{
  "index": "1968-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-4. Given \\( n \\) points on the sphere \\( \\left\\{(x, y, z): x^{2}+y^{2}+z^{2}=1\\right\\} \\), demonstrate that the sum of the squares of the distances between them does not exceed \\( n^{2} \\).",
  "solution": "A-4 The \\( n \\) points can be represented by vectors \\( v_{i}(i=1, \\cdots, n) \\) with \\( \\left|v_{i}\\right|=1 \\). Expanding \"the sum of the squares of the distances between them\" in the case \\( n=3 \\) suggests the following general identities:\n\\[\n\\begin{aligned}\n\\sum_{1 \\leq i<j \\leq n}\\left|v_{i}-v_{j}\\right|^{2} & =\\sum_{1 \\leq i<j \\leq n}\\left(v_{i}-v_{j}\\right) \\cdot\\left(v_{i}-v_{j}\\right) \\\\\n& =(n-1) \\sum_{1 \\leq i \\leq n} v_{i} \\cdot v_{i}-2 \\sum_{1 \\leq i<j \\leq n} v_{i} \\cdot v_{j} \\\\\n& =n \\sum_{1 \\leq i \\leq n} v_{i} \\cdot v_{i}-\\left[\\sum_{1 \\leq i \\leq n} v_{i} \\cdot v_{i}+2 \\sum_{1 \\leq i<j \\leq n} v_{i} \\cdot v_{j}\\right] \\\\\n& =n^{2}-\\left(\\sum_{1 \\leq i \\leq n} v_{i}\\right) \\cdot\\left(\\sum_{1 \\leq i \\leq n} v_{i}\\right) .\n\\end{aligned}\n\\]\n\nThus the result follows and, in addition, equality exists if and only if \\( \\sum_{1 \\leq i \\leq n} v_{i}=0 \\).",
  "vars": [
    "x",
    "y",
    "z",
    "v_i",
    "v_j",
    "i",
    "j"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoordn",
        "y": "ycoordn",
        "z": "zcoordn",
        "v_i": "vectpointi",
        "v_j": "vectpointj",
        "i": "indexi",
        "j": "indexj",
        "n": "pointcount"
      },
      "question": "A-4. Given \\( pointcount \\) points on the sphere \\( \\left\\{(xcoordn, ycoordn, zcoordn): xcoordn^{2}+ycoordn^{2}+zcoordn^{2}=1\\right\\} \\), demonstrate that the sum of the squares of the distances between them does not exceed \\( pointcount^{2} \\).",
      "solution": "A-4 The \\( pointcount \\) points can be represented by vectors \\( vectpointi(indexi=1, \\cdots, pointcount) \\) with \\( \\left|vectpointi\\right|=1 \\). Expanding \"the sum of the squares of the distances between them\" in the case \\( pointcount=3 \\) suggests the following general identities:\n\\[\n\\begin{aligned}\n\\sum_{1 \\leq indexi<indexj \\leq pointcount}\\left|vectpointi-vectpointj\\right|^{2} & =\\sum_{1 \\leq indexi<indexj \\leq pointcount}\\left(vectpointi-vectpointj\\right) \\cdot\\left(vectpointi-vectpointj\\right) \\\\\n& =(pointcount-1) \\sum_{1 \\leq indexi \\leq pointcount} vectpointi \\cdot vectpointi-2 \\sum_{1 \\leq indexi<indexj \\leq pointcount} vectpointi \\cdot vectpointj \\\\\n& =pointcount \\sum_{1 \\leq indexi \\leq pointcount} vectpointi \\cdot vectpointi-\\left[\\sum_{1 \\leq indexi \\leq pointcount} vectpointi \\cdot vectpointi+2 \\sum_{1 \\leq indexi<indexj \\leq pointcount} vectpointi \\cdot vectpointj\\right] \\\\\n& =pointcount^{2}-\\left(\\sum_{1 \\leq indexi \\leq pointcount} vectpointi\\right) \\cdot\\left(\\sum_{1 \\leq indexi \\leq pointcount} vectpointi\\right) .\n\\end{aligned}\n\\]\n\nThus the result follows and, in addition, equality exists if and only if \\( \\sum_{1 \\leq indexi \\leq pointcount} vectpointi=0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "galaxyone",
        "y": "rainforest",
        "z": "moonlight",
        "v_i": "vectorlake",
        "v_j": "vectorcave",
        "i": "sunflower",
        "j": "thunderfog",
        "n": "crystalnet"
      },
      "question": "A-4. Given \\( crystalnet \\) points on the sphere \\( \\left\\{(galaxyone, rainforest, moonlight): galaxyone^{2}+rainforest^{2}+moonlight^{2}=1\\right\\} \\), demonstrate that the sum of the squares of the distances between them does not exceed \\( crystalnet^{2} \\).",
      "solution": "A-4 The \\( crystalnet \\) points can be represented by vectors \\( vectorlake(sunflower=1, \\cdots, crystalnet) \\) with \\( \\left|vectorlake\\right|=1 \\). Expanding \"the sum of the squares of the distances between them\" in the case \\( crystalnet=3 \\) suggests the following general identities:\n\\[\n\\begin{aligned}\n\\sum_{1 \\leq sunflower<thunderfog \\leq crystalnet}\\left|vectorlake-vectorcave\\right|^{2} & =\\sum_{1 \\leq sunflower<thunderfog \\leq crystalnet}\\left(vectorlake-vectorcave\\right) \\cdot\\left(vectorlake-vectorcave\\right) \\\\\n& =(crystalnet-1) \\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake \\cdot vectorlake-2 \\sum_{1 \\leq sunflower<thunderfog \\leq crystalnet} vectorlake \\cdot vectorcave \\\\\n& =crystalnet \\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake \\cdot vectorlake-\\left[\\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake \\cdot vectorlake+2 \\sum_{1 \\leq sunflower<thunderfog \\leq crystalnet} vectorlake \\cdot vectorcave\\right] \\\\\n& =crystalnet^{2}-\\left(\\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake\\right) \\cdot\\left(\\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake\\right) .\n\\end{aligned}\n\\]\n\nThus the result follows and, in addition, equality exists if and only if \\( \\sum_{1 \\leq sunflower \\leq crystalnet} vectorlake=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "steadystate",
        "y": "fixedpoint",
        "z": "constantval",
        "v_i": "scalarentity",
        "v_j": "scalarobject",
        "i": "constantindex",
        "j": "staticcounter",
        "n": "singularcount"
      },
      "question": "A-4. Given \\( singularcount \\) points on the sphere \\( \\left\\{(steadystate, fixedpoint, constantval): steadystate^{2}+fixedpoint^{2}+constantval^{2}=1\\right\\} \\), demonstrate that the sum of the squares of the distances between them does not exceed \\( singularcount^{2} \\).",
      "solution": "A-4 The \\( singularcount \\) points can be represented by vectors \\( scalarentity(constantindex=1, \\cdots, singularcount) \\) with \\( \\left|scalarentity\\right|=1 \\). Expanding \"the sum of the squares of the distances between them\" in the case \\( singularcount=3 \\) suggests the following general identities:\n\\[\n\\begin{aligned}\n\\sum_{1 \\leq constantindex<staticcounter \\leq singularcount}\\left|scalarentity-scalarobject\\right|^{2} & =\\sum_{1 \\leq constantindex<staticcounter \\leq singularcount}\\left(scalarentity-scalarobject\\right) \\cdot\\left(scalarentity-scalarobject\\right) \\\\\n& =(singularcount-1) \\sum_{1 \\leq constantindex \\leq singularcount} scalarentity \\cdot scalarentity-2 \\sum_{1 \\leq constantindex<staticcounter \\leq singularcount} scalarentity \\cdot scalarobject \\\\\n& =singularcount \\sum_{1 \\leq constantindex \\leq singularcount} scalarentity \\cdot scalarentity-\\left[\\sum_{1 \\leq constantindex \\leq singularcount} scalarentity \\cdot scalarentity+2 \\sum_{1 \\leq constantindex<staticcounter \\leq singularcount} scalarentity \\cdot scalarobject\\right] \\\\\n& =singularcount^{2}-\\left(\\sum_{1 \\leq constantindex \\leq singularcount} scalarentity\\right) \\cdot\\left(\\sum_{1 \\leq constantindex \\leq singularcount} scalarentity\\right) .\n\\end{aligned}\n\\]\n\nThus the result follows and, in addition, equality exists if and only if \\( \\sum_{1 \\leq constantindex \\leq singularcount} scalarentity=0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mtcprkvo",
        "v_i": "ngqpwlrs",
        "v_j": "pyxtrmau",
        "i": "fkljdwne",
        "j": "awzqpsld",
        "n": "lxvwtaeq"
      },
      "question": "A-4. Given \\( lxvwtaeq \\) points on the sphere \\( \\left\\{(qzxwvtnp, hjgrksla, mtcprkvo): qzxwvtnp^{2}+hjgrksla^{2}+mtcprkvo^{2}=1\\right\\} \\), demonstrate that the sum of the squares of the distances between them does not exceed \\( lxvwtaeq^{2} \\).",
      "solution": "A-4 The \\( lxvwtaeq \\) points can be represented by vectors \\( ngqpwlrs(fkljdwne=1, \\cdots, lxvwtaeq) \\) with \\( \\left|ngqpwlrs\\right|=1 \\). Expanding \"the sum of the squares of the distances between them\" in the case \\( lxvwtaeq=3 \\) suggests the following general identities:\n\\[\n\\begin{aligned}\n\\sum_{1 \\leq fkljdwne<awzqpsld \\leq lxvwtaeq}\\left|ngqpwlrs-pyxtrmau\\right|^{2} & =\\sum_{1 \\leq fkljdwne<awzqpsld \\leq lxvwtaeq}\\left(ngqpwlrs-pyxtrmau\\right) \\cdot\\left(ngqpwlrs-pyxtrmau\\right) \\\\\n& =(lxvwtaeq-1) \\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs \\cdot ngqpwlrs-2 \\sum_{1 \\leq fkljdwne<awzqpsld \\leq lxvwtaeq} ngqpwlrs \\cdot pyxtrmau \\\\\n& =lxvwtaeq \\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs \\cdot ngqpwlrs-\\left[\\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs \\cdot ngqpwlrs+2 \\sum_{1 \\leq fkljdwne<awzqpsld \\leq lxvwtaeq} ngqpwlrs \\cdot pyxtrmau\\right] \\\\\n& =lxvwtaeq^{2}-\\left(\\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs\\right) \\cdot\\left(\\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs\\right) .\n\\end{aligned}\n\\]\n\nThus the result follows and, in addition, equality exists if and only if \\( \\sum_{1 \\leq fkljdwne \\leq lxvwtaeq} ngqpwlrs=0 \\)."
    },
    "kernel_variant": {
      "question": "Let k ≥ 2 and let U be a real inner–product space of dimension k.  \nFix a radius r > 0 and put  \n\n  S := {x ∈ U : ‖x‖ = r},\n\nthe (k – 1)-sphere of radius r centred at the origin.  \nChoose an integer n ≥ k+2 and pick points  \n\n  P₁ , … , Pₙ ∈ S.                                                                                            (★)\n\nFor these points introduce the n × n distance–squared matrix  \n\n  D = (D_{ij})_{1≤i,j≤n},  D_{ij} := ‖P_i − P_j‖².  \n\nPut 𝟙 := (1,…,1)ᵀ and 𝟙^{⊥} := {α ∈ ℝⁿ : αᵀ𝟙 = 0}.  \nFor α ∈ 𝟙^{⊥} define the quadratic form  \n\n  Q(α) := ½ αᵀDα   (so Q(α) ≤ 0 by part (a)).                                                       (0)\n\nFor an orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} (1 ≤ m ≤ k) set  \n\n  Σ_m := −½ ∑_{s=1}^{m} α^{(s)ᵀ}D α^{(s)}                                                   (†)\n\n     = ∑_{s=1}^{m}‖∑_{i=1}^{n}α^{(s)}_i P_i‖²  ≥ 0.\n\nMoreover put  \n\n  C : 𝟙^{⊥} → U,  C(α) := ∑_{i=1}^{n}α_i P_i,                                             (1)\n\nand denote by T the positive semidefinite operator  \n\n  T := −½ D↾_{𝟙^{⊥}} = CᵀC.                                                                        (2)\n\n(a) Prove that for every α ∈ 𝟙^{⊥}  \n\n  αᵀD α = −2‖Cα‖²,\n\nand hence that D is negative semidefinite on 𝟙^{⊥}.\n\n(b) Show rank T = rank C ≤ k and that the non-zero eigenvalues of T are the squares σ₁² ≥ … ≥ σ_{rank T}² of the singular values σ₁�≥… ≥ σ_{rank T} of the k × (n−1) matrix representing C in orthonormal bases.  Prove further that  \n\n  ‖C‖ ≤ r√n and spec (T) ⊂ [0, n r²].\n\n(c) Put  \n\n  c := (1/n)∑_{i=1}^{n}P_i.\n\nShow that  \n\n  tr T = n r² − n‖c‖².\n\nDeduce 0 ≤ tr T ≤ n r² and determine when the extreme values occur.\n\n(d) From now on suppose that the points (★) span U and satisfy the centroid condition ∑_{i=1}^{n}P_i = 0 (so tr T = n r²).  \nLet λ₁ ≥ … ≥ λ_k > 0 be the (necessarily k) positive eigenvalues of T, and let Π denote the orthogonal projection onto Im T.\n\n(i) Prove the determinant bound  \n\n  det T ≤ (n r² / k)^k.                                                                                (6′)\n\n(ii) Show that equality in (6′) holds iff  \n\n  T Π = (n r² / k) Π,                                                                              (7′)\n\ni.e. T acts as a scalar multiple of the identity on its k-dimensional image.  \nEquivalently, this is the case iff  \n\n    P Pᵀ = (n r² / k) Id_U,\n\nso {P_i / r} is a unit-norm tight frame for U (regular k-simplex when n = k+1, regular cross-polytope when n = 2k, etc.).\n\n(iii) For every m = 1,…,k and every orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} prove  \n\n  Σ_m ≤ n r²,                                                                                    (8′)\n\nand show that equality holds iff m = k and  \n\n  Im T ⊂ span{α^{(1)},…,α^{(k)}}.                                                   (9′)\n\nIn particular, condition (7′) is sufficient but not necessary for equality in (8′).\n\n───────────────────────────────────────────────────────────────────────",
      "solution": "Step 0.  Notation.  \nWrite  \n\n  P := [P₁ … Pₙ] ∈ U^{k×n}, e := 𝟙/√n, M := Iₙ − e eᵀ,                        (10)\n\nso that M is the orthogonal projection onto 𝟙^{⊥}.  Euclidean inner products are used throughout ℝⁿ.\n\n────────────────────────────────  \n(a)  Identity on 𝟙^{⊥}.  \nFor i ≠ j the cosine rule with ‖P_i‖ = ‖P_j‖ = r gives  \n\n  D_{ij} = 2r² − 2 P_i·P_j, whereas D_{ii} = 0.\n\nWith the Gram matrix G := PᵀP we get  \n\n  D = 2r² 𝟙𝟙ᵀ − 2G.                                                                        (11)\n\nBecause α ∈ 𝟙^{⊥}, αᵀ𝟙 = 0, so αᵀ𝟙𝟙ᵀα = 0 and therefore  \n\n  αᵀDα = −2 αᵀGα = −2‖Pα‖² = −2‖Cα‖² ≤ 0,\n\nas required; thus D is negative semidefinite on 𝟙^{⊥}.\n\n────────────────────────────────  \n(b)  The operator T.  \nEquation (11) with (2) gives T = −½ D|_{𝟙^{⊥}} = CᵀC, whence rank T = rank C ≤ k.  \nIn orthonormal bases of 𝟙^{⊥} (dimension n−1) and U (dimension k) the matrix of C is k × (n−1); its non-zero singular values are σ₁,…,σ_{rank T}, and the non-zero eigenvalues of T are σ_j².\n\nFor the norm, take α ∈ 𝟙^{⊥} with ‖α‖ = 1:\n\n  ‖Cα‖ = ‖∑_{i}α_i P_i‖ ≤ ∑_{i}|α_i|‖P_i‖ ≤ r ∑_{i}|α_i| ≤ r√n,        (12)\n\nso ‖C‖ ≤ r√n and every eigenvalue of T lies in [0, n r²].\n\n────────────────────────────────  \n(c)  Trace identity.  \nBecause C = P M,  \n\n  T = (P M)ᵀ(P M) = M G M.                                                        (13)\n\nTaking traces and using tr M = n−1 and P e = c√n,\n\n  tr T = tr(M G) = tr G − tr(e eᵀ G)  \n     = ∑_{i=1}^{n}‖P_i‖² − ‖∑_{i=1}^{n}P_i‖² / n  \n     = n r² − n‖c‖².                                                        (14)\n\nHence 0 ≤ tr T ≤ n r²; tr T = 0 iff P₁ = … = Pₙ, and tr T = n r² iff ∑P_i = 0.\n\n────────────────────────────────  \n(d)  Centred, spanning case (∑P_i = 0, rank T = k).  \nNow c = 0, so tr T = n r² and T has exactly k positive eigenvalues λ₁≥…≥λ_k>0.  \nLet Π be the projection onto Im T (rank k).\n\n(d-i)  Determinant bound (6′).  \nBy AM–GM,\n\n  (det T)^{1/k} ≤ (tr T)/k = n r²/k,\n\nso det T ≤ (n r²/k)^k.                                                            (15)\n\n(d-ii)  Equality case for (6′).  \nEquality in (15) occurs iff λ₁ = … = λ_k = n r²/k, i.e.  \n\n  T Π = (n r²/k) Π.                                                        (16)\n\nBecause C has rank k and T = CᵀC, multiplying (16) on the left by C gives  \n\n  (C T)Π = (n r²/k) C Π.  \n\nSince C Π = C, this is equivalent to  \n\n  P Pᵀ = (n r²/k) Id_U.                                                (17)\n\nConversely, if (17) holds then C Cᵀ = (n r²/k) Id_U, so the non-zero singular values of C (and hence the positive eigenvalues of T) are all equal to √(n r²/k); consequently (16) and (15) are equalities.  \nThus (6′) is sharp precisely when (16) (equivalently (17)) holds; geometrically {P_i / r} is a unit-norm tight frame for U.\n\n(d-iii)  The Σ_m inequality (8′).  \nFor an orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} set  \n\n  S := ∑_{s=1}^{m}α^{(s)}α^{(s)ᵀ};  S is the orthogonal projection onto W := span{α^{(1)},…,α^{(m)}}.  \nBecause 0 ≤ S ≤ I_{𝟙^{⊥}},\n\n  Σ_m = tr(T S) ≤ tr T = n r².                                           (18)\n\nEquality, Σ_m = n r², means tr(T(I−S)) = 0, hence (I−S)T = 0, i.e.  \n\n  Im T ⊂ Im S = W.                                                     (19)\n\nSince dim Im T = k and m ≤ k, equality forces m = k and W ⊇ Im T.  \nConversely, if m = k and W ⊇ Im T, then S Π = Π and (18) becomes an equality.  \nTherefore (8′) is sharp precisely when m = k and condition (9′) holds.  \nNote that (17) automatically implies (9′) for any orthonormal basis of 𝟙^{⊥}, but (9′) can hold without (17).\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.576555",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher mathematical structures.  \n  The problem now involves the distance matrix D, its restriction to the codimension-1 subspace 𝟙^{⊥}, singular-value considerations for the column operator C and spectral estimates (parts (b),(d)).  None of these appear in the original statement.\n\n•  Additional interacting concepts.  \n  One must combine geometry on the sphere, the Gram matrix, operator inequalities, singular-value decomposition, trace bounds and characterisation of equality via regular simplices.\n\n•  Deeper theoretical requirements.  \n  The solver must know and use the polar decomposition, properties of the spectrum of CᵀC and of projections, trace techniques, and the connection between extremal spectra and regular simplices.\n\n•  Multiple steps instead of a single computation.  \n  The solution proceeds through: (i) a non-trivial identity; (ii) an operator factorisation and rank argument; (iii) a trace–spectral inequality with an equality characterisation; (iv) optimisation of a Hilbert–Schmidt norm.  Each layer builds on the previous ones.\n\n•  Strictly harder than previous variants.  \n  The earlier kernel variant asked for a single scalar inequality with one weight vector; our variant demands the full spectral description of a matrix operator, multiple weight vectors simultaneously, and identification of extremal geometric configurations.  Simple pattern matching or the original algebraic manipulation is no longer sufficient—the solver must mobilise linear-algebraic machinery and a fine understanding of regular simplices."
      }
    },
    "original_kernel_variant": {
      "question": "Let k ≥ 2 and let U be a real inner–product space of dimension k.  \nFix a radius r > 0 and put  \n\n  S := {x ∈ U : ‖x‖ = r},\n\nthe (k – 1)-sphere of radius r centred at the origin.  \nChoose an integer n ≥ k+2 and pick points  \n\n  P₁ , … , Pₙ ∈ S.                                                                                            (★)\n\nFor these points introduce the n × n distance–squared matrix  \n\n  D = (D_{ij})_{1≤i,j≤n},  D_{ij} := ‖P_i − P_j‖².  \n\nPut 𝟙 := (1,…,1)ᵀ and 𝟙^{⊥} := {α ∈ ℝⁿ : αᵀ𝟙 = 0}.  \nFor α ∈ 𝟙^{⊥} define the quadratic form  \n\n  Q(α) := ½ αᵀDα   (so Q(α) ≤ 0 by part (a)).                                                       (0)\n\nFor an orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} (1 ≤ m ≤ k) set  \n\n  Σ_m := −½ ∑_{s=1}^{m} α^{(s)ᵀ}D α^{(s)}                                                   (†)\n\n     = ∑_{s=1}^{m}‖∑_{i=1}^{n}α^{(s)}_i P_i‖²  ≥ 0.\n\nMoreover put  \n\n  C : 𝟙^{⊥} → U,  C(α) := ∑_{i=1}^{n}α_i P_i,                                             (1)\n\nand denote by T the positive semidefinite operator  \n\n  T := −½ D↾_{𝟙^{⊥}} = CᵀC.                                                                        (2)\n\n(a) Prove that for every α ∈ 𝟙^{⊥}  \n\n  αᵀD α = −2‖Cα‖²,\n\nand hence that D is negative semidefinite on 𝟙^{⊥}.\n\n(b) Show rank T = rank C ≤ k and that the non-zero eigenvalues of T are the squares σ₁² ≥ … ≥ σ_{rank T}² of the singular values σ₁�≥… ≥ σ_{rank T} of the k × (n−1) matrix representing C in orthonormal bases.  Prove further that  \n\n  ‖C‖ ≤ r√n and spec (T) ⊂ [0, n r²].\n\n(c) Put  \n\n  c := (1/n)∑_{i=1}^{n}P_i.\n\nShow that  \n\n  tr T = n r² − n‖c‖².\n\nDeduce 0 ≤ tr T ≤ n r² and determine when the extreme values occur.\n\n(d) From now on suppose that the points (★) span U and satisfy the centroid condition ∑_{i=1}^{n}P_i = 0 (so tr T = n r²).  \nLet λ₁ ≥ … ≥ λ_k > 0 be the (necessarily k) positive eigenvalues of T, and let Π denote the orthogonal projection onto Im T.\n\n(i) Prove the determinant bound  \n\n  det T ≤ (n r² / k)^k.                                                                                (6′)\n\n(ii) Show that equality in (6′) holds iff  \n\n  T Π = (n r² / k) Π,                                                                              (7′)\n\ni.e. T acts as a scalar multiple of the identity on its k-dimensional image.  \nEquivalently, this is the case iff  \n\n    P Pᵀ = (n r² / k) Id_U,\n\nso {P_i / r} is a unit-norm tight frame for U (regular k-simplex when n = k+1, regular cross-polytope when n = 2k, etc.).\n\n(iii) For every m = 1,…,k and every orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} prove  \n\n  Σ_m ≤ n r²,                                                                                    (8′)\n\nand show that equality holds iff m = k and  \n\n  Im T ⊂ span{α^{(1)},…,α^{(k)}}.                                                   (9′)\n\nIn particular, condition (7′) is sufficient but not necessary for equality in (8′).\n\n───────────────────────────────────────────────────────────────────────",
      "solution": "Step 0.  Notation.  \nWrite  \n\n  P := [P₁ … Pₙ] ∈ U^{k×n}, e := 𝟙/√n, M := Iₙ − e eᵀ,                        (10)\n\nso that M is the orthogonal projection onto 𝟙^{⊥}.  Euclidean inner products are used throughout ℝⁿ.\n\n────────────────────────────────  \n(a)  Identity on 𝟙^{⊥}.  \nFor i ≠ j the cosine rule with ‖P_i‖ = ‖P_j‖ = r gives  \n\n  D_{ij} = 2r² − 2 P_i·P_j, whereas D_{ii} = 0.\n\nWith the Gram matrix G := PᵀP we get  \n\n  D = 2r² 𝟙𝟙ᵀ − 2G.                                                                        (11)\n\nBecause α ∈ 𝟙^{⊥}, αᵀ𝟙 = 0, so αᵀ𝟙𝟙ᵀα = 0 and therefore  \n\n  αᵀDα = −2 αᵀGα = −2‖Pα‖² = −2‖Cα‖² ≤ 0,\n\nas required; thus D is negative semidefinite on 𝟙^{⊥}.\n\n────────────────────────────────  \n(b)  The operator T.  \nEquation (11) with (2) gives T = −½ D|_{𝟙^{⊥}} = CᵀC, whence rank T = rank C ≤ k.  \nIn orthonormal bases of 𝟙^{⊥} (dimension n−1) and U (dimension k) the matrix of C is k × (n−1); its non-zero singular values are σ₁,…,σ_{rank T}, and the non-zero eigenvalues of T are σ_j².\n\nFor the norm, take α ∈ 𝟙^{⊥} with ‖α‖ = 1:\n\n  ‖Cα‖ = ‖∑_{i}α_i P_i‖ ≤ ∑_{i}|α_i|‖P_i‖ ≤ r ∑_{i}|α_i| ≤ r√n,        (12)\n\nso ‖C‖ ≤ r√n and every eigenvalue of T lies in [0, n r²].\n\n────────────────────────────────  \n(c)  Trace identity.  \nBecause C = P M,  \n\n  T = (P M)ᵀ(P M) = M G M.                                                        (13)\n\nTaking traces and using tr M = n−1 and P e = c√n,\n\n  tr T = tr(M G) = tr G − tr(e eᵀ G)  \n     = ∑_{i=1}^{n}‖P_i‖² − ‖∑_{i=1}^{n}P_i‖² / n  \n     = n r² − n‖c‖².                                                        (14)\n\nHence 0 ≤ tr T ≤ n r²; tr T = 0 iff P₁ = … = Pₙ, and tr T = n r² iff ∑P_i = 0.\n\n────────────────────────────────  \n(d)  Centred, spanning case (∑P_i = 0, rank T = k).  \nNow c = 0, so tr T = n r² and T has exactly k positive eigenvalues λ₁≥…≥λ_k>0.  \nLet Π be the projection onto Im T (rank k).\n\n(d-i)  Determinant bound (6′).  \nBy AM–GM,\n\n  (det T)^{1/k} ≤ (tr T)/k = n r²/k,\n\nso det T ≤ (n r²/k)^k.                                                            (15)\n\n(d-ii)  Equality case for (6′).  \nEquality in (15) occurs iff λ₁ = … = λ_k = n r²/k, i.e.  \n\n  T Π = (n r²/k) Π.                                                        (16)\n\nBecause C has rank k and T = CᵀC, multiplying (16) on the left by C gives  \n\n  (C T)Π = (n r²/k) C Π.  \n\nSince C Π = C, this is equivalent to  \n\n  P Pᵀ = (n r²/k) Id_U.                                                (17)\n\nConversely, if (17) holds then C Cᵀ = (n r²/k) Id_U, so the non-zero singular values of C (and hence the positive eigenvalues of T) are all equal to √(n r²/k); consequently (16) and (15) are equalities.  \nThus (6′) is sharp precisely when (16) (equivalently (17)) holds; geometrically {P_i / r} is a unit-norm tight frame for U.\n\n(d-iii)  The Σ_m inequality (8′).  \nFor an orthonormal family {α^{(1)},…,α^{(m)}} ⊂ 𝟙^{⊥} set  \n\n  S := ∑_{s=1}^{m}α^{(s)}α^{(s)ᵀ};  S is the orthogonal projection onto W := span{α^{(1)},…,α^{(m)}}.  \nBecause 0 ≤ S ≤ I_{𝟙^{⊥}},\n\n  Σ_m = tr(T S) ≤ tr T = n r².                                           (18)\n\nEquality, Σ_m = n r², means tr(T(I−S)) = 0, hence (I−S)T = 0, i.e.  \n\n  Im T ⊂ Im S = W.                                                     (19)\n\nSince dim Im T = k and m ≤ k, equality forces m = k and W ⊇ Im T.  \nConversely, if m = k and W ⊇ Im T, then S Π = Π and (18) becomes an equality.  \nTherefore (8′) is sharp precisely when m = k and condition (9′) holds.  \nNote that (17) automatically implies (9′) for any orthonormal basis of 𝟙^{⊥}, but (9′) can hold without (17).\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.466880",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher mathematical structures.  \n  The problem now involves the distance matrix D, its restriction to the codimension-1 subspace 𝟙^{⊥}, singular-value considerations for the column operator C and spectral estimates (parts (b),(d)).  None of these appear in the original statement.\n\n•  Additional interacting concepts.  \n  One must combine geometry on the sphere, the Gram matrix, operator inequalities, singular-value decomposition, trace bounds and characterisation of equality via regular simplices.\n\n•  Deeper theoretical requirements.  \n  The solver must know and use the polar decomposition, properties of the spectrum of CᵀC and of projections, trace techniques, and the connection between extremal spectra and regular simplices.\n\n•  Multiple steps instead of a single computation.  \n  The solution proceeds through: (i) a non-trivial identity; (ii) an operator factorisation and rank argument; (iii) a trace–spectral inequality with an equality characterisation; (iv) optimisation of a Hilbert–Schmidt norm.  Each layer builds on the previous ones.\n\n•  Strictly harder than previous variants.  \n  The earlier kernel variant asked for a single scalar inequality with one weight vector; our variant demands the full spectral description of a matrix operator, multiple weight vectors simultaneously, and identification of extremal geometric configurations.  Simple pattern matching or the original algebraic manipulation is no longer sufficient—the solver must mobilise linear-algebraic machinery and a fine understanding of regular simplices."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}