2cos C).Also, since (alpha) and (beta) are angles of a triangle, we have:[alpha + beta = pi - C]And from (tan alpha tan beta = 1/3).Let me denote (x = alpha) and (y = beta). Then:[x + y = pi - C][tan x tan y = frac{1}{3}]We need to express (tan C) in terms of x and y, or in terms of C.Alternatively, we can use the following identity:Given (x + y = S) and (tan x tan y = k), then (tan x + tan y = frac{sin S}{cos x cos y - k}).But not sure. Alternatively, since (x + y = pi - C), then:[tan(x + y) = tan(pi - C) = -tan C]But,[tan(x + y) = frac{tan x + tan y}{1 - tan x tan y} = frac{tan x + tan y}{1 - frac{1}{3}} = frac{3}{2}(tan x + tan y)]Therefore,[-tan C = frac{3}{2}(tan x + tan y)]So,[tan x + tan y = -frac{2}{3}tan C]But we need to relate this to angle C. Let's use the identity for (sin(x + y)):[sin(x + y) = sin(pi - C) = sin C]Also,[sin(x + y) = sin x cos y + cos x sin y]Divide both sides by (cos x cos y):[frac{sin(x + y)}{cos x cos y} = tan x + tan y]But,[frac{sin C}{cos x cos y} = tan x + tan y = -frac{2}{3}tan C]Therefore,[sin C = -frac{2}{3}tan C cos x cos y]But from earlier, we have (cos x cos y = 3 sin x sin y), and also (sin x sin y = -frac{1}{2}cos C). Substitute:[cos x cos y = 3 left( -frac{1}{2}cos C right ) = -frac{3}{2}cos C]Therefore,[sin C = -frac{2}{3}tan C left( -frac{3}{2}cos C right )]Simplify:Left side: (sin C)Right side: (-frac{2}{3} tan C cdot -frac{3}{2} cos C = tan C cos C = sin C)Therefore, we get:[sin C = sin C]Again, a tautology. Therefore, this approach doesn't help.Alternative method: Express everything in terms of C.We have:From (tan A tan B = frac{1}{3}), and (A + B = pi - C). Let me use variables (A) and (B), and express (tan A tan B = frac{1}{3}).Let me set (A = pi - C - B). Then,[tan(pi - C - B) tan B = frac{1}{3}]But (tan(pi - C - B) = tan(pi - (C + B)) = -tan(C + B)), but this seems complicated.Alternatively, consider writing (A = frac{pi - C + D}{2}) and (B = frac{pi - C - D}{2}), where (D = A - B). Then,[tan A tan B = frac{1}{3}]Express (tan A tan B) in terms of D and C.But this might be messy. Alternatively, express in terms of sine and cosine.From (tan A tan B = frac{sin A sin B}{cos A cos B} = frac{1}{3}), and from earlier:[cos A cos B = 3 sin A sin B]Therefore,[cos A cos B - 3 sin A sin B = 0]Which can be written as:[cos(A + B) + 2sin A sin B = 0]But not sure.Alternatively, recall that:[cos(A + B) = cos A cos B - sin A sin B = 3 sin A sin B - sin A sin B = 2 sin A sin B]But (cos(A + B) = -cos C), so:[-cos C = 2 sin A sin B]Therefore,[sin A sin B = -frac{1}{2}cos C]But (A) and (B) are angles in a triangle, so they are between 0 and π. If C is obtuse, then (cos C) is negative, so the right-hand side is positive, which matches since (sin A sin B) is positive.Also, using the identity:[cos(A - B) = cos A cos B + sin A sin B = 3 sin A sin B + sin A sin B = 4 sin A sin B = 4(-frac{1}{2}cos C) = -2cos C]But earlier we had (cos(A - B) = -2cos C), which is the same as here.Thus, from (cos(A - B) = -2cos C), and since (|cos(A - B)| leq 1), then:[| -2cos C | leq 1 implies | cos C | leq frac{1}{2}]Therefore, (cos C leq frac{1}{2}) or (cos C geq -frac{1}{2}). But since C is obtuse ((pi/2 < C < pi)), (cos C) is negative, so:[-2cos C leq 1 implies cos C geq -frac{1}{2}]Therefore, (-frac{1}{2} leq cos C < 0), which implies that:[frac{pi}{2} < C leq frac{2pi}{3}]Because (cos^{-1}(-1/2) = (2π/3). So C is between π/2 and 2π/3.Therefore, the maximum value of (tan C) would occur when C is as large as possible (approaching 2π/3), but since (tan) is increasing in (π/2, π), but actually, (tan C) is negative and becomes more negative as C approaches π. However, since C is at most 2π/3, the minimal value of (tan C) (most negative) would be at C = 2π/3. But the problem asks for the maximum value of (tan C). Since (tan C) is negative in this interval, the maximum (least negative) would be when C is smallest, i.e., approaching π/2. But as C approaches π/2 from above, (tan C) approaches negative infinity. Wait, no. Wait, (tan(π/2)) is undefined, approaching +infty from the left and -infty from the right. However, in our case, C is greater than π/2 but less than or equal to 2π/3. Therefore, the maximum value of (tan C) (closest to zero, i.e., least negative) occurs at C = 2π/3, where (tan(2π/3) = tan(π - π/3) = -tan(π/3) = -sqrt{3}). So the maximum value is -√3, but since it's negative, if we consider the maximum in terms of algebraic value, it's -√3, but if considering the maximum magnitude, it's √3. However, the problem says "Find the maximum value of (tan C)", so likely considering the algebraic maximum, which would be the least negative value, which is -√3 when C = 2π/3. But this seems counterintuitive.Wait, but earlier, we derived that C ≤ 2π/3. Therefore, the maximum value of (tan C) (closest to zero) occurs at C = 2π/3, where (tan C = -sqrt{3}). However, this is the least negative value. But if we are to find the maximum value (in terms of positive numbers), then it's not applicable since (tan C) is negative. But the problem doesn't specify. Let's check the problem statement again."Find the maximum value of (tan C) and determine the shape of the triangle at this time."Since in mathematics, unless specified otherwise, maximum value refers to the greatest real value, so even if negative, the maximum is the largest (closest to positive infinity). However, in this case, as C approaches π/2 from above, (tan C) approaches negative infinity, so there's no maximum. But we have a constraint that (cos C geq -1/2), which implies C ≤ 2π/3. Therefore, C is in (π/2, 2π/3]. Therefore, (tan C) is in (-∞, -√3]. Therefore, the maximum value of (tan C) is -√3, achieved when C = 2π/3. Hence, the maximum value is -√3, and the triangle is such that C = 2π/3, which makes it an obtuse triangle with angles A and B satisfying (tan A tan B = 1/3). When C = 2π/3, then A + B = π/3. Since (tan A tan B = 1/3), and A + B = π/3, so A and B are acute angles summing to π/3. In this case, the triangle would be an obtuse triangle with angles A, B, and C = 2π/3. The shape would be such that angle C is 120 degrees, and angles A and B are 30 degrees each, but wait, if A + B = π/3 and (tan A tan B = 1/3), are they equal?Let me check. Suppose A = B, then A = B = π/6. Then, (tan A tan B = (tan π/6)^2 = (1/√3)^2 = 1/3). So yes, if A = B = π/6, then C = 2π/3, which satisfies the condition. Therefore, the triangle is an isosceles triangle with angles π/6, π/6, and 2π/3. However, is this the only possibility? Suppose A ≠ B, but A + B = π/3 and (tan A tan B = 1/3). For example, if A = π/12 and B = π/4, then A + B = π/3. Let's compute (tan A tan B = tan(π/12)tan(π/4)). (tan(π/12) = 2 - √3), (tan(π/4) = 1), so their product is 2 - √3 ≈ 0.2679, which is less than 1/3 ≈ 0.3333. Therefore, such angles would not satisfy the condition. Alternatively, suppose A approaches π/3 and B approaches 0. Then, (tan A tan B) approaches 0. So, the maximum occurs when A = B = π/6, leading to (tan A tan B = 1/3), which fits. Therefore, the maximum value of (tan C) is -√3, achieved when C = 2π/3, and the triangle is isosceles with angles 30°, 30°, 120°.But the problem asks for the maximum value of (tan C). Since (tan C = -sqrt{3}) is the least negative value (as C approaches 2π/3 from above), but actually, as C increases from π/2 to 2π/3, (tan C) decreases from -infty to -√3. Therefore, the maximum value (i.e., the largest value) of (tan C) is -√3, achieved when C = 2π/3. However, this is the largest value in the interval (π/2, 2π/3]. Therefore, the answer should be -√3, but since the problem might expect a positive maximum, perhaps there's an error in the analysis.Wait, going back to the equation:We had (cos(A - B) = -2cos C). Since (cos(A - B)) has a range of [-1, 1], and since C is obtuse, (cos C) is in [-1/2, 0). Therefore, -2cos C is in (0, 1]. Therefore, (cos(A - B)) is in (0, 1], which implies that (A - B) is between -π/2 and π/2, but since angles A and B are positive and sum to π - C < π/2 (if C is obtuse), wait no: if C is obtuse, then A + B = π - C < π - π/2 = π/2. Wait, if C is greater than π/2, then A + B is less than π/2. Wait, this contradicts earlier statements. Wait, if C is obtuse (π/2 < C < π), then A + B = π - C, which would be between 0 and π/2. Therefore, angles A and B are both acute and sum to less than π/2. Therefore, A and B are both less than π/2, and their sum is less than π/2. Therefore, A and B are both acute angles less than π/2, and their sum is less than π/2.But in this case, both A and B are less than π/2, so their tangents are positive. Then, according to the earlier equation (tan C = -frac{3}{2}(tan A + tan B)), which is negative. Therefore, (tan C) is negative, which aligns with C being obtuse.But since A + B < π/2, which implies C > π/2. But wait, earlier steps suggested that C is in (π/2, 2π/3], but how?Wait, if A + B = π - C < π/2, then C > π - π/2 = π/2. But also, from the equation (cos(A - B) = -2cos C), and since (cos(A - B) leq 1), then:[-2cos C leq 1 implies cos C geq -1/2 implies C leq 2π/3]Therefore, C is in (π/2, 2π/3].Therefore, the maximum value of (tan C) (i.e., the least negative) occurs at C = 2π/3, where (tan C = tan(2π/3) = -sqrt{3}).Therefore, the maximum value is -√3, and the triangle at this time has angles A = B = π/6, C = 2π/3, making it an isosceles triangle with two angles of 30° and one angle of 120°.However, typically, problems like this might expect the answer in positive terms, but according to the analysis, it's negative. Let me check again.If we consider that the problem asks for the maximum value of (tan C), and C is in (π/2, 2π/3], then (tan C) is in (-∞, -√3]. Therefore, the maximum value is -√3, and the minimum value is negative infinity. But since C cannot be less than π/2 (as C is obtuse here), the maximum value of (tan C) is indeed -√3.However, in the problem statement, they might have a different approach. Let me think alternatively.Suppose we express (tan C) in terms of A and B. Given that (tan A tan B = 1/3) and (A + B < pi/2). Wait, no, if C is obtuse, (A + B = pi - C < pi/2). So A + B is acute, and A and B are both acute angles.Let me use the AM-GM inequality on (tan A + tan B). The sum (tan A + tan B) is minimized when A = B due to the equality condition of AM-GM. Since (tan A tan B = 1/3), the minimum of (tan A + tan B) occurs when A = B, so:If A = B, then (tan A = tan B = sqrt{1/3} = 1/sqrt{3}), so A = B = π/6. Therefore, (tan A + tan B = 2/sqrt{3}). Therefore, (tan C = -frac{3}{2} times frac{2}{sqrt{3}} = -sqrt{3}). This is the minimum of (tan A + tan B), leading to the maximum (least negative) value of (tan C), which is -√3.Therefore, the maximum value of (tan C) is -√3, achieved when A = B = π/6, C = 2π/3, and the triangle is an isosceles triangle with angles 30°, 30°, and 120°.Therefore, the answer to part (2) is the maximum value of (tan C) is -√3, and the triangle is an isosceles triangle with angles 30°, 30°, 120°.But let me confirm if this is correct. If A and B are both π/6, then C = 2π/3, which is 120 degrees. Then, compute the vector (overrightarrow{a}):[overrightarrow{a} = sqrt{2}cosfrac{A+B}{2}overrightarrow{i} + sinfrac{A-B}{2}overrightarrow{j}]Since A = B = π/6, then A + B = π/3, so (frac{A + B}{2} = π/6), and A - B = 0, so (sinfrac{A - B}{2} = 0).Therefore, the vector becomes:[sqrt{2}cos(pi/6)overrightarrow{i} + 0overrightarrow{j} = sqrt{2} times (sqrt{3}/2) = sqrt{6}/2 overrightarrow{i}]Therefore, the magnitude is (sqrt{6}/2), which matches the given condition. Therefore, this is a valid case. Hence, the maximum value of (tan C) is indeed -√3, and the triangle is isosceles with angles π/6, π/6, 2π/3.However, in terms of the problem statement, sometimes maximum of a trigonometric function might refer to the maximum absolute value, but in this case, the problem specifies "maximum value", so the algebraic maximum is -√3.But since the problem is in Chinese, maybe the translation is accurate. Alternatively, the problem might expect the magnitude, but according to mathematical terminology, it should be -√3.Therefore, the answers are:(1) Yes, (tan A cdot tan B = frac{1}{3}).(2) The maximum value of (tan C) is (-sqrt{3}), achieved when the triangle is isosceles with angles 30°, 30°, and 120°.But in some contexts, the problem might expect the magnitude, but unless specified, the answer is -√3. However, in the problem statement, since they ask for the maximum value, and C is an interior angle, which can be obtuse, and the maximum value of (tan C) is indeed -√3. Therefore, the answer is as above.Final Answer(1) (boxed{dfrac{1}{3}})(2) The maximum value of (tan C) is (boxed{-sqrt{3}}), and the triangle is isosceles with angles (30^circ), (30^circ), and (120^circ).</think>Given that A and B are two interior angles of (triangle ABC), and (overrightarrow{a} = sqrt{2}cosfrac{A+B}{2}overrightarrow{i} + sinfrac{A-B}{2}overrightarrow{j}) with (|overrightarrow{a}| = frac{sqrt{6}}{2}):(1) To determine if (tan A cdot tan B) is a constant value, we start by calculating the magnitude of (overrightarrow{a}):[|overrightarrow{a}|^2 = left( sqrt{2}cosfrac{A+B}{2} right)^2 + left( sinfrac{A-B}{2} right)^2 = 2cos^2left(frac{A+B}{2}right) + sin^2left(frac{A-B}{2}right)]Given (|overrightarrow{a}| = frac{sqrt{6}}{2}), we have:[2cos^2left(frac{A+B}{2}right) + sin^2left(frac{A-B}{2}right) = frac{3}{2}]Using trigonometric identities and simplifying, we find:[cos(A - B) = 2cos(A + B)]Expanding both sides using trigonometric identities and simplifying, we get:[cos A cos B = 3sin A sin B]Dividing both sides by (cos A cos B):[1 = 3tan A tan B implies tan A tan B = frac{1}{3}]Thus, (tan A cdot tan B) is a constant value (boxed{frac{1}{3}}).(2) To find the maximum value of (tan C) and determine the shape of the triangle, we use the result from part (1) and the identity (tan C = -frac{3}{2}(tan A + tan B)). Given (tan A tan B = frac{1}{3}), we use the AM-GM inequality to find that the minimum value of (tan A + tan B) occurs when (A = B). This leads to (A = B = frac{pi}{6}) and (C = frac{2pi}{3}).At this point, (tan C = tanleft(frac{2pi}{3}right) = -sqrt{3}). The triangle is isosceles with angles (30^circ), (30^circ), and (120^circ).Thus, the maximum value of (tan C) is (boxed{-sqrt{3}}), and the triangle is isosceles with angles (30^circ), (30^circ), and (120^circ).,<think>Okay, let's try to tackle this problem step by step. First, let me understand what's given and what's being asked.We have triangle ABC, and A and B are two interior angles of this triangle. The vector a is given as:a = √2 cos[(A+B)/2] i + sin[(A-B)/2] jAnd the magnitude of this vector |a| is √6 / 2.There are two parts to the problem:1. Determine if tan A · tan B is a constant value. If yes, find it; if not, explain why.2. Find the maximum value of tan C and determine the shape of the triangle at that time.Starting with part (1). Let's recall that in any triangle, the sum of angles A + B + C = π. So, A + B = π - C. Therefore, (A + B)/2 = (π - C)/2. Also, since A and B are angles in a triangle, they must each be between 0 and π, and their sum must be less than π (since C is positive). But let's see how that plays into the vector's components.Given the vector a has components in i and j directions. The magnitude squared is the sum of the squares of the components. So:|a|² = (√2 cos[(A+B)/2])² + (sin[(A - B)/2])² = (6/4) = 3/2.Calculating the left side:(√2 cos[(A+B)/2])² = 2 cos²[(A+B)/2]And (sin[(A - B)/2])² remains as is. Therefore:2 cos²[(A+B)/2] + sin²[(A - B)/2] = 3/2.Let me write that equation again:2 cos²[(π - C)/2] + sin²[(A - B)/2] = 3/2.But since A + B = π - C, we can express everything in terms of C and (A - B). Maybe that's a way to relate the angles. But we need to find tan A · tan B. Let's see if we can express tan A tan B in terms of other trigonometric functions.First, let me recall that in a triangle, tan A + tan B + tan C = tan A tan B tan C. But I'm not sure if that's helpful here. Alternatively, since A + B = π - C, so tan(A + B) = tan(π - C) = -tan C. But tan(A + B) = [tan A + tan B] / [1 - tan A tan B]. Therefore:[tan A + tan B] / [1 - tan A tan B] = -tan CBut tan C is positive since C is an acute angle (assuming the triangle is acute, but maybe not necessarily). But perhaps we can express tan C in terms of tan A and tan B. Let's note that:tan C = tan(π - A - B) = -tan(A + B) = [tan A + tan B] / [tan A tan B - 1]But this might complicate things. Alternatively, maybe we can use the equation given from the vector's magnitude to find a relationship between A and B.Let me focus back on the equation:2 cos²[(A + B)/2] + sin²[(A - B)/2] = 3/2.Since A + B = π - C, but we can also express C as π - (A + B). Let me denote D = (A + B)/2, so that D = (π - C)/2. Then, the equation becomes:2 cos² D + sin²[(A - B)/2] = 3/2.But A - B = 2E, let me let E = (A - B)/2, so that:2 cos² D + sin² E = 3/2.But D = (π - C)/2, and E = (A - B)/2. However, since A + B = π - C, we can express A as D + E and B as D - E. Therefore, A = D + E and B = D - E. Then, since D = (π - C)/2, and C is another angle. But maybe this substitution can help.Alternatively, perhaps express the equation 2 cos² D + sin² E = 3/2 in terms of D and E. Let's note that:cos² D = (1 + cos 2D)/2sin² E = (1 - cos 2E)/2But substituting that:2 * [ (1 + cos 2D)/2 ] + [ (1 - cos 2E)/2 ] = 3/2Simplify:(1 + cos 2D) + (1 - cos 2E)/2 = 3/2Multiply through by 2 to eliminate denominators:2(1 + cos 2D) + (1 - cos 2E) = 3Expand:2 + 2 cos 2D + 1 - cos 2E = 3Combine constants:3 + 2 cos 2D - cos 2E = 3Subtract 3:2 cos 2D - cos 2E = 0Therefore:2 cos 2D = cos 2EBut D = (A + B)/2 = (π - C)/2, so 2D = π - C. Therefore, cos 2D = cos(π - C) = -cos C.Similarly, E = (A - B)/2, so 2E = A - B. So cos 2E = cos(A - B).Thus, substituting back:2 (-cos C) = cos(A - B)So:-2 cos C = cos(A - B)So the equation simplifies to:cos(A - B) = -2 cos CBut since A + B + C = π, C = π - A - B. Therefore:cos(A - B) = -2 cos(π - A - B) = -2 (-cos(A + B)) = 2 cos(A + B)Therefore, we have:cos(A - B) = 2 cos(A + B)So this is a trigonometric equation relating angles A and B. Let's write that:cos(A - B) = 2 cos(A + B)This seems like a key equation. Let me see if I can manipulate this equation to find a relation between tan A and tan B.First, expand both sides using trigonometric identities.Left side: cos(A - B) = cos A cos B + sin A sin BRight side: 2 cos(A + B) = 2 (cos A cos B - sin A sin B)Therefore:cos A cos B + sin A sin B = 2 cos A cos B - 2 sin A sin BBring all terms to the left side:cos A cos B + sin A sin B - 2 cos A cos B + 2 sin A sin B = 0Combine like terms:(-cos A cos B) + 3 sin A sin B = 0So:- cos A cos B + 3 sin A sin B = 0Divide both sides by cos A cos B (assuming cos A cos B ≠ 0, which is true since A and B are angles in a triangle, so they are between 0 and π, but not 0 or π, so cosines are not zero except if angle is π/2. But even if one is π/2, cos would be zero. Wait, but maybe we can proceed by dividing:-1 + 3 tan A tan B = 0Therefore:3 tan A tan B = 1Hence:tan A tan B = 1/3So, that is a constant value. Therefore, the answer to part (1) is yes, and tan A · tan B = 1/3.Wait, let me check the steps again to make sure.Starting from |a| = √6 / 2, so |a|² = 6/4 = 3/2.Then, expanding the components:(√2 cos[(A+B)/2])² = 2 cos²[(A+B)/2]Plus (sin[(A-B)/2])².Set equal to 3/2:2 cos²[(A+B)/2] + sin²[(A - B)/2] = 3/2.Then we let D = (A + B)/2, E = (A - B)/2.Then equation becomes 2 cos² D + sin² E = 3/2.Then we converted to double angles:2*(1 + cos 2D)/2 + (1 - cos 2E)/2 = 3/2.Wait, maybe that step had an error. Let me verify:cos² D = (1 + cos 2D)/2, so 2 cos² D = (1 + cos 2D). Then sin² E = (1 - cos 2E)/2.Therefore, 2 cos² D + sin² E = (1 + cos 2D) + (1 - cos 2E)/2.So:Left side: (1 + cos 2D) + (1 - cos 2E)/2Set equal to 3/2.Multiply both sides by 2:2(1 + cos 2D) + (1 - cos 2E) = 3Which becomes:2 + 2 cos 2D + 1 - cos 2E = 3So 3 + 2 cos 2D - cos 2E = 3Subtract 3: 2 cos 2D - cos 2E = 0.Then, substituting 2D = π - C and 2E = A - B, so cos 2D = cos(π - C) = -cos C, cos 2E = cos(A - B).So:2*(-cos C) - cos(A - B) = 0 => -2 cos C - cos(A - B) = 0 => cos(A - B) = -2 cos C.But since C = π - A - B, cos C = -cos(A + B). Therefore:cos(A - B) = -2*(-cos(A + B)) = 2 cos(A + B)Thus, cos(A - B) = 2 cos(A + B). Then expanding both sides:cos A cos B + sin A sin B = 2 (cos A cos B - sin A sin B)Bring all terms to the left:cos A cos B + sin A sin B - 2 cos A cos B + 2 sin A sin B = 0Simplify:(-cos A cos B) + 3 sin A sin B = 0Then, divide by cos A cos B (assuming they are not zero):-1 + 3 tan A tan B = 0 => 3 tan A tan B = 1 => tan A tan B = 1/3.Yes, that seems correct. Therefore, tan A · tan B is indeed a constant value, 1/3. So part (1) is answered.Now moving to part (2): Find the maximum value of tan C and determine the shape of the triangle at this time.We need to find the maximum value of tan C, given that in triangle ABC, angles A and B satisfy tan A tan B = 1/3, and C = π - A - B.So, since tan A tan B = 1/3, we can express tan C in terms of A and B, or maybe find a relationship.First, recall that in a triangle, A + B + C = π, so C = π - A - B.Therefore, tan C = tan(π - A - B) = -tan(A + B).But tan(A + B) = [tan A + tan B] / [1 - tan A tan B] = [tan A + tan B] / [1 - 1/3] = [tan A + tan B] / (2/3) = (3/2)(tan A + tan B)Therefore, tan C = -tan(A + B) = -(3/2)(tan A + tan B)But since C is an angle in a triangle, 0 < C < π, so tan C is defined only if C ≠ π/2. However, if C is acute, tan C is positive; if C is obtuse, tan C is negative. But since in a triangle, only one angle can be obtuse. But if C is obtuse, then A and B must be acute. However, given that tan A tan B = 1/3, which is positive, so both tan A and tan B are positive (since angles A and B are between 0 and π, and their product is positive, so they are both acute angles, so tan A and tan B are positive). Therefore, A and B are acute, so C could be acute or obtuse.But in the equation tan C = -(3/2)(tan A + tan B), if C is acute, then tan C is positive, but the right side is negative unless (tan A + tan B) is negative, but tan A and tan B are positive. Therefore, there's a contradiction here. Wait, this suggests that maybe we have to consider the sign properly.Wait, let's double-check. Since A and B are angles in a triangle, they are each between 0 and π. Since A + B = π - C, and C is between 0 and π. If C is acute, then A + B = π - C > π/2, so A and B could be anything as long as their sum is greater than π/2. If C is right angle, then A + B = π/2. If C is obtuse, then A + B < π/2. But A and B are both positive angles, so in that case, both A and B must be less than π/2.But given that tan A tan B = 1/3, both tan A and tan B are positive, so A and B are acute angles. Therefore, C can be either acute or obtuse, but given that A and B are both acute, if C is obtuse, then A + B < π/2, but since A and B are both acute, their sum would be less than π, but for C to be obtuse, A + B < π/2. Wait, that's impossible because if A and B are both positive angles adding to less than π/2, but in a triangle, the sum of all three angles is π. So if C is obtuse (greater than π/2), then A + B = π - C < π - π/2 = π/2. Therefore, A + B < π/2, so both A and B are less than π/2, which is okay because they are acute. But in this case, tan C would be negative because C is obtuse (since tangent of an angle between π/2 and π is negative). But in the equation above, we had tan C = -(3/2)(tan A + tan B). Since tan A and tan B are positive, then tan C is negative, which matches the case when C is obtuse. But the problem asks for the maximum value of tan C. Since tan C can be positive or negative, but the maximum value would be when tan C is as large as possible. However, when C is acute (tan C positive), how is this possible?Wait, there seems to be confusion here. Let me re-examine the equation.We have:tan C = -tan(A + B) = - [ (tan A + tan B) / (1 - tan A tan B) ]But since tan A tan B = 1/3, so:tan C = - [ (tan A + tan B) / (1 - 1/3) ] = - [ (tan A + tan B) / (2/3) ] = - (3/2)(tan A + tan B)Therefore, tan C is equal to negative three-halves times the sum of tan A and tan B. Since tan A and tan B are positive (as angles A and B are acute), tan C is negative. But tan C is negative only if angle C is obtuse (since tangent is negative in the second quadrant). Therefore, C is obtuse, and tan C is negative. However, the problem asks for the maximum value of tan C. Since tan C is negative, its maximum value (closest to zero) would be the least negative value. Alternatively, perhaps the problem is referring to the maximum absolute value? But the question says "maximum value of tan C", which, if taken literally, since tan C is negative, the maximum value would be the least negative one, i.e., closest to zero. However, maybe there is a miscalculation here.Wait, perhaps I made a mistake in deriving tan C. Let me check again.We have C = π - A - B. Therefore, tan C = tan(π - A - B) = -tan(A + B). Then, tan(A + B) = [tan A + tan B] / [1 - tan A tan B]. Therefore:tan C = - [ (tan A + tan B) / (1 - tan A tan B) ]But we know tan A tan B = 1/3. Therefore:tan C = - [ (tan A + tan B) / (1 - 1/3) ] = - [ (tan A + tan B) / (2/3) ] = - (3/2)(tan A + tan B)Therefore, tan C is indeed negative, implying angle C is obtuse. However, the problem asks for the maximum value of tan C. Since tan C is negative, its maximum value is the least negative value, which occurs when tan A + tan B is minimized. Alternatively, if we consider the absolute value, the maximum |tan C| would occur when tan A + tan B is maximized. But the problem says "maximum value of tan C", so as a real number, the highest value on the real line, which for negative numbers would be the one closest to zero. Therefore, to maximize tan C (i.e., make it less negative), we need to minimize (tan A + tan B).Alternatively, maybe the problem expects the maximum in magnitude, but the wording is unclear. However, since in the problem statement it's mentioned as "maximum value", and since in mathematics, unless specified otherwise, maximum value refers to the greatest real value, which for tan C would be approaching zero from below. But that seems counterintuitive. Alternatively, perhaps there's a mistake in the reasoning here.Wait, let's think differently. If C is acute, then tan C is positive. However, according to our previous result, tan C is negative, which would imply that C is obtuse. But how can that be? If we have A and B such that A + B < π/2, then C = π - A - B > π/2, making it obtuse. But in that case, A + B must be less than π/2, which would require A and B both to be quite small. However, given that tan A tan B = 1/3, which is a fixed product, perhaps there's a way to have C vary.Wait, maybe there's a mistake in assuming that C must be obtuse. Let's verify with an example. Suppose A = B. Then tan A tan B = tan² A = 1/3, so tan A = 1/√3, so A = π/6. Then B = π/6, so A + B = π/3, so C = π - π/3 = 2π/3, which is obtuse. Therefore, tan C = tan(2π/3) = -√3. So in this case, tan C = -√3.But if A and B are such that A approaches 0, then B approaches (π - C), but tan A approaches 0, so tan B must approach infinity to maintain tan A tan B = 1/3. However, B cannot approach π/2 because A approaches 0, then A + B approaches B, which would need to be less than π/2 (if C is obtuse). But if B approaches π/2, then A approaches (π - C - B), but this seems conflicting. Maybe the maximum value of |tan C| is unbounded? Wait, but in reality, A and B are constrained by the triangle. Let's consider.Given that A and B are both positive angles, with A + B = π - C < π/2 (since C is obtuse). Also, tan A tan B = 1/3.We can model this as a constraint optimization problem: maximize tan C, which is equivalent to minimizing (tan A + tan B) because tan C = - (3/2)(tan A + tan B). So to maximize tan C (which is negative), we need to minimize (tan A + tan B). Conversely, to minimize tan C (make it more negative), we need to maximize (tan A + tan B). So the problem is to find the minimum of (tan A + tan B) given that tan A tan B = 1/3.This is a classic optimization problem. Let me recall that for positive numbers x and y with a fixed product, the sum x + y is minimized when x = y. This follows from the AM ≥ GM inequality. Since tan A and tan B are positive (since A and B are acute), and their product is fixed as 1/3, the minimum of tan A + tan B occurs when tan A = tan B = sqrt(1/3) = 1/√3. Therefore, the minimum sum is 2/√3. Thus, the minimum value of (tan A + tan B) is 2/√3, achieved when tan A = tan B = 1/√3. Therefore, the maximum value of tan C is - (3/2)*(2/√3) = - (3/2)*(2/√3) = - 3/√3 = -√3.However, if we consider the maximum value of |tan C|, then it would be unbounded as tan A + tan B can go to infinity. But in reality, in the triangle, angles A and B cannot be arbitrary. Let's see if that's possible.Given that A and B are both acute angles with A + B < π/2 (since C is obtuse). Let's suppose A approaches 0, then B must approach (π - C) - A, but since C is obtuse, π - C < π/2. Wait, if A approaches 0, then B = π - C - A ≈ π - C. But since C is obtuse, π - C < π/2. Therefore, B approaches π - C, which is less than π/2. But with A approaching 0, tan A approaches 0, and tan B approaches tan(π - C - 0) = tan(π - C) = -tan C. But tan B must satisfy tan A tan B = 1/3, so as A approaches 0, tan A approaches 0, so tan B must approach infinity to maintain the product 1/3. Therefore, B approaches π/2, but π - C approaches B, so π - C approaches π/2, hence C approaches π - π/2 = π/2. But C is supposed to be obtuse, so this leads to a contradiction. Therefore, B cannot approach π/2 because then C would approach π/2, which is not obtuse. Therefore, there must be a lower bound on how small A can be, hence a lower bound on B, hence preventing tan B from approaching infinity. Wait, this is confusing. Let's model it more carefully.Suppose A approaches 0. Then B = π - C - A. Since C is obtuse, π - C < π/2. Therefore, B = π - C - A < π/2 - A. As A approaches 0, B approaches π - C, which is less than π/2. Therefore, B approaches some value less than π/2. But if A approaches 0, then tan A approaches 0, so to maintain tan A tan B = 1/3, tan B must approach infinity. But if B approaches π/2, then π - C approaches π/2, so C approaches π/2, which is not obtuse. Therefore, this is impossible. Therefore, there's a contradiction here. So perhaps the maximum of tan A + tan B is unbounded, but due to the triangle constraints, it's actually bounded.Wait, let me consider the following. Let's set tan A = t, then tan B = 1/(3t). We need to find the possible values of t given that angles A and B satisfy A + B < π/2 (since C is obtuse). So A and B are both less than π/2, and A + B < π/2.Expressing A and B in terms of t:A = arctan(t)B = arctan(1/(3t))Then, the condition A + B < π/2.We need to find the domain of t such that arctan(t) + arctan(1/(3t)) < π/2.Recall that arctan(a) + arctan(b) < π/2 if and only if ab < 1. So:t * (1/(3t)) = 1/3 < 1Therefore, arctan(t) + arctan(1/(3t)) < π/2 is always true. Therefore, for all t > 0, this condition holds. Therefore, angles A and B can vary such that tan A = t and tan B = 1/(3t) for any t > 0, and C will always be obtuse.Therefore, tan A + tan B = t + 1/(3t)We need to find the minimum and maximum of this expression for t > 0.As t approaches 0, tan A approaches 0, tan B approaches infinity, so tan A + tan B approaches infinity.As t approaches infinity, tan A approaches infinity, tan B approaches 0, so tan A + tan B approaches infinity.The function f(t) = t + 1/(3t) has a minimum at some finite t. To find the minimum, take derivative:f'(t) = 1 - 1/(3t²)Set to zero:1 - 1/(3t²) = 0 => 1 = 1/(3t²) => 3t² = 1 => t² = 1/3 => t = 1/√3Thus, the minimum of f(t) is at t = 1/√3:f(1/√3) = 1/√3 + 1/(3*(1/√3)) = 1/√3 + (1/3)*√3 = 1/√3 + √3/3 = 2/√3Therefore, the minimum value of tan A + tan B is 2/√3, achieved when t = 1/√3. Therefore, the maximum value of tan C is - (3/2)*(2/√3) = -√3.However, the maximum value of tan C would be the least negative value, which is -√3. But if we consider the magnitude, |tan C| is √3, which is the maximum magnitude. However, since tan C is negative, the problem states "maximum value of tan C", which is -√3. But -√3 is actually the minimum value of tan C because as t increases or decreases, tan C becomes more negative (i.e., decreasing). But the problem says "maximum value", so in terms of real numbers, -√3 is the maximum (since it's the highest value on the real line for tan C, which is negative). For example, between -√3 (-1.732) and -infinity, the maximum is -√3.Therefore, the maximum value of tan C is -√3, and this occurs when tan A = tan B = 1/√3, so A = B = arctan(1/√3) = π/6. Therefore, angles A and B are both π/6, and angle C = π - π/6 - π/6 = 2π/3. Therefore, the triangle is an isosceles triangle with angles π/6, π/6, and 2π/3. However, in terms of sides, since angles A and B are equal, the sides opposite them are equal. Therefore, the triangle is isosceles with two sides equal and the third side different. However, given the angles are π/6, π/6, 2π/3, the triangle is an obtuse isosceles triangle.But let me verify this. If A = B = π/6, then sides opposite these angles (a and b) would be equal. The side opposite angle C (which is 2π/3) would be the longest side. Therefore, the triangle is indeed isosceles with sides a = b and angle C = 2π/3. So the shape is an isosceles triangle with apex angle 120 degrees.Alternatively, if we consider the triangle with angles 30°, 30°, 120°, which is an isosceles triangle.Therefore, the maximum value of tan C is -√3, and the triangle is an isosceles triangle with angles 30°, 30°, 120°.But the problem asks for the shape, so probably stating it's an isosceles triangle or specifically a 30-30-120 triangle. However, in terms of side lengths, since angles are π/6, π/6, 2π/3, using the Law of Sines:a / sin A = b / sin B = c / sin CSince A = B = π/6, sin A = sin B = 1/2, and sin C = sin(2π/3) = √3/2. Therefore, the sides opposite A and B are equal (a = b), and the side opposite C is c = a * (sin C / sin A) = a * (√3/2 / 1/2) = a * √3. Therefore, the sides are in the ratio 1 : 1 : √3, making it an isosceles triangle with two sides equal and the third side √3 times longer.Therefore, the maximum value of tan C is -√3, and the triangle is an isosceles triangle with angles 30°, 30°, and 120°.But let me confirm if this is indeed the case. Suppose A = B = π/6, then tan A = tan π/6 = 1/√3, so tan A tan B = 1/3, which matches the given condition. Then, C = 2π/3, tan C = tan(2π/3) = -√3, which is the value we found. If we take another example, say tan A = 1, then tan B = 1/3, so A = π/4, B = arctan(1/3). Then, A + B = π/4 + arctan(1/3). Let's compute this:arctan(1/3) ≈ 0.32175 radians ≈ 18.43 degrees.Therefore, A + B ≈ 45° + 18.43° ≈ 63.43°, so C ≈ 116.57°, which is still obtuse. tan C ≈ tan(116.57°) ≈ tan(π - 63.43°) ≈ -tan(63.43°) ≈ -2.000. So tan C ≈ -2.0, which is greater than -√3 ≈ -1.732. Wait, but -2 is less than -√3. Wait, this contradicts our previous conclusion. Wait, tan C in this case is -2, which is less than -√3 ≈ -1.732. So tan C is more negative here. But according to our earlier conclusion, the maximum value (least negative) is -√3. So in this example, tan C is -2, which is smaller than -√3. That suggests that our previous conclusion was incorrect.Wait, this is confusing. Let's calculate properly.If tan A = 1, so A = π/4, then tan B = 1/3, so B = arctan(1/3). Then, A + B = π/4 + arctan(1/3). Let's compute angle C = π - (A + B). Then:tan C = -tan(A + B) = - [ (tan A + tan B) / (1 - tan A tan B) ] = - [ (1 + 1/3) / (1 - 1*1/3) ] = - [ (4/3) / (2/3) ] = - [ 4/3 * 3/2 ] = -2So tan C = -2, which is indeed more negative than -√3. However, according to our earlier analysis, the minimum value of (tan A + tan B) is 2/√3 ≈ 1.1547, so tan C = - (3/2)(tan A + tan B) ≈ - (3/2)(1.1547) ≈ -1.732 ≈ -√3. So in this case, when tan A = tan B = 1/√3, tan C is -√3 ≈ -1.732. Whereas, when tan A = 1 and tan B = 1/3, tan C is -2, which is smaller. So the earlier conclusion was that the minimum of (tan A + tan B) is 2/√3, leading to the maximum (least negative) value of tan C as -√3. However, the example shows that tan C can be more negative than that. Therefore, there must be a mistake in the reasoning.Wait, so where is the error?Earlier, we considered that given tan A tan B = 1/3, tan A + tan B is minimized when tan A = tan B = sqrt(1/3). However, in the triangle, angles A and B are subject to the constraint that A + B < π/2 (since C is obtuse). However, when we set tan A = tan B = 1/√3, angles A and B are π/6 each, so A + B = π/3, which is less than π/2. Therefore, C = 2π/3, which is obtuse, and this is valid. But in this case, tan C = -√3. However, when we take tan A = 1 and tan B = 1/3, angles A and B are π/4 and arctan(1/3) ≈ 0.32175 radians, so their sum is ≈ 0.785 + 0.32175 ≈ 1.1068 radians ≈ 63.43 degrees, which is less than π/2 ≈ 1.5708 radians ≈ 90 degrees. So C ≈ 116.57 degrees, which is obtuse. Therefore, this is also a valid case, and in this case, tan C ≈ -2. So in this case, tan C is more negative (smaller) than -√3.But according to the calculus approach, the minimum of tan A + tan B is 2/√3, which would correspond to maximum tan C of -√3. But this example shows that tan A + tan B can be larger (1 + 1/3 ≈ 1.333), which is larger than 2/√3 ≈ 1.1547, leading to a smaller (more negative) tan C.Therefore, there's a mistake in assuming that tan A + tan B is minimized when tan A = tan B = 1/√3. Wait, but according to the AM ≥ GM inequality, for positive numbers, the sum is minimized when the numbers are equal, given a fixed product. Indeed, tan A * tan B = 1/3, so the minimum sum occurs when tan A = tan B = sqrt(1/3). Therefore, tan A + tan B = 2/sqrt(3) ≈ 1.1547. However, in our example, tan A + tan B = 1 + 1/3 ≈ 1.333, which is larger. Therefore, the minimum sum is indeed 2/sqrt(3), and larger sums are possible. Therefore, tan C can take values from -√3 down to negative infinity? But that contradicts the example where tan C = -2, which is a finite value.Wait, but how can tan C approach negative infinity? If tan A approaches infinity, then tan B approaches zero such that tan A tan B = 1/3. But if tan A approaches infinity, then angle A approaches π/2, but in our case, A + B < π/2, so if A approaches π/2, then B must approach a negative angle, which is impossible. Therefore, there must be a bound on how large tan A can get.Wait, this is key. Earlier, we considered that angles A and B must satisfy A + B < π/2 (since C is obtuse). Therefore, both A and B are less than π/2, and their sum is less than π/2. Therefore, if A approaches π/2, then B must approach π/2 - A, which would go to a negative number as A approaches π/2 from below, which is impossible. Therefore, angle A cannot approach π/2. So there is an upper bound on angle A. Let's compute the maximum possible value of angle A.Given that A + B < π/2, and B > 0, so A < π/2 - B < π/2. Similarly, B < π/2 - A. Therefore, angles A and B are each less than π/2, and their sum is less than π/2. Therefore, the maximum possible value for A is approaching π/2, but in reality, since B must remain positive, A < π/2 - B, so if A approaches π/2, B approaches zero. But tan A tan B = 1/3. If B approaches zero, then tan B approaches zero, so tan A must approach infinity to maintain the product 1/3. But angle A is approaching π/2, so tan A approaches infinity. But angle A cannot exceed π/2, so this suggests that as A approaches π/2, B approaches zero, with tan A approaching infinity and tan B approaching zero such that their product remains 1/3. Therefore, in this limit, tan C = - (3/2)(tan A + tan B) approaches - (3/2)(infty + 0) = -infty. Therefore, tan C can approach negative infinity, meaning there is no maximum value (since it can be made as negative as desired). But this contradicts our earlier example where tan C = -2. Therefore, where is the mistake here?Ah, the key is that in reality, angles A and B cannot be such that A approaches π/2 while B approaches zero, because A + B must remain less than π/2. But if A approaches π/2, then B must approach π/2 - A, but if A approaches π/2 from below, then π/2 - A approaches zero. Therefore, B approaches zero, so tan B approaches zero, but then tan A must approach infinity to keep their product 1/3. Therefore, this is possible in the limit, which would imply that tan C approaches negative infinity. However, in reality, angles in a triangle are bounded, but mathematically, in this setup, with the given conditions, tan C can indeed approach negative infinity. However, in practical terms, angles cannot be zero or π/2, but can approach those limits.But this contradicts the example where tan C = -2. So what's the resolution here? The problem states that A and B are interior angles of triangle ABC, so they must be greater than 0 and less than π, and their sum must be less than π (since C is positive). But in the case where tan A approaches infinity, angle A approaches π/2, angle B approaches zero, their sum approaches π/2, angle C approaches π/2, but C must be obtuse, which is greater than π/2. Therefore, there's a contradiction here. If A approaches π/2, then C approaches π - (A + B) ≈ π - π/2 = π/2. But C must be obtuse, so C > π/2. Therefore, angle C cannot approach π/2 from above; it must be strictly greater than π/2. Therefore, there is a lower bound on how close C can be to π/2, which implies an upper bound on how close A + B can be to π/2.Wait, this seems like a loop. Let me think carefully.Given that C is obtuse, then A + B = π - C < π/2. Therefore, A + B must be less than π/2. Therefore, A and B are both less than π/2, and their sum is less than π/2. Therefore, both angles are acute and their sum is less than π/2. So for example, if A approaches π/2, then B must approach π/2 - A, but since A + B < π/2, if A approaches π/2 from below, then B must approach a negative angle, which is impossible. Therefore, angle A cannot approach π/2. Therefore, there must be a maximum possible value for angle A such that B remains positive.To find the maximum possible angle A, we set B approaches 0, so A + B approaches A, which must be less than π/2. Therefore, angle A can approach π/2 but cannot reach it. So in reality, angle A can approach π/2 - ε, and angle B approaches ε, where ε approaches zero. In this case, tan A approaches tan(π/2 - ε) = cot ε ≈ 1/ε, and tan B ≈ tan ε ≈ ε. Therefore, tan A tan B ≈ (1/ε)(ε) = 1, but in our problem, tan A tan B = 1/3. Therefore, this is a contradiction. Therefore, angle A cannot approach π/2 in this problem. Therefore, there must be a bound on angle A.Wait, here's the key: in our problem, tan A tan B = 1/3. If angle A approaches some maximum value, then tan A approaches a corresponding maximum value, and tan B approaches a value such that their product is 1/3. So perhaps we can express this.Let me denote angle A = α, angle B = β, with α + β < π/2.We have tan α tan β = 1/3.We need to find the range of α and β.Express β in terms of α: β = arctan(1/(3 tan α)).Then, the condition α + β < π/2 becomes:α + arctan(1/(3 tan α)) < π/2.We need to find the allowable range of α.Let me consider α in (0, π/2), but such that α + arctan(1/(3 tan α)) < π/2.This seems complex to solve analytically. Let's consider specific values.When α = arctan(1/√3) = π/6, then β = arctan(1/(3*(1/√3))) = arctan(1/√3) = π/6. So α + β = π/3 < π/2, which is valid.When α increases, say α = π/4, then tan α = 1, so β = arctan(1/3) ≈ 0.32175 radians, so α + β ≈ 0.785 + 0.32175 ≈ 1.10675 radians ≈ 63.43 degrees < π/2 ≈ 90 degrees. Still valid.If α increases further, say α approaches π/3 (60 degrees), then tan α = √3 ≈ 1.732, so β = arctan(1/(3*1.732)) ≈ arctan(0.19245) ≈ 0.190 radians ≈ 10.89 degrees. Then, α + β ≈ 60 + 10.89 ≈ 70.89 degrees < 90 degrees. Still valid.If α increases to, say, 80 degrees (1.396 radians), then tan α ≈ 5.671, so β = arctan(1/(3*5.671)) ≈ arctan(0.0588) ≈ 0.0588 radians ≈ 3.37 degrees. Then α + β ≈ 83.37 degrees < 90 degrees. Still valid.If α approaches 90 degrees (π/2), tan α approaches infinity, so β = arctan(1/(3*infty)) = arctan(0) = 0. But α + β approaches 90 degrees + 0 = 90 degrees, which is not less than π/2. Therefore, α cannot reach 90 degrees. However, as α approaches 90 degrees from below, β approaches 0, and α + β approaches 90 degrees, which violates the condition α + β < π/2. Therefore, there must be a maximum α for which α + β = π/2. Let's find this α.Set α + β = π/2. Then β = π/2 - α. Then:tan α tan β = tan α tan(π/2 - α) = tan α cot α = 1.But in our problem, tan α tan β = 1/3. Therefore, 1 = 1/3, which is impossible. Therefore, there's no solution when α + β = π/2. Therefore, angles α and β can approach α + β = π/2 but with tan α tan β = 1/3. Let's find the limit as α approaches some value such that α + β approaches π/2.Let me set α + β = π/2 - ε, where ε approaches 0+.Then, β = π/2 - α - ε.But tan α tan β = tan α tan(π/2 - α - ε) = tan α cot(α + ε) = tan α / tan(α + ε).Set this equal to 1/3:tan α / tan(α + ε) = 1/3.As ε approaches 0, tan(α + ε) ≈ tan α + ε sec² α.Therefore:tan α / (tan α + ε sec² α) ≈ 1/3.Multiply both sides by denominator:tan α ≈ (1/3)(tan α + ε sec² α)Multiply both sides by 3:3 tan α ≈ tan α + ε sec² αSubtract tan α:2 tan α ≈ ε sec² αTherefore:ε ≈ 2 tan α / sec² α = 2 sin α cos α.As ε approaches 0, either sin α or cos α approaches 0. But if α approaches 0, then cos α approaches 1, so ε ≈ 2*0*1 = 0. If α approaches π/2, then sin α approaches 1, cos α approaches 0, so ε ≈ 2*1*0 = 0. But in between, ε is maximized.But this seems to indicate that as we vary α, ε can be made small. Therefore, angles α and β can approach α + β = π/2 as ε approaches 0, but maintaining tan α tan β = 1/3. However, in this limit, we can see that tan α tan β approaches 1/3, but as α + β approaches π/2, we have tan α approaches sqrt(1/3), but this seems inconsistent. Wait, perhaps another approach.Let me set α + β = π/2 - ε, and try to express tan α tan β = 1/3 in terms of ε.Using the identity:tan(α + β) = tan(π/2 - ε) = cot ε.But tan(α + β) = (tan α + tan β) / (1 - tan alpha tan beta) = (tan alpha + tan beta) / (1 - 1/3) = (tan alpha + tan beta) / (2/3).Therefore:(tan alpha + tan beta) / (2/3) = cot epsilon=> tan alpha + tan beta = (2/3) cot epsilon.But as epsilon approaches 0, cot epsilon approaches infinity, so tan alpha + tan beta approaches infinity. Therefore, in this limit, tan C = - (3/2)(tan alpha + tan beta) approaches -infty.Therefore, as alpha + beta approaches pi/2 (i.e., epsilon approaches 0), tan C approaches -infty. Therefore, tan C can be made arbitrarily small (more negative), meaning there is no maximum value. However, this contradicts our earlier example where tan C was finite.But in the problem statement, it's asked to find the maximum value of tan C, which suggests that such a maximum exists. Therefore, there must be a constraint that limits how small tan C can be.But according to the previous analysis, tan C can be made arbitrarily small (approaching -infty). However, in the context of a triangle, angles are bounded. If tan C approaches -infty, then angle C approaches pi/2 from above, but C must be obtuse, greater than pi/2. Wait, tan C approaches -infty when C approaches pi/2 from above. Because tan(pi/2 + x) approaches -infty as x approaches 0 from the positive side. Therefore, angle C approaches pi/2 (but stays greater than pi/2), making tan C approach -infty. But angle C can't be exactly pi/2 or greater than pi, but between pi/2 and pi.But in reality, as C approaches pi/2 from above, angle A + B approaches pi/2 from below. In this case, angles A and B are approaching two angles whose sum is pi/2, but such that tan A tan B = 1/3. As C approaches pi/2, angles A and B approach certain limits. However, this suggests that C can approach pi/2, making tan C approach -infty, but in reality, in a triangle, angles must be positive. However, if C approaches pi, then A + B approaches 0, but with tan A tan B = 1/3, which is impossible since A and B would both approach 0, making tan A and tan B approach 0, contradicting the product 1/3. Therefore, the only possible way for tan C to approach -infty is for C to approach pi/2 from above, making angles A and B approach a sum of pi/2, with their product of tangents being 1/3. But this seems mathematically possible, leading to tan C approaching -infty.However, the problem asks for the maximum value of tan C, implying that such a maximum exists. This contradiction suggests that there might be a mistake in the problem's translation or my understanding. Alternatively, perhaps the problem assumes that C is acute, but the earlier analysis showed that C must be obtuse.Wait, let me re-examine the problem statement:"A and B are two interior angles of triangle ABC..."The vector is given with magnitude sqrt(6)/2. From part (1), we derived that tan A tan B = 1/3. In part (2), we are to find the maximum value of tan C. But according to our analysis, tan C is negative and can be made arbitrarily small (more negative). However, the problem might expect the maximum value in magnitude, but the wording is "maximum value of tan C". If we consider the maximum value as in the least negative, then it would be -sqrt(3). However, in the example with angles pi/6, pi/6, 2pi/3, we get tan C = -sqrt(3), and in another example with angles pi/4, arctan(1/3), we get tan C = -2, which is more negative. But the first case gives the least negative value of tan C. Therefore, the maximum value of tan C (as the highest real number) is -sqrt(3), achieved when A = B = pi/6, and C = 2pi/3. This must be the answer expected by the problem, even though mathematically, tan C can be made more negative. Perhaps there's a constraint that I missed.Wait, perhaps when maximizing tan C, we should consider C as acute. But according to our previous deduction, tan C is negative, implying C is obtuse. But if we consider C acute, then tan C would be positive, but this contradicts the equation tan C = - (3/2)(tan A + tan B). However, if we revisit the earlier step:We had tan C = - (3/2)(tan A + tan B)But tan C is negative, meaning angle C is obtuse. Therefore, C is obtuse, and tan C is negative. Therefore, the maximum value of tan C is the least negative value, which occurs when tan A + tan B is minimized. As previously found using AM-GM inequality, the minimum of tan A + tan B is 2/sqrt(3), leading to tan C = -sqrt(3). Therefore, the answer is that the maximum value of tan C is -sqrt(3), achieved when A = B = pi/6, making the triangle isosceles.But the example with angles pi/4 and arctan(1/3) gives a more negative tan C, so why is this not considered? It must be because in that case, the sum tan A + tan B is larger, leading to a smaller (more negative) tan C. Therefore, the minimum of tan A + tan B gives the maximum (least negative) tan C. Therefore, the correct maximum value of tan C is -sqrt(3), achieved when tan A = tan B = 1/sqrt(3). Therefore, the answer is:(1) Yes, tan A · tan B is a constant, 1/3.(2) The maximum value of tan C is -sqrt(3), achieved when the triangle is an isosceles triangle with angles 30°, 30°, and 120°.But why does the example with tan C = -2 exist? Because when tan A and tan B are not equal, the sum tan A + tan B is larger, leading to a more negative tan C. Therefore, the minimum sum corresponds to the maximum tan C (least negative), and there's no lower bound on tan C (it can approach -infty). However, in a triangle, angles are constrained, but according to the mathematical model, as long as A + B < pi/2 and tan A tan B = 1/3, angles A and B can vary such that tan A + tan B can be made arbitrarily large, making tan C arbitrarily negative. However, this is possible only if A and B can be such that one approaches 0 and the other approaches pi/2 - 0 = pi/2, but in reality, as one angle approaches 0, the other approaches a value less than pi/2, but their product of tangents must remain 1/3. This suggests that tan A can approach infinity only if angle A approaches pi/2, but in that case, angle B approaches pi/2 - A, which approaches a negative angle, which is impossible. Therefore, there must be a bound.The confusion arises because while mathematically, the equations allow tan A to approach infinity, geometrically, in a triangle, angles are bounded. Therefore, there must be a maximum possible value for tan A, which would impose a lower bound on tan C. However, the exact bound requires more careful analysis.Let me consider the relationship between angles A and B. Given that A + B < pi/2 and tan A tan B = 1/3, we can express B as a function of A.Let A be an angle in (0, pi/2), and B = arctan(1/(3 tan A)). Then, A + B < pi/2.Let's find the maximum possible A:A + arctan(1/(3 tan A)) < pi/2.Let me attempt to solve this equation for A.Let’s denote x = A, then:x + arctan(1/(3 tan x)) < pi/2.Let’s find when equality holds:x + arctan(1/(3 tan x)) = pi/2.Let’s set y = arctan(1/(3 tan x)).Then, tan y = 1/(3 tan x).From the equation x + y = pi/2, we have y = pi/2 - x.Therefore, tan(pi/2 - x) = cot x = 1/(3 tan x).Thus:cot x = 1/(3 tan x) => cot x = (1/3) cot x => 3 cot x = cot x => 2 cot x = 0 => cot x = 0 => x = pi/2.But x = pi/2 is not allowed in the equation tan x, so there is no solution. Therefore, the equality x + arctan(1/(3 tan x)) = pi/2 is never achieved, confirming that A + B is always less than pi/2.Therefore, there's no upper bound on A other than approaching pi/2, but as A approaches pi/2, B approaches zero, and tan A approaches infinity, leading to tan C approaching negative infinity. Therefore, mathematically, tan C can be made as small as desired, implying there's no maximum value. However, the problem asks to find the maximum value, so there must be a constraint I missed.Looking back at the problem statement:"A and B are two interior angles of triangle ABC..."Ah, wait, the problem doesn't specify whether the triangle is acute, obtuse, or right-angled. But from the derived condition, since tan A tan B = 1/3, both A and B are acute angles, and C must be obtuse. Therefore, the triangle is obtuse.But in the problem statement part (2), it says "find the maximum value of tan C and determine the shape of the triangle at this time." Since C is obtuse, tan C is negative, and the maximum value would be the least negative, which corresponds to the case when tan C is closest to zero. However, tan C = - (3/2)(tan A + tan B). To make tan C closest to zero, we need to minimize (tan A + tan B). As established before, the minimum of tan A + tan B is 2/sqrt(3), leading to tan C = -sqrt(3). Therefore, even though tan C can be made more negative, the "maximum" value (i.e., the least negative) is -sqrt(3), and this occurs when A = B = pi/6, making the triangle isosceles.Therefore, the answer expected is:(1) tan A · tan B is a constant, 1/3.(2) The maximum value of tan C is -sqrt(3), achieved when the triangle is isosceles with angles 30°, 30°, and 120°.The confusion arises because the term "maximum value" can be ambiguous for negative numbers, but in the context of the problem, since C is an obtuse angle, and the problem asks for the maximum value of tan C, the intended answer is the least negative value, which is -sqrt(3).Final Answer(1) boxed{dfrac{1}{3}}(2) The maximum value of tan C is boxed{-sqrt{3}}, and the triangle is isosceles with angles 30^circ, 30^circ, and 120^circ.</think>Given that A and B are two interior angles of triangle ABC, and the vector (overrightarrow{a} = sqrt{2}cosfrac{A+B}{2}overrightarrow{i} + sinfrac{A-B}{2}overrightarrow{j}) has a magnitude of (frac{sqrt{6}}{2}):(1) To determine if (tan A cdot tan B) is a constant value, we start by calculating the magnitude of the vector (overrightarrow{a}):[|overrightarrow{a}|^2 = (sqrt{2} cos frac{A+B}{2})^2 + (sin frac{A-B}{2})^2 = frac{3}{2}]Expanding and simplifying, we get:[2 cos^2 frac{A+B}{2} + sin^2 frac{A-B}{2} = frac{3}{2}]Using trigonometric identities and substituting (A + B = pi - C), we derive:[2 cos^2 left(frac{pi - C}{2}right) + sin^2 left(frac{A - B}{2}right) = frac{3}{2}]This simplifies to:[2 cos^2 D + sin^2 E = frac{3}{2}]where (D = frac{pi - C}{2}) and (E = frac{A - B}{2}). Further simplification leads to:[cos(A - B) = 2 cos(A + B)]Expanding both sides and simplifying, we find:[- cos A cos B + 3 sin A sin B = 0]Dividing by (cos A cos B) gives:[3 tan A tan B = 1 implies tan A cdot tan B = frac{1}{3}]Thus, (tan A cdot tan B) is a constant value (boxed{frac{1}{3}}).(2) To find the maximum value of (tan C) and determine the shape of the triangle, we use the relationship (tan C = -frac{3}{2}(tan A + tan B)). Given (tan A cdot tan B = frac{1}{3}), we minimize (tan A + tan B) using the AM-GM inequality:[tan A + tan B geq 2 sqrt{tan A tan B} = frac{2}{sqrt{3}}]The minimum value of (tan A + tan B) is (frac{2}{sqrt{3}}), leading to:[tan C = -frac{3}{2} cdot frac{2}{sqrt{3}} = -sqrt{3}]This occurs when (tan A = tan B = frac{1}{sqrt{3}}), implying (A = B = 30^circ) and (C = 120^circ). The triangle is isosceles with angles (30^circ), (30^circ), and (120^circ).The maximum value of (tan C) is (boxed{-sqrt{3}}), and the triangle is isosceles with angles (30^circ), (30^circ), and (120^circ).]

🔑:(1) We have |overrightarrow{a}|^2 = 2cos^2frac{A+B}{2} + sin^2frac{A-B}{2} = frac{3}{2},1 + cos(A+B) + frac{1-cos(A-B)}{2} = frac{3}{2}cos Acos B - sin Asin B - frac{cos Acos B + sin Asin B}{2} = 0frac{1}{2} - frac{3tan Atan B}{2} = 0 then tan Atan B = frac{1}{3}So, boxed{tan Atan B = frac{1}{3}} is a constant value.(2) From (1), we know that A and B are acute angles.tan C = -tan(B+A) = -frac{tan A + tan B}{1 - tan Atan B} = -frac{3(tan A + tan B)}{2} leq -3sqrt{tan Atan B} = -sqrt{3}Therefore, the maximum value of tan C is boxed{-sqrt{3}}.At this time, triangle ABC is an obtuse triangle.

🧠:[<think>Okay, so I need to find the sum of the third, fourth, and fifth terms of a geometric sequence. The first term is 3, and the sum of the first three terms is 21. Let me recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, the nth term of a geometric sequence can be written as a_n = a_1 * r^(n-1).Given that a_1 is 3, the first term is 3. Then the second term a_2 would be 3*r, and the third term a_3 would be 3*r^2. The problem states that the sum of the first three terms, S_3, is 21. So, S_3 = a_1 + a_2 + a_3 = 3 + 3r + 3r^2 = 21.I need to solve for r here. Let me write that equation out:3 + 3r + 3r^2 = 21.First, I can subtract 21 from both sides to set the equation to zero:3r^2 + 3r + 3 - 21 = 0.Simplifying the constants:3r^2 + 3r - 18 = 0.Hmm, this is a quadratic equation. To make it simpler, I can divide all terms by 3:r^2 + r - 6 = 0.Now, solving for r. The quadratic equation is r^2 + r - 6 = 0. Let's try to factor it. I need two numbers that multiply to -6 and add to 1. Let me think: 3 and -2. Yes, 3 * (-2) = -6 and 3 + (-2) = 1. So, the factors are (r + 3)(r - 2) = 0.Setting each factor equal to zero gives r + 3 = 0 or r - 2 = 0. So, r = -3 or r = 2. However, the problem states that all terms are positive. If r were negative, say -3, then the terms would alternate in sign. For example, a_2 would be 3*(-3) = -9, which is negative, but the problem says all terms are positive. Therefore, we discard r = -3, and take r = 2.So, the common ratio is 2. Now, we need to find the sum of a_3 + a_4 + a_5. Let's compute each term:a_3 = 3 * r^(3-1) = 3 * r^2. Since r is 2, a_3 = 3*4 = 12.a_4 = 3 * r^(4-1) = 3 * r^3 = 3*8 = 24.a_5 = 3 * r^(4) = 3*16 = 48.Adding them together: 12 + 24 + 48 = 84. So, the sum of a_3 + a_4 + a_5 is 84.Wait, let me verify the steps again to make sure I didn't make any mistakes. First, S_3 is 21. The formula for the sum of the first n terms in a geometric sequence is S_n = a_1*(1 - r^n)/(1 - r) when r ≠ 1. Let me try using that formula instead to see if I get the same result.Given a_1 = 3, n = 3, S_3 = 21. So,21 = 3*(1 - r^3)/(1 - r)Divide both sides by 3:7 = (1 - r^3)/(1 - r)Simplify the numerator: 1 - r^3 can be factored as (1 - r)(1 + r + r^2). So,7 = (1 - r)(1 + r + r^2)/(1 - r) => 7 = 1 + r + r^2.So, the equation becomes:1 + r + r^2 = 7Which simplifies to r^2 + r - 6 = 0, which is the same quadratic equation as before. So, that's consistent. Then, solving gives r = 2 or r = -3, and since all terms are positive, r must be 2. Then, the terms a_3, a_4, a_5 are 12, 24, 48, summing to 84. So that seems correct.Alternatively, perhaps I can compute the sum a_3 + a_4 + a_5 without calculating each term individually. Since it's a geometric sequence, the sum from a_3 to a_5 is a geometric series with first term a_3 = 12 and common ratio 2, for 3 terms. The sum would be a_3*(1 - r^3)/(1 - r). Wait, but since there are 3 terms, it's a_3 + a_4 + a_5 = a_3*(1 - r^3)/(1 - r). Wait, no, actually, the formula for the sum of the first n terms is S_n = a_1*(1 - r^n)/(1 - r). But in this case, the first term of this sub-series is a_3, so the sum would be S_3' = a_3*(1 - r^3)/(1 - r). Let's check that.Alternatively, since we need the sum of the 3rd, 4th, and 5th terms, which is equivalent to the sum from term 3 to term 5. The number of terms is 5 - 3 + 1 = 3 terms. Alternatively, you can think of it as the sum of the first 5 terms minus the sum of the first 2 terms.Sum from a_3 to a_5 = S_5 - S_2.But we can compute S_5 and S_2 if we know r. Since r is 2, let's compute S_5 and S_2:S_2 = a_1 + a_2 = 3 + 6 = 9.S_5 = 3*(1 - 2^5)/(1 - 2) = 3*(1 - 32)/(-1) = 3*(-31)/(-1) = 3*31 = 93.Therefore, S_5 - S_2 = 93 - 9 = 84. Which matches the previous result.Alternatively, if I use the formula for the sum starting at a_3:The sum of a_3 + a_4 + a_5 is a geometric series with first term a_3 = 12, ratio r = 2, and 3 terms. So the sum is 12*(1 - 2^3)/(1 - 2) = 12*(1 - 8)/(-1) = 12*(-7)/(-1) = 12*7 = 84. Same answer.So, multiple methods lead to the same answer, 84. Therefore, I can be confident that the sum of a_3 + a_4 + a_5 is 84.But let me just double-check once more with the original terms. If r = 2, then:a_1 = 3a_2 = 3*2 = 6a_3 = 6*2 = 12a_4 = 12*2 = 24a_5 = 24*2 = 48Thus, a_3 + a_4 + a_5 = 12 + 24 + 48 = 84. Correct.Alternatively, if I had a different ratio, but since we concluded that r must be 2 because the terms are all positive, this is the only possible ratio. So, the answer is 84.Final Answerboxed{84}</think>Given a geometric sequence ({a_n}) with all terms positive, where (a_1 = 3) and (S_3 = 21), we need to find the sum of (a_3 + a_4 + a_5).First, we recognize that the sum of the first three terms (S_3) is given by:[S_3 = a_1 + a_2 + a_3 = 3 + 3r + 3r^2 = 21]Simplifying this equation:[3 + 3r + 3r^2 = 21 implies 3r^2 + 3r - 18 = 0 implies r^2 + r - 6 = 0]Factoring the quadratic equation:[(r + 3)(r - 2) = 0]Since all terms are positive, the common ratio (r) must be positive. Thus, (r = 2).Next, we calculate the terms (a_3), (a_4), and (a_5):[a_3 = 3 cdot r^2 = 3 cdot 2^2 = 12][a_4 = 3 cdot r^3 = 3 cdot 2^3 = 24][a_5 = 3 cdot r^4 = 3 cdot 2^4 = 48]Summing these terms:[a_3 + a_4 + a_5 = 12 + 24 + 48 = 84]Thus, the sum of (a_3 + a_4 + a_5) is (boxed{84}).,<think>Okay, so I need to find the sum of the third, fourth, and fifth terms of a geometric sequence. The problem says that the first term a₁ is 3, and the sum of the first three terms S₃ is 21. All terms are positive. Hmm, let me recall what I know about geometric sequences.A geometric sequence is one where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, if a₁ is the first term, then the second term a₂ is a₁ * r, the third term a₃ is a₂ * r = a₁ * r², and so on. So in general, the nth term is a₁ * r^(n-1).Given that a₁ = 3, so the first three terms would be:a₁ = 3a₂ = 3 * ra₃ = 3 * r²Then, the sum S₃ is a₁ + a₂ + a₃ = 3 + 3r + 3r². According to the problem, this sum is 21. So I can set up the equation:3 + 3r + 3r² = 21Hmm, let me simplify that equation. First, divide both sides by 3 to make it easier:1 + r + r² = 7So the equation becomes:r² + r + 1 = 7Subtract 7 from both sides:r² + r + 1 - 7 = 0Which simplifies to:r² + r - 6 = 0Now, I need to solve this quadratic equation for r. Let me check if I can factor it. The quadratic is r² + r - 6. Let's see, I need two numbers that multiply to -6 and add to 1. Let's think: 3 and -2. 3 * (-2) = -6, and 3 + (-2) = 1. Yes, that works. So the factors would be (r + 3)(r - 2) = 0. Wait, let me check:(r + 3)(r - 2) = r² - 2r + 3r -6 = r² + r -6. Perfect. So the solutions are r = -3 or r = 2.But the problem states that all terms are positive. Since the common ratio is r, if r were negative, then the terms would alternate in sign. However, since all terms are positive, r must be positive. Therefore, r = 2 is the only valid solution. So the common ratio is 2.Now that I know r is 2, I can find the terms a₃, a₄, and a₅. Let's compute them:a₁ = 3a₂ = 3 * 2 = 6a₃ = 6 * 2 = 12a₄ = 12 * 2 = 24a₅ = 24 * 2 = 48So the sum a₃ + a₄ + a₅ = 12 + 24 + 48. Let's compute that:12 + 24 = 3636 + 48 = 84Therefore, the sum is 84. Wait, let me verify everything again to make sure I didn't make a mistake.First, confirming S₃: 3 + 6 + 12 = 21. Yes, that's correct because 3 + 6 is 9, plus 12 is 21. So that's right.Then, the common ratio is 2, so each subsequent term is multiplied by 2. So a₃ is 12, a₄ is 24, a₅ is 48. Adding those gives 12 + 24 + 48 = 84. Yep, that seems correct.Alternatively, maybe there's a formulaic way to compute the sum of terms from a₃ to a₅ without calculating each term individually. Let's see.In a geometric sequence, the sum of the first n terms is S_n = a₁(r^n - 1)/(r - 1). However, here we need the sum from a₃ to a₅. So that's the sum from term 3 to term 5. Which is equivalent to S₅ - S₂.First, let's compute S₅ and S₂ using the formula.S₅ = 3*(2^5 - 1)/(2 - 1) = 3*(32 - 1)/1 = 3*31 = 93S₂ = 3*(2^2 - 1)/(2 - 1) = 3*(4 - 1)/1 = 3*3 = 9So S₅ - S₂ = 93 - 9 = 84. Which matches the previous result. So that's good.Alternatively, another way to think about it is that a₃ is the third term, which is a₁*r². Then the terms a₃, a₄, a₅ form their own geometric sequence with first term a₃ = 3*r² and common ratio r. So the sum of these three terms is a₃ + a₄ + a₅ = a₃*(1 + r + r²). Since we know r = 2, substituting:a₃ = 3*2² = 12Sum = 12*(1 + 2 + 4) = 12*7 = 84. Same result. That's reassuring.So whether I compute each term individually, use the sum formula for the first five terms minus the first two, or treat the sum from a₃ to a₅ as a new geometric sequence starting at a₃, I get the same answer of 84.Therefore, I'm confident that the correct answer is 84.Final AnswerThe sum of (a_3 + a_4 + a_5) is boxed{84}.</think>Given a geometric sequence ({a_n}) with (a_1 = 3) and (S_3 = 21), we need to find the sum of (a_3 + a_4 + a_5).First, we know the sum of the first three terms of the geometric sequence is given by:[S_3 = a_1 + a_2 + a_3 = 3 + 3r + 3r^2 = 21]Simplifying this equation:[3 + 3r + 3r^2 = 21 implies 1 + r + r^2 = 7 implies r^2 + r - 6 = 0]Solving the quadratic equation (r^2 + r - 6 = 0), we factorize it as:[(r + 3)(r - 2) = 0]Since all terms are positive, the common ratio (r) must be positive, giving (r = 2).Next, we calculate the terms (a_3), (a_4), and (a_5):[begin{align*}a_1 &= 3, a_2 &= 3 times 2 = 6, a_3 &= 6 times 2 = 12, a_4 &= 12 times 2 = 24, a_5 &= 24 times 2 = 48.end{align*}]The sum (a_3 + a_4 + a_5) is:[12 + 24 + 48 = 84]We can also verify this using the sum formula for a geometric sequence. The sum from (a_3) to (a_5) can be found by:[S_5 - S_2 = left(3 frac{2^5 - 1}{2 - 1}right) - left(3 frac{2^2 - 1}{2 - 1}right) = 93 - 9 = 84]Thus, the sum of (a_3 + a_4 + a_5) is (boxed{84}).]

🔑:Given that a_1 = 3 and the sum of the first three terms S_3 = 21, we can find the common ratio and then calculate the sum of a_3 + a_4 + a_5.However, the detailed steps are omitted here. The final answer is boxed{84}.

🧠:[<think>Okay, so I need to find all continuous functions ( f: mathbb{R} rightarrow mathbb{R} ) that satisfy the functional equation:[f(xy) = fleft( frac{x^2 + y^2}{2} right) + (x - y)^2]for all real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but since the problem mentions continuity, that might help. Let me start by trying some standard approaches.First, maybe I can plug in specific values for ( x ) and ( y ) to simplify the equation and see if I can spot a pattern or figure out the form of ( f ).Let me try letting ( x = y ). Then the equation becomes:[f(x^2) = fleft( frac{x^2 + x^2}{2} right) + (x - x)^2]Simplifying the right-hand side:[f(x^2) = fleft( frac{2x^2}{2} right) + 0 implies f(x^2) = f(x^2)]Hmm, that just gives me an identity, which doesn't provide new information. Okay, maybe try different substitutions. What if I set ( y = 0 )? Let's see:Substituting ( y = 0 ):Left-hand side: ( f(x cdot 0) = f(0) )Right-hand side: ( fleft( frac{x^2 + 0^2}{2} right) + (x - 0)^2 = fleft( frac{x^2}{2} right) + x^2 )So we get:[f(0) = fleft( frac{x^2}{2} right) + x^2]Let me denote ( t = frac{x^2}{2} ), which implies ( x^2 = 2t ). Since ( x^2 ) is non-negative, ( t ) is non-negative. But since ( f ) is defined for all real numbers, I need to check if this substitution covers all possible ( t ), but ( t ) here is non-negative. However, maybe later I can extend it. Anyway, substituting ( t ):[f(0) = f(t) + 2t]Therefore, solving for ( f(t) ):[f(t) = f(0) - 2t]But this holds for ( t geq 0 ), since ( t = frac{x^2}{2} ). So for non-negative real numbers, ( f(t) = f(0) - 2t ).But the original functional equation requires ( f ) to be defined on all real numbers. So what about negative arguments? Let's see if we can find an expression for ( f ) when the input is negative.Another substitution: perhaps set ( x = 0 ). Let's try that.Set ( x = 0 ):Left-hand side: ( f(0 cdot y) = f(0) )Right-hand side: ( fleft( frac{0^2 + y^2}{2} right) + (0 - y)^2 = fleft( frac{y^2}{2} right) + y^2 )Again, similar to the previous substitution, we get:[f(0) = fleft( frac{y^2}{2} right) + y^2]Which is the same as before. So again, for ( t = frac{y^2}{2} geq 0 ), ( f(t) = f(0) - 2t ). So this doesn't help with negative arguments. Hmm.So far, we know that for all ( t geq 0 ), ( f(t) = f(0) - 2t ). But what about ( t < 0 )? How can we express ( f(t) ) for negative ( t )?Let me think. The original functional equation involves ( xy ) and ( frac{x^2 + y^2}{2} ). Note that ( frac{x^2 + y^2}{2} ) is always non-negative, as it's the average of squares. Therefore, the argument of ( f ) on the right-hand side is always non-negative. However, the left-hand side is ( f(xy) ), which can be negative if ( x ) and ( y ) have opposite signs.Therefore, to find ( f(z) ) for negative ( z ), we need to set ( xy = z ) where ( z ) is negative. So suppose ( z ) is a negative real number. Let me set ( x = sqrt{-2z} ) and ( y = -sqrt{-2z} ), but wait, let me check.Wait, perhaps a better approach. Let me pick ( x = a ) and ( y = -b ), so that ( xy = -ab ). Then the equation becomes:[f(-ab) = fleft( frac{a^2 + b^2}{2} right) + (a + b)^2]But we already know that ( fleft( frac{a^2 + b^2}{2} right) = f(0) - 2 cdot frac{a^2 + b^2}{2} = f(0) - (a^2 + b^2) )Therefore, substituting into the equation:[f(-ab) = [f(0) - (a^2 + b^2)] + (a + b)^2]Simplify the right-hand side:[f(-ab) = f(0) - (a^2 + b^2) + a^2 + 2ab + b^2 = f(0) + 2ab]But the left-hand side is ( f(-ab) ), so:[f(-ab) = f(0) + 2ab]Let me set ( z = -ab ). Then ( ab = -z ), so substituting:[f(z) = f(0) + 2(-z) = f(0) - 2z]Wait, but this is for ( z = -ab ). But ( a ) and ( b ) can be any real numbers, so ( ab ) can take any real value (since for any real number ( c ), set ( a = c ), ( b = 1 ), then ( ab = c )). Therefore, ( z = -ab ) can be any real number. Therefore, for any real ( z ), ( f(z) = f(0) - 2z ).But wait, let me check that again. If ( z = -ab ), then ( ab = -z ). But ( a ) and ( b ) can be any real numbers, so ( z ) can indeed be any real number. Therefore, the expression ( f(z) = f(0) - 2z ) holds for all real numbers ( z ).But wait, earlier when we considered ( t geq 0 ), we had ( f(t) = f(0) - 2t ), which matches this. So this suggests that for all real numbers ( z ), ( f(z) = f(0) - 2z ). Therefore, ( f ) is a linear function of the form ( f(z) = C - 2z ), where ( C = f(0) ).But we need to verify that this function actually satisfies the original equation. Let's check that.Let ( f(z) = C - 2z ). Then substitute into the functional equation:Left-hand side: ( f(xy) = C - 2xy )Right-hand side: ( fleft( frac{x^2 + y^2}{2} right) + (x - y)^2 = [C - 2 cdot frac{x^2 + y^2}{2}] + (x^2 - 2xy + y^2) )Simplify the right-hand side:[C - (x^2 + y^2) + x^2 - 2xy + y^2 = C - x^2 - y^2 + x^2 - 2xy + y^2 = C - 2xy]Which equals the left-hand side. Therefore, the function ( f(z) = C - 2z ) satisfies the equation for any constant ( C ).But wait, in our earlier steps, we found that ( f(z) = f(0) - 2z ), so ( C = f(0) ). Therefore, the constant term is determined by the value of ( f ) at 0. However, we need to check if there are any constraints on ( C ).Wait, let's substitute back into the equation when we set ( y = 0 ):From earlier, when ( y = 0 ), we had:[f(0) = fleft( frac{x^2}{2} right) + x^2]But if ( f(z) = C - 2z ), then:Left-hand side: ( f(0) = C - 0 = C )Right-hand side: ( fleft( frac{x^2}{2} right) + x^2 = [C - 2 cdot frac{x^2}{2}] + x^2 = C - x^2 + x^2 = C )So this holds true. Therefore, there's no restriction on ( C ); it's simply ( f(0) ). However, since we have already derived that ( f(z) = C - 2z ) for all ( z ), and this satisfies the functional equation for any constant ( C ), the general solution is ( f(z) = C - 2z ).But wait, is that possible? Let me test with specific numbers. Let's pick ( C = 0 ), so ( f(z) = -2z ).Check the equation:Left-hand side: ( f(xy) = -2xy )Right-hand side: ( fleft( frac{x^2 + y^2}{2} right) + (x - y)^2 = -2 cdot frac{x^2 + y^2}{2} + (x - y)^2 = -(x^2 + y^2) + x^2 - 2xy + y^2 = -x^2 - y^2 + x^2 - 2xy + y^2 = -2xy )Which matches. If ( C = 5 ), then:Left-hand side: ( f(xy) = 5 - 2xy )Right-hand side: ( 5 - 2 cdot frac{x^2 + y^2}{2} + (x - y)^2 = 5 - (x^2 + y^2) + x^2 - 2xy + y^2 = 5 - 2xy )Which also matches. So indeed, any constant ( C ) works. Therefore, the general solution is ( f(z) = C - 2z ), where ( C ) is a constant.However, wait a second. Let me check another substitution to ensure there isn't a hidden constraint. Let's pick ( x = 1 ) and ( y = 1 ):Left-hand side: ( f(1 cdot 1) = f(1) = C - 2(1) = C - 2 )Right-hand side: ( fleft( frac{1 + 1}{2} right) + (1 - 1)^2 = f(1) + 0 = C - 2(1) = C - 2 ). So that's consistent.Another test: ( x = 2 ), ( y = 3 ):Left-hand side: ( f(6) = C - 12 )Right-hand side: ( fleft( frac{4 + 9}{2} right) + (2 - 3)^2 = f(6.5) + 1 = (C - 13) + 1 = C - 12 ). Which matches.So it seems that regardless of the constant ( C ), the equation holds. Therefore, the general continuous solution is ( f(x) = C - 2x ).But wait, the problem says "find all continuous functions", so maybe the constant ( C ) can be any real number. However, let me recall that when we derived the function, we had ( f(z) = f(0) - 2z ), so ( C = f(0) ). Therefore, the constant term is just the value of the function at 0, which can indeed be any real number. Therefore, the family of solutions is ( f(x) = C - 2x ) for some constant ( C in mathbb{R} ).But let me verify once more if there are any other possible functions. Since we used substitutions that led us directly to this solution and verified it works, and since the problem states that ( f ) is continuous, I don't see any other possible solutions. Usually, in functional equations, without continuity, there might be pathological solutions, but with continuity, the solution is unique up to constants determined by the equation. However, in this case, the constant isn't determined by the equation, which is a bit strange. Wait, but when I set ( y = 0 ), we found ( f(0) = f(0) - 2 cdot 0 + 0 ), which is an identity, so it doesn't impose any restrictions on ( f(0) ). Therefore, ( C ) can indeed be any real number.Wait, but hold on, in the step where I set ( x = a ), ( y = -b ), and derived that ( f(z) = f(0) - 2z ) for all real ( z ), which includes all real numbers, both positive and negative. So this suggests that the only possible functions are linear functions of the form ( f(z) = C - 2z ). Therefore, the general continuous solution is indeed ( f(z) = C - 2z ).But let me check another substitution where ( x ) and ( y ) are such that ( xy ) is negative. For example, take ( x = 1 ), ( y = -1 ):Left-hand side: ( f(-1) = C - 2(-1) = C + 2 )Right-hand side: ( fleft( frac{1 + 1}{2} right) + (1 - (-1))^2 = f(1) + (2)^2 = (C - 2(1)) + 4 = C - 2 + 4 = C + 2 ). Which matches.Another test case: ( x = 3 ), ( y = -2 ):Left-hand side: ( f(-6) = C - 2(-6) = C + 12 )Right-hand side: ( fleft( frac{9 + 4}{2} right) + (3 - (-2))^2 = f(6.5) + 25 = (C - 13) + 25 = C + 12 ). Perfect.Therefore, all test cases seem to validate that ( f(z) = C - 2z ) satisfies the functional equation for any constant ( C ). Since the problem asks for all continuous functions, and we derived that the solution must be linear of this form, I think that's the complete solution. There are no other continuous functions that satisfy the equation because the functional equation essentially forces ( f ) to be linear, and the continuity ensures there are no other pathological solutions.So, putting it all together, the continuous solutions are all linear functions with slope -2, i.e., ( f(x) = C - 2x ), where ( C ) is a real constant. However, the problem might expect a specific form. Let me check if ( C ) can be determined. But in our analysis, ( C ) remained arbitrary because substituting back into the equation didn't impose any restrictions on ( C ). For example, when we set ( x = y = 0 ), let's see:If ( x = y = 0 ):Left-hand side: ( f(0 cdot 0) = f(0) = C )Right-hand side: ( fleft( frac{0 + 0}{2} right) + (0 - 0)^2 = f(0) + 0 = C ). So again, it's consistent for any ( C ).Therefore, ( C ) is indeed arbitrary. Hence, the final answer is that all continuous functions satisfying the equation are of the form ( f(x) = C - 2x ), where ( C ) is a constant. However, since the problem asks to "find all continuous functions", we can write the general solution as ( f(x) = -2x + C ), or equivalently ( f(x) = -2x + K ), where ( K ) is a constant.But let me check once more for any possible oversight. Suppose there exists another function different from ( -2x + C ) that satisfies the equation. But since through substitution we derived that ( f(z) = f(0) - 2z ), which is linear, and with continuity, there can't be any non-linear solutions. Hence, the conclusion holds.Final AnswerThe continuous functions satisfying the equation are all linear functions of the form boxed{f(x) = -2x + C} where ( C ) is a real constant.</think>To find all continuous functions ( f : mathbb{R} rightarrow mathbb{R} ) that satisfy the functional equation:[f(xy) = fleft( frac{x^2 + y^2}{2} right) + (x - y)^2]we start by substituting specific values for ( x ) and ( y ).1. Substituting ( y = 0 ): [ f(0) = fleft( frac{x^2}{2} right) + x^2 ] Let ( t = frac{x^2}{2} ), then ( x^2 = 2t ). This gives: [ f(0) = f(t) + 2t implies f(t) = f(0) - 2t quad text{for } t geq 0 ]2. Substituting ( x = 0 ): [ f(0) = fleft( frac{y^2}{2} right) + y^2 ] Similarly, this gives the same result for ( t geq 0 ).3. Considering negative arguments: Set ( x = a ) and ( y = -b ), then: [ f(-ab) = fleft( frac{a^2 + b^2}{2} right) + (a + b)^2 ] Substituting ( fleft( frac{a^2 + b^2}{2} right) = f(0) - (a^2 + b^2) ): [ f(-ab) = f(0) - (a^2 + b^2) + a^2 + 2ab + b^2 = f(0) + 2ab ] Let ( z = -ab ), then ( ab = -z ): [ f(z) = f(0) - 2z quad text{for all } z in mathbb{R} ]4. Verification: The function ( f(z) = C - 2z ) (where ( C = f(0) )) satisfies the original equation. Substituting ( f(z) = C - 2z ) into both sides of the equation confirms this.Thus, the continuous functions satisfying the equation are all linear functions of the form:[boxed{f(x) = -2x + C}]where ( C ) is a real constant.,<think>Okay, so I need to find all continuous functions f from the real numbers to the real numbers that satisfy the functional equation f(xy) = f((x² + y²)/2) + (x - y)² for any real numbers x and y. Hmm, functional equations can be tricky, but maybe I can find some substitutions or specific values for x and y that simplify things.First, let's see if plugging in some particular values for x and y helps. Maybe setting x = y? If I do that, then the left-hand side becomes f(x²), and the right-hand side becomes f((x² + x²)/2) + (x - x)², which simplifies to f(x²) + 0. So that gives me f(x²) = f(x²). Well, that doesn't tell me anything new. Maybe that wasn't helpful.What if I set y = 0? Let's try that. Then the equation becomes f(0) = f((x² + 0)/2) + (x - 0)², so f(0) = f(x²/2) + x². Hmm, interesting. Let me write that down:f(0) = f(x²/2) + x² for all real x.This seems like it relates f at 0 to f at x²/2. Maybe I can express f in terms of x² here. Let's rearrange the equation:f(x²/2) = f(0) - x².But x²/2 can be any non-negative real number, since x² is non-negative. So perhaps for any z ≥ 0, we can write z = x²/2, so x² = 2z, and then:f(z) = f(0) - 2z.So this suggests that for all non-negative z, f(z) = f(0) - 2z. But the functional equation is supposed to hold for all real numbers x and y, so maybe f is defined for all real numbers, but the substitution y=0 gives us information only about non-negative arguments. Wait, but the original problem states that f is defined from R to R, so f needs to be defined for all real numbers, not just non-negative. But with y=0, we only get information about f(x²/2), which is non-negative. So maybe we need to figure out the behavior for negative arguments as well.Alternatively, maybe f is a quadratic function. Let me make a guess. Suppose f(t) = at² + bt + c. Then maybe we can substitute this into the functional equation and see what the coefficients must be. Let's try that.Assume f(t) = at² + bt + c. Then the left-hand side of the equation is f(xy) = a(xy)² + b(xy) + c.The right-hand side is f((x² + y²)/2) + (x - y)² = a[(x² + y²)/2]² + b[(x² + y²)/2] + c + (x - y)².Let's expand both sides:Left-hand side (LHS):a x² y² + b x y + c.Right-hand side (RHS):a [(x² + y²)^2 /4] + b (x² + y²)/2 + c + x² - 2xy + y².First, expand (x² + y²)^2:(x² + y²)^2 = x^4 + 2x² y² + y^4.So RHS becomes:a (x^4 + 2x² y² + y^4)/4 + b (x² + y²)/2 + c + x² - 2xy + y².Let me write that out term by term:- For the first term: (a/4)x^4 + (a/2)x² y² + (a/4)y^4- Second term: (b/2)x² + (b/2)y²- Third term: c- Fourth term: x² - 2xy + y²So combining all terms:RHS = (a/4)x^4 + (a/2)x² y² + (a/4)y^4 + (b/2)x² + (b/2)y² + c + x² - 2xy + y².Now let's collect like terms:- x^4: (a/4)- y^4: (a/4)- x² y²: (a/2)- x²: (b/2) + 1- y²: (b/2) + 1- xy: -2- Constants: cComparing LHS and RHS:LHS: a x² y² + b x y + c.RHS: (a/4)x^4 + (a/4)y^4 + (a/2)x² y² + [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy + c.So, to have equality for all x and y, the coefficients of corresponding terms must be equal.But on the LHS, there are no x^4 or y^4 terms, so their coefficients must be zero. Similarly, coefficients for x², y², and xy must match, as well as constants.So, first, set the coefficients of x^4 and y^4 to zero:(a/4) = 0 ⇒ a = 0.But if a = 0, then the x² y² term on the LHS would be 0, but on the RHS, the x² y² term is (a/2)x² y² = 0. So that's okay.Wait, but if a = 0, then f(t) = bt + c. So f is a linear function.Wait, but let's check. If a = 0, then the LHS is 0 + bxy + c.The RHS is 0 + 0 + 0 + [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy + c.So LHS: bxy + c.RHS: [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy + c.Therefore, setting LHS = RHS:bxy + c = [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy + c.Subtract c from both sides:bxy = [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy.Now, collect like terms:Left side: bxy.Right side: [(b/2) + 1]x² + [(b/2) + 1]y² - 2xy.So, for this equality to hold for all x and y, the coefficients of x², y², and xy must match on both sides. But the left side has no x² or y² terms, so their coefficients on the right must be zero. Similarly, the coefficient of xy on the left is b, and on the right, it's -2.Therefore:Coefficients of x² and y²:[(b/2) + 1] = 0.So (b/2) + 1 = 0 ⇒ b/2 = -1 ⇒ b = -2.Coefficient of xy:Left side: b.Right side: -2.So b = -2. Which matches the previous result.Therefore, with a = 0 and b = -2, we have:f(t) = -2t + c.Now, check if this works in the original equation. Let's verify.Let f(t) = -2t + c. Then:Left-hand side: f(xy) = -2xy + c.Right-hand side: f((x² + y²)/2) + (x - y)^2 = -2*( (x² + y²)/2 ) + c + x² - 2xy + y².Simplify:-2*( (x² + y²)/2 ) = -(x² + y²).So right-hand side becomes:-(x² + y²) + c + x² - 2xy + y² = (-x² - y² + x² + y²) + c - 2xy = 0 + c - 2xy = -2xy + c.Which matches the left-hand side. So indeed, f(t) = -2t + c satisfies the equation for any constant c. However, the problem states that f is defined for all real numbers, and we need to check if there are any constraints on c.Wait, but earlier when we substituted y = 0, we had f(0) = f(x²/2) + x². If f(t) = -2t + c, then:Left side: f(0) = -2*0 + c = c.Right side: f(x²/2) + x² = (-2*(x²/2) + c) + x² = (-x² + c) + x² = c.So this holds for any x. So that's consistent.But wait, the functional equation is supposed to hold for all real numbers x and y. If our solution is f(t) = -2t + c, which works, then that seems to be a valid solution. But the problem says "find all continuous functions," so maybe there are other solutions?But wait, in our initial substitution, when we set y = 0, we found that f(z) = f(0) - 2z for z ≥ 0, which matches the linear function. However, for z < 0, we don't have any constraints from that substitution.But the original functional equation must hold for all real x and y. Let's see if we can get information about negative arguments. For example, suppose we take x and y such that xy is negative. Then (x² + y²)/2 is still non-negative. Hmm. So maybe f is determined only on non-negative numbers by the equation, but for negative numbers, we need to see if the function can be extended.Wait, but if x and y are real numbers, then xy can be any real number (positive, negative, or zero), depending on x and y. Similarly, (x² + y²)/2 is always non-negative. So perhaps for any real number t, we can write t as xy for some x and y, but depending on t's sign.For example, if t is negative, then we can write t = x y where one of x or y is negative. Then the functional equation would relate f(t) to f((x² + y²)/2) + (x - y)^2.But (x² + y²)/2 is still non-negative, so f(t) would be expressed in terms of f evaluated at a non-negative argument, plus some quadratic term.But we already have f(z) = -2 z + c for z ≥ 0. So for t negative, we can express f(t) as f((x² + y²)/2) + (x - y)^2 where xy = t.But since (x² + y²)/2 is non-negative, f((x² + y²)/2) = -2*( (x² + y²)/2 ) + c = - (x² + y² ) + c.Therefore, substituting into the functional equation:f(t) = [ - (x² + y² ) + c ] + (x - y)^2.But (x - y)^2 = x² - 2xy + y², so:f(t) = -x² - y² + c + x² - 2xy + y² = (-x² + x²) + (- y² + y²) + c - 2xy = 0 + 0 + c - 2xy = c - 2xy.But t = xy, so f(t) = c - 2t.Which is the same as f(t) = -2t + c. Therefore, even for negative t, f(t) = -2t + c. Therefore, the function is linear everywhere, with slope -2 and intercept c.But earlier, when we considered the substitution y = 0, we found that for z ≥ 0, f(z) = f(0) - 2z. So f(0) = c, which is consistent. Thus, the function is linear everywhere, f(t) = -2t + c, which can be written as f(t) = -2t + c, where c is a constant.But wait, the problem says "find all continuous functions," so maybe this is the only solution? However, we derived that f must be linear, and specifically with slope -2. So the only solutions are linear functions f(t) = -2t + c, where c is a constant.But let's check if this works for all x and y. Let's verify with a specific example. Let x = 1, y = 1. Then:Left-hand side: f(1*1) = f(1) = -2(1) + c = -2 + c.Right-hand side: f((1 + 1)/2) + (1 - 1)^2 = f(1) + 0 = -2 + c. So that's okay.Another example: x = 2, y = 3.Left-hand side: f(2*3) = f(6) = -12 + c.Right-hand side: f((4 + 9)/2) + (2 - 3)^2 = f(6.5) + 1 = (-13 + c) + 1 = -12 + c. That works.Negative arguments: Let x = -1, y = 1. Then xy = -1.Left-hand side: f(-1) = -2*(-1) + c = 2 + c.Right-hand side: f((1 + 1)/2) + (-1 -1)^2 = f(1) + 4 = (-2 + c) + 4 = 2 + c. That works too.Another one: x = -2, y = 3. Then xy = -6.Left-hand side: f(-6) = -2*(-6) + c = 12 + c.Right-hand side: f((4 + 9)/2) + (-2 -3)^2 = f(6.5) + 25 = (-13 + c) + 25 = 12 + c. Perfect.So seems like the linear function f(t) = -2t + c satisfies the equation for any constant c. But wait, let me check if c can be any constant or if there's a restriction. When we considered y=0, we had f(0) = f(x²/2) + x². Plugging f(t) = -2t + c into that gives c = (-2*(x²/2) + c) + x² => c = (-x² + c) + x² => c = c. So no restriction on c, it can be any constant.Therefore, the general continuous solution is f(t) = -2t + c for some constant c. But the problem says "find all continuous functions," and we've derived that this linear function is the only solution. Since linear functions are continuous, these are all the solutions. So the answer should be f(x) = -2x + c for some constant c. But let me check once more if there could be non-linear solutions.Wait, in the beginning, I assumed f was quadratic and found that a=0, leading to a linear function. But maybe there's a non-polynomial solution? However, the substitution with y=0 gave us f(z) = f(0) - 2z for z >=0. Since the problem states that f is continuous everywhere, and we have f(z) = -2z + c for z >=0, then for z <0, since f is continuous, we can extend it in such a way that f(z) must also be -2z + c. Because if there's a different behavior for z <0, the function would have a discontinuity at z=0 unless it also follows the same linear rule. Therefore, continuity forces f(z) = -2z + c for all real z.Therefore, the only continuous solutions are linear functions of the form f(x) = -2x + c, where c is a real constant.Final AnswerThe continuous functions satisfying the equation are all linear functions of the form boxed{f(x) = -2x + C} where ( C ) is a real constant.</think>To find all continuous functions ( f : mathbb{R} rightarrow mathbb{R} ) that satisfy the functional equation ( f(xy) = fleft(frac{x^2 + y^2}{2}right) + (x - y)^2 ) for any real numbers ( x ) and ( y ), we start by examining specific substitutions for ( x ) and ( y ).1. Substitution ( y = 0 ): [ f(0) = fleft(frac{x^2}{2}right) + x^2 ] Rearranging gives: [ fleft(frac{x^2}{2}right) = f(0) - x^2 ] Letting ( z = frac{x^2}{2} ), we get ( x^2 = 2z ), so: [ f(z) = f(0) - 2z quad text{for} quad z geq 0 ]2. Assuming a linear form: We hypothesize ( f(t) = -2t + c ) where ( c ) is a constant. Substituting this into the functional equation, we check both sides: - Left-hand side (LHS): [ f(xy) = -2xy + c ] - Right-hand side (RHS): [ fleft(frac{x^2 + y^2}{2}right) + (x - y)^2 = -2left(frac{x^2 + y^2}{2}right) + c + x^2 - 2xy + y^2 ] Simplifying the RHS: [ - (x^2 + y^2) + c + x^2 - 2xy + y^2 = -2xy + c ] Both sides match, confirming that ( f(t) = -2t + c ) satisfies the equation.3. Continuity consideration: Since ( f ) is continuous and we have determined ( f(z) = -2z + c ) for ( z geq 0 ), continuity at ( z = 0 ) and for all real numbers implies ( f ) must be linear everywhere. Thus, ( f(t) = -2t + c ) for all ( t in mathbb{R} ).The continuous functions satisfying the given functional equation are all linear functions of the form:[boxed{f(x) = -2x + C}]where ( C ) is a real constant.]

🔑:1. Define a new function ( g(x) ): Let ( g(x) = f(x) + 2x ). Then the given functional equation ( f(xy) = fleft(frac{x^2 + y^2}{2}right) + (x - y)^2 ) can be rewritten in terms of ( g ): [ f(xy) = g(xy) - 2xy ] and [ fleft(frac{x^2 + y^2}{2}right) = gleft(frac{x^2 + y^2}{2}right) - x^2 - y^2 ] Substituting these into the original equation, we get: [ g(xy) - 2xy = gleft(frac{x^2 + y^2}{2}right) - x^2 - y^2 + (x - y)^2 ] Simplifying the right-hand side: [ g(xy) - 2xy = gleft(frac{x^2 + y^2}{2}right) - x^2 - y^2 + x^2 - 2xy + y^2 ] [ g(xy) - 2xy = gleft(frac{x^2 + y^2}{2}right) - 2xy ] Therefore, we have: [ g(xy) = gleft(frac{x^2 + y^2}{2}right) ]2. Analyze the functional equation ( g(xy) = gleft(frac{x^2 + y^2}{2}right) ): - Set ( y = 0 ): [ g(0) = gleft(frac{x^2 + 0^2}{2}right) = gleft(frac{x^2}{2}right) ] This implies that ( g ) is constant for all non-negative arguments. Let ( g(x) = c ) for ( x geq 0 ). - Set ( y = -x ): [ g(-x^2) = gleft(frac{x^2 + (-x)^2}{2}right) = g(x^2) ] Since ( g(x^2) = c ) for ( x geq 0 ), we have ( g(-x^2) = c ). Therefore, ( g(x) = c ) for all ( x in mathbb{R} ).3. Determine the form of ( f(x) ): Since ( g(x) = c ) for all ( x in mathbb{R} ), we have: [ f(x) + 2x = c ] Solving for ( f(x) ): [ f(x) = c - 2x ] where ( c ) is a real constant.Conclusion:The function ( f(x) = c - 2x ) satisfies the given functional equation for any real constant ( c ).The final answer is ( boxed{ f(x) = c - 2x } ) for any real constant ( c ).

🧠:[<think>Okay, so I need to find all integer solutions that satisfy both inequalities in the system. Let me start by writing down the two inequalities again to make sure I have them right. The first one is 2x + 4 > 0, and the second one is 1 + x ≥ 2x - 1. Alright, so I need to solve each inequality separately and then find the intersection of their solution sets, right? Since we’re dealing with integers, the final answer will be a list of integers that satisfy both conditions.Starting with the first inequality: 2x + 4 > 0. Let me solve this step by step. First, subtract 4 from both sides. That gives me 2x > -4. Then, divide both sides by 2. So x > -2. Hmm, okay. So the first inequality simplifies to x being greater than -2. Since we’re dealing with integers, that means x can be -1, 0, 1, 2, and so on.Now moving on to the second inequality: 1 + x ≥ 2x - 1. Let me solve this one as well. First, I want to get all the x terms on one side. Let me subtract x from both sides. That gives me 1 ≥ x - 1. Then, add 1 to both sides to isolate x. So 1 + 1 ≥ x, which simplifies to 2 ≥ x, or equivalently, x ≤ 2. So the second inequality tells me that x has to be less than or equal to 2. Since we’re dealing with integers, that means x can be 2, 1, 0, -1, -2, etc., all the way to negative infinity. But wait, the first inequality said x > -2, so x has to be greater than -2. So combining these two, x must be greater than -2 and less than or equal to 2.But let me double-check my steps to make sure I didn't make a mistake. For the first inequality: 2x + 4 > 0. Subtract 4: 2x > -4. Divide by 2: x > -2. Yup, that seems right. For the second inequality: 1 + x ≥ 2x -1. Subtract x: 1 ≥ x -1. Add 1: 2 ≥ x. Yes, that's correct. So x ≤ 2. So combining both, x must be in the interval (-2, 2]. But since x has to be an integer, the possible integer values are -1, 0, 1, 2. Wait, but let me check if that's right.Wait, x has to be greater than -2, so the smallest integer greater than -2 is -1. Then, x can be -1, 0, 1, 2. Because 2 is included since x ≤ 2. So the integer solutions are -1, 0, 1, 2. Let me check each of these in the original inequalities to make sure they work.First, x = -1:Check first inequality: 2*(-1) + 4 = -2 + 4 = 2 > 0. True.Check second inequality: 1 + (-1) = 0 ≥ 2*(-1) -1 = -2 -1 = -3. 0 ≥ -3 is true. So -1 is valid.x = 0:First inequality: 2*0 +4 = 4 >0. True.Second inequality: 1 +0 =1 ≥ 2*0 -1 = -1. 1 ≥ -1. True. So 0 is valid.x =1:First inequality: 2*1 +4 =6>0. True.Second inequality:1 +1=2 ≥ 2*1 -1=2-1=1. 2≥1. True. Valid.x=2:First inequality:2*2 +4=8>0. True.Second inequality:1 +2=3 ≥2*2 -1=4 -1=3. 3≥3. True. So 2 is valid.So all these integers work. Now, let me check the next integer, 3, just to make sure it's not included. x=3:First inequality:2*3 +4=10>0. True.Second inequality:1 +3=4 ≥2*3 -1=6 -1=5. 4 ≥5? No, that's false. So 3 is not included. Similarly, if x=-2, which was excluded by the first inequality:x=-2:First inequality:2*(-2) +4=-4 +4=0>0? No, 0 is not greater than 0. So x=-2 is invalid. So indeed, the integer solutions are -1, 0, 1, 2. Therefore, the set of integer solutions is {-1, 0, 1, 2}.Wait, but the problem says "using the listing method". I think that means we need to list all the elements in the set. Since the integers are -1, 0, 1, 2, that's the list. So putting them in order from smallest to largest, separated by commas, enclosed in braces. So the answer should be boxed{{-1, 0, 1, 2}}.But let me just confirm once again. The first inequality: x > -2. Integers greater than -2 start at -1. The second inequality: x ≤ 2. So the integers from -1 up to 2 inclusive. So yes, exactly those four numbers. So I think that's correct. No mistakes in my reasoning. All test cases worked.Final AnswerThe set of integer solutions is boxed{{-1, 0, 1, 2}}.</think>To find the set of integer solutions for the system of inequalities:[begin{cases}2x + 4 > 0 1 + x geq 2x - 1end{cases}]we solve each inequality separately and then find the intersection of their solution sets.1. Solving the first inequality (2x + 4 > 0): [ 2x + 4 > 0 2x > -4 x > -2 ] This means (x) must be greater than (-2).2. Solving the second inequality (1 + x geq 2x - 1): [ 1 + x geq 2x - 1 1 + x - x geq 2x - 1 - x 1 geq x - 1 1 + 1 geq x 2 geq x ] This means (x) must be less than or equal to (2).Combining both inequalities, we get:[-2 < x leq 2]Since (x) must be an integer, the valid integer solutions are (-1, 0, 1, 2). Checking each of these values in the original inequalities confirms they satisfy both conditions.Thus, the set of integer solutions is (boxed{{-1, 0, 1, 2}}).,<think>Okay, let's see. I need to find all integer solutions that satisfy both inequalities in the system. The system is:1. 2x + 4 > 02. 1 + x ≥ 2x - 1Hmm, starting with the first inequality: 2x + 4 > 0. Let me solve for x. Subtract 4 from both sides? Wait, no, first step is to isolate the term with x. So, 2x + 4 > 0. Let me subtract 4 from both sides:2x > -4Then divide both sides by 2:x > -2Okay, so the first inequality tells me that x must be greater than -2.Now the second inequality: 1 + x ≥ 2x - 1. Let me solve this step by step as well. First, maybe bring all the x terms to one side. Subtract x from both sides:1 ≥ x - 1Wait, that's not right. Let me check again. Starting with 1 + x ≥ 2x - 1.If I subtract x from both sides, I get:1 ≥ x - 1Then, add 1 to both sides:1 + 1 ≥ xSo, 2 ≥ x, which means x ≤ 2.Wait, so the second inequality simplifies to x ≤ 2.So putting both inequalities together, x must be greater than -2 and less than or equal to 2. So the solution set in real numbers would be (-2, 2]. But since we need integer solutions, x has to be an integer within that interval.Let me list the integers greater than -2 and less than or equal to 2. So starting from the next integer after -2, which is -1, then 0, 1, 2. Let me check each integer:-1: Is -1 > -2? Yes. And -1 ≤ 2? Yes. So -1 is included.0: 0 > -2? Yes. 0 ≤ 2? Yes.1: Similarly, yes.2: 2 > -2? Yes. 2 ≤ 2? Yes.So the integer solutions are -1, 0, 1, 2.Wait, but hold on. Let me double-check the inequalities with each of these values to make sure.For x = -1:First inequality: 2*(-1) + 4 = -2 + 4 = 2 > 0. True.Second inequality: 1 + (-1) = 0 ≥ 2*(-1) -1 = -2 -1 = -3. 0 ≥ -3. True.x=0:First: 0 + 4 = 4 >0. True.Second: 1 + 0 =1 ≥ -1. True.x=1:First: 2 + 4 =6 >0. True.Second: 1 +1 =2 ≥ 2*1 -1 =2 -1=1. 2 ≥1. True.x=2:First: 4 +4=8>0. True.Second: 1 +2=3 ≥ 4 -1=3. 3 ≥3. True.So all those work. Are there any other integers in that interval?Wait, between -2 and 2, the integers are -1, 0, 1, 2. Because x has to be greater than -2, so -2 is excluded, and the next integer is -1. Then up to 2, inclusive.So the list is indeed -1, 0, 1, 2.But let me check if there are any other integers outside that range that might satisfy both inequalities. For example, x=-2:First inequality: 2*(-2) +4= -4 +4=0. Which is not greater than 0. So x=-2 is excluded.x=3:Second inequality: 3 ≤2? No, so x=3 is excluded. Similarly, x= -1.5 is not integer, so we don't care. So yes, the integer solutions are -1, 0, 1, 2.Therefore, using the listing method, the set is {-1, 0, 1, 2}.Final AnswerThe set of integer solutions is boxed{{-1, 0, 1, 2}}.</think>To find the set of integer solutions for the system of inequalities:[begin{cases}2x + 4 > 0 1 + x geq 2x - 1end{cases}]we start by solving each inequality separately.1. For the first inequality (2x + 4 > 0): [ 2x + 4 > 0 implies 2x > -4 implies x > -2 ]2. For the second inequality (1 + x geq 2x - 1): [ 1 + x geq 2x - 1 implies 1 + x - 2x geq -1 implies 1 - x geq -1 implies -x geq -2 implies x leq 2 ]Combining both inequalities, we get:[-2 < x leq 2]The integer solutions within this interval are (-1, 0, 1, 2). Verifying each integer:- For (x = -1): Both inequalities hold true.- For (x = 0): Both inequalities hold true.- For (x = 1): Both inequalities hold true.- For (x = 2): Both inequalities hold true.Thus, the set of integer solutions is ({-1, 0, 1, 2}).[boxed{{-1, 0, 1, 2}}]]

🔑:Solve the system of inequalities begin{cases} left.begin{matrix}2x+4>0 1+x≥2x-1end{matrix}right.end{cases} to get -20 1+x≥2x-1end{matrix}right.end{cases} is {-1, 0, 1, 2}Hence, the answer is: boxed{{-1, 0, 1, 2}}Solve the inequalities to find the range of x, and then list the set of integer solutions accordingly.This question tests the method of solving a system of inequalities and the way to express sets, and it is a basic question.