🧠:Okay, let me try to figure out this problem step by step. So, we have a 2x2 real matrix A with determinant d, which is not zero. And we know that the determinant of A plus d times its adjugate (A*) is zero. We need to prove that the determinant of A minus d times its adjugate is 4. First, let me recall some properties of 2x2 matrices. For a 2x2 matrix A, the adjugate matrix A* (which is sometimes called the classical adjoint) has a specific form. If A is [[a, b], [c, d]], then A* is [[d, -b], [-c, a]]. Also, for 2x2 matrices, we know that A* = det(A) * A^{-1} when A is invertible. Since det(A) = d ≠ 0, A is invertible, so A* = d * A^{-1}. Wait, no, hold on. Actually, the adjugate matrix satisfies A * A* = det(A) * I. So, if A is invertible, then A* = det(A) * A^{-1}. Therefore, in this case, A* = d * A^{-1}. That seems important. Let me verify that. If A is invertible, then yes, multiplying both sides of A * A* = det(A) * I by A^{-1} gives A* = det(A) * A^{-1}. So that's correct. So, A* = d * A^{-1}. Therefore, the expression A + d A* can be rewritten using this. Let's substitute A* with d * A^{-1}:A + d A* = A + d*(d * A^{-1}) = A + d^2 A^{-1}Similarly, A - d A* = A - d*(d * A^{-1}) = A - d^2 A^{-1}So, the problem states that det(A + d^2 A^{-1}) = 0, and we need to prove det(A - d^2 A^{-1}) = 4.Hmm. Maybe we can use some properties of determinants here. Let's denote B = A + d^2 A^{-1}, and we know det(B) = 0. Also, note that for 2x2 matrices, the determinant of a matrix plus a scalar multiple of its inverse can sometimes be simplified. But I'm not sure. Let me think.Alternatively, since A is a 2x2 matrix, maybe we can express everything in terms of the trace and determinant. For a 2x2 matrix, the characteristic polynomial is λ^2 - tr(A) λ + det(A) = 0. Also, the adjugate matrix A* has trace equal to the trace of A, since for a 2x2 matrix, the adjugate's diagonal entries are the same as the original matrix but swapped. Wait, no. If A is [[a, b], [c, d]], then A* is [[d, -b], [-c, a]], so the trace of A* is d + a, which is the same as the trace of A, which is a + d. So tr(A*) = tr(A). Also, determinant of A* is det(A*) = (a d - b c) = det(A) = d. Wait, no. Wait, det(A*) for a 2x2 matrix. Wait, if A* is [[d, -b], [-c, a]], then det(A*) is d*a - (-b)(-c) = a d - b c = det(A). So det(A*) = det(A). So for 2x2 matrices, det(A*) = det(A). Interesting. So det(A*) = d as well. But let's get back to the problem. Let's try to compute det(A + d A*). Let's express this in terms of A's trace and determinant. Let me recall that for any 2x2 matrix M, det(M) = (tr(M)^2 - tr(M^2)) / 2. But not sure if that helps here. Alternatively, for 2x2 matrices, if we have two matrices, maybe we can use the identity det(X + Y) = det X + det Y + tr(X) tr(Y) - tr(XY). Wait, is that correct? Let me check. No, actually, the determinant of a sum is not simply the sum of determinants. There is a formula for 2x2 matrices, which is det(A + B) = det A + det B + tr(A) tr(B) - tr(AB). Let me verify this with a simple example. Let A be [[1, 0], [0, 0]] and B be [[0, 0], [0, 1]]. Then det(A + B) = det([[1, 0], [0, 1]]) = 1. det A = 0, det B = 0, tr(A) = 1, tr(B) = 1, tr(AB) = 0. So 0 + 0 + 1*1 - 0 = 1, which matches det(A + B). So yes, this formula seems to hold: det(A + B) = det A + det B + tr(A) tr(B) - tr(AB). Therefore, maybe we can apply this formula to compute det(A + d A*). Let's let B = d A*. So det(A + B) = det A + det B + tr(A) tr(B) - tr(AB). According to the problem, this equals zero. Let's compute each term.First, det A = d. Then det B = det(d A*) = d^2 det(A*) = d^2 * d = d^3, since determinant is multiplicative and scalar multiplication by d in 2x2 matrix raises determinant by d^2. Wait, wait: det(k M) = k^n det(M) for n x n matrix. So since we're in 2x2, det(d A*) = d^2 det(A*) = d^2 * det(A) = d^2 * d = d^3. Wait, det(A*) is equal to det(A) for 2x2 matrices? Wait, earlier we thought that det(A*) = det(A). Let me check again. If A is invertible, then A* = det(A) A^{-1}, so det(A*) = det(det(A) A^{-1}) = (det(A))^2 det(A^{-1}) = (det(A))^2 * (1/det(A)) ) = det(A). Therefore, yes, det(A*) = det(A). So det(A*) = d. Therefore, det(d A*) = d^2 * d = d^3. So det B = d^3. Then tr(A) is, let's say, tr(A) = t. Then tr(B) = tr(d A*) = d tr(A*) = d tr(A), since tr(A*) = tr(A). So tr(B) = d t. Then tr(AB) = tr(A * d A*) = d tr(A A*). But A A* = det(A) I = d I, so tr(A A*) = tr(d I) = 2 d. Therefore, tr(AB) = d * 2 d = 2 d^2. Putting it all together: det(A + B) = d + d^3 + t * (d t) - 2 d^2. According to the problem, this is zero. So:d + d^3 + d t^2 - 2 d^2 = 0We can factor out a d:d (1 + d^2 + t^2 - 2 d) = 0But d ≠ 0, so:1 + d^2 + t^2 - 2 d = 0So t^2 + d^2 - 2 d + 1 = 0Which can be written as t^2 + (d - 1)^2 = 0Wait, t is the trace of A, which is a real number, and d is a real number. The sum of squares equals zero only if each term is zero. Therefore:t = 0 and d - 1 = 0, so d = 1Wait, but the problem states that d ≠ 0, but here we get d = 1. So if d must be 1, then maybe in the problem d is given as d ≠ 0, but from the equation t^2 + (d - 1)^2 = 0, since t and d are real numbers, this equation implies both t = 0 and d = 1. Therefore, under the problem's conditions, d must equal 1. But the problem says "det(A) = d ≠ 0", so perhaps d can be any non-zero real number? But in our deduction, we get that t^2 + (d - 1)^2 = 0, which requires t = 0 and d = 1. Therefore, under the given conditions, the only possibility is d = 1 and tr(A) = 0. Therefore, perhaps the problem is only valid when d = 1? Wait, but the problem doesn't specify d = 1, just d ≠ 0. Hmm. This suggests that either there's a mistake in the reasoning, or the problem has a specific setup. Wait, but the problem says "Let A ∈ M2(R) such that det(A) = d ≠ 0 and det(A + d A*) = 0. Prove that det(A - d A*) = 4." So according to our deduction, the conditions force d = 1 and tr(A) = 0, and then the determinant of (A - d A*) would be det(A - A*) with d = 1, and then we need to compute that. Let's see.If d = 1, then A has determinant 1, trace 0. So A is a 2x2 real matrix with trace 0 and determinant 1. Then A* is the adjugate of A, which for a 2x2 matrix with determinant 1 is equal to A^{-1} because A* = det(A) A^{-1} = 1 * A^{-1} = A^{-1}. So in this case, A* = A^{-1}. So the expression A - d A* becomes A - A^{-1}. So we need to compute det(A - A^{-1}) and show that it's 4. But wait, if A has trace 0 and determinant 1, then its characteristic polynomial is λ^2 - tr(A) λ + det(A) = λ^2 + 1. Therefore, the eigenvalues of A are i and -i. But A is a real matrix, so it can't have real eigenvalues; it must be a rotation matrix or something similar. For example, a matrix like [[0, 1], [-1, 0]] which has trace 0 and determinant 1. Let's compute det(A - A^{-1}) for this specific matrix. Let A = [[0, 1], [-1, 0]]. Then A^{-1} = [[0, -1], [1, 0]]. So A - A^{-1} = [[0 - 0, 1 - (-1)], [-1 - 1, 0 - 0]] = [[0, 2], [-2, 0]]. The determinant of this matrix is (0)(0) - (2)(-2) = 4. So det(A - A^{-1}) = 4, which matches the problem's conclusion. But in this case, d was 1. But the problem allows d to be any non-zero real number. However, according to our previous deduction, under the given conditions, d must equal 1 and tr(A) must be 0. Therefore, perhaps the problem implicitly requires d = 1, even though it's stated as d ≠ 0. Wait, maybe there's an error in my earlier steps. Let me check again.Starting from the equation det(A + d A*) = 0. Then, using the determinant formula for the sum of two matrices:det(A + d A*) = det A + det(d A*) + tr(A) tr(d A*) - tr(A (d A*)) = 0As before, det A = d, det(d A*) = d^2 det(A*) = d^2 * d = d^3 (since det(A*) = det A = d). Then tr(A) tr(d A*) = tr(A) * d tr(A*) = tr(A) * d tr(A) (since tr(A*) = tr(A)). So that term is d (tr A)^2. Then tr(A (d A*)) = d tr(A A*). But A A* = det(A) I = d I, so tr(A A*) = tr(d I) = 2 d. Therefore, tr(A (d A*)) = d * 2 d = 2 d^2. Putting it all together:det(A + d A*) = d + d^3 + d (tr A)^2 - 2 d^2 = 0Factor out d:d [1 + d^2 + (tr A)^2 - 2 d] = 0Since d ≠ 0, we have:1 + d^2 + (tr A)^2 - 2 d = 0Which rearranged is:(tr A)^2 + d^2 - 2 d + 1 = 0Which is:(tr A)^2 + (d - 1)^2 = 0Since both terms are squares of real numbers, each must be zero. Hence:tr A = 0 and d - 1 = 0 ⇒ d = 1So indeed, the only solution is d = 1 and tr A = 0. Therefore, even though the problem states det A = d ≠ 0, the additional condition det(A + d A*) = 0 forces d = 1. So the problem is essentially saying: Let A be a 2x2 real matrix with determinant 1 and trace 0, and such that det(A + A*) = 0. Prove that det(A - A*) = 4. Wait, but if d = 1, then A* = A^{-1}, since A* = det(A) A^{-1} = 1 * A^{-1} = A^{-1}. Therefore, A + A* = A + A^{-1}, and det(A + A^{-1}) = 0. But for the example I took earlier, A = [[0,1],[-1,0]], A + A^{-1} = [[0,1],[-1,0]] + [[0,-1],[1,0]] = [[0,0],[0,0]], which has determinant 0. So that works. So in this case, det(A - A*) = det(A - A^{-1}) = 4. So the problem seems to hinge on d being 1, even though it's presented as arbitrary non-zero. But according to our deduction, the only way for det(A + d A*) = 0 is if d = 1 and tr A = 0. Therefore, the problem must assume that d = 1, even if not explicitly stated, otherwise there's no solution. Therefore, we can proceed under the conclusion that d = 1 and tr A = 0.Therefore, given that d = 1 and tr A = 0, we need to compute det(A - A*). Since A* = A^{-1}, this is det(A - A^{-1}).Given that A is a 2x2 real matrix with trace 0 and determinant 1. Let's denote A as [[a, b], [c, -a]], since the trace is 0. Then the determinant is -a^2 - b c = 1. Therefore, -a^2 - b c = 1 ⇒ a^2 + b c = -1.Now, A* (the adjugate) is [[-a, -b], [-c, a]]. Wait, original matrix A is [[a, b], [c, -a]], so swapping the diagonal entries gives [[-a, b], [c, a]]? Wait, no. Wait, the adjugate matrix is the transpose of the cofactor matrix. For a 2x2 matrix [[p, q], [r, s]], the adjugate is [[s, -q], [-r, p]]. So for our matrix A = [[a, b], [c, -a]], the adjugate should be [[-a, -b], [-c, a]]. Because:- The (1,1) entry is the determinant of the submatrix obtained by removing row 1 and column 1, which is (-a).- The (1,2) entry is the determinant of the submatrix removing row 1, column 2: [[c]], determinant is c, multiplied by (-1)^{1+2} = -1, so -c. Wait, wait. Maybe I should compute it properly.Wait, the cofactor matrix is computed as follows: for entry (i,j), it's (-1)^{i+j} times the determinant of the submatrix obtained by deleting row i and column j.So for matrix A = [[a, b], [c, -a]]:Cofactor at (1,1): (-1)^{1+1} * det([[-a]]) = 1 * (-a) = -aCofactor at (1,2): (-1)^{1+2} * det([[c]]) = -1 * c = -cCofactor at (2,1): (-1)^{2+1} * det([[b]]) = -1 * b = -bCofactor at (2,2): (-1)^{2+2} * det([[a]]) = 1 * a = aTherefore, the cofactor matrix is [[-a, -c], [-b, a]]. Then the adjugate matrix is the transpose of the cofactor matrix, so [[-a, -b], [-c, a]]. Therefore, A* = [[-a, -b], [-c, a]].Now, let's compute A - A*:A - A* = [[a, b], [c, -a]] - [[-a, -b], [-c, a]] = [[a - (-a), b - (-b)], [c - (-c), -a - a]] = [[2a, 2b], [2c, -2a]]So the matrix A - A* is [[2a, 2b], [2c, -2a]]. The determinant of this matrix is (2a)(-2a) - (2b)(2c) = -4a^2 - 4b c. Factor out -4: -4(a^2 + b c). But from earlier, we know that a^2 + b c = -1. Therefore, determinant is -4*(-1) = 4. Hence, det(A - A*) = 4. Therefore, proved.So the key steps were:1. Using the determinant formula for the sum of two matrices to express det(A + d A*) in terms of d and tr(A), leading to the equation (tr A)^2 + (d - 1)^2 = 0, which forces tr A = 0 and d = 1.2. With d = 1 and tr A = 0, writing A in terms of its entries, computing A*, then finding A - A* and its determinant, which simplifies to 4 due to the condition a^2 + b c = -1 from the determinant of A.Therefore, despite the problem initially seeming to allow any non-zero d, the condition det(A + d A*) = 0 actually constraints d to be 1, and then the rest follows from there.Final Answerboxed{4}

✅:Given a 2x2 real matrix ( A ) with (det(A) = d neq 0) and (det(A + dA^*) = 0), we need to prove that (det(A - dA^*) = 4).1. Key Properties and Substitutions: - For a 2x2 matrix ( A ), the adjugate matrix ( A^* ) satisfies ( A^* = det(A) cdot A^{-1} ). Since (det(A) = d ), we have ( A^* = d cdot A^{-1} ). - Substitute ( A^* ) in the given conditions: ( A + dA^* = A + d^2A^{-1} ) and ( A - dA^* = A - d^2A^{-1} ).2. Determinant Condition Analysis: - Using the determinant formula for the sum of two matrices ( det(A + B) = det A + det B + tr(A)tr(B) - tr(AB) ): [ det(A + dA^*) = d + d^3 + d(tr(A))^2 - 2d^2 = 0 ] - Factoring out ( d ): [ d(1 + d^2 + (tr(A))^2 - 2d) = 0 ] - Since ( d neq 0 ), we get: [ 1 + d^2 + (tr(A))^2 - 2d = 0 implies (tr(A))^2 + (d - 1)^2 = 0 ] - This implies ( tr(A) = 0 ) and ( d = 1 ).3. Matrix Properties with ( d = 1 ): - For ( d = 1 ), ( A ) has (det(A) = 1) and (tr(A) = 0). Thus, ( A ) is a 2x2 real matrix with trace 0 and determinant 1. - The adjugate matrix ( A^* = A^{-1} ).4. Determinant Calculation: - Express ( A ) as (begin{bmatrix} a & b c & -a end{bmatrix}) with ( a^2 + bc = -1 ). - Compute ( A - A^* ): [ A - A^* = begin{bmatrix} a & b c & -a end{bmatrix} - begin{bmatrix} -a & -b -c & a end{bmatrix} = begin{bmatrix} 2a & 2b 2c & -2a end{bmatrix} ] - The determinant of ( A - A^* ): [ detleft( begin{bmatrix} 2a & 2b 2c & -2a end{bmatrix} right) = (2a)(-2a) - (2b)(2c) = -4a^2 - 4bc = -4(a^2 + bc) ] - Using ( a^2 + bc = -1 ): [ det(A - A^*) = -4(-1) = 4 ]Thus, the determinant of ( A - dA^* ) is (boxed{4}).

🔑:1. Given ( A in mathcal{M}_2(mathbb{R}) ) such that (det(A) = d neq 0) and (det(A + dA^*) = 0), we need to prove that (det(A - dA^*) = 4).2. Let (text{tr}(A) = t). The adjugate of (A), denoted (text{adj}(A)), satisfies the relation (A cdot text{adj}(A) = det(A) cdot I = d cdot I). For a (2 times 2) matrix, (text{adj}(A) = tI - A).3. Given (det(A + d text{adj}(A)) = 0), we substitute (text{adj}(A) = tI - A): [ det(A + d(tI - A)) = det(A + dtI - dA) = det(dtI - (d-1)A) = 0 ]4. If (d neq 1), we can factor out ((d-1)): [ detleft[(d-1) left(frac{dt}{d-1}I - Aright)right] = (d-1)^2 detleft(frac{dt}{d-1}I - Aright) = 0 ] This implies: [ detleft(frac{dt}{d-1}I - Aright) = 0 ]5. The above equation means that (frac{dt}{d-1}) is an eigenvalue of (A). The characteristic polynomial of (A) is: [ phi(x) = x^2 - tx + d ] Substituting (frac{dt}{d-1}) into the characteristic polynomial: [ phileft(frac{dt}{d-1}right) = left(frac{dt}{d-1}right)^2 - tleft(frac{dt}{d-1}right) + d = 0 ]6. Simplifying the above equation: [ left(frac{dt}{d-1}right)^2 - tleft(frac{dt}{d-1}right) + d = 0 ] [ frac{d^2 t^2}{(d-1)^2} - frac{d t^2}{d-1} + d = 0 ] Multiplying through by ((d-1)^2): [ d^2 t^2 - d t^2 (d-1) + d (d-1)^2 = 0 ] [ d^2 t^2 - d t^2 d + d t^2 + d^2 - 2d + 1 = 0 ] [ d^2 t^2 - d^2 t^2 + d t^2 + d^2 - 2d + 1 = 0 ] [ d t^2 + d^2 - 2d + 1 = 0 ]7. Since (d neq 0), we can solve for (t): [ d t^2 + d^2 - 2d + 1 = 0 ] [ t^2 + (d-1)^2 = 0 ] This implies: [ t^2 = -(d-1)^2 ] Since (t^2) is non-negative, the only solution is: [ t = 0 quad text{and} quad d-1 = 0 implies d = 1 ]8. Therefore, (d = 1). Substituting (d = 1) back into the original determinant condition: [ det(A + text{adj}(A)) = 0 ] Since (text{adj}(A) = -A) when (t = 0): [ det(A - A) = det(0) = 0 ]9. Finally, we need to find (det(A - d text{adj}(A))): [ det(A - text{adj}(A)) = det(A - (-A)) = det(2A) = 2^2 det(A) = 4 cdot 1 = 4 ](blacksquare)The final answer is ( boxed{ 4 } )

🧠:Okay, so I have this geometry problem here. Let me try to parse it step by step. The problem is about triangle ABC with incenter I. The sides are given as AB = 10, BC = 21, and CA = 17. Points D and E are on side BC such that BD = 4, DE = 6, and EC = 11. Then, the circumcircles of triangles BIE and CID intersect again at point P (other than I). Line IP meets the altitude from A to BC at point X. We need to find (DX · EX)^2.First, let me recall some basics. The incenter I is the intersection of the angle bisectors of triangle ABC. Since the sides are given, maybe I can compute coordinates for the points. Coordinates might help here because the problem involves various constructions like circumcircles, intersections, and altitudes. Let me consider placing the triangle in a coordinate system to make calculations easier.Let me set side BC on the x-axis with B at (0, 0) and C at (21, 0). Then, the altitude from A to BC will be vertical if BC is horizontal. Let me denote the coordinates of A as (d, h), where h is the height. Then, using the given lengths AB = 10, AC = 17, and BC = 21, I can compute d and h.Wait, let's check. If B is at (0, 0) and C is at (21, 0), then the coordinates of A are (d, h). The distance from A to B is 10, so sqrt((d - 0)^2 + (h - 0)^2) = 10, so d^2 + h^2 = 100. Similarly, the distance from A to C is 17, so sqrt((d - 21)^2 + (h - 0)^2) = 17, so (d - 21)^2 + h^2 = 289. Subtract the first equation from the second: (d - 21)^2 + h^2 - (d^2 + h^2) = 289 - 100 => (d^2 - 42d + 441) - d^2 = 189 => -42d + 441 = 189 => -42d = -252 => d = 6. Then, substituting back into d^2 + h^2 = 100, we get 36 + h^2 = 100 => h^2 = 64 => h = 8. So, point A is at (6, 8).Alright, so coordinates:- B: (0, 0)- C: (21, 0)- A: (6, 8)Now, points D and E are on BC. BD = 4, DE = 6, EC = 11. Since BC is from (0,0) to (21,0), the coordinates of D and E can be determined. Since BD = 4, D is at (4, 0). Then DE = 6, so E is at (4 + 6, 0) = (10, 0). EC is 11, which is 21 - 10 = 11, so that checks out.So D is (4, 0) and E is (10, 0).Next, we need to find the incenter I of triangle ABC. The incenter can be found using the formula that weights the coordinates by the lengths of the sides. The formula is:I = (aA + bB + cC)/(a + b + c)Wait, no, actually, the incenter coordinates are given by ( (a x_A + b x_B + c x_C)/ (a + b + c), (a y_A + b y_B + c y_C)/ (a + b + c) ), where a, b, c are the lengths of the sides opposite to vertices A, B, C. Wait, in triangle ABC, the sides are AB = 10, BC = 21, CA = 17. Wait, in standard notation, a is BC, b is AC, c is AB. So a = 21, b = 17, c = 10. Then the incenter coordinates would be ( (a x_A + b x_B + c x_C)/(a + b + c), (a y_A + b y_B + c y_C)/(a + b + c) )Wait, let me confirm: the formula for the incenter is ( (a x_A + b x_B + c x_C ) / (a + b + c), same for y). But actually, in standard terms, the incenter formula uses the lengths opposite to the angles. So in standard triangle notation, a is BC, which is opposite to A, b is AC opposite to B, c is AB opposite to C. So in this case:a = BC = 21 (opposite A)b = AC = 17 (opposite B)c = AB = 10 (opposite C)Therefore, the incenter coordinates would be:x = (a x_A + b x_B + c x_C)/(a + b + c)y = (a y_A + b y_B + c y_C)/(a + b + c)Given:- A is (6, 8)- B is (0, 0)- C is (21, 0)- a = 21, b = 17, c = 10So:x = (21*6 + 17*0 + 10*21)/(21 + 17 + 10)y = (21*8 + 17*0 + 10*0)/(21 + 17 + 10)Compute numerator for x:21*6 = 12617*0 = 010*21 = 210Total x numerator: 126 + 0 + 210 = 336Denominator: 21 + 17 + 10 = 48So x = 336 / 48 = 7For y:21*8 = 16817*0 = 010*0 = 0Total y numerator: 168y = 168 / 48 = 3.5 = 7/2Thus, the incenter I is at (7, 7/2).So coordinates of I: (7, 3.5)Alright, moving on. Now, we need the circumcircles of triangles BIE and CID. Let me note the coordinates again:- B is (0,0)- I is (7, 3.5)- E is (10, 0)- C is (21, 0)- D is (4, 0)First, let's find the circumcircle of triangle BIE. Points B (0,0), I (7, 3.5), E (10,0). Similarly, the circumcircle of triangle CID: points C (21,0), I (7, 3.5), D (4,0).To find the circumcircle of three points, we can use the circumcircle equation. Let's first handle triangle BIE.Circumcircle of BIE:Points B (0,0), I (7, 3.5), E (10, 0). Let's find the equation of the circle passing through these three points.General equation of a circle: x² + y² + Dx + Ey + F = 0.Plugging in B (0,0):0 + 0 + 0 + 0 + F = 0 => F = 0.So equation becomes x² + y² + Dx + Ey = 0.Now plug in I (7, 3.5):7² + (3.5)^2 + D*7 + E*3.5 = 049 + 12.25 + 7D + 3.5E = 061.25 + 7D + 3.5E = 0 --> equation (1)Plug in E (10,0):10² + 0 + D*10 + E*0 = 0100 + 10D = 0 => 10D = -100 => D = -10.So from E, D = -10. Now substitute D into equation (1):61.25 + 7*(-10) + 3.5E = 061.25 - 70 + 3.5E = 0-8.75 + 3.5E = 0 => 3.5E = 8.75 => E = 8.75 / 3.5 = 2.5.Therefore, the equation of the circumcircle of BIE is x² + y² -10x + 2.5y = 0.Alternatively, we can write it in standard form by completing the squares:x² -10x + y² + 2.5y = 0x² -10x +25 + y² +2.5y + (2.5/2)^2 = 25 + (2.5/2)^2Wait, completing the square for x: coefficient of x is -10, so half of that is -5, square is 25.For y: coefficient of y is 2.5, half is 1.25, square is (5/4)^2 = 25/16 = 1.5625.So left side becomes (x - 5)^2 + (y + 1.25)^2 = 25 + 1.5625 = 26.5625Which is 26.5625 = (5.15)^2 approximately, but exact value is 26.5625 = 85/3.2... Wait, 26.5625 * 16 = 425, so 26.5625 = 425/16. Hmm, 25 + 25/16 = (400 +25)/16 = 425/16. Yes, so radius squared is 425/16, and center is (5, -1.25) or (5, -5/4).Wait, the standard form would be (x - 5)^2 + (y + 5/4)^2 = (sqrt(425)/4)^2. But maybe we don't need the standard form. Anyway, the equation is x² + y² -10x + 2.5y = 0.Now, let's find the circumcircle of triangle CID. Points C (21,0), I (7, 3.5), D (4,0).Again, general equation: x² + y² + Gx + Hy + K = 0.Plugging in C (21,0):21² + 0 + G*21 + H*0 + K = 0 => 441 + 21G + K = 0 --> equation (2)Plugging in I (7, 3.5):7² + (3.5)^2 + G*7 + H*3.5 + K = 049 + 12.25 + 7G + 3.5H + K = 061.25 + 7G + 3.5H + K = 0 --> equation (3)Plugging in D (4,0):4² + 0 + G*4 + H*0 + K = 0 => 16 + 4G + K = 0 --> equation (4)Now, we have three equations:(2): 441 + 21G + K = 0(3): 61.25 + 7G + 3.5H + K = 0(4): 16 + 4G + K = 0Let me subtract equation (4) from equation (2):(441 +21G + K) - (16 +4G + K) = 0 - 0 => 425 +17G = 0 => 17G = -425 => G = -25.Then, substitute G = -25 into equation (4):16 +4*(-25) + K = 0 => 16 -100 + K = 0 => K = 84.Now, substitute G = -25 and K = 84 into equation (3):61.25 +7*(-25) +3.5H +84 = 061.25 -175 +84 +3.5H = 0(61.25 +84) -175 +3.5H = 0145.25 -175 +3.5H = 0-29.75 +3.5H = 0 => 3.5H =29.75 => H =29.75 /3.5 =8.5.Thus, H = 8.5 = 17/2.Therefore, the equation of the circumcircle of CID is x² + y² -25x + (17/2)y +84 =0.Again, maybe convert to standard form:x² -25x + y² + (17/2)y = -84Complete the squares:For x: coefficient of x is -25, half is -12.5, square is 156.25For y: coefficient of y is 17/2, half is 17/4, square is (17/4)^2 = 289/16So add 156.25 and 289/16 to both sides:Left side: (x - 12.5)^2 + (y +17/4)^2Right side: -84 +156.25 +289/16Convert all to fractions:-84 = -84/1156.25 = 625/4289/16 remains.So total right side:-84 +625/4 +289/16 = (-84*16 +625*4 +289)/16Compute:-84*16 = -1344625*4 = 2500289 = 289Total: (-1344 +2500 +289)/16 = (2500 -1344 +289)/16 = (1156 +289)/16 =1445/16So radius squared is 1445/16, center at (12.5, -17/4). But perhaps not necessary.Anyway, the equation is x² + y² -25x + (17/2)y +84 =0.Now, we need to find the other intersection point P of the two circumcircles (other than I). So we have two circles:1. BIE's circle: x² + y² -10x + 2.5y =02. CID's circle: x² + y² -25x + (17/2)y +84 =0Subtract equation 1 from equation 2 to find the radical axis:(x² + y² -25x + (17/2)y +84) - (x² + y² -10x + 2.5y) =0Simplify:(-25x + (17/2)y +84) - (-10x + 2.5y) =0-25x + (17/2)y +84 +10x -2.5y =0(-15x) + (17/2 - 5/2)y +84 =0-15x + (12/2)y +84 =0-15x +6y +84 =0Divide through by -3:5x -2y -28 =0So radical axis is 5x -2y -28 =0. The line IP intersects the radical axis at P. Wait, but points I and P lie on both circles, so the line connecting them is the radical axis? Wait, no. The radical axis is the set of points with equal power regarding both circles, which is the line we found: 5x -2y -28 =0. Since both I and P lie on both circles, they lie on the radical axis, so the line IP is the radical axis. Wait, but radical axis is a line, so IP must be the radical axis. Therefore, the line IP is 5x -2y -28 =0.But we need to find point P, which is the other intersection. Since I is on both circles, P is the other point. So we can parametrize the line 5x -2y -28 =0 and find its intersection with either circle.Alternatively, since we know I is on both circles, we can solve the two circle equations and find the other intersection.Let me solve the system:1. x² + y² -10x + 2.5y =02. x² + y² -25x + (17/2)y +84 =0We already subtracted them to get the radical axis: 5x -2y -28 =0, so y = (5x -28)/2.Now, substitute y = (5x -28)/2 into equation 1:x² + [(5x -28)/2]^2 -10x +2.5*(5x -28)/2 =0Let me compute each term.First, x².Second, [(5x -28)/2]^2 = (25x² -280x +784)/4Third, -10x.Fourth, 2.5*(5x -28)/2 = (5/2)*(5x -28)/2 = (25x -140)/4So putting all together:x² + (25x² -280x +784)/4 -10x + (25x -140)/4 =0Multiply all terms by 4 to eliminate denominators:4x² +25x² -280x +784 -40x +25x -140 =0Combine like terms:(4x² +25x²) + (-280x -40x +25x) + (784 -140) =029x² -295x +644 =0So quadratic equation 29x² -295x +644 =0Let me compute discriminant D:D = 295² -4*29*644Compute 295²: 295*295. Let's compute:(300 -5)^2 = 300² -2*300*5 +5² =90000 -3000 +25=870254*29*644: 4*29=116; 116*644. Let's compute:644*100=64400644*16=10304So 644*116=64400 +10304=74704Thus, D=87025 -74704=12321Square root of 12321: 111*111=12321, so sqrt(D)=111Therefore, x=(295 ±111)/(2*29)Compute numerator:295 +111=406295 -111=184Thus, x=406/58=203/29≈7.0 and x=184/58=92/29≈3.1724But we know point I is at x=7, so 203/29≈7.0 (since 29*7=203). Therefore, x=203/29=7, which is point I. The other solution is x=92/29≈3.1724. So the other intersection point P has x=92/29. Then, y=(5x -28)/2. Plugging in x=92/29:y=(5*(92/29) -28)/2 = (460/29 -28)/2 = (460/29 - 812/29)/2 = (-352/29)/2 = -176/29Thus, point P is at (92/29, -176/29)So coordinates of P: (92/29, -176/29)Now, we need to find line IP. Since I is at (7, 3.5) and P is at (92/29, -176/29). Let me convert 3.5 to 7/2 = 203/58? Wait, 3.5 = 7/2 = 101.5/29? Wait, maybe better to use fractions.Coordinates of I: (7, 7/2) = (203/29, 7/2). Wait, but 7 = 203/29? 29*7=203, yes. Similarly, 7/2 = 101.5/29? Wait, that might complicate things. Let me just compute the parametric equations or the slope.Slope of IP:(y_P - y_I)/(x_P - x_I) = (-176/29 - 7/2)/(92/29 -7) = (-176/29 - 101.5/29)/(92/29 -203/29) = (-277.5/29)/(-111/29)= (-277.5)/(-111)=277.5/111=2.5=5/2.Wait, 277.5 divided by 111: 111*2=222, 111*2.5=277.5, yes. So slope is 2.5 or 5/2.Therefore, the equation of line IP is y - y_I = (5/2)(x -x_I)Given I is (7, 7/2), so:y -7/2 = (5/2)(x -7)Multiply both sides by 2:2y -7 =5(x -7)=> 5x -2y -35 +7=0Wait, let me check:Starting from y -7/2 = (5/2)(x -7)Multiply both sides by 2: 2y -7 =5x -35Bring all terms to left: 5x -2y -28=0Wait, that's the same as the radical axis equation! Which makes sense because IP is the radical axis. So line IP is 5x -2y -28=0, which we already knew. So that's consistent.Now, the next part: line IP meets the altitude from A to BC at X. We need to find point X, which is the intersection of IP and the altitude from A.Given that A is (6,8), and the altitude from A to BC is the vertical line? Wait, BC is horizontal from (0,0) to (21,0). The altitude from A to BC is the vertical line, since in our coordinate system, we found that A is at (6,8), and BC is on the x-axis. The altitude from A to BC is the vertical line x=6, because the foot of the altitude is at (6,0). Wait, actually, the foot of the altitude is the projection of A onto BC. Since BC is horizontal, the projection is indeed (6,0). So the altitude is the line from (6,8) to (6,0), which is vertical line x=6.Therefore, the altitude from A to BC is x=6. Then, line IP is 5x -2y -28=0. To find their intersection X, set x=6 in the line equation:5*6 -2y -28=0 =>30 -2y -28=0 =>2 -2y=0 =>2y=2 => y=1.Thus, point X is at (6,1).Now, we need to compute DX and EX, then find (DX · EX)^2.Points D (4,0), E (10,0), and X (6,1).Compute distance DX:DX = sqrt( (6 -4)^2 + (1 -0)^2 ) = sqrt(4 +1)=sqrt(5)Similarly, EX = sqrt( (10 -6)^2 + (0 -1)^2 )= sqrt(16 +1)=sqrt(17)Thus, DX · EX = sqrt(5) * sqrt(17) = sqrt(85)Therefore, (DX · EX)^2 = (sqrt(85))^2 =85.Wait, so is it 85? Hmm, but let me double-check the calculations.Coordinates of X: (6,1). Correct, since the altitude is x=6, and solving 5x -2y =28 gives at x=6, 5*6 -2y =28 =>30 -2y=28 =>2y=2 =>y=1. Correct.Distance from D (4,0) to X (6,1):Difference in x: 6-4=2, difference in y:1-0=1. So sqrt(2² +1²)=sqrt(5). Correct.Distance from E (10,0) to X (6,1):Difference in x:10-6=4, difference in y:0-1=-1. So sqrt(4² + (-1)^2)=sqrt(16 +1)=sqrt(17). Correct.Product DX * EX = sqrt(5)*sqrt(17)=sqrt(85). Then square is 85. So answer is 85. Therefore, (DX · EX)^2=85.But wait, the problem says "Find (DX · EX)^2". So the answer is boxed{85}.Hmm, but I should check if all steps are correct.First, coordinates of triangle ABC: yes, using distance formula, A is at (6,8). Correct.Incenter calculation: used formula correctly. a=21, b=17, c=10. Computations led to (7, 3.5). Correct.Points D and E: BD=4, so D at (4,0); DE=6, so E at (10,0). Correct.Circumcircle of BIE: solved equations and found center and equation. Then for CID, same. Subtraction gave radical axis 5x -2y -28=0. Solving the system found point P at (92/29, -176/29). Then line IP parametrized, found intersection with altitude x=6 gives X at (6,1). Distances computed. All steps seem correct.Therefore, the answer is 85.Final Answerboxed{85}

✅:Given triangle (ABC) with incenter (I), sides (AB = 10), (BC = 21), and (CA = 17). Points (D) and (E) lie on side (BC) such that (BD = 4), (DE = 6), and (EC = 11). The circumcircles of triangles (BIE) and (CID) intersect again at point (P), and line (IP) meets the altitude from (A) to (BC) at (X). We need to find ((DX cdot EX)^2).1. Coordinates of Points: - Place (B) at ((0, 0)) and (C) at ((21, 0)). - Using the distance formula, coordinates of (A) are found to be ((6, 8)). - Points (D) and (E) are at ((4, 0)) and ((10, 0)) respectively.2. Incenter Calculation: - Using the formula for the incenter, (I) is calculated as ((7, frac{7}{2})).3. Circumcircles of Triangles (BIE) and (CID): - Circumcircle of (BIE): Equation found to be (x^2 + y^2 - 10x + 2.5y = 0). - Circumcircle of (CID): Equation found to be (x^2 + y^2 - 25x + frac{17}{2}y + 84 = 0).4. Radical Axis and Intersection Point (P): - Subtracting the equations of the circumcircles gives the radical axis (5x - 2y - 28 = 0). - Solving the system of equations for the circumcircles, the other intersection point (P) is found to be (left(frac{92}{29}, -frac{176}{29}right)).5. Intersection of Line (IP) with Altitude from (A): - The altitude from (A) to (BC) is the vertical line (x = 6). - Solving the radical axis equation (5x - 2y - 28 = 0) at (x = 6) gives (y = 1). Thus, point (X) is ((6, 1)).6. Distances (DX) and (EX): - (DX = sqrt{(6 - 4)^2 + (1 - 0)^2} = sqrt{5}) - (EX = sqrt{(10 - 6)^2 + (0 - 1)^2} = sqrt{17}) - Product ((DX cdot EX)^2 = (sqrt{5} cdot sqrt{17})^2 = 85).Thus, the final answer is (boxed{85}).

🔑:1. Determine the coordinates of points (A), (B), and (C): - Given (AB = 10), (BC = 21), and (CA = 17), we can place (B) at ((0,0)) and (C) at ((21,0)) on the coordinate plane. - To find the coordinates of (A), we use the distance formula: [ AB = 10 implies sqrt{(x-0)^2 + (y-0)^2} = 10 implies x^2 + y^2 = 100 ] [ AC = 17 implies sqrt{(x-21)^2 + y^2} = 17 implies (x-21)^2 + y^2 = 289 ] Solving these equations simultaneously, we get: [ x^2 + y^2 = 100 ] [ (x-21)^2 + y^2 = 289 implies x^2 - 42x + 441 + y^2 = 289 implies 100 - 42x + 441 = 289 implies -42x + 541 = 289 implies x = 6 ] [ y^2 = 100 - 6^2 = 100 - 36 = 64 implies y = 8 text{ or } y = -8 ] Thus, (A) can be ((6, 8)) or ((6, -8)). We choose ((6, 8)) for simplicity.2. Determine the coordinates of points (D) and (E): - Given (BD = 4), (DE = 6), and (EC = 11), we place (D) at ((4,0)) and (E) at ((10,0)).3. Find the incenter (I): - The incenter (I) of (triangle ABC) can be found using the formula: [ I = left( frac{aA_x + bB_x + cC_x}{a+b+c}, frac{aA_y + bB_y + cC_y}{a+b+c} right) ] where (a = BC = 21), (b = CA = 17), and (c = AB = 10): [ I = left( frac{21 cdot 6 + 17 cdot 0 + 10 cdot 21}{21+17+10}, frac{21 cdot 8 + 17 cdot 0 + 10 cdot 0}{21+17+10} right) = left( frac{126 + 210}{48}, frac{168}{48} right) = left( frac{336}{48}, frac{168}{48} right) = left( 7, frac{7}{2} right) ]4. Find the circumcircles of (triangle BIE) and (triangle CID): - The equation of the circumcircle of (triangle BIE) can be written as: [ (x-7)^2 + left(y - frac{7}{2}right)^2 = r^2 ] Similarly, the equation of the circumcircle of (triangle CID) can be written as: [ (x-7)^2 + left(y - frac{7}{2}right)^2 = r^2 ] These circles intersect at points (I) and (P).5. Find the coordinates of point (P): - The intersection of the two circles gives us point (P). Solving the system of equations for the circles, we find (P).6. Find the intersection of line (IP) with the altitude from (A) to (BC): - The altitude from (A) to (BC) is a vertical line through (A) at (x = 6). - The line (IP) intersects this altitude at point (X).7. Calculate ((DX cdot EX)^2): - Using the coordinates of (D), (E), and (X), we calculate: [ DX = sqrt{(6-4)^2 + (1-0)^2} = sqrt{4 + 1} = sqrt{5} ] [ EX = sqrt{(10-6)^2 + (1-0)^2} = sqrt{16 + 1} = sqrt{17} ] [ (DX cdot EX)^2 = (sqrt{5} cdot sqrt{17})^2 = 5 cdot 17 = 85 ]The final answer is (boxed{85})

🧠:Okay, so there's this game between Ani and Budi involving three buckets and n marbles. Ani starts by distributing all n marbles into the three buckets, each having at least one marble. Then Budi and Ani take turns, starting with Budi, where each player can take 1, 2, or 3 marbles from a single bucket. The person who takes the last marble wins. We need to find all n ≥ 3 where Ani has a winning strategy, and also describe her initial distribution for those n.First, let me recall similar games. This seems like a variant of the Nim game, but with three heaps (buckets) and a restriction that each move is taking 1-3 marbles from one heap. In standard Nim, the winning strategy involves binary representations and XOR operations, but this might be different due to the move restriction (only 1-3 marbles can be taken each turn). So maybe the Grundy numbers concept applies here? Grundy numbers are used in combinatorial game theory to determine the winning positions.Let me start by understanding the game mechanics. Since it's a normal play game (last marble wins), the key is to determine the losing positions, known as P-positions. If Ani can distribute the marbles into three buckets such that the resulting position is a P-position, then Budi has no winning move, and Ani can win with proper play.But first, let's recall how Grundy numbers work. Each position (heap size) in a game has a Grundy number, which is the minimum excludant (mex) of the Grundy numbers of positions reachable in one move. For a single heap where you can take 1, 2, or 3 marbles, the Grundy numbers would be periodic. Let's compute them:For heap size 0: not possible here since each bucket must have at least 1 marble initially. But in general, G(0) = 0.For heap size 1: possible moves lead to 0 marbles. So mex{G(0)} = mex{0} = 1.Heap size 2: moves lead to 1 or 0. So mex{G(1), G(0)} = mex{1, 0} = 2.Heap size 3: moves lead to 2, 1, or 0. mex{2, 1, 0} = 3.Heap size 4: moves lead to 3, 2, 1. mex{3, 2, 1} = 0.Heap size 5: moves lead to 4, 3, 2. mex{0, 3, 2} = 1.Heap size 6: moves lead to 5, 4, 3. mex{1, 0, 3} = 2.Heap size 7: moves lead to 6, 5, 4. mex{2, 1, 0} = 3.Heap size 8: moves lead to 7, 6, 5. mex{3, 2, 1} = 0.So the Grundy numbers for a single heap in this game cycle every 4: 1, 2, 3, 0, 1, 2, 3, 0, etc.Therefore, the Grundy number for a heap of size k is G(k) = k mod 4. If k mod 4 is 0, then G(k) = 0, otherwise G(k) = k mod 4.But wait, when k is 4, mex{3, 2, 1} is 0. So yes, the pattern is 1, 2, 3, 0 repeating every 4.Therefore, in this game, each heap's Grundy number is k mod 4 if k mod 4 ≠ 0, else 0. So the Grundy number for a heap of size k is equivalent to k modulo 4, except when k is a multiple of 4, in which case it's 0.Then, for the game as a whole, which is the XOR (nimber sum) of the Grundy numbers of each heap. So the position is a P-position (losing for the player about to move) if the XOR of the Grundy numbers is 0.Therefore, Ani wants to distribute the marbles into three heaps (a, b, c) such that G(a) XOR G(b) XOR G(c) = 0. If she can do that, then Budi is faced with a P-position and Ani can win with correct play.So the key is to find n such that n can be partitioned into a, b, c ≥1 with a + b + c = n and G(a) XOR G(b) XOR G(c) = 0.But since each G(k) is k mod 4 (except when k mod 4 = 0, then G(k)=0), we can model this as:G(a) XOR G(b) XOR G(c) = (a mod 4) XOR (b mod 4) XOR (c mod 4) = 0, considering that if any of a, b, c is divisible by 4, their G is 0, which is same as (k mod 4).Wait, but in general, G(k) is k mod 4 if k mod 4 ≠0, else 0. So actually, G(k) is equal to k mod 4. Wait, let's verify:For k=4, G(4) is mex{G(3), G(2), G(1)} = mex{3, 2, 1} = 0. 4 mod 4 is 0.For k=5, G(5)= mex{G(4), G(3), G(2)}= mex{0,3,2}=1. 5 mod 4 is 1.Similarly, k=6, G(6)= mex{G(5), G(4), G(3)}= mex{1,0,3}=2. 6 mod4=2.k=7, mex{2,1,0}=3. 7 mod4=3.k=8, mex{3,2,1}=0. 8 mod4=0.So actually, G(k) = k mod4. So regardless, G(k) = k mod4. So even if k is a multiple of 4, G(k) is 0, which is 4 mod4=0. So in general, G(k) = k mod4.Therefore, the Grundy number for each heap is simply k mod4. Then the total Grundy number is the XOR of (a mod4, b mod4, c mod4). So Ani needs to partition n into a,b,c ≥1 such that (a mod4) XOR (b mod4) XOR (c mod4) =0.Therefore, for Ani to have a winning strategy, such a partition must exist. Therefore, the problem reduces to finding all n≥3 for which there exists a partition of n into three positive integers a,b,c with a+b+c=n and (a mod4) XOR (b mod4) XOR (c mod4)=0.So our task is to find all n≥3 where such a partition exists.Alternatively, for each n, check whether there exists a triple (a,b,c) of positive integers summing to n such that the XOR of (a mod4, b mod4, c mod4) is 0.First, let's note that since a,b,c ≥1 and sum to n, and n ≥3.Alternatively, maybe we can characterize n modulo some number. Let's see.Alternatively, let's consider the possible residues of n modulo 4, and for each residue, see if such a partition exists.But maybe we need to consider the possible triples (x,y,z) where x,y,z are in {1,2,3,0} (since mod4) and x XOR y XOR z=0, and then check if there exists such a triple that sums to n modulo4? Wait, but sum modulo4 is not directly connected to the sum n. Because even if the sum modulo4 is equal to n modulo4, but the actual sum is n. So the triples (x,y,z) must satisfy x + y + z ≡n mod4, but also x XOR y XOR z=0.But this seems complicated. Maybe a better approach is to consider all possible triples (x,y,z) where x,y,z are in {1,2,3,0}, x XOR y XOR z=0, and then for each such triple, find the possible n where there exists a partition a,b,c of n with a ≡x, b≡y, c≡z mod4.Alternatively, since the residues mod4 matter, perhaps n must be congruent to (x + y + z) mod4. But since x, y, z are residues, then x + y + z ≡ n mod4. But also, x XOR y XOR z=0. So for each such triple (x,y,z) where x XOR y XOR z=0 and x+y+z ≡n mod4, we need to see if n can be written as x + y + z + 4k for some k≥0, but with a,b,c ≥1. Wait, but x, y, z are residues mod4, but a,b,c are each at least 1, so each of a,b,c must be at least 1, meaning that x, y, z can be 1,2,3, or 0 (but if x=0, then a must be 4, 8, etc.)But this is getting a bit tangled. Let's step back.First, list all possible triples (x,y,z) where x,y,z ∈ {0,1,2,3} (mod4) and x XOR y XOR z =0. Then for each such triple, check if there exists a partition of n into a,b,c such that a ≡x, b≡y, c≡z mod4, and a,b,c ≥1.The key is that the sum a + b + c =n, so x + y + z ≡n mod4. Therefore, for each triple (x,y,z) with x XOR y XOR z=0, n must ≡x+y+z mod4. However, x, y, z can be 0,1,2,3, but when considering a,b,c, each must be at least 1. So if x=0, then a must be 4,8,12,... etc. Similarly for y and z.So to find possible n, we need to consider all triples (x,y,z) with x XOR y XOR z=0, and x+y+z ≡n mod4, and such that there exists a,b,c ≥1 with a ≡x, b≡y, c≡z mod4 and a + b + c =n.Alternatively, for each triple (x,y,z) with x XOR y XOR z=0, the minimal n for which such a partition exists is x + y + z (if x,y,z are all at least 1) or higher. But since a,b,c must be at least 1, if x,y,z are 0 mod4, then the actual a,b,c would be 4, 8, etc. So the minimal n for each triple can vary.This seems complicated. Maybe instead of trying to list all possible triples, let's look for patterns.First, note that the XOR of three numbers is 0 if either all three are equal (since x XOR x XOR x = x XOR x =0 XOR x =x, which is 0 only if x=0), or if two of them are equal and the third is 0. Wait, no, let's think:In general, for three numbers x,y,z:x XOR y XOR z =0 iff z = x XOR y.But then, substituting back, x XOR y XOR (x XOR y)=0, which is always true. So the possible triples are those where z =x XOR y.But z must be in {0,1,2,3}, as must x and y. So for each x and y in {0,1,2,3}, compute z=x XOR y, and check if z is in {0,1,2,3}. Well, since x and y are 0-3, their XOR is also 0-3, so yes.Therefore, all possible triples (x,y,z) with x XOR y XOR z=0 are exactly those where z=x XOR y. So we can generate all such triples.Let me list all possible (x,y,z):For x,y in 0-3:(0,0,0), (0,1,1), (0,2,2), (0,3,3),(1,0,1), (1,1,0), (1,2,3), (1,3,2),(2,0,2), (2,1,3), (2,2,0), (2,3,1),(3,0,3), (3,1,2), (3,2,1), (3,3,0).Now, considering that in the initial distribution, each bucket must have at least 1 marble. Therefore, a,b,c ≥1, so their residues mod4 can be 1,2,3, but if the residue is 0, that means the actual number is a multiple of 4 (like 4,8, etc.), which is allowed as long as it's ≥1.Wait, 0 mod4 corresponds to numbers like 4,8,12,... which are ≥4, so they are allowed since the buckets must have at least 1 marble. So residues can be 0,1,2,3, but the actual number must be at least 1. So a residue of 0 corresponds to a number 4,8, etc.So when we consider triples (x,y,z), even if x,y,z are 0, as long as the actual numbers a,b,c are 4 or more, it's okay.Therefore, for each triple (x,y,z) where x,y,z ∈{0,1,2,3}, x XOR y XOR z=0, we need to check if there exists a way to write n = (4k + x) + (4m + y) + (4p + z) where k,m,p are non-negative integers (since x,y,z are residues, so a=4k +x, etc.), and 4k +x ≥1, 4m + y ≥1, 4p + z ≥1.But since x,y,z ∈{0,1,2,3}, and a,b,c ≥1, the constraints are:If x=0, then 4k +0 ≥1 ⇒k≥1 (since 4*0 +0=0 <1), so k≥1.Similarly, if y=0, m≥1; if z=0, p≥1.But this complicates things because for triples where any of x,y,z is 0, the minimal contribution to n is higher.Let me try to categorize the possible triples (x,y,z) based on whether they contain zeros or not.Case 1: All residues x,y,z are non-zero (i.e., 1,2,3). Then, the minimal n for such a triple is x + y + z (since k=m=p=0). So for example, if (x,y,z)=(1,1,0), but wait, no, in this case z=0, which would require p≥1. Wait, no, in the previous list, (1,1,0) is a triple where z=0. So in such a case, to have z=0, the actual c would have to be 4p +0 where p≥1, so c≥4. Therefore, the minimal n for this triple would be a=1 (x=1), b=1 (y=1), c=4 (z=0), totaling n=6.Similarly, for the triple (0,0,0), all residues are 0, so each bucket must have at least 4 marbles, so minimal n=12.But this seems messy. Maybe a better approach is to first find for each residue class of n mod4, whether there exists a triple (x,y,z) with x XOR y XOR z=0 and x + y + z ≡n mod4, and such that n can be expressed as the sum of three numbers congruent to x,y,z mod4, each at least 1.Alternatively, let's consider the possible values of n mod4 and see which ones allow such triples.Possible n mod4: 0,1,2,3.Let's check for each residue:Case 1: n ≡0 mod4.We need a triple (x,y,z) where x XOR y XOR z=0 and x + y + z ≡0 mod4.Looking at the list of triples:For example, (0,0,0): x+y+z=0 mod4, and x XOR y XOR z=0. The sum is 0+0+0=0 mod4. To get n=0 mod4, n=4k. But to have a triple (0,0,0), each bucket must have at least 4 marbles (since x=0 ⇒a=4p, etc.). So minimal n=12 (4+4+4). So n=12,16,20,...But also, there might be other triples. For example, (1,1,2): 1+1+2=4≡0 mod4, and 1 XOR1 XOR2= (0) XOR2=2≠0. Wait, no, wait (1,1,2) is not a valid triple since z=x XOR y=1 XOR1=0, not 2. So scratch that.Looking back at the list of valid triples:For example, (1,2,3): 1+2+3=6≡2 mod4. Not 0.(0,0,0): sum 0 mod4.(2,2,0): sum 4≡0 mod4. 2+2+0=4≡0. But z=0. So minimal n for this triple would be a=2, b=2, c=4 (since z=0 requires c≥4). So sum=8. Similarly, (3,3,0): sum=6≡2 mod4. No.Wait, let's check the triples:Looking at the list generated earlier:Take (2,2,0): sum=4≡0 mod4. So if we use this triple, the minimal n is a=2, b=2, c=4 ⇒n=8. Then n=8,12,16,... (adding multiples of 4). Similarly, (1,1,0): sum=2≡2 mod4. Not 0. (3,3,0): sum=6≡2 mod4. (1,3,2): sum=6≡2. (0,0,0): sum=0.So the only triples that give sum≡0 mod4 are (0,0,0) and (2,2,0), (0,2,2), etc. Wait, let's check:Wait, in the list of valid triples:(2,2,0) is there (since x=2, y=2, z=0). Similarly, (2,0,2) and (0,2,2). All these have sum 2+2+0=4≡0 mod4. So any triple where two are 2 and one is 0. Similarly, (1,3,2): sum=1+3+2=6≡2 mod4. Not 0.Also, (3,3,0) sum=6≡2 mod4.So the possible triples for sum≡0 mod4 are:- (0,0,0): sum=0.- Triples with two 2s and one 0: (2,2,0), (2,0,2), (0,2,2): sum=4≡0.Similarly, is there any other triple?(1,1,0): sum=2≡2.(1,2,3): sum=6≡2.(3,3,0): sum=6≡2.(0,1,1): sum=2≡2.(0,3,3): sum=6≡2.So only (0,0,0) and the (2,2,0) type triples give sum≡0 mod4.Now, for n≡0 mod4, we need to see if n can be expressed as either:1. 4k + 4m + 4p =4(k + m + p), i.e., n=4t, t≥3 (since each bucket must have at least 1 marble, so k,m,p≥1 ⇒t=k+m+p≥3). So n=12,16,20,...2. Or 2 + 2 +4p, where p≥1 (since z=0 ⇒c=4p). So n=2+2+4p=4(p+1). So p≥1 ⇒n=8,12,16,...Similarly, other permutations: 2+4p+2= same as above.So for n≡0 mod4, n≥8 can be expressed as either 4t (t≥3 ⇒n≥12) or 4(p+1) (p≥1 ⇒n≥8). Therefore, the minimal n for case 1 (sum≡0) is 8 (from the second type). For example, n=8 can be partitioned as 2,2,4. Then check if the Grundy numbers XOR to 0: 2 mod4=2, 2 mod4=2, 4 mod4=0. So 2 XOR2 XOR0=0. Yes. So n=8 is possible.Similarly, n=12 can be partitioned as 4,4,4 (Grundy numbers 0,0,0 XOR=0) or 2,2,8 (2,2,0 XOR=0). So both work.Similarly, n=4t for t≥2: wait, n=8=4*2, but t=2, but in the first case (0,0,0), each bucket needs at least 4 marbles, so t must be at least 3 (since 4+4+4=12). But in the second case, n=8=2+2+4.Therefore, for n≡0 mod4:If n≥8, Ani can distribute as (2,2,n-4), provided that n-4 is ≡0 mod4 (since n is 0 mod4, n-4 is also 0 mod4). So n-4 must be ≥4 (since each bucket must have at least 1, but n-4=0 would mean bucket c has 4 marbles. Wait, n=8: 2,2,4. 4 is allowed. So yes, for n≥8 and n≡0 mod4, Ani can split into (2,2,n-4), where n-4 is a multiple of 4 and ≥4. Wait, but n=8: n-4=4, which is allowed. n=12: 2,2,8 (8 is allowed). So yes.But also, for n=4t where t≥3, she can split into (4,4,4(t-2)), but 4(t-2) must be ≥4 ⇒t-2≥1 ⇒t≥3 ⇒n≥12.So for n≡0 mod4:- If n≥8, Ani can win by splitting into (2,2,n-4). For n=8, 2,2,4. For n=12, 2,2,8, etc.- For n=4, which is not allowed since n≥3 marbles, but also, Ani cannot split 4 into three buckets each with at least 1, as the minimal split is 1+1+2=4. For n=4, the split would be (1,1,2). Let's check the Grundy numbers: 1,1,2. XOR=1 XOR1 XOR2=0 XOR2=2≠0. So Budi can win. Hence, n=4 is not a winning position for Ani.Similarly, n=0 mod4 but n=8 and above: Ani can win. So n≡0 mod4 and n≥8.Case 2: n≡1 mod4.We need triples (x,y,z) with x XOR y XOR z=0 and x + y + z ≡1 mod4.Looking at the list:Possible triples:For example, (1,0,1): sum=2≡2.(1,1,0): sum=2≡2.(1,2,3): sum=6≡2.(3,3,0): sum=6≡2.(0,0,1): not in the list. Wait, the valid triples are those where z=x XOR y.Looking for triples where sum≡1 mod4.Looking through the earlier list:(1,2,3): sum=6≡2.(1,1,0): sum=2≡2.(0,3,3): sum=6≡2.(2,3,1): sum=6≡2.(3,2,1): same.(3,1,2): same.(2,2,0): sum=4≡0.(0,0,0): sum=0.(0,1,1): sum=2.(0,2,2): sum=4≡0.(1,3,2): sum=6≡2.(2,0,2): sum=4≡0.(3,0,3): sum=6≡2.(1,0,1): sum=2.(2,1,3): sum=6≡2.(3,3,0): sum=6≡2.Hmm, none of the triples sum to 1 mod4. Wait, is there any triple in the list with sum≡1 mod4?Wait, let's check:Take triple (1,0,1). Wait, z=1. But how is that generated? If x=1, y=0, then z=x XOR y=1 XOR0=1. So triple (1,0,1). Sum=1+0+1=2≡2.Another triple: (3,3,0). Sum=6≡2.What about (0,1,1)? Wait, no, (0,1,1) is x=0, y=1, z=0 XOR1=1. Sum=0+1+1=2.Similarly, (0,2,2): sum=4.So all triples either sum to 0,2, or 3 mod4?Wait, what about triple (3,0,3): x=3, y=0, z=3. Sum=6≡2.Wait, is there any triple in the list with sum≡1 or 3 mod4?Looking at the list:Take (3,3,0): sum=6≡2.(1,3,2): sum=6≡2.(3,1,2): same.(2,3,1): same.(2,1,3): same.(1,2,3): same.What about (3,2,1): sum=6≡2.(0,0,0): sum0.(2,2,0): sum4≡0.(0,2,2): sum4≡0.(1,1,0): sum2.(0,1,1): sum2.(2,0,2): sum4≡0.(3,0,3): sum6≡2.(1,0,1): sum2.Wait, so no triples sum to 1 or 3 mod4. Only 0,2 mod4.Wait, is that possible? That seems odd. Wait, maybe I made a mistake in generating the triples.Wait, when we generate triples where z=x XOR y, for x and y in 0-3:For example, x=1, y=2: z=3. Sum=1+2+3=6≡2 mod4.x=3, y=3: z=0. Sum=6≡2.x=3, y=1: z=2. Sum=3+1+2=6≡2.x=2, y=3: z=1. Sum=2+3+1=6≡2.x=1, y=1: z=0. Sum=2≡2.x=2, y=2: z=0. Sum=4≡0.x=0, y=0: z=0. Sum=0.x=0, y=1: z=1. Sum=0+1+1=2.x=0, y=2: z=2. Sum=0+2+2=4.x=0, y=3: z=3. Sum=0+3+3=6≡2.So indeed, all possible triples either sum to 0,2, or 4≡0, etc., but modulo4, the sums are only 0 or 2. Therefore, there are no triples where the sum is ≡1 or3 mod4. That seems strange but according to this, it's impossible.Therefore, for n≡1 mod4 or n≡3 mod4, there are no triples (x,y,z) with x XOR y XOR z=0 and sum x+y+z ≡n mod4. Therefore, Ani cannot create a P-position in these cases, so Budi has a winning strategy.Wait, but this can't be right. Because for example, take n=3 marbles. Ani must distribute them as 1,1,1. The Grundy numbers are 1,1,1. XOR=1 XOR1 XOR1=1≠0. So Budi can win. Similarly, n=5≡1 mod4. If Ani splits into 1,1,3. Grundy numbers 1,1,3. XOR=1 XOR1 XOR3=3≠0. Budi can win.But wait, is there any way for Ani to force a win for n≡1 or3 mod4?But according to our analysis, there's no possible triple (x,y,z) with XOR=0 and sum≡1 or3 mod4. Therefore, Ani cannot create a P-position, so Budi can always win regardless of how Ani splits the marbles. Therefore, Ani cannot win for n≡1 or3 mod4.Similarly, for n≡2 mod4:We need triples (x,y,z) with sum≡2 mod4 and x XOR y XOR z=0.Looking at our earlier list, several triples sum≡2 mod4:For example, (1,1,0): sum=2.(1,0,1): sum=2.(0,1,1): sum=2.(3,3,0): sum=6≡2.Similarly, (1,3,2): sum=6≡2.So there are triples with sum≡2 mod4.So for n≡2 mod4, Ani can try to split into such triples.For example, the triple (1,1,0). To make this into actual numbers:a=1 +4k, b=1 +4m, c=0 +4p (with p≥1 since c≥1). So sum=1+1+4p +4(k + m). So n=2 +4(p +k +m). So n≡2 mod4. The minimal n is when k=m=0, p=1: a=1, b=1, c=4 ⇒n=6.Similarly, another triple: (3,3,0). Sum=6≡2 mod4. Minimal n=3+3+4=10.Another triple: (1,3,2). Sum=6≡2 mod4. Minimal n=1+3+2=6. But here, a=1, b=3, c=2. All ≥1. So n=6. The Grundy numbers: 1,3,2. XOR=1 XOR3 XOR2=0.Wait, 1 XOR3=2, then XOR2=0. Yes. So n=6 can be split into (1,3,2), which gives a Grundy sum of 0.Similarly, n=6 can also be split into (1,1,4): Grundy numbers 1,1,0. XOR=0. Yes. So n=6 can be split both ways.So for n≡2 mod4:If n≥6, Ani can split into (1,1,n-2) where n-2≡0 mod4 and n-2≥4. For example, n=6: 1,1,4. n=10:1,1,8. Also, she can split into (1,3,2) if n=6. For higher n≡2 mod4, like n=10, can she split into (1,3,6)? Let's check: 1,3,6. Grundy numbers 1,3,2 (since6≡2). XOR=1 XOR3 XOR2=0. Yes. So n=10: (1,3,6). Also, (3,3,4): 3,3,0 (Grundy numbers 3,3,0). XOR=0. So yes.Therefore, for n≡2 mod4 and n≥6, Ani can split the marbles into such a triple. For n=2, it's impossible since n must be ≥3. For n=6,10,14,..., Ani can win.But wait, what about n=2 mod4 and n=6,10,14,...Wait, n=2 mod4: n=2,6,10,14,...But n≥3, so starting from n=6.Wait, n=2 is not allowed, but n=6 is the first in this class.Similarly, for n≡0 mod4 and n≥8.Now, checking n=4: it's 0 mod4 but Ani cannot win because the only split is 1,1,2 which gives XOR=1+1+2=4≡0 mod4, but wait, wait. Wait, n=4: split must be three buckets each ≥1. So possible splits are:1,1,2: Grundy numbers 1,1,2. XOR=1 XOR1 XOR2=2≠0. So Budi can win.1,2,1: same as above.2,1,1: same.So n=4 is losing for Ani.Similarly, n=8: split as 2,2,4. Grundy numbers 2,2,0. XOR=0. Ani can win.Similarly, n=12: 4,4,4: Grundy 0,0,0. XOR=0.So for n≡0 mod4, n≥8.For n≡2 mod4, n≥6.For n≡1 or3 mod4, Ani cannot win.Therefore, the answer is all n≡0 mod4 with n≥8 and all n≡2 mod4 with n≥6. But wait, let's verify with small n:n=3: 1,1,1. XOR=1+1+1=3≠0. Budi wins.n=4: 1,1,2. XOR=2≠0. Budi.n=5: possible splits like 1,1,3 (XOR=3), 1,2,2 (XOR=1+2+2=5≡1). Not 0. So Budi.n=6: split into 1,1,4 (XOR=1+1+0=0). Ani wins.n=7: possible splits. If she splits into 1,2,4. XOR=1+2+0=3≠0. If she splits into 3,3,1. XOR=3+3+1=7≡3≠0. So Budi.n=8: split into 2,2,4. XOR=0. Ani wins.n=9: splits. Try 1,1,7: XOR=1+1+3=5≡1. No. 3,3,3: XOR=3+3+3=3≠0. Budi.n=10: split into 1,1,8: XOR=1+1+0=0. Ani wins.n=11: splits. 1,2,8: XOR=1+2+0=3. No. Budi.n=12: split into 4,4,4: XOR=0. Ani wins.So yes, the pattern holds. Therefore, the solution is:All integers n ≥6 where n is even but not congruent to 0 mod4. Wait, no. Wait, n=6≡2 mod4, n=8≡0 mod4, n=10≡2 mod4, n=12≡0 mod4. So it's all even n≥6. But wait, n=6≡2 mod4, n=8≡0, n=10≡2, etc. So actually, it's all even numbers n≥6, regardless of being 0 or2 mod4? Wait no, because in our analysis, for n≡0 mod4, n≥8, and n≡2 mod4, n≥6. So the union is all even n≥6. Because:- Even n≥6: if n≡2 mod4 (6,10,14,...) and n≡0 mod4 (8,12,16,...).Yes, so combining both cases, all even n≥6.But wait, let's check:n=6: yes.n=8: yes.n=10: yes.n=12: yes.n=14: yes.But according to our earlier analysis, for n≡0 mod4, starting from 8, and n≡2 mod4 starting from 6. So union is even n≥6.Therefore, the answer is all even integers n≥6. Ani can win for even n≥6 by distributing the marbles such that two buckets have 1 marble each and the third has n-2 marbles (which is even and ≥4), making the Grundy numbers 1,1,0 (since n-2≡0 mod4), thus XOR=0. Alternatively, for n≡0 mod4, she can split into two 2s and the rest, like 2,2,n-4 (n-4≡0 mod4). Both strategies work.Wait, but when n=6, splitting into 1,1,4: Grundy numbers 1,1,0. XOR=0.For n=8: 2,2,4: Grundy 2,2,0. XOR=0.Similarly, n=10:1,1,8: Grundy 1,1,0. XOR=0.Yes, so whether n is 0 or2 mod4, as long as it's even and ≥6, Ani can split into either (1,1,n-2) or (2,2,n-4), whichever makes the third bucket a multiple of4.Thus, the final answer is all even n≥6. However, let's check for n=6:Ani's move: (1,1,4). Correct.n=8: (2,2,4).n=10: (1,1,8) or (3,3,4) since 3+3+4=10, and Grundy numbers 3,3,0. XOR=0.Yes, either way.Therefore, all even numbers n≥6. So the answer is that Ani can win for all even integers n≥6, and her first move is to distribute the marbles into two buckets with 1 marble each and the third with n-2 marbles (if n≡2 mod4), or into two buckets with 2 marbles each and the third with n-4 marbles (if n≡0 mod4). Alternatively, any distribution that results in Grundy numbers XORing to zero.But to specify the answer as per the question: "Find all n such that Ani has a winning strategy, including what Ani's first move should be for these n."Thus, the answer is all even integers n≥6. Ani's first move is to divide the marbles into three buckets such that two of them are 1 each and the third is n−2 (if n ≡2 mod4), or two of them are 2 each and the third is n−4 (if n ≡0 mod4). Both distributions result in Grundy numbers XORing to zero.But wait, let's confirm:For n=6 (≡2 mod4): two 1s and one 4. Yes, 1+1+4=6.For n=8 (≡0 mod4): two 2s and one 4. Yes.For n=10 (≡2 mod4): two 1s and one 8. Yes.For n=12 (≡0 mod4): two 2s and one 8. Yes.Alternatively, she can also split into other distributions. For example, for n=6, she can do (3,3,0) but 0 isn't allowed. Wait, no, 0 mod4 corresponds to 4 marbles. Wait, no, (3,3,0) would mean the third bucket has 0 marbles, which is invalid. So she has to have all buckets with at least1. So for n=6, she can't do (3,3,0) but can do (1,1,4).Alternatively, (1,3,2) for n=6: 1+3+2=6. Grundy numbers 1,3,2. XOR=0. So yes, that's another possible distribution.Similarly, for n=10: (1,3,6). Grundy numbers 1,3,2 (since6≡2 mod4). XOR=0.So there can be multiple winning distributions. However, the question asks to include what Ani's first move should be. So we need to specify at least one distribution for each n.Therefore, summarizing:All even integers n≥6. For such n, Ani can win by splitting the marbles into either:- Two buckets with 1 marble each and the third with n−2 marbles (if n ≡2 mod4), since n−2 will be ≡0 mod4.- Two buckets with 2 marbles each and the third with n−4 marbles (if n ≡0 mod4), since n−4 will be ≡0 mod4.Alternatively, she can distribute them in other ways that achieve a Grundy XOR of zero, but the above are standard methods.Thus, the final answer is all even n≥6, with the specified distributions.

✅:All even integers ( n geq 6 ). For these ( n ), Ani can distribute the marbles as follows:- If ( n equiv 0 pmod{4} ): Split the marbles into two buckets with 2 marbles each and the third bucket with ( n - 4 ) marbles. This ensures the Grundy numbers (2, 2, 0) modulo 4, whose XOR is 0. - If ( n equiv 2 pmod{4} ): Split the marbles into two buckets with 1 marble each and the third bucket with ( n - 2 ) marbles. This results in Grundy numbers (1, 1, 0) modulo 4, whose XOR is 0.Final Answer: Ani has a winning strategy for all even integers ( n geq 6 ). She should distribute the marbles as ( (2, 2, n - 4) ) when ( n equiv 0 pmod{4} ) and as ( (1, 1, n - 2) ) when ( n equiv 2 pmod{4} ). Thus, the solution is all even ( n ) with ( n geq 6 ), and the required distributions are:[boxed{n text{ is even and } n geq 6}]

🔑:1. Initial Setup and Strategy for Ani: - Ani distributes the marbles such that each bucket has at least one marble. - For even ( n ge 6 ), Ani can distribute the marbles as follows: - Place 4 marbles in the first bucket. - Place (frac{n-4}{2}) marbles in each of the other two buckets. - This ensures that the total number of marbles is ( n ) and each bucket has at least one marble.2. Ani's Winning Strategy: - After the initial distribution, Ani can use a mirroring strategy: - If Budi takes ( x ) marbles from the second bucket, Ani takes ( x ) marbles from the third bucket. - If Budi takes ( x ) marbles from the third bucket, Ani takes ( x ) marbles from the second bucket. - This strategy ensures that the number of marbles in the second and third buckets remains equal after each pair of turns.3. Characterizing Losing Triples: - A triple ((a, b, c)) is losing if: 1. Two of ( a, b, c ) are congruent modulo 4, and the other one is 0 modulo 4. 2. ( a, b, c ) are congruent to 1, 2, 3 modulo 4 in some order. - These conditions ensure that the next player to move will lose if they encounter such a configuration.4. Proof of Losing Triples: - If ((a, b, c)) is golden (satisfies one of the losing conditions), removing 1, 2, or 3 marbles from any bucket will make it not golden. - If ((a, b, c)) is not golden, it is always possible to remove 1, 2, or 3 marbles from some bucket to make it golden. - These observations imply that the player encountering a golden configuration will lose if the opponent plays optimally.5. Odd ( n ) and ( n = 4 ): - For odd ( n ), the sum ( a + b + c ) is odd, so the configuration cannot be losing. - For ( n = 4 ), the possible distributions are ( (2, 1, 1) ) in some order. Budi can take both marbles from the bucket with 2 marbles, ensuring a win.The final answer is all even ( boxed{ n ge 6 } ).

🧠:Okay, let me try to figure out how to solve this problem. So, we need to find the value of x² + y² + z² given that x + y + z = 13, xyz = 72, and 1/x + 1/y + 1/z = 3/4. Hmm, I remember that when dealing with symmetric expressions like this, it's often useful to use Vieta's formulas, which relate the sums and products of variables to the coefficients of a polynomial. Maybe I can consider x, y, z as roots of a cubic equation. Let's see.First, let me recall Vieta's formulas for a cubic equation. If x, y, z are roots of the equation t³ - at² + bt - c = 0, then:a = x + y + z,b = xy + yz + zx,c = xyz.Given that we know a = 13 and c = 72, but we don't know b. If I can find the value of b, then I can use the identity x² + y² + z² = (x + y + z)² - 2(xy + yz + zx) = a² - 2b. So, if I can find b, then I can compute the desired expression.So, how do I find b? We also have the information that 1/x + 1/y + 1/z = 3/4. Let's try to express this in terms of the known values. The sum of reciprocals can be written as (xy + yz + zx)/xyz. So, (xy + yz + zx)/xyz = 3/4. Substituting the known value of xyz, we get (b)/72 = 3/4. Solving for b, we multiply both sides by 72: b = (3/4)*72. Let's compute that. 72 divided by 4 is 18, so 3*18 = 54. Therefore, b = 54.Now that we have b = 54, we can substitute into the expression for x² + y² + z². So, x² + y² + z² = a² - 2b = 13² - 2*54. Calculating 13 squared: 13*13 is 169. Then, 2*54 is 108. Subtracting, 169 - 108 = 61. So, the value should be 61.Wait, let me check my steps again to make sure I didn't make a mistake. First, Vieta's formula: correct. Sum of reciprocals: (xy + yz + zx)/(xyz) = 3/4. So, substituting the known xyz = 72: (b)/72 = 3/4. Multiply both sides by 72: b = (3/4)*72. 3 divided by 4 is 0.75, 0.75*72: let's compute that. 72*0.75 is 54. So, yes, b = 54.Then, x² + y² + z² = (x + y + z)² - 2(xy + yz + zx) = 13² - 2*54. 13 squared is 169, 2*54 is 108, 169 - 108 is 61. That seems right.But maybe I should verify this another way to be thorough. Let's see. Suppose x, y, z are integers. Since xyz = 72, maybe they are integers. Let's list the factors of 72 and see if three numbers add up to 13 and multiply to 72.Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Let's try to find three numbers among these that multiply to 72 and add up to 13.Start with the smaller numbers. Let's try 2, 3, 12: 2+3+12=17, which is too big. 2, 4, 9: 2+4+9=15. Still too big. 3, 4, 6: 3+4+6=13. Perfect! And 3*4*6=72. So the numbers are 3, 4, and 6. Let's check the sum of reciprocals: 1/3 + 1/4 + 1/6 = (4 + 3 + 2)/12 = 9/12 = 3/4. That matches the given condition. So, indeed, x, y, z are 3, 4, 6 in some order. Then, x² + y² + z² = 9 + 16 + 36 = 61. Yep, that's correct.So, even if I didn't use the Vieta's method, by testing possible integers, I also get the same result. Therefore, the answer is 61.But just to ensure that there are no other possible triplet of numbers (non-integer perhaps) that could satisfy the conditions, even though in this case we found integer solutions. Let's think: If the variables are not integers, could there be another solution? Let's consider the cubic equation constructed from Vieta's formulas: t³ - 13t² + 54t -72 = 0. If we factor this, let's check if it factors as (t - 3)(t - 4)(t - 6). Multiply them out: (t - 3)(t - 4) = t² -7t +12, then multiply by (t - 6): t³ -6t² -7t² +42t +12t -72 = t³ -13t² +54t -72. Yes, that's correct. So the equation factors into (t -3)(t -4)(t -6), so the roots are indeed 3, 4, 6. Therefore, there are no other roots, so the only solutions are these integers, so x² + y² + z² must be 61.Therefore, the answer is 61. All steps check out. No mistakes detected in the process. The answer is confirmed through multiple methods: Vieta's formula leading to the calculation, checking integer solutions, and verifying the polynomial factors. All confirm the result is 61.Final AnswerThe value of the expression is boxed{61}.

✅:Given the equations ( x + y + z = 13 ), ( xyz = 72 ), and ( frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{3}{4} ), we need to determine the value of ( x^2 + y^2 + z^2 ).First, we use the identity for the sum of squares:[x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx)]We know ( x + y + z = 13 ), so we need to find ( xy + yz + zx ). Using the given sum of reciprocals:[frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{xy + yz + zx}{xyz} = frac{3}{4}]Substituting ( xyz = 72 ):[frac{xy + yz + zx}{72} = frac{3}{4} implies xy + yz + zx = frac{3}{4} times 72 = 54]Now, substituting ( x + y + z = 13 ) and ( xy + yz + zx = 54 ) into the sum of squares identity:[x^2 + y^2 + z^2 = 13^2 - 2 times 54 = 169 - 108 = 61]To verify, we check if ( x, y, z ) can be integers. The factors of 72 that add up to 13 are 3, 4, and 6. These values satisfy all given conditions:[3 + 4 + 6 = 13, quad 3 times 4 times 6 = 72, quad frac{1}{3} + frac{1}{4} + frac{1}{6} = frac{3}{4}]Calculating the sum of squares for these values:[3^2 + 4^2 + 6^2 = 9 + 16 + 36 = 61]Thus, the value of ( x^2 + y^2 + z^2 ) is (boxed{61}).

🔑:1. Given the equations: [ x + y + z = 13, ] [ xyz = 72, ] [ frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{3}{4}, ] we need to determine the value of ( x^2 + y^2 + z^2 ).2. First, we use the identity for the sum of the reciprocals: [ frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{xy + yz + zx}{xyz}. ] Substituting the given values, we have: [ frac{xy + yz + zx}{72} = frac{3}{4}. ]3. Solving for ( xy + yz + zx ): [ xy + yz + zx = 72 cdot frac{3}{4} = 54. ]4. Next, we use the identity for the sum of squares: [ x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx). ] Substituting the known values: [ x^2 + y^2 + z^2 = 13^2 - 2 cdot 54. ]5. Calculating the right-hand side: [ 13^2 = 169, ] [ 2 cdot 54 = 108, ] [ x^2 + y^2 + z^2 = 169 - 108 = 61. ]6. Therefore, the value of the expression ( x^2 + y^2 + z^2 ) is: [ boxed{61}. ]

🧠:Okay, let me try to figure out this problem step by step. So, we have an array with 2*(2^n - 1) rows and k columns. The coloring is 2-color, which I assume means each element is either one color or the other, like red or blue. The key condition here is that any two columns must agree on less than 2^n - 1 entries on the same row. Wait, hold on, the wording says "less than 2^n -1 entries on the same row." Hmm, that part is a bit confusing. Let me parse it again."Any two columns agree on less than 2^n -1 entries on the same row." So, for each row, if we look at two columns, the number of entries where they have the same color on that row is less than 2^n -1. Wait, but each row has k entries, right? Each column is a vector of length 2*(2^n -1), with each entry colored either color 1 or color 2.Wait, maybe I need to clarify the problem. The array has 2*(2^n -1) rows and k columns. Each element is colored with two colors. The coloring is acceptable if any two columns agree on less than 2^n -1 entries on the same row. So, for any pair of columns, when you look at any single row, the number of positions (entries) in that row where both columns have the same color is less than 2^n -1.Wait, but each row is part of the columns. So, each column is a vector of length 2*(2^n -1). So, each column has an entry for each row. So, when we compare two columns, they have 2*(2^n -1) entries, each of which can agree or disagree. But the problem states that they agree on less than 2^n -1 entries on the same row. Wait, maybe "on the same row" is a mistranslation or misunderstanding.Wait, maybe the problem is that for any two columns, when restricted to the same row, the number of entries where they agree is less than 2^n -1. But each row has k columns, so each row is a vector of k entries. If we take two columns, they are two vectors of length 2*(2^n -1). The agreement between two columns is the number of rows where their entries are the same. But the problem says "agree on less than 2^n -1 entries on the same row." Wait, maybe the translation is off? Maybe it's supposed to say "on the same row" or "in the same row."Alternatively, perhaps the problem is that any two columns must not agree on more than 2^n -2 entries in any single row. Since they have to agree on less than 2^n -1 entries per row.But each row has k entries, so when they say "any two columns agree on less than 2^n -1 entries on the same row," does that mean that, for any two columns, the number of rows where they have the same color in that row is less than 2^n -1? Or does it mean that, within a single row, the number of columns where they agree is less than 2^n -1?Wait, the original problem states: "any two columns agree on less than 2^n -1 entries on the same row." So, "on the same row" – so, for each row, when you look at the entries of two different columns in that row, how many times do they agree? Wait, but each column has an entry for each row, so each column is a vertical vector. Then two columns would have two entries per row. Wait, maybe the problem is that in each row, there are k entries (since the array is 2*(2^n -1) rows by k columns). So, each row has k elements, each colored. Then, two columns (each column being a vertical vector of 2*(2^n -1) elements) would correspond to two different vertical vectors. Then, when they say "agree on less than 2^n -1 entries on the same row," perhaps that is a mistranslation or misphrasing. Maybe it's supposed to be "agree on less than 2^n -1 rows for each entry"? Wait, maybe the problem is that the number of rows where the two columns have the same color in that row is less than 2^n -1. Because otherwise, "entries on the same row" doesn't quite make sense.Wait, let's rephrase the problem statement again to make sure. The array has 2*(2^n -1) rows and k columns. A 2-coloring is acceptable if any two columns agree on less than 2^n -1 entries on the same row. Hmm. So, for any two columns, and for any row, the number of entries in that row where they agree is less than 2^n -1. But each row has k entries, so each entry is part of a column. Wait, maybe the problem is that each row has k entries, each entry being part of a different column. So, for a given row, each column has one entry in that row. Then, two columns would each have one entry in that row. So, for each row, two columns can either agree or disagree on that row. Then, the total number of rows where two columns agree is the number of rows where their entries are the same. So, maybe the problem is that this number (the number of rows where two columns agree) must be less than 2^n -1. But the problem says "entries on the same row," which is confusing. If two columns agree on an entry on the same row, that would be one entry. So, perhaps the problem is phrased incorrectly? Maybe it should be "agree on less than 2^n -1 rows" instead of "entries on the same row."Alternatively, maybe the problem is that in each row, the number of columns where they agree in that row is less than 2^n -1. But that would be per row. If two columns are compared, in each row, they can either agree or disagree. So, for each row, the two entries (one from each column) either match or don't. So, the number of rows where they agree is the Hamming distance or something. But the problem says "agree on less than 2^n -1 entries on the same row." Wait, but "entries on the same row" – for a single row, two columns only have one entry each in that row. So, they can either agree or disagree on that row. So, how can two columns agree on multiple entries on the same row? Unless each column has multiple entries per row. But the array is 2*(2^n -1) rows by k columns, so each column has one entry per row. Therefore, in each row, two columns have one entry each, so they can either agree or disagree on that single entry. Therefore, the number of entries where they agree on the same row is either 0 or 1. But 2^n -1 is a much larger number. So, this suggests that perhaps the problem is misstated or I am misunderstanding.Wait, maybe there's a misinterpretation here. Let's check again: the array has 2*(2^n -1) rows and k columns. Each element is colored with two colors. The coloring is acceptable if any two columns agree on less than 2^n -1 entries on the same row. Hmm. If each column is a vector of length 2*(2^n -1), with entries colored in two colors, then two columns agree on an entry if, for a particular row (i.e., position in the vector), they have the same color. Therefore, the number of entries they agree on is the number of rows where their colors match. Therefore, the problem might actually be saying that for any two columns, the number of rows where they agree is less than 2^n -1. That is, the Hamming distance between any two columns is greater than 2*(2^n -1) - (2^n -2) = 2*(2^n -1) - 2^n + 2 = 2*2^n - 2 - 2^n + 2 = 2^n. Wait, but the Hamming distance is the number of positions where they differ. If the number of agreements is less than 2^n -1, then the number of disagreements is more than 2*(2^n -1) - (2^n -2) = 2*2^n -2 -2^n +2 = 2^n. So, the Hamming distance is greater than 2^n. But in coding theory, codes with certain Hamming distances have bounds on their maximum size. Maybe this is related to error-correcting codes?Alternatively, maybe the problem is about a different kind of structure. Let me think again. If the array has 2*(2^n -1) rows and k columns, each entry colored with two colors. The condition is that any two columns agree on less than 2^n -1 entries. So, if we think of each column as a binary vector (since it's two-colored) of length 2*(2^n -1), then the maximum number of agreements between any two vectors is less than 2^n -1. The question is to determine the maximum k for which such a coloring exists.Alternatively, maybe the problem is in combinatorial design. For example, each pair of columns must not have too many agreements. The question is, given the length of the vectors (2*(2^n -1)), what's the maximum number of vectors (k) such that any two vectors agree in fewer than 2^n -1 positions.This seems similar to a code with certain distance properties. In coding theory, the maximum number of codewords with a certain minimum distance is a classic problem. However, here the condition is not on the minimum distance (which is about the number of disagreements), but on the maximum number of agreements. So, instead of requiring that any two codewords differ in at least d positions, we require that any two codewords agree in at most t positions.This is a different kind of code, sometimes called a "code with limited pairwise coincidence" or a "code with bounded correlation." The problem then is to find the maximum size of such a code.Alternatively, in combinatorics, this is similar to a set system where the intersection of any two sets is limited. But here, instead of sets, we have vectors, and their "intersection" is the number of coordinates where they agree.Another perspective is that each column is a function from the set of rows to {0,1} (assuming the two colors are 0 and 1). The condition is that for any two functions, the number of inputs (rows) where they agree is less than 2^n -1. So, the maximum k such that there exist k functions with this property.Alternatively, we can model this as a graph problem where each column is a vertex, and edges represent pairs that agree on too many entries. Then, we're looking for the maximum independent set in this graph. But this is probably not directly helpful.Alternatively, perhaps we can use probabilistic methods or algebraic constructions to bound k. However, given the problem structure, it might be related to combinatorial designs or finite geometries.Given that the number of rows is 2*(2^n -1), which is twice the number of non-zero vectors in an n-dimensional vector space over GF(2). Wait, 2^n -1 is the number of non-zero vectors in GF(2)^n. So, 2*(2^n -1) might be related to pairing each non-zero vector with its complement or something. Hmm, maybe the rows are indexed by pairs (v, -v) where v is a non-zero vector in GF(2)^n. But I need to check.Alternatively, consider that in coding theory, certain codes have parameters related to 2^n. For example, Reed-Muller codes, or Reed-Solomon codes over GF(2^n). But I need to think carefully.Alternatively, since the number of rows is 2*(2^n -1), perhaps we can consider each row as corresponding to a hyperplane or a co-dimension 1 subspace in GF(2)^n. But I'm not sure.Wait, here's another thought. If we have 2*(2^n -1) rows, perhaps each row corresponds to a pair of a non-zero vector and its complement in GF(2)^n. For example, for each non-zero vector v in GF(2)^n, we have two rows: one for v and one for -v (but in GF(2), -v is the same as v). Wait, that might not make sense.Alternatively, maybe the rows correspond to the non-zero vectors of GF(2)^n and their complements, but since in GF(2), complementing a vector is equivalent to adding the all-ones vector. However, the number of non-zero vectors in GF(2)^n is 2^n -1, so doubling that would give 2*(2^n -1) rows. Maybe each row is duplicated? But why?Alternatively, think of each row as corresponding to a pair consisting of a non-zero vector and a non-zero vector with a sign? But over GF(2), signs don't matter. Hmm.Alternatively, maybe the rows are structured such that each original row from a set of 2^n -1 rows is duplicated, leading to 2*(2^n -1) rows. If that's the case, perhaps the problem is related to a design where each element is repeated twice.But perhaps this is overcomplicating. Let's step back.We need to construct a binary matrix with 2*(2^n -1) rows and k columns, such that any two columns agree in less than 2^n -1 positions. What is the maximum k?Alternatively, since each column is a vector in {0,1}^(2*(2^n -1)), and any two vectors share less than 2^n -1 ones in common positions.Wait, but the agreement is on the same color. So, if two columns have the same color in a row, that counts as an agreement. So, the number of agreements is the number of rows where the two columns have the same bit (if we think of the colors as bits).So, for two vectors u and v in {0,1}^m, where m = 2*(2^n -1), the number of coordinates where u_i = v_i must be less than 2^n -1. So, the inner product (using standard inner product) would be related to the number of agreements. Wait, actually, the number of agreements is equal to (m + inner product(u, v)) / 2 if we consider 0 and 1 as bits. Wait, if we map the colors to { -1, 1 }, then the inner product would be the sum of u_i * v_i, which is equal to the number of agreements minus the number of disagreements. So, agreements = (inner product + m)/2. So, if the number of agreements is less than t, then inner product + m < 2t, so inner product < 2t - m. But here, t is 2^n -1, so inner product < 2*(2^n -1) - m. But m is 2*(2^n -1). So, inner product < 2*(2^n -1) - 2*(2^n -1) = 0. Therefore, inner product < 0. But if we use ±1 vectors, the inner product being negative means that the two vectors are negatively correlated. However, in coding theory, this is a bit different. Alternatively, using 0/1 vectors, the number of agreements is the dot product. So, if u and v are 0/1 vectors, then the dot product is the number of positions where both are 1. But the total number of agreements is the number of positions where u_i = v_i, which is equal to the dot product plus the number of positions where both are 0. So, if we let A = dot product, and B = number of positions where both are 0, then total agreements = A + B. If the vectors are in {0,1}^m, then the total agreements is (m - (number of positions where u ≠ v)). So, the number of agreements is m - d_H(u, v), where d_H is the Hamming distance. Therefore, the condition that agreements < 2^n -1 is equivalent to m - d_H(u, v) < 2^n -1, which implies that d_H(u, v) > m - (2^n -1). Since m = 2*(2^n -1), this gives d_H(u, v) > 2*(2^n -1) - (2^n -1) = 2^n -1. Therefore, the Hamming distance between any two columns must be greater than 2^n -1. So, this is equivalent to a code with length m = 2*(2^n -1), minimal Hamming distance greater than 2^n -1, and we need the maximal k (size of the code). Ah! Now this makes sense. So, the problem is equivalent to finding the maximum number of codewords in a binary code of length 2*(2^n -1) with minimal Hamming distance at least 2^n. Because if the Hamming distance is greater than 2^n -1, then it's at least 2^n. So, the question reduces to: What is the maximum size k of a binary code with length 2*(2^n -1), minimal distance 2^n?In coding theory, there are bounds on the maximum size of codes given their length and minimal distance. The most relevant bound here might be the Singleton Bound, which states that for a code of length m and minimal distance d, the maximum size is at most 2^{m - d +1}. However, the Singleton Bound is generally for codes over arbitrary alphabets, and for binary codes, the Hamming Bound or the Johnson Bound might be tighter. However, in some cases, the Plotkin Bound gives a tighter upper limit.The Plotkin Bound is useful when the minimal distance is a significant fraction of the code length. The Plotkin theorem states that if the minimal distance d satisfies d > (m/2), then the maximum number of codewords is bounded by floor(d / (2d - m)). In our case, the code length is m = 2*(2^n -1), and the minimal distance d = 2^n. Let's check if d > m/2:d = 2^nm/2 = (2*(2^n -1))/2 = 2^n -1So, d = 2^n > m/2 = 2^n -1. Yes, so the Plotkin Bound applies here.Applying Plotkin's theorem: If d > m/2, then the maximum number of codewords k satisfies k <= floor( d / (2d - m) )Plugging in d = 2^n and m = 2*(2^n -1):2d - m = 2*(2^n) - 2*(2^n -1) = 2*2^n - 2*2^n + 2 = 2Therefore, k <= floor( 2^n / 2 ) = 2^{n -1}But wait, Plotkin's Bound says that if d > m/2, then A(m,d) <= floor( d / (2d - m) )But in our case, 2d - m = 2*2^n - 2*(2^n -1) = 2*2^n - 2*2^n + 2 = 2. So, A(m,d) <= floor( 2^n / 2 ) = 2^{n-1}However, Plotkin's theorem also has another condition: when d is even, or in some cases, but perhaps in the binary case, the formula is slightly different. Wait, let me check the exact statement of Plotkin's theorem.Plotkin's theorem states that for a code of length m over an alphabet of size q, if the minimal distance d satisfies d > (q -1)m / q, then the maximum number of codewords is floor( d / (d - (q -1)m / q ) )But in our case, q = 2, so (q -1)m / q = (1)*m /2 = m/2. So, if d > m/2, then the maximum number of codewords is floor( d / (d - m/2) )Wait, in our case, d = 2^n, m = 2*(2^n -1). So, d - m/2 = 2^n - (2*(2^n -1))/2 = 2^n - (2^n -1) = 1. Therefore, k <= floor( d / 1 ) = d = 2^n.Wait, that contradicts my previous calculation. Wait, maybe I had the formula wrong before.Let me re-derive Plotkin's Bound for binary codes.For binary codes, if d > m/2, then the maximum number of codewords k satisfies k <= floor( d / (2d - m) )Wait, let's check with our parameters:2d - m = 2*2^n - 2*(2^n -1) = 2^{n+1} - 2^{n+1} + 2 = 2So, k <= floor( 2^n / 2 ) = 2^{n -1}But another source says that if d > m/2, then k <= floor( d / (2d - m) )Wait, but according to the calculation, if we use this formula, we get k <= 2^{n}/2 = 2^{n-1}But when d - m/2 = 1, as above, then the formula k <= floor( d / (d - m/2) ) would be floor(2^n /1) = 2^n. But this seems different.Wait, maybe there's a different version of Plotkin's Bound. Let me check.According to Wikipedia, Plotkin's theorem states:"If C is a code of length n over an alphabet of size q with minimum distance d, then if d > (1 - 1/q) n, then |C| ≤ floor( d / (d - (1 - 1/q) n ) )"For binary codes, q=2, so (1 - 1/q)n = n/2. So, if d > n/2, then |C| ≤ floor( d / (d - n/2 ) )In our case, n (length) = 2*(2^n -1), d = 2^n. Then,d - n/2 = 2^n - (2*(2^n -1))/2 = 2^n - (2^n -1) = 1Therefore, |C| ≤ floor( d /1 ) = d = 2^nBut this contradicts the other version. Hmm.Wait, so according to this version of Plotkin's theorem, the maximum k is at most 2^n. But in our case, when n=1, for example, let's test.Wait, let's take n=1. Then the problem parameters:Number of rows = 2*(2^1 -1) = 2*1 = 2 rows.We need a 2-coloring (binary) of a 2x k array where any two columns agree in less than 2^1 -1 = 1 entry. So, they must agree in 0 entries. So, any two columns must be different in both rows. Since there are two rows, each column is a pair of bits. To have all columns pairwise differing in both positions. The maximum such k is 2, since there are 2^2 =4 possible pairs, but any two must differ in both bits. The possible pairs are 00, 11 (which agree with themselves but not with others). Wait, actually, if we pick 00 and 11, they differ in both positions. Similarly, 01 and 10. So, maximum k is 2.But according to Plotkin's theorem (the second version), with n=2 (wait, no, n in the theorem is the length). Wait, in our case, for n=1 (original problem's n=1), the code length m=2*(2^1 -1)=2, minimal distance d=2^1=2. Then, according to Plotkin's theorem, if d > m/2, which is 2 > 1, yes. Then |C| <= floor( d / (d - m/2) ) = floor(2 / (2 -1)) = 2. Which matches the example. So, maximum k=2. Which is correct.But according to the original problem's question, given n, determine the maximum k. So, when n=1, k=2. What does 2^{n} give? 2^1=2, which matches. If we use the second version of Plotkin's theorem, we get |C| <=2^n, which in this case is 2, which is correct.But wait, for n=2. Let's check. For n=2, original problem's n=2. Then m=2*(2^2 -1)=2*3=6 rows. The minimal distance d=2^2=4. According to Plotkin's theorem, |C| <= floor(4 / (4 - 6/2)) = floor(4 / (4 -3))=floor(4/1)=4. So maximum k=4. Let's see if that's achievable.Is there a binary code of length 6, minimal distance 4, with 4 codewords? For example, the repetition code: 000000, 111111, 000111, 111000. Let's check the distances. Between 000000 and 111111: distance 6. Between 000000 and 000111: distance 3. Which is less than 4. So that's no good. Alternatively, take the codewords as 000000, 001111, 110011, 111100. Let's compute pairwise distances. Between 000000 and 001111: distance 4. Between 000000 and 110011: distance 4. Between 000000 and 111100: distance 4. Between 001111 and 110011: compare 001111 vs 110011. They differ in positions 1,2,5,6: distance 4. Similarly, 001111 vs 111100: differ in positions 1,2,5,6: distance 4. 110011 vs 111100: differ in positions 3,4,5,6: distance 4. So yes, this code has 4 codewords with minimal distance 4. So, it is achievable. Therefore, Plotkin's theorem gives a tight bound here.Thus, in general, according to Plotkin's theorem, for our parameters, the maximum k is 2^n. But in the n=2 case, we achieved k=4=2^2, which matches. For n=1, k=2=2^1. So, the general formula seems to be k=2^n. But wait, the Plotkin bound says it's at most 2^n. But is this achievable for all n?Wait, for example, take n=3. Then m=2*(2^3 -1)=14 rows. Minimal distance d=8. Then, according to Plotkin, |C| <=8. But can we actually construct such a code with 8 codewords?Well, in coding theory, there is the Reed-Muller code. Reed-Muller codes of order m have parameters related to the length, dimension, and minimal distance. For example, the first-order Reed-Muller code of length 2^m has minimal distance 2^{m-1}. However, our code has length 14, which is not a power of two. Alternatively, perhaps using a Simplex code. Wait, Simplex codes are codes where the codewords are the vertices of a simplex in Euclidean space, and they have maximal minimal distance. But for binary codes, Simplex codes have parameters [2^r -1, r, 2^{r-1}]. But again, our code length is 14, which is 2*(2^3 -1). Not sure.Alternatively, maybe using a code formed by concatenating two codes. Alternatively, since m=2*(2^n -1), which is twice the number of non-zero vectors in GF(2)^n. Maybe there's a connection to the Simplex code, which has codewords corresponding to the non-zero vectors in GF(2)^n.Wait, in the Simplex code, each codeword is a vector of length 2^n -1, where each position corresponds to a non-zero vector in GF(2)^n, and the codeword is the evaluation of a linear function at those vectors. The Simplex code has minimal distance 2^{n-1}, which is half the length. But our code needs minimal distance 2^n, which is more than half the length (since m=2*(2^n -1), half the length is 2^n -1, and 2^n is greater than that). So, perhaps this is a different code.Alternatively, maybe the code is constructed by taking two copies of a Simplex code. For example, if we take each codeword and duplicate it, resulting in a code of length 2*(2^n -1). Then, the minimal distance would be doubled? Wait, no. If you duplicate each bit, the minimal distance would also be doubled. So, if the original code has minimal distance d, the duplicated code would have minimal distance 2d. So, if we take the Simplex code of length 2^n -1, minimal distance 2^{n-1}, duplicating each bit would give a code of length 2*(2^n -1), minimal distance 2*2^{n-1}=2^n. Which is exactly what we need. Then, the number of codewords in the Simplex code is 2^n. So, duplicating each codeword would result in 2^n codewords of length 2*(2^n -1) with minimal distance 2^n. Therefore, such a code exists, achieving k=2^n. Since the Plotkin bound gives k <=2^n, this is optimal.Therefore, the maximum k is 2^n. Therefore, the answer should be boxed{2^n}But let me verify this again. The Simplex code is a linear code with parameters [2^n -1, n, 2^{n-1}]. If we duplicate each coordinate, we get a code with parameters [2*(2^n -1), n, 2*2^{n-1}]=[2*(2^n -1),n,2^n]. Therefore, this code has 2^n codewords (since the dimension is n, the number of codewords is 2^n). The minimal distance is 2^n, which satisfies our requirement that the Hamming distance is greater than 2^n -1. Therefore, such a code exists, and by Plotkin's bound, this is the maximum possible. Therefore, the maximum k is 2^n.Hence, the answer is boxed{2^n}

✅:Given positive integers ( k ) and ( n ), we need to determine the maximum ( k ) for which an acceptable 2-coloring exists in an array of ( 2(2^n - 1) ) rows and ( k ) columns. An acceptable 2-coloring requires that any two columns agree on less than ( 2^n - 1 ) entries on the same row.Key Steps:1. Problem Transformation: - The problem can be transformed into a coding theory problem where each column is a binary vector of length ( 2(2^n - 1) ). - The requirement translates to any two vectors (columns) having a Hamming distance greater than ( 2^n - 1 ), meaning they must differ in more than ( 2^n - 1 ) positions.2. Plotkin Bound Application: - The Plotkin Bound is used to determine the maximum number of codewords (columns) in a binary code with given length and minimal distance. - For code length ( m = 2(2^n - 1) ) and minimal distance ( d = 2^n ), the Plotkin Bound states that the maximum number of codewords ( k ) is ( leftlfloor frac{d}{2d - m} rightrfloor ). - Substituting ( d = 2^n ) and ( m = 2(2^n - 1) ), we find ( 2d - m = 2 ), leading to ( k leq 2^n ).3. Code Construction: - Using the Simplex code (a linear code with parameters ([2^n - 1, n, 2^{n-1}])) and duplicating each coordinate, we construct a code with parameters ([2(2^n - 1), n, 2^n]). - This construction achieves the required minimal distance and matches the Plotkin Bound, confirming the maximum ( k ) is ( 2^n ).Conclusion:The maximum value of ( k ) for which an acceptable 2-coloring exists is (boxed{2^n}).

🔑:To determine the maximum value of ( k ) for an acceptable ( 2 )-coloring to exist, we need to ensure that any two columns agree on fewer than ( 2^n - 1 ) entries. Let's break down the solution step by step.1. Establish the upper bound for ( k ): - Consider the array with ( 2(2^n - 1) ) rows and ( k ) columns. - The condition requires each pair of columns to have at least ( 2^n ) disagreements. - The total number of disagreements for ( k ) columns is at least ( binom{k}{2} cdot 2^n ). - This can be written as: [ frac{k(k-1)}{2} cdot 2^n ] - If we look at rows instead of columns, a row with ( a ) white squares and ( b ) black squares has ( ab ) disagreements, where ( a + b = k ). - Summing over all rows, the total number of disagreements is: [ sum ab ] - Since ( ab leq frac{(a+b)^2}{4} = frac{k^2}{4} ), we have: [ sum ab leq 2(2^n - 1) cdot frac{k^2}{4} ] - Therefore, we get: [ frac{k(k-1)}{2} cdot 2^n leq 2(2^n - 1) cdot frac{k^2}{4} ] - Simplifying this inequality: [ k(k-1)2^n leq (2^n - 1)k^2 ] [ k(k-1) leq frac{(2^n - 1)k^2}{2^n} ] [ k - 1 leq frac{(2^n - 1)k}{2^n} ] [ 2^n(k - 1) leq (2^n - 1)k ] [ 2^n k - 2^n leq 2^n k - k ] [ k leq 2^n ]2. Construct an example to show ( k = 2^n ) is achievable: - We will use induction to construct an acceptable ( 2 )-coloring. - Base case (( n = 1 )): - For ( n = 1 ), we have ( 2(2^1 - 1) = 2 ) rows and ( k = 2 ) columns. - A valid ( 2 )-coloring is: [ begin{array}{cc} 0 & 1 1 & 0 end{array} ] - Here, each pair of columns disagrees on both entries, satisfying the condition. - Inductive step: - Assume we have a valid ( 2 )-coloring for ( n ) with ( 2(2^n - 1) ) rows and ( k = 2^n ) columns. - Construct the ( 2 )-coloring for ( n+1 ) as follows: - Create a top row with ( 2^{n+1} ) columns, where the first ( 2^n ) entries are ( 0 ) and the next ( 2^n ) entries are ( 1 ). - Divide the remaining ( 2(2^{n+1} - 1) - 1 = 2(2^n - 1) times 2 ) rows into four ( (2^n - 1) times 2^n ) rectangles. - Fill three of these rectangles with the ( n )-example and the fourth with the ( n )-example ( +1 pmod{2} ).By induction, we can construct an acceptable ( 2 )-coloring for ( k = 2^n ).The final answer is ( boxed{ k = 2^n } ).

🧠:Okay, let's see. So the problem is about points X, Y, Z being collinear in that order, with XY = 1 and YZ = 3. There's a point W such that YW = 5. We need to find the minimum possible length of the segment O₁O₂, where O₁ and O₂ are the circumcenters of triangles WXY and WYZ respectively. Hmm.First, let me visualize this. X, Y, Z are on a straight line, right? So X----Y----Z, with XY = 1 and YZ = 3. So the total distance from X to Z is 4. Then there's a point W such that YW = 5. Since Y is fixed on the line, W is somewhere in the plane 5 units away from Y. The circumcenters O₁ and O₂ are the centers of the circumcircles of triangles WXY and WYZ. The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. So for each triangle, we need to find the perpendicular bisectors and then their intersection point, which is the circumcenter. Then, once we have O₁ and O₂, we need to find the minimum distance between them as W moves around such that YW = 5.Let me break this down step by step.First, maybe I should set up a coordinate system to make things easier. Let's place point Y at the origin (0, 0). Then, since X, Y, Z are collinear in that order with XY = 1 and YZ = 3, we can place X at (-1, 0), Y at (0, 0), and Z at (3, 0). Wait, hold on. If they are in order X, Y, Z, and XY = 1, YZ = 3, then from X to Y is 1, Y to Z is 3. So if Y is at (0, 0), then X is at (-1, 0) and Z is at (3, 0). Wait, but from Y to Z is 3, so if Y is at (0,0), then Z is at (3,0). Hmm, yes. That seems correct.Now, point W is such that YW = 5. Since Y is at (0,0), W lies somewhere on the circle centered at Y with radius 5. So the coordinates of W can be parameterized as (5 cos θ, 5 sin θ) for some angle θ. That might be helpful.Now, we need to find the circumcenters of triangles WXY and WYZ. Let's start with triangle WXY. The circumcenter O₁ is the intersection of the perpendicular bisectors of the sides of WXY. Similarly, O₂ is the intersection of the perpendicular bisectors of WYZ.Let me recall that the perpendicular bisector of a segment is the line perpendicular to the segment at its midpoint. So, for triangle WXY, the sides are WX, WY, and XY. Similarly for triangle WYZ, the sides are WY, WZ, and YZ.But maybe it's easier to focus on two perpendicular bisectors for each triangle to find their circumcenters.Starting with triangle WXY. Let's find the perpendicular bisectors of two sides, say WX and XY.First, the side XY: since X is at (-1, 0) and Y is at (0, 0), the midpoint of XY is at (-0.5, 0). The segment XY is horizontal, so its perpendicular bisector is vertical, passing through (-0.5, 0). Therefore, the perpendicular bisector of XY is the line x = -0.5.Next, let's take another side, say WX. Let's find the midpoint of WX and the slope of WX to determine the perpendicular bisector.Coordinates of W: (5 cos θ, 5 sin θ)Coordinates of X: (-1, 0)Midpoint of WX: [(5 cos θ + (-1))/2, (5 sin θ + 0)/2] = [(5 cos θ - 1)/2, (5 sin θ)/2]Slope of WX: [5 sin θ - 0]/[5 cos θ - (-1)] = (5 sin θ)/(5 cos θ + 1)Therefore, the slope of the perpendicular bisector of WX is the negative reciprocal, which is -(5 cos θ + 1)/(5 sin θ)Thus, the equation of the perpendicular bisector of WX is:y - (5 sin θ)/2 = [-(5 cos θ + 1)/(5 sin θ)] (x - (5 cos θ - 1)/2 )Similarly, the perpendicular bisector of XY is x = -0.5. So, the intersection point of these two lines is the circumcenter O₁.Similarly, for triangle WYZ, let's find O₂.Points W(5 cos θ, 5 sin θ), Y(0,0), Z(3, 0)Again, let's find the perpendicular bisectors of two sides. Let's take YZ and WY.First, side YZ: from Y(0,0) to Z(3,0). Midpoint is (1.5, 0), and since YZ is horizontal, the perpendicular bisector is vertical: x = 1.5.Next, perpendicular bisector of WY.Midpoint of WY: [(5 cos θ + 0)/2, (5 sin θ + 0)/2] = (2.5 cos θ, 2.5 sin θ)Slope of WY: [5 sin θ - 0]/[5 cos θ - 0] = (5 sin θ)/(5 cos θ) = tan θThus, the slope of the perpendicular bisector is -cot θTherefore, the equation of the perpendicular bisector of WY is:y - 2.5 sin θ = -cot θ (x - 2.5 cos θ )Simplify this equation:y = -cot θ (x - 2.5 cos θ ) + 2.5 sin θ= -cot θ x + 2.5 cos θ cot θ + 2.5 sin θBut cot θ = cos θ / sin θ, so:= - (cos θ / sin θ) x + 2.5 cos θ (cos θ / sin θ) + 2.5 sin θ= - (cos θ / sin θ) x + (2.5 cos² θ)/ sin θ + 2.5 sin θCombine the terms:= - (cos θ / sin θ) x + [2.5 cos² θ + 2.5 sin² θ]/ sin θ= - (cos θ / sin θ) x + 2.5 (cos² θ + sin² θ)/ sin θSince cos² θ + sin² θ = 1:= - (cos θ / sin θ) x + 2.5 / sin θSo, the equation becomes:y = - (cos θ / sin θ) x + 2.5 / sin θNow, the perpendicular bisectors for triangle WYZ are x = 1.5 and the above line. So the intersection point O₂ is found by plugging x = 1.5 into the equation:y = - (cos θ / sin θ) * 1.5 + 2.5 / sin θ= (-1.5 cos θ + 2.5)/ sin θSo, coordinates of O₂ are (1.5, (-1.5 cos θ + 2.5)/ sin θ )Wait, but let me check that again. Wait, x = 1.5 is one perpendicular bisector, so plugging x = 1.5 into the other perpendicular bisector's equation gives the y-coordinate. So:y = - (cos θ / sin θ) * 1.5 + 2.5 / sin θ= ( -1.5 cos θ + 2.5 ) / sin θTherefore, O₂ is at (1.5, ( -1.5 cos θ + 2.5 ) / sin θ )Similarly, for O₁. Let's go back to triangle WXY. The perpendicular bisector of XY is x = -0.5, and the perpendicular bisector of WX is:y - (5 sin θ)/2 = [ - (5 cos θ + 1 ) / (5 sin θ ) ] ( x - (5 cos θ - 1 ) / 2 )This seems a bit complicated. Let's try to simplify this equation.Let me denote m = slope = - (5 cos θ + 1 ) / (5 sin θ )So the equation is:y = m (x - (5 cos θ - 1)/2 ) + (5 sin θ)/2Plugging x = -0.5 (since the other perpendicular bisector is x = -0.5) into this equation will give the y-coordinate of O₁.Therefore, substituting x = -0.5:y = m ( -0.5 - (5 cos θ - 1)/2 ) + (5 sin θ)/2First, let's compute the term inside the parentheses:-0.5 - (5 cos θ - 1)/2 = - (1/2) - (5 cos θ - 1)/2 = [ -1 -5 cos θ + 1 ] / 2 = ( -5 cos θ ) / 2Therefore:y = m ( -5 cos θ / 2 ) + (5 sin θ)/2Substituting m = - (5 cos θ + 1 ) / (5 sin θ )Thus:y = [ - (5 cos θ + 1 ) / (5 sin θ ) ] * ( -5 cos θ / 2 ) + (5 sin θ)/2Compute first term:[ (5 cos θ + 1 ) * 5 cos θ ) / (5 sin θ * 2 ) ] = [ (5 cos θ + 1 ) * cos θ ) / (2 sin θ ) ]Therefore:y = [ (5 cos² θ + cos θ ) / (2 sin θ ) ] + (5 sin θ ) / 2Combine the terms:= (5 cos² θ + cos θ + 5 sin² θ ) / (2 sin θ )Since 5 cos² θ + 5 sin² θ = 5 (cos² θ + sin² θ ) = 5So:= (5 + cos θ ) / (2 sin θ )Therefore, coordinates of O₁ are ( -0.5, (5 + cos θ ) / (2 sin θ ) )So, summarizing:O₁ is at ( -0.5, (5 + cos θ ) / (2 sin θ ) )O₂ is at ( 1.5, ( -1.5 cos θ + 2.5 ) / sin θ )Now, to find the distance O₁O₂, we can use the distance formula.First, let's compute the difference in x-coordinates: 1.5 - (-0.5) = 2. So the horizontal distance is 2 units.The vertical distance is [ ( -1.5 cos θ + 2.5 ) / sin θ ] - [ (5 + cos θ ) / (2 sin θ ) ]Let me compute this:= [ (-1.5 cos θ + 2.5 ) - (5 + cos θ ) / 2 ] / sin θFirst, simplify the numerator:Let's compute (-1.5 cos θ + 2.5 ) - (5 + cos θ ) / 2= -1.5 cos θ + 2.5 - 2.5 - 0.5 cos θWait, let me do this step by step.First, expand the numerator:(-1.5 cos θ + 2.5 ) - (5 + cos θ ) / 2= (-1.5 cos θ + 2.5 ) - 5/2 - (cos θ)/2= (-1.5 cos θ - 0.5 cos θ ) + (2.5 - 2.5 )= (-2 cos θ ) + 0= -2 cos θWait, wait, let's check:Wait, 2.5 is 5/2. So,= (-1.5 cos θ + 5/2 ) - (5/2 + cos θ / 2 )= -1.5 cos θ + 5/2 -5/2 - 0.5 cos θ= (-1.5 cos θ - 0.5 cos θ ) + (5/2 -5/2 )= (-2 cos θ ) + 0= -2 cos θYes, that's correct. So the numerator is -2 cos θ, so the vertical difference is -2 cos θ / sin θ = -2 cot θ.But the vertical distance is the absolute value, but since we are squaring it later in the distance formula, maybe we can proceed.Therefore, the vertical component is (-2 cos θ ) / sin θ = -2 cot θTherefore, the vertical distance between O₁ and O₂ is | -2 cot θ |, but since distance is positive, but actually, in the distance formula, we take the square of the difference, so:Distance O₁O₂ = sqrt[ (2)^2 + ( (-2 cos θ ) / sin θ )^2 ] = sqrt[4 + (4 cos² θ ) / sin² θ ]= sqrt[4 + 4 cot² θ ] = sqrt[4(1 + cot² θ ) ] = sqrt[4 csc² θ ] = 2 |csc θ | = 2 / |sin θ |.Wait, that's interesting. So the distance between O₁ and O₂ is 2 / |sin θ |. But this seems surprising. Let me check the steps again to ensure there's no mistake.So we had:O₁ at (-0.5, (5 + cos θ)/(2 sin θ )) and O₂ at (1.5, ( -1.5 cos θ + 2.5 ) / sin θ )Compute vertical difference:( (-1.5 cos θ + 2.5 ) / sin θ ) - ( (5 + cos θ ) / (2 sin θ ) )= [ (-1.5 cos θ + 2.5 ) - (5 + cos θ ) / 2 ] / sin θLet me compute the numerator:First, write -1.5 cos θ as -3/2 cos θ, and 2.5 as 5/2.So numerator:[ -3/2 cos θ + 5/2 - (5 + cos θ ) / 2 ]= [ (-3/2 cos θ + 5/2 ) - 5/2 - (cos θ)/2 ]= (-3/2 cos θ + 5/2 -5/2 - cos θ /2 )= (-3/2 cos θ - cos θ /2 ) + (5/2 -5/2 )= (-4/2 cos θ ) + 0= -2 cos θYes, so numerator is -2 cos θ, denominator is sin θ. So vertical difference is (-2 cos θ)/ sin θ = -2 cot θ.Therefore, the vertical component is -2 cot θ. So the distance between O₁O₂ is sqrt( (2)^2 + (-2 cot θ )^2 )= sqrt(4 + 4 cot² θ )= 2 sqrt(1 + cot² θ )But 1 + cot² θ = csc² θ, so sqrt(csc² θ ) = |csc θ |, so total distance is 2 |csc θ | = 2 / |sin θ |.So, the distance between O₁ and O₂ is 2 / |sin θ |. Therefore, to minimize this distance, we need to maximize |sin θ |. Since |sin θ | ≤ 1, the maximum value is 1, so the minimal distance is 2 / 1 = 2. But wait, that seems too straightforward. Is there a mistake here?Wait, but hold on. The coordinate system was set up with Y at (0,0), X at (-1,0), Z at (3,0). The point W is parameterized as (5 cos θ, 5 sin θ ). But does θ range over all angles? Or are there constraints? Since W can be anywhere such that YW = 5, so W lies on a circle of radius 5 centered at Y. So θ can be from 0 to 2π. However, the triangles WXY and WYZ must be non-degenerate. So W cannot be colinear with X, Y, or Y, Z, otherwise the circumradius would be undefined or infinite. So as long as W is not on the line XYZ, the triangles are non-degenerate. If W is on the line XYZ, then the circumcenters would be at infinity. So we need to exclude the cases where W is colinear with X, Y, Z. Since X, Y, Z are colinear, if W is also on that line, then the triangles would be degenerate. So θ cannot be 0 or π, since that would place W on the x-axis (the line XYZ). So θ ∈ (0, π) ∪ (π, 2π). Therefore, sin θ can be positive or negative, but |sin θ | is between 0 and 1. So the minimal distance O₁O₂ is 2, achieved when |sin θ | = 1, i.e., θ = π/2 or 3π/2. Therefore, when W is at (0,5) or (0,-5), the points O₁ and O₂ are 2 units apart.But let's check this. Let me take θ = π/2. Then W is at (5 cos π/2, 5 sin π/2 ) = (0,5). Let's compute O₁ and O₂.For triangle WXY with W(0,5), X(-1,0), Y(0,0).Circumcenter O₁: perpendicular bisectors of XY and WX.Perpendicular bisector of XY is x = -0.5 as before.Perpendicular bisector of WX: midpoint of WX is [ (-1 + 0)/2, (0 + 5)/2 ] = (-0.5, 2.5 )Slope of WX: (5 - 0)/(0 - (-1)) = 5/1 = 5. Therefore, perpendicular slope is -1/5.Equation of perpendicular bisector: y - 2.5 = -1/5 (x + 0.5 )To find O₁, intersect this with x = -0.5.Substituting x = -0.5:y - 2.5 = -1/5 ( -0.5 + 0.5 ) = -1/5 (0) = 0Thus, y = 2.5. Therefore, O₁ is at (-0.5, 2.5 )Similarly, for triangle WYZ with W(0,5), Y(0,0), Z(3,0).Circumcenter O₂: perpendicular bisectors of YZ and WY.Perpendicular bisector of YZ is x = 1.5.Perpendicular bisector of WY: midpoint of WY is (0, 2.5). Slope of WY is (5 - 0)/(0 - 0) undefined, which is vertical. Therefore, the perpendicular bisector is horizontal line through midpoint. Wait, WY is vertical, so its perpendicular bisector is horizontal line through midpoint (0, 2.5). So the perpendicular bisector is y = 2.5.Therefore, intersection with x = 1.5 is (1.5, 2.5 ). Therefore, O₂ is at (1.5, 2.5 )Therefore, distance between O₁(-0.5, 2.5 ) and O₂(1.5, 2.5 ) is sqrt( (1.5 - (-0.5))² + (2.5 - 2.5 )² ) = sqrt( (2)^2 + 0 ) = 2. So yes, that gives distance 2.Similarly, if θ = 3π/2, W is at (0, -5 ). Let's check.For W(0, -5):Triangle WXY: points W(0, -5), X(-1,0), Y(0,0). Midpoint of WX is (-0.5, -2.5 ). Slope of WX: ( -5 - 0 ) / ( 0 - (-1) ) = -5 /1 = -5. Perpendicular slope is 1/5. Perpendicular bisector equation: y + 2.5 = (1/5)(x + 0.5 )Intersecting with x = -0.5, gives y + 2.5 = (1/5)(0) => y = -2.5. So O₁ is at (-0.5, -2.5 )For triangle WYZ: W(0, -5), Y(0,0), Z(3,0). Perpendicular bisector of YZ is x = 1.5. Perpendicular bisector of WY: midpoint of WY is (0, -2.5 ). Slope of WY is vertical, so perpendicular bisector is horizontal line y = -2.5. Intersection with x = 1.5 is (1.5, -2.5 )Distance between O₁(-0.5, -2.5 ) and O₂(1.5, -2.5 ) is again 2. So that checks out.Therefore, in these cases, the distance is indeed 2, which is the minimal possible. Since our general formula gives the distance as 2 / |sin θ |, which is minimized when |sin θ | is maximized, i.e., at 1. Therefore, the minimal distance is 2.But wait, the problem says "the minimum possible length of segment O₁O₂". So the answer should be 2. But let me make sure that there are no other positions of W where |sin θ | could be larger than 1, but since |sin θ | ≤ 1, the minimal distance is indeed 2.But wait, is there a case when the distance could be less than 2? For example, if W is somewhere else?Wait, let's take θ such that sin θ is approaching 0. Then the distance O₁O₂ approaches infinity. So the minimal distance is achieved when sin θ is maximum. So yes, the minimal possible distance is 2.Alternatively, let's consider another angle. Let's take θ = π/2, which we already did, giving distance 2. What if θ is some other angle where sin θ is less than 1? For example, θ = π/4. Then sin θ = √2 / 2 ≈ 0.707. Then the distance would be 2 / (√2 / 2 ) = 2 * 2 / √2 = 2√2 ≈ 2.828, which is larger than 2. Similarly, θ = π/6, sin θ = 0.5, distance is 2 / 0.5 = 4. So yes, indeed, as sin θ decreases, the distance increases. Therefore, 2 is indeed the minimal possible length.Therefore, the answer is 2.But wait, the problem mentions "the minimum possible length of segment O₁O₂". But wait, in our coordinate system, we fixed Y at (0,0), X at (-1,0), Z at (3,0). But is there a different coordinate system or configuration where the distance could be smaller?Wait, no. Because the problem is about collinear points X, Y, Z with XY=1, YZ=3. The position of W is such that YW=5. So regardless of the coordinate system, the minimal distance should be the same. By setting up coordinates, we converted the problem into a coordinate system where calculations are easier, but the result is general.Alternatively, is there a way to see this result geometrically without coordinates?Let me think. The circumcenters O₁ and O₂ are the centers of the circles passing through W, X, Y and W, Y, Z respectively.Since both triangles share the side WY, which has length 5. The circumradius of each triangle can be found, but maybe that's not helpful here.Alternatively, note that O₁ is the intersection of the perpendicular bisectors of WX and WY, while O₂ is the intersection of the perpendicular bisectors of WY and WZ. So both O₁ and O₂ lie on the perpendicular bisector of WY. Wait, is that true?Wait, O₁ lies on the perpendicular bisector of WX and the perpendicular bisector of XY (which is fixed). Similarly, O₂ lies on the perpendicular bisector of WZ and the perpendicular bisector of YZ (which is fixed). Wait, but if we consider that both O₁ and O₂ lie on the perpendicular bisector of WY? Let me check.Wait, for O₁: the perpendicular bisectors of WX and XY. For O₂: the perpendicular bisectors of WZ and YZ. So unless WY is involved, they don't necessarily lie on the same perpendicular bisector. Wait, maybe not. Wait, O₁ is the intersection of perpendicular bisectors of WX and XY, which is different from the perpendicular bisector of WY. Similarly, O₂ is the intersection of perpendicular bisectors of WZ and YZ. Therefore, they are not necessarily on the same line.But in our coordinate calculation, we found that the vertical distance between O₁ and O₂ depends on cot θ, but when θ = π/2, the points O₁ and O₂ lie on the same horizontal line, so their vertical distance is zero, and horizontal distance is 2. So in that case, the distance is 2. In other cases, they are separated vertically as well as horizontally.So, in that case, the minimal distance occurs when the vertical component is zero, which happens when cot θ = 0, i.e., when cos θ = 0, which is θ = π/2 or 3π/2. Thus, the minimal distance is 2.Therefore, the answer is boxed{2}.Final AnswerThe minimum possible length of segment (overline{O_1O_2}) is (boxed{2}).

✅:Given collinear points (X), (Y), and (Z) with (XY = 1) and (YZ = 3), we place (Y) at the origin ((0, 0)), (X) at ((-1, 0)), and (Z) at ((3, 0)). Point (W) lies on a circle centered at (Y) with radius 5, parameterized as ((5 cos theta, 5 sin theta)).To find the circumcenters (O_1) and (O_2) of triangles (triangle WXY) and (triangle WYZ), we use the perpendicular bisectors of the sides of these triangles.For (triangle WXY):- The perpendicular bisector of (XY) is (x = -0.5).- The perpendicular bisector of (WX) intersects (x = -0.5) at (O_1).For (triangle WYZ):- The perpendicular bisector of (YZ) is (x = 1.5).- The perpendicular bisector of (WY) intersects (x = 1.5) at (O_2).The coordinates of (O_1) and (O_2) are found to be:- (O_1) at (left(-0.5, frac{5 + cos theta}{2 sin theta}right))- (O_2) at (left(1.5, frac{-1.5 cos theta + 2.5}{sin theta}right))The distance between (O_1) and (O_2) is calculated as:[sqrt{left(2right)^2 + left(frac{-2 cos theta}{sin theta}right)^2} = sqrt{4 + 4 cot^2 theta} = 2 csc theta]The minimum distance occurs when (|sin theta| = 1), i.e., (theta = frac{pi}{2}) or (theta = frac{3pi}{2}), giving the minimum distance as (2).Thus, the minimum possible length of segment (overline{O_1O_2}) is (boxed{2}).

🔑:1. Assign coordinates to points: Let ( Y ) be at the origin ((0,0)). Then, ( X ) is at ((-1,0)) and ( Z ) is at ((3,0)). Point ( W ) is at ((a, b)) such that ( YW = 5 ).2. Calculate coordinates of ( W ): Since ( YW = 5 ), we have: [ sqrt{a^2 + b^2} = 5 implies a^2 + b^2 = 25 ]3. Determine the circumcenters ( O_1 ) and ( O_2 ): - The circumcenter ( O_1 ) of (triangle WXY) is the intersection of the perpendicular bisectors of ( WX ) and ( XY ). - The circumcenter ( O_2 ) of (triangle WYZ) is the intersection of the perpendicular bisectors of ( WY ) and ( YZ ).4. Find the perpendicular bisectors: - The perpendicular bisector of ( WX ) passes through the midpoint of ( WX ) and is perpendicular to ( WX ). - The perpendicular bisector of ( XY ) passes through the midpoint of ( XY ) and is perpendicular to ( XY ).5. Calculate the coordinates of ( O_1 ): - The midpoint of ( WX ) is (left(frac{a-1}{2}, frac{b}{2}right)). - The slope of ( WX ) is (frac{b}{a+1}), so the slope of the perpendicular bisector is (-frac{a+1}{b}). - The equation of the perpendicular bisector of ( WX ) is: [ y - frac{b}{2} = -frac{a+1}{b} left(x - frac{a-1}{2}right) ] - The midpoint of ( XY ) is (left(-frac{1}{2}, 0right)). - The slope of ( XY ) is (0), so the perpendicular bisector is a vertical line ( x = -frac{1}{2} ).6. Calculate the coordinates of ( O_2 ): - The midpoint of ( WY ) is (left(frac{a}{2}, frac{b}{2}right)). - The slope of ( WY ) is (frac{b}{a}), so the slope of the perpendicular bisector is (-frac{a}{b}). - The equation of the perpendicular bisector of ( WY ) is: [ y - frac{b}{2} = -frac{a}{b} left(x - frac{a}{2}right) ] - The midpoint of ( YZ ) is (left(frac{3}{2}, 0right)). - The slope of ( YZ ) is (0), so the perpendicular bisector is a vertical line ( x = frac{3}{2} ).7. Find the intersection points: - Solve the equations of the perpendicular bisectors to find the coordinates of ( O_1 ) and ( O_2 ).8. Calculate the distance ( O_1O_2 ): - Use the distance formula to find the distance between ( O_1 ) and ( O_2 ): [ O_1O_2 = sqrt{left(frac{3}{2} - left(-frac{1}{2}right)right)^2 + left(frac{frac{5}{2} - frac{3cos{theta}}{2}}{sin{theta}} - frac{frac{5}{2} + frac{cos{theta}}{2}}{sin{theta}}right)^2} ] - Simplify the expression to find the minimum distance.9. Conclusion: - After simplification, we find that the minimum possible length of segment ( overline{O_1O_2} ) is ( 2 ).The final answer is ( boxed{2} ).