{
  "index": "1951-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. In the plane, what is the locus of points the sum of the squares of whose distances from \\( n \\) fixed points is a constant? What restrictions, stated in geometric terms, must be put on the constant so that the locus is non-null?",
  "solution": "Solution. Let the given points be \\( \\left\\{\\left(x_{i}, y_{i}\\right)\\right\\}, i=1,2, \\ldots, n \\). Let a point on the locus be \\( (x, y) \\). Then the required condition is\n\\[\n\\Sigma\\left[\\left(x-x_{i}\\right)^{2}+\\left(y-y_{i}\\right)^{2}\\right]=C\n\\]\nwhere \\( C \\) is a constant. This can be rewritten as \\( n x^{2}-2\\left(\\Sigma x_{i}\\right) x+\\Sigma x_{i}{ }^{2}+ \\) \\( n y^{2}-2\\left(\\Sigma y_{i}\\right) y+\\Sigma y_{i}{ }^{2}=C \\) and by completing the squares in the form\n\\[\n\\left(x-\\frac{\\Sigma x_{i}}{n}\\right)^{2}+\\left(y-\\frac{\\Sigma y_{i}}{n}\\right)^{2}=\\frac{C}{n}+\\left(\\frac{\\Sigma x_{i}}{n}\\right)^{2}+\\left(\\frac{\\Sigma y_{i}}{n}\\right)^{2}-\\frac{\\Sigma x_{i}^{2}}{n}-\\frac{\\Sigma y_{i}^{2}}{n}\n\\]\n\nThe locus is therefore empty if the right member is negative; a single point (namely the centroid \\( \\left((1 / n) \\Sigma x_{i},(1 / n) \\Sigma y_{i}\\right) \\) of the given points) if the right member is zero; and a circle with center at the centroid if the right member is positive. If coordinates are chosen with the origin at the centroid of the given points, the condition for a real non-null locus then becomes\n\\[\nC \\geq \\Sigma\\left(x_{i}^{2}+y_{i}^{2}\\right)=\\Sigma r_{i}^{2}\n\\]\nwhere \\( r_{i} \\) is the distance of the \\( i \\) th point from the new center. That is, \\( C \\) must be at least as large as the sum of the squares of the distances from the centroid. This last formulation is a special instance of the general fact that a planar mass has its minimal moment of inertia about an axis perpendicular to the plane through the centroid of the mass.",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "x_i",
    "y_i",
    "C",
    "r_i",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "x_i": "pointabsc",
        "y_i": "pointordi",
        "C": "squaretotal",
        "r_i": "pointdist",
        "n": "pointcount"
      },
      "question": "2. In the plane, what is the locus of points the sum of the squares of whose distances from \\( pointcount \\) fixed points is a constant? What restrictions, stated in geometric terms, must be put on the constant so that the locus is non-null?",
      "solution": "Solution. Let the given points be \\( \\left\\{\\left(pointabsc, pointordi\\right)\\right\\}, i=1,2, \\ldots, pointcount \\). Let a point on the locus be \\( (abscissa, ordinate) \\). Then the required condition is\n\\[\n\\Sigma\\left[\\left(abscissa-pointabsc\\right)^{2}+\\left(ordinate-pointordi\\right)^{2}\\right]=squaretotal\n\\]\nwhere \\( squaretotal \\) is a constant. This can be rewritten as \\( pointcount\\, abscissa^{2}-2\\left(\\Sigma pointabsc\\right) abscissa+\\Sigma pointabsc^{2}+ pointcount\\, ordinate^{2}-2\\left(\\Sigma pointordi\\right) ordinate+\\Sigma pointordi^{2}=squaretotal \\) and by completing the squares in the form\n\\[\n\\left(abscissa-\\frac{\\Sigma pointabsc}{pointcount}\\right)^{2}+\\left(ordinate-\\frac{\\Sigma pointordi}{pointcount}\\right)^{2}=\\frac{squaretotal}{pointcount}+\\left(\\frac{\\Sigma pointabsc}{pointcount}\\right)^{2}+\\left(\\frac{\\Sigma pointordi}{pointcount}\\right)^{2}-\\frac{\\Sigma pointabsc^{2}}{pointcount}-\\frac{\\Sigma pointordi^{2}}{pointcount}\n\\]\n\nThe locus is therefore empty if the right member is negative; a single point (namely the centroid \\( \\left((1 / pointcount) \\Sigma pointabsc,(1 / pointcount) \\Sigma pointordi\\right) \\) of the given points) if the right member is zero; and a circle with center at the centroid if the right member is positive. If coordinates are chosen with the origin at the centroid of the given points, the condition for a real non-null locus then becomes\n\\[\nsquaretotal \\geq \\Sigma\\left(pointabsc^{2}+pointordi^{2}\\right)=\\Sigma pointdist^{2}\n\\]\nwhere \\( pointdist \\) is the distance of the \\( i \\) th point from the new center. That is, \\( squaretotal \\) must be at least as large as the sum of the squares of the distances from the centroid. This last formulation is a special instance of the general fact that a planar mass has its minimal moment of inertia about an axis perpendicular to the plane through the centroid of the mass."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "y": "hazelnut",
        "x_i": "pendulum",
        "y_i": "sapphire",
        "C": "butterfly",
        "r_i": "pineapple",
        "n": "pavilion"
      },
      "question": "In the plane, what is the locus of points the sum of the squares of whose distances from \\( pavilion \\) fixed points is a constant? What restrictions, stated in geometric terms, must be put on the constant so that the locus is non-null?",
      "solution": "Solution. Let the given points be \\( \\left\\{\\left(pendulum, sapphire\\right)\\right\\}, i=1,2, \\ldots, pavilion \\). Let a point on the locus be \\( (marigold, hazelnut) \\). Then the required condition is\n\\[\n\\Sigma\\left[\\left(marigold-pendulum\\right)^{2}+\\left(hazelnut-sapphire\\right)^{2}\\right]=butterfly\n\\]\nwhere \\( butterfly \\) is a constant. This can be rewritten as \\( pavilion\\, marigold^{2}-2\\left(\\Sigma pendulum\\right) marigold+\\Sigma pendulum^{2}+ \\) \\( pavilion\\, hazelnut^{2}-2\\left(\\Sigma sapphire\\right) hazelnut+\\Sigma sapphire^{2}=butterfly \\) and by completing the squares in the form\n\\[\n\\left(marigold-\\frac{\\Sigma pendulum}{pavilion}\\right)^{2}+\\left(hazelnut-\\frac{\\Sigma sapphire}{pavilion}\\right)^{2}=\\frac{butterfly}{pavilion}+\\left(\\frac{\\Sigma pendulum}{pavilion}\\right)^{2}+\\left(\\frac{\\Sigma sapphire}{pavilion}\\right)^{2}-\\frac{\\Sigma pendulum^{2}}{pavilion}-\\frac{\\Sigma sapphire^{2}}{pavilion}\n\\]\n\nThe locus is therefore empty if the right member is negative; a single point (namely the centroid \\( \\left((1 / pavilion) \\Sigma pendulum,(1 / pavilion) \\Sigma sapphire\\right) \\) of the given points) if the right member is zero; and a circle with center at the centroid if the right member is positive. If coordinates are chosen with the origin at the centroid of the given points, the condition for a real non-null locus then becomes\n\\[\nbutterfly \\geq \\Sigma\\left(pendulum^{2}+sapphire^{2}\\right)=\\Sigma pineapple^{2}\n\\]\nwhere \\( pineapple \\) is the distance of the \\( i \\) th point from the new center. That is, \\( butterfly \\) must be at least as large as the sum of the squares of the distances from the centroid. This last formulation is a special instance of the general fact that a planar mass has its minimal moment of inertia about an axis perpendicular to the plane through the centroid of the mass."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "x_i": "ycomponents",
        "y_i": "xcomponents",
        "C": "variabletotal",
        "r_i": "tangentlen",
        "n": "infinitecount"
      },
      "question": "In the plane, what is the locus of points the sum of the squares of whose distances from \\( infinitecount \\) fixed points is a constant? What restrictions, stated in geometric terms, must be put on the constant so that the locus is non-null?",
      "solution": "Solution. Let the given points be \\( \\left\\{\\left(ycomponents_{i}, xcomponents_{i}\\right)\\right\\}, i=1,2, \\ldots, infinitecount \\). Let a point on the locus be \\( (verticalaxis, horizontalaxis) \\). Then the required condition is\n\\[\n\\Sigma\\left[\\left(verticalaxis-ycomponents_{i}\\right)^{2}+\\left(horizontalaxis-xcomponents_{i}\\right)^{2}\\right]=variabletotal\n\\]\nwhere \\( variabletotal \\) is a constant. This can be rewritten as \\( infinitecount verticalaxis^{2}-2\\left(\\Sigma ycomponents_{i}\\right) verticalaxis+\\Sigma ycomponents_{i}{ }^{2}+ infinitecount horizontalaxis^{2}-2\\left(\\Sigma xcomponents_{i}\\right) horizontalaxis+\\Sigma xcomponents_{i}{ }^{2}=variabletotal \\) and by completing the squares in the form\n\\[\n\\left(verticalaxis-\\frac{\\Sigma ycomponents_{i}}{infinitecount}\\right)^{2}+\\left(horizontalaxis-\\frac{\\Sigma xcomponents_{i}}{infinitecount}\\right)^{2}=\\frac{variabletotal}{infinitecount}+\\left(\\frac{\\Sigma ycomponents_{i}}{infinitecount}\\right)^{2}+\\left(\\frac{\\Sigma xcomponents_{i}}{infinitecount}\\right)^{2}-\\frac{\\Sigma ycomponents_{i}^{2}}{infinitecount}-\\frac{\\Sigma xcomponents_{i}^{2}}{infinitecount}\n\\]\n\nThe locus is therefore empty if the right member is negative; a single point (namely the centroid \\( \\left((1 / infinitecount) \\Sigma ycomponents_{i},(1 / infinitecount) \\Sigma xcomponents_{i}\\right) \\) of the given points) if the right member is zero; and a circle with center at the centroid if the right member is positive. If coordinates are chosen with the origin at the centroid of the given points, the condition for a real non-null locus then becomes\n\\[\nvariabletotal \\geq \\Sigma\\left(ycomponents_{i}^{2}+xcomponents_{i}^{2}\\right)=\\Sigma tangentlen_{i}^{2}\n\\]\nwhere \\( tangentlen_{i} \\) is the distance of the \\( i \\) th point from the new center. That is, \\( variabletotal \\) must be at least as large as the sum of the squares of the distances from the centroid. This last formulation is a special instance of the general fact that a planar mass has its minimal moment of inertia about an axis perpendicular to the plane through the centroid of the mass."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "x_i": "vbxchzqe",
        "y_i": "klmpdfru",
        "C": "snrtqfwa",
        "r_i": "plqmsdzn",
        "n": "gzrtplke"
      },
      "question": "2. In the plane, what is the locus of points the sum of the squares of whose distances from \\( gzrtplke \\) fixed points is a constant?",
      "solution": "Solution. Let the given points be \\( \\left\\{\\left(vbxchzqe, klmpdfru\\right)\\right\\}, i=1,2, \\ldots, gzrtplke \\). Let a point on the locus be \\( (qzxwvtnp, hjgrksla) \\). Then the required condition is\n\\[\n\\Sigma\\left[\\left(qzxwvtnp-vbxchzqe\\right)^{2}+\\left(hjgrksla-klmpdfru\\right)^{2}\\right]=snrtqfwa\n\\]\nwhere \\( snrtqfwa \\) is a constant. This can be rewritten as \\( gzrtplke qzxwvtnp^{2}-2\\left(\\Sigma vbxchzqe\\right) qzxwvtnp+\\Sigma vbxchzqe{ }^{2}+ \\) \\( gzrtplke hjgrksla^{2}-2\\left(\\Sigma klmpdfru\\right) hjgrksla+\\Sigma klmpdfru{ }^{2}=snrtqfwa \\) and by completing the squares in the form\n\\[\n\\left(qzxwvtnp-\\frac{\\Sigma vbxchzqe}{gzrtplke}\\right)^{2}+\\left(hjgrksla-\\frac{\\Sigma klmpdfru}{gzrtplke}\\right)^{2}=\\frac{snrtqfwa}{gzrtplke}+\\left(\\frac{\\Sigma vbxchzqe}{gzrtplke}\\right)^{2}+\\left(\\frac{\\Sigma klmpdfru}{gzrtplke}\\right)^{2}-\\frac{\\Sigma vbxchzqe^{2}}{gzrtplke}-\\frac{\\Sigma klmpdfru^{2}}{gzrtplke}\n\\]\n\nThe locus is therefore empty if the right member is negative; a single point (namely the centroid \\( \\left((1 / gzrtplke) \\Sigma vbxchzqe,(1 / gzrtplke) \\Sigma klmpdfru\\right) \\) of the given points) if the right member is zero; and a circle with center at the centroid if the right member is positive. If coordinates are chosen with the origin at the centroid of the given points, the condition for a real non-null locus then becomes\n\\[\nsnrtqfwa \\geq \\Sigma\\left(vbxchzqe^{2}+klmpdfru^{2}\\right)=\\Sigma plqmsdzn^{2}\n\\]\nwhere \\( plqmsdzn \\) is the distance of the \\( i \\) th point from the new center. That is, \\( snrtqfwa \\) must be at least as large as the sum of the squares of the distances from the centroid. This last formulation is a special instance of the general fact that a planar mass has its minimal moment of inertia about an axis perpendicular to the plane through the centroid of the mass."
    },
    "kernel_variant": {
      "question": "Fix integers d≥2 and N≥2.  \nLet P₁,…,P_N be pairwise distinct points of the Euclidean space ℝᵈ with position vectors p₁,…,p_N, and let w₁,…,w_N be positive real “masses”.  \nFix a real symmetric positive–definite d×d matrix A (so ⟨x,y⟩_A := xᵀAy is an anisotropic inner product).  \nLet S be an s–dimensional affine subspace (1≤s≤d) given in Cartesian form by  \n  Bq = c,                                                                (1)  \nwhere B is an (d–s)×d matrix of rank d–s and c∈ℝ^{d–s}.  \nFor q∈S set  \n\n  Φ(q) = Σ_{j=1}^{N} w_j‖q−p_j‖_A²   with ‖x‖_A² := xᵀAx.                (2)\n\nGiven a real constant K, study the locus  \n\n  𝓛(K) = { q∈S : Φ(q)=K }.                                     (3)\n\nTasks  \n(i) Derive an explicit Cartesian equation of 𝓛(K).  \n(ii) Give a geometric description of 𝓛(K) (empty set / single point / (s–1)-dimensional ellipsoid) according to K.  \n(iii) Express, purely in geometric terms, the necessary and sufficient condition on K for 𝓛(K)≠∅.  \n(iv) Compute the unique point q₀∈S minimising Φ and the minimal value Φ_min.  \n(v) Determine the principal semi–axes of 𝓛(K) (when it is an ellipsoid) and their lengths.",
      "solution": "Throughout let  \n\n  W := Σ_{j=1}^{N} w_j  > 0,        μ := (1/W) Σ_{j=1}^{N} w_j p_j   (the weighted centroid).        (4)\n\nStep 1.  Re-express Φ(q) as a single quadratic term.  \nUsing (2):\n\n Φ(q)=Σ w_j[ qᵀAq − 2qᵀAp_j + p_jᵀAp_j ]  \n     = W qᵀAq − 2W qᵀAμ + Σ w_j p_jᵀAp_j                                     (5)\n\n   = W (q−μ)ᵀA(q−μ) − W μᵀAμ + Σ w_j p_jᵀAp_j                             (6)\n\nSet the “structural constant”\n\n C₀ := Σ w_j p_jᵀAp_j − W μᵀAμ = Σ w_j ‖p_j−μ‖_A²  (≥0).                  (7)\n\nHence\n\n Φ(q) = W (q−μ)ᵀA(q−μ) + C₀.                                              (8)\n\nStep 2.  Restrict to the affine subspace S.  \nA point q lies in S iff (1) holds.  \nIntroduce Lagrange multipliers λ∈ℝ^{d−s} and minimise (q−μ)ᵀA(q−μ) subject to Bq=c.\n\nThe stationarity equations are  \n\n 2A(q−μ) + Bᵀλ = 0,         Bq = c.                                        (9)\n\nSolve for q.  Since A is positive–definite, A⁻¹ exists.  \nFrom the first equation q = μ − ½ A⁻¹Bᵀλ.  Substitute into Bq=c:\n\n Bμ − ½ B A⁻¹ Bᵀ λ = c  \n ⟹ λ = 2 (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                                          (10)\n\n(The matrix in parentheses is invertible because B has full rank and A⁻¹ is positive–definite.)\n\nHence the unique minimiser is  \n\n q₀ = μ − A⁻¹ Bᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                                   (11)\n\nStep 3.  Minimal value of Φ on S.  \nPut h := q₀ − μ = −A⁻¹ Bᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).  \n\nThen, by (8),\n\n Φ_min := Φ(q₀) = W‖h‖_A² + C₀.                                          (12)\n\nCompute  \n\n ‖h‖_A² = hᵀA h  \n           = (Bμ − c)ᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                            (13)\n\nStep 4.  Equation of the locus.  \nFor any q∈S write q = q₀ + y where y lies in the linear subspace  \n\n T := { y∈ℝᵈ : By = 0 }.                                                  (14)\n\nBecause B has rank d−s, T is s–dimensional.  \nInsert q = q₀ + y into (8):\n\n Φ(q) = W (y + h)ᵀA(y + h) + C₀  \n       = W(‖h‖_A² + 2⟨h,y⟩_A + ‖y‖_A²) + C₀  \n       = Φ_min + W( 2⟨h,y⟩_A + ‖y‖_A² ).                                (15)\n\nBut ⟨h,y⟩_A = hᵀAy = 0, because Ay lies in im(A) and y∈T ⇒ By=0 ⇒ B A⁻¹ Ay=0, whence h is A⁻orthogonal to T.  (A short direct check is routine.)  Therefore the cross term vanishes and\n\n Φ(q) = Φ_min + W‖y‖_A².                                                (16)\n\nThus the level set Φ(q)=K inside S becomes\n\n W‖y‖_A² = K − Φ_min.                                                    (17)\n\nExplicit Cartesian form: choose any basis of T, assemble its columns into an d×s matrix U of full rank with BU=0.  Write y=Uξ (ξ∈ℝˢ).  Set G:=UᵀAU (positive–definite).  Then (17) is\n\n W ξᵀG ξ = K − Φ_min.                                                    (18)\n\nStep 5.  Geometric classification & condition on K.  \nBecause G is positive–definite, ξᵀGξ>0 unless ξ=0.\n\n• If K < Φ_min  ⟹ RHS<0  ⟹ no real ξ satisfy (18)  ⟹ 𝓛(K)=∅.  \n• If K = Φ_min  ⟹ RHS=0  ⟹ ξ=0 only         ⟹ 𝓛(K)={q₀}.  \n• If K > Φ_min  ⟹ RHS>0  ⟹ (18) is a non-degenerate quadratic equation in ξ, hence an (s−1)-dimensional ellipsoid in S with centre q₀.  \n\nNecessary and sufficient condition:\n\n 𝓛(K)≠∅ ⇔ K ≥ Φ_min = C₀ + W‖h‖_A²,                                  (19)\n\ni.e. the prescribed constant must not be smaller than the A-moment of inertia of the weighted system of points about the affine subspace S.\n\nStep 6.  Principal axes and semi-axis lengths.  \nDiagonalise G: choose an orthogonal matrix R with RᵀGR=diag(γ₁,…,γ_s) (all γ_i>0).  \nUnder the coordinate change ξ = Rη equation (18) becomes\n\n Σ_{i=1}^{s} γ_i η_i² = (K − Φ_min)/W.                                  (20)\n\nHence the semi-axis lengths are  \n\n a_i = √((K − Φ_min)/(W γ_i) ),  i=1,…,s.                              (21)\n\nTheir directions in ℝᵈ are the A-orthonormal vectors U r_i (columns of UR).\n\nThis completes all requested items (i)–(v).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.435974",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional & anisotropic metric: The problem moves from ordinary Euclidean space to ℝᵈ endowed with an arbitrary positive-definite matrix A, forcing the solver to handle quadratic forms, inner products depending on A, and coordinate changes via eigen–decomposition.\n\n2. Additional constraints: The point q is not free in ℝᵈ but must satisfy Bq=c, introducing an affine subspace and the need for Lagrange multipliers/projection techniques.\n\n3. Weighted sums: Unequal positive weights w_j turn the centroid into a weighted centroid, complicating constant-term calculations.\n\n4. Multi-concept interaction: The solution uses linear algebra (Moore–Penrose inversion of BA⁻¹Bᵀ), classical analytic geometry (ellipsoids), optimisation (constrained minima), and spectral theory (diagonalisation of a symmetric positive-definite matrix).\n\n5. Deeper theoretical requirements: Identifying that h is A-orthogonal to T and proving uniqueness of the minimiser demand familiarity with projections in inner-product spaces not necessarily orthogonal in the usual sense.\n\n6. More steps: Compared with the original “circle/sphere” picture, one must (a) rewrite Φ with weights, (b) carry out a constrained optimisation, (c) detect and eliminate cross terms, (d) transform to principal axes, and (e) derive explicit semi–axis lengths.\n\nAltogether these layers raise the problem well beyond simple pattern recognition, making it significantly harder than both the original question and the earlier 3-dimensional kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix integers d≥2 and N≥2.  \nLet P₁,…,P_N be pairwise distinct points of the Euclidean space ℝᵈ with position vectors p₁,…,p_N, and let w₁,…,w_N be positive real “masses”.  \nFix a real symmetric positive–definite d×d matrix A (so ⟨x,y⟩_A := xᵀAy is an anisotropic inner product).  \nLet S be an s–dimensional affine subspace (1≤s≤d) given in Cartesian form by  \n  Bq = c,                                                                (1)  \nwhere B is an (d–s)×d matrix of rank d–s and c∈ℝ^{d–s}.  \nFor q∈S set  \n\n  Φ(q) = Σ_{j=1}^{N} w_j‖q−p_j‖_A²   with ‖x‖_A² := xᵀAx.                (2)\n\nGiven a real constant K, study the locus  \n\n  𝓛(K) = { q∈S : Φ(q)=K }.                                     (3)\n\nTasks  \n(i) Derive an explicit Cartesian equation of 𝓛(K).  \n(ii) Give a geometric description of 𝓛(K) (empty set / single point / (s–1)-dimensional ellipsoid) according to K.  \n(iii) Express, purely in geometric terms, the necessary and sufficient condition on K for 𝓛(K)≠∅.  \n(iv) Compute the unique point q₀∈S minimising Φ and the minimal value Φ_min.  \n(v) Determine the principal semi–axes of 𝓛(K) (when it is an ellipsoid) and their lengths.",
      "solution": "Throughout let  \n\n  W := Σ_{j=1}^{N} w_j  > 0,        μ := (1/W) Σ_{j=1}^{N} w_j p_j   (the weighted centroid).        (4)\n\nStep 1.  Re-express Φ(q) as a single quadratic term.  \nUsing (2):\n\n Φ(q)=Σ w_j[ qᵀAq − 2qᵀAp_j + p_jᵀAp_j ]  \n     = W qᵀAq − 2W qᵀAμ + Σ w_j p_jᵀAp_j                                     (5)\n\n   = W (q−μ)ᵀA(q−μ) − W μᵀAμ + Σ w_j p_jᵀAp_j                             (6)\n\nSet the “structural constant”\n\n C₀ := Σ w_j p_jᵀAp_j − W μᵀAμ = Σ w_j ‖p_j−μ‖_A²  (≥0).                  (7)\n\nHence\n\n Φ(q) = W (q−μ)ᵀA(q−μ) + C₀.                                              (8)\n\nStep 2.  Restrict to the affine subspace S.  \nA point q lies in S iff (1) holds.  \nIntroduce Lagrange multipliers λ∈ℝ^{d−s} and minimise (q−μ)ᵀA(q−μ) subject to Bq=c.\n\nThe stationarity equations are  \n\n 2A(q−μ) + Bᵀλ = 0,         Bq = c.                                        (9)\n\nSolve for q.  Since A is positive–definite, A⁻¹ exists.  \nFrom the first equation q = μ − ½ A⁻¹Bᵀλ.  Substitute into Bq=c:\n\n Bμ − ½ B A⁻¹ Bᵀ λ = c  \n ⟹ λ = 2 (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                                          (10)\n\n(The matrix in parentheses is invertible because B has full rank and A⁻¹ is positive–definite.)\n\nHence the unique minimiser is  \n\n q₀ = μ − A⁻¹ Bᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                                   (11)\n\nStep 3.  Minimal value of Φ on S.  \nPut h := q₀ − μ = −A⁻¹ Bᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).  \n\nThen, by (8),\n\n Φ_min := Φ(q₀) = W‖h‖_A² + C₀.                                          (12)\n\nCompute  \n\n ‖h‖_A² = hᵀA h  \n           = (Bμ − c)ᵀ (B A⁻¹ Bᵀ)⁻¹ (Bμ − c).                            (13)\n\nStep 4.  Equation of the locus.  \nFor any q∈S write q = q₀ + y where y lies in the linear subspace  \n\n T := { y∈ℝᵈ : By = 0 }.                                                  (14)\n\nBecause B has rank d−s, T is s–dimensional.  \nInsert q = q₀ + y into (8):\n\n Φ(q) = W (y + h)ᵀA(y + h) + C₀  \n       = W(‖h‖_A² + 2⟨h,y⟩_A + ‖y‖_A²) + C₀  \n       = Φ_min + W( 2⟨h,y⟩_A + ‖y‖_A² ).                                (15)\n\nBut ⟨h,y⟩_A = hᵀAy = 0, because Ay lies in im(A) and y∈T ⇒ By=0 ⇒ B A⁻¹ Ay=0, whence h is A⁻orthogonal to T.  (A short direct check is routine.)  Therefore the cross term vanishes and\n\n Φ(q) = Φ_min + W‖y‖_A².                                                (16)\n\nThus the level set Φ(q)=K inside S becomes\n\n W‖y‖_A² = K − Φ_min.                                                    (17)\n\nExplicit Cartesian form: choose any basis of T, assemble its columns into an d×s matrix U of full rank with BU=0.  Write y=Uξ (ξ∈ℝˢ).  Set G:=UᵀAU (positive–definite).  Then (17) is\n\n W ξᵀG ξ = K − Φ_min.                                                    (18)\n\nStep 5.  Geometric classification & condition on K.  \nBecause G is positive–definite, ξᵀGξ>0 unless ξ=0.\n\n• If K < Φ_min  ⟹ RHS<0  ⟹ no real ξ satisfy (18)  ⟹ 𝓛(K)=∅.  \n• If K = Φ_min  ⟹ RHS=0  ⟹ ξ=0 only         ⟹ 𝓛(K)={q₀}.  \n• If K > Φ_min  ⟹ RHS>0  ⟹ (18) is a non-degenerate quadratic equation in ξ, hence an (s−1)-dimensional ellipsoid in S with centre q₀.  \n\nNecessary and sufficient condition:\n\n 𝓛(K)≠∅ ⇔ K ≥ Φ_min = C₀ + W‖h‖_A²,                                  (19)\n\ni.e. the prescribed constant must not be smaller than the A-moment of inertia of the weighted system of points about the affine subspace S.\n\nStep 6.  Principal axes and semi-axis lengths.  \nDiagonalise G: choose an orthogonal matrix R with RᵀGR=diag(γ₁,…,γ_s) (all γ_i>0).  \nUnder the coordinate change ξ = Rη equation (18) becomes\n\n Σ_{i=1}^{s} γ_i η_i² = (K − Φ_min)/W.                                  (20)\n\nHence the semi-axis lengths are  \n\n a_i = √((K − Φ_min)/(W γ_i) ),  i=1,…,s.                              (21)\n\nTheir directions in ℝᵈ are the A-orthonormal vectors U r_i (columns of UR).\n\nThis completes all requested items (i)–(v).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.376779",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional & anisotropic metric: The problem moves from ordinary Euclidean space to ℝᵈ endowed with an arbitrary positive-definite matrix A, forcing the solver to handle quadratic forms, inner products depending on A, and coordinate changes via eigen–decomposition.\n\n2. Additional constraints: The point q is not free in ℝᵈ but must satisfy Bq=c, introducing an affine subspace and the need for Lagrange multipliers/projection techniques.\n\n3. Weighted sums: Unequal positive weights w_j turn the centroid into a weighted centroid, complicating constant-term calculations.\n\n4. Multi-concept interaction: The solution uses linear algebra (Moore–Penrose inversion of BA⁻¹Bᵀ), classical analytic geometry (ellipsoids), optimisation (constrained minima), and spectral theory (diagonalisation of a symmetric positive-definite matrix).\n\n5. Deeper theoretical requirements: Identifying that h is A-orthogonal to T and proving uniqueness of the minimiser demand familiarity with projections in inner-product spaces not necessarily orthogonal in the usual sense.\n\n6. More steps: Compared with the original “circle/sphere” picture, one must (a) rewrite Φ with weights, (b) carry out a constrained optimisation, (c) detect and eliminate cross terms, (d) transform to principal axes, and (e) derive explicit semi–axis lengths.\n\nAltogether these layers raise the problem well beyond simple pattern recognition, making it significantly harder than both the original question and the earlier 3-dimensional kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}