{
  "index": "2013-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "For $m \\geq 3$, a list of $\\binom{m}{3}$ real numbers $a_{ijk}$ ($1 \\leq i < < j < k \\leq m$) is said to be \\emph{area definite} for $\\mathbb{R}^n$ if the inequality\n\\[\n\\sum_{1 \\leq i < j < k \\leq m} a_{ijk} \\cdot \\mathrm{Area}(\\Delta A_i A_j A_k) \\geq 0\n\\]\nholds for every choice of $m$ points $A_1,\\dots,A_m$ in $\\mathbb{R}^n$.\nFor example, the list of four numbers $a_{123} = a_{124} = a_{134} = 1$, $a_{234} = -1$ is area definite for $\\mathbb{R}^2$. Prove that if a list of $\\binom{m}{3}$ numbers is area definite for $\\mathbb{R}^2$,\nthen it is area definite for $\\mathbb{R}^3$.",
  "solution": "Let $A_1,\\ldots,A_m$ be points in $\\mathbb{R}^3$, and let $\\hat{n}_{ijk}$ denote a unit vector normal to $\\Delta A_iA_jA_k$ (unless $A_i,A_j,A_k$ are collinear, there are two possible choices for $\\hat{n}_{ijk}$). If $\\hat{n}$ is a unit vector in $\\mathbb{R}^3$, and $\\Pi_{\\hat{n}}$ is a plane perpendicular to $\\hat{n}$, then the area of the orthogonal projection of $\\Delta A_iA_jA_k$ onto $\\Pi_{\\hat{n}}$ is $\\text{Area}(\\Delta A_iA_jA_k) |\\hat{n}_{ijk} \\cdot \\hat{n}|$. Thus if $\\{a_{ijk}\\}$ is area definite for $\\mathbb{R}^2$, then for any $\\hat{n}$,\n\\[\n\\sum a_{ijk} \\text{Area}(\\Delta A_iA_jA_k) |\\hat{n}_{ijk} \\cdot \\hat{n}| \\geq 0.\n\\]\nNote that integrating $|\\hat{n}_{ijk} \\cdot \\hat{n}|$ over $\\hat{n} \\in S^2$, the unit sphere in $\\mathbb{R}^3$, with respect to the natural measure on $S^2$ gives a positive number $c$, which is independent of $\\hat{n}_{ijk}$ since the measure on $S^2$ is rotation-independent. Thus integrating the above inequality over $\\hat{n}$ gives\n$c \\sum a_{ijk} \\text{Area}(\\Delta A_iA_jA_k) \\geq 0$. It follows that $\\{a_{ijk}\\}$ is area definite for $\\mathbb{R}^3$, as desired. \n\n\\noindent\n\\textbf{Remark:}\nIt is not hard to check (e.g., by integration in spherical coordinates) that the constant $c$ occurring above is equal to $2\\pi$. It follows that for any convex body $C$ in $\\mathbb{R}^3$, the average over $\\hat{n}$ of the area of the projection of $C$ onto $\\Pi_{\\hat{n}}$ equals $1/4$ of the surface area of $C$. \n\nMore generally, let $C$ be a convex body in $\\mathbb{R}^n$.\nFor $\\hat{n}$ a unit vector, let $\\Pi_{\\hat{n}}$ denote the hyperplane through the origin perpendicular to $\\hat{n}$. Then the average over $\\hat{n}$ of the volume of the projection of $C$ onto $\\Pi_{\\hat{n}}$ equals a constant (depending only on $n$) times the $(n-1)$-dimensional surface area of $C$. \n\nStatements of this form inhabit the field of \\emph{inverse problems}, in which one attempts to reconstruct information about a geometric object from low-dimensional samples. This field has important applications in imaging and tomography.",
  "vars": [
    "a_ijk",
    "i",
    "j",
    "k",
    "A_i",
    "\\\\hat{n}",
    "\\\\hat{n}_ijk",
    "\\\\Pi_\\\\hat{n}",
    "C"
  ],
  "params": [
    "m",
    "n",
    "c",
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_ijk": "areacoef",
        "i": "firstix",
        "j": "secondix",
        "k": "thirdix",
        "A_i": "pointvertex",
        "\\hat{n}": "unitnormal",
        "\\hat{n}_ijk": "facenormal",
        "\\Pi_\\hat{n}": "orthoplane",
        "C": "convexbody",
        "m": "pointcount",
        "n": "dimension",
        "c": "constantval",
        "S": "unitsphere"
      },
      "question": "For $pointcount \\geq 3$, a list of $\\binom{pointcount}{3}$ real numbers $areacoef$ ($1 \\leq firstix < < secondix < thirdix \\leq pointcount$) is said to be \\emph{area definite} for $\\mathbb{R}^{dimension}$ if the inequality\n\\[\n\\sum_{1 \\leq firstix < secondix < thirdix \\leq pointcount} areacoef \\cdot \\mathrm{Area}(\\Delta pointvertex_{firstix} pointvertex_{secondix} pointvertex_{thirdix}) \\geq 0\n\\]\nholds for every choice of $pointcount$ points $pointvertex_1,\\dots,pointvertex_{pointcount}$ in $\\mathbb{R}^{dimension}$.\nFor example, the list of four numbers $areacoef_{123} = areacoef_{124} = areacoef_{134} = 1$, $areacoef_{234} = -1$ is area definite for $\\mathbb{R}^2$. Prove that if a list of $\\binom{pointcount}{3}$ numbers is area definite for $\\mathbb{R}^2$, then it is area definite for $\\mathbb{R}^3$.",
      "solution": "Let pointvertex_{firstix},\\ldots,pointvertex_{pointcount} be points in $\\mathbb{R}^3$, and let facenormal denote a unit vector normal to $\\Delta pointvertex_{firstix}pointvertex_{secondix}pointvertex_{thirdix}$ (unless pointvertex_{firstix},pointvertex_{secondix},pointvertex_{thirdix} are collinear, there are two possible choices for facenormal). If unitnormal is a unit vector in $\\mathbb{R}^3$, and orthoplane is a plane perpendicular to unitnormal, then the area of the orthogonal projection of $\\Delta pointvertex_{firstix}pointvertex_{secondix}pointvertex_{thirdix}$ onto orthoplane is $\\text{Area}(\\Delta pointvertex_{firstix}pointvertex_{secondix}pointvertex_{thirdix}) |facenormal \\cdot unitnormal|$. Thus if $\\{areacoef\\}$ is area definite for $\\mathbb{R}^2$, then for any unitnormal,\n\\[\n\\sum areacoef \\text{Area}(\\Delta pointvertex_{firstix}pointvertex_{secondix}pointvertex_{thirdix}) |facenormal \\cdot unitnormal| \\geq 0.\n\\]\nNote that integrating $|facenormal \\cdot unitnormal|$ over unitnormal $\\in unitsphere^2$, the unit sphere in $\\mathbb{R}^3$, with respect to the natural measure on unitsphere^2 gives a positive number constantval, which is independent of facenormal since the measure on unitsphere^2 is rotation-independent. Thus integrating the above inequality over unitnormal gives\n$constantval \\sum areacoef \\text{Area}(\\Delta pointvertex_{firstix}pointvertex_{secondix}pointvertex_{thirdix}) \\geq 0$. It follows that $\\{areacoef\\}$ is area definite for $\\mathbb{R}^3$, as desired. \n\nRemark:\nIt is not hard to check (e.g., by integration in spherical coordinates) that the constant constantval occurring above is equal to $2\\pi$. It follows that for any convexbody in $\\mathbb{R}^3$, the average over unitnormal of the area of the projection of convexbody onto orthoplane equals $1/4$ of the surface area of convexbody. \n\nMore generally, let convexbody be a convex body in $\\mathbb{R}^{dimension}$. For unitnormal a unit vector, let orthoplane denote the hyperplane through the origin perpendicular to unitnormal. Then the average over unitnormal of the volume of the projection of convexbody onto orthoplane equals a constant (depending only on dimension) times the $(dimension-1)$-dimensional surface area of convexbody. \n\nStatements of this form inhabit the field of \\emph{inverse problems}, in which one attempts to reconstruct information about a geometric object from low-dimensional samples. This field has important applications in imaging and tomography."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_ijk": "blueberry",
        "i": "seashell",
        "j": "butterfly",
        "k": "lemonade",
        "A_i": "sandstone",
        "\\hat{n}": "lighthouse",
        "\\hat{n}_ijk": "monolithic",
        "\\Pi_\\hat{n}": "pineapple",
        "C": "waterfall",
        "m": "rainstorm",
        "n": "snowflake",
        "c": "turquoise",
        "S": "raspberry"
      },
      "question": "For $rainstorm \\geq 3$, a list of $\\binom{rainstorm}{3}$ real numbers blueberry ($1 \\leq seashell < < butterfly < lemonade \\leq rainstorm$) is said to be \\emph{area definite} for $\\mathbb{R}^{snowflake}$ if the inequality\n\\[\n\\sum_{1 \\leq seashell < butterfly < lemonade \\leq rainstorm} blueberry \\cdot \\mathrm{Area}(\\Delta sandstone A_j A_k) \\geq 0\n\\]\nholds for every choice of $rainstorm$ points $A_1,\\dots,A_{rainstorm}$ in $\\mathbb{R}^{snowflake}$. For example, the list of four numbers $a_{123} = a_{124} = a_{134} = 1$, $a_{234} = -1$ is area definite for $\\mathbb{R}^2$. Prove that if a list of $\\binom{rainstorm}{3}$ numbers is area definite for $\\mathbb{R}^2$, then it is area definite for $\\mathbb{R}^3$.",
      "solution": "Let $A_1,\\ldots,A_{rainstorm}$ be points in $\\mathbb{R}^3$, and let monolithic denote a unit vector normal to $\\Delta sandstone A_jA_k$ (unless sandstone, $A_j$, and $A_k$ are collinear, there are two possible choices for monolithic). If lighthouse is a unit vector in $\\mathbb{R}^3$, and pineapple is a plane perpendicular to lighthouse, then the area of the orthogonal projection of $\\Delta sandstone A_jA_k$ onto pineapple is $\\text{Area}(\\Delta sandstone A_jA_k) |monolithic \\cdot lighthouse|$. Thus if $\\{blueberry\\}$ is area definite for $\\mathbb{R}^2$, then for any lighthouse,\n\\[\n\\sum blueberry \\text{Area}(\\Delta sandstone A_jA_k) |monolithic \\cdot lighthouse| \\geq 0.\n\\]\nNote that integrating $|monolithic \\cdot lighthouse|$ over lighthouse $\\in raspberry^2$, the unit sphere in $\\mathbb{R}^3$, with respect to the natural measure on raspberry^2 gives a positive number turquoise, which is independent of monolithic since the measure on raspberry^2 is rotation-independent. Thus integrating the above inequality over lighthouse gives\n$turquoise \\sum blueberry \\text{Area}(\\Delta sandstone A_jA_k) \\geq 0$. It follows that $\\{blueberry\\}$ is area definite for $\\mathbb{R}^3$, as desired.\n\n\\noindent\\textbf{Remark:} It is not hard to check (e.g., by integration in spherical coordinates) that the constant turquoise occurring above is equal to $2\\pi$. It follows that for any convex body waterfall in $\\mathbb{R}^3$, the average over lighthouse of the area of the projection of waterfall onto pineapple equals $1/4$ of the surface area of waterfall.\n\nMore generally, let waterfall be a convex body in $\\mathbb{R}^{snowflake}$. For lighthouse a unit vector, let pineapple denote the hyperplane through the origin perpendicular to lighthouse. Then the average over lighthouse of the volume of the projection of waterfall onto pineapple equals a constant (depending only on snowflake) times the $(snowflake-1)$-dimensional surface area of waterfall.\n\nStatements of this form inhabit the field of \\emph{inverse problems}, in which one attempts to reconstruct information about a geometric object from low-dimensional samples. This field has important applications in imaging and tomography."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_{ijk}": "nullifier",
        "i": "entirety",
        "j": "wholeness",
        "k": "completeness",
        "A_i": "hypersolid",
        "\\hat{n}": "antidirect",
        "\\hat{n}_{ijk}": "antinormal",
        "\\Pi_{\\hat{n}}": "voidline",
        "C": "emptiness",
        "m": "scarcity",
        "n": "nothingness",
        "c": "variability",
        "S": "flatplane"
      },
      "question": "For $scarcity \\geq 3$, a list of $\\binom{scarcity}{3}$ real numbers nullifier ($1 \\leq entirety < < wholeness < completeness \\leq scarcity$) is said to be \\emph{area definite} for $\\mathbb{R}^{nothingness}$ if the inequality\n\\[\n\\sum_{1 \\leq entirety < wholeness < completeness \\leq scarcity} nullifier \\cdot \\mathrm{Area}(\\Delta hypersolid A_{wholeness} A_{completeness}) \\geq 0\n\\]\nholds for every choice of $scarcity$ points $A_1,\\dots,A_{scarcity}$ in $\\mathbb{R}^{nothingness}$. For example, the list of four numbers $a_{123} = a_{124} = a_{134} = 1$, $a_{234} = -1$ is area definite for $\\mathbb{R}^2$. Prove that if a list of $\\binom{scarcity}{3}$ numbers is area definite for $\\mathbb{R}^2$, then it is area definite for $\\mathbb{R}^3$.",
      "solution": "Let $A_1,\\ldots,A_{scarcity}$ be points in $\\mathbb{R}^3$, and let $antinormal$ denote a unit vector normal to $\\Delta hypersolid A_{wholeness} A_{completeness}$ (unless $hypersolid,A_{wholeness},A_{completeness}$ are collinear, there are two possible choices for $antinormal$). If $antidirect$ is a unit vector in $\\mathbb{R}^3$, and $voidline$ is a plane perpendicular to $antidirect$, then the area of the orthogonal projection of $\\Delta hypersolid A_{wholeness} A_{completeness}$ onto $voidline$ is $\\text{Area}(\\Delta hypersolid A_{wholeness} A_{completeness}) |antinormal \\cdot antidirect|$. Thus if $\\{nullifier\\}$ is area definite for $\\mathbb{R}^2$, then for any $antidirect$,\n\\[\n\\sum nullifier \\text{Area}(\\Delta hypersolid A_{wholeness} A_{completeness}) |antinormal \\cdot antidirect| \\geq 0.\n\\]\nNote that integrating $|antinormal \\cdot antidirect|$ over $antidirect \\in flatplane^{2}$, the unit sphere in $\\mathbb{R}^3$, with respect to the natural measure on $flatplane^{2}$ gives a positive number $variability$, which is independent of $antinormal$ since the measure on $flatplane^{2}$ is rotation-independent. Thus integrating the above inequality over $antidirect$ gives $variability \\sum nullifier \\text{Area}(\\Delta hypersolid A_{wholeness} A_{completeness}) \\geq 0$. It follows that $\\{nullifier\\}$ is area definite for $\\mathbb{R}^3$, as desired.\n\n\\noindent\\textbf{Remark:} It is not hard to check (e.g., by integration in spherical coordinates) that the constant $variability$ occurring above is equal to $2\\pi$. It follows that for any convex body $emptiness$ in $\\mathbb{R}^3$, the average over $antidirect$ of the area of the projection of $emptiness$ onto $voidline$ equals $1/4$ of the surface area of $emptiness$.\n\nMore generally, let $emptiness$ be a convex body in $\\mathbb{R}^{nothingness}$. For $antidirect$ a unit vector, let $voidline$ denote the hyperplane through the origin perpendicular to $antidirect$. Then the average over $antidirect$ of the volume of the projection of $emptiness$ onto $voidline$ equals a constant (depending only on $nothingness$) times the $(nothingness-1)$-dimensional surface area of $emptiness$.\n\nStatements of this form inhabit the field of \\emph{inverse problems}, in which one attempts to reconstruct information about a geometric object from low-dimensional samples. This field has important applications in imaging and tomography."
    },
    "garbled_string": {
      "map": {
        "a_ijk": "phqvzctu",
        "i": "lxadrope",
        "j": "fsygmkzn",
        "k": "qbvrejhi",
        "A_i": "eumlscvo",
        "\\hat{n}": "jpawrcgy",
        "\\hat{n}_ijk": "hvnesylq",
        "\\Pi_{\\hat{n}}": "xgodflkb",
        "C": "rozybten",
        "m": "krtuzsba",
        "n": "wqovipher",
        "c": "mldoynve",
        "S": "tqsprnzl"
      },
      "question": "For $krtuzsba \\geq 3$, a list of $\\binom{krtuzsba}{3}$ real numbers $phqvzctu_{ijk}$ ($1 \\leq lxadrope < < fsygmkzn < qbvrejhi \\leq krtuzsba$) is said to be \\emph{area definite} for $\\mathbb{R}^{wqovipher}$ if the inequality\n\\[\n\\sum_{1 \\leq lxadrope < fsygmkzn < qbvrejhi \\leq krtuzsba} phqvzctu_{ijk} \\cdot \\mathrm{Area}(\\Delta eumlscvo_{lxadrope} eumlscvo_{fsygmkzn} eumlscvo_{qbvrejhi}) \\geq 0\n\\]\nholds for every choice of $krtuzsba$ points $eumlscvo_1,\\dots,eumlscvo_{krtuzsba}$ in $\\mathbb{R}^{wqovipher}$.  For example, the list of four numbers $phqvzctu_{123} = phqvzctu_{124} = phqvzctu_{134} = 1$, $phqvzctu_{234} = -1$ is area definite for $\\mathbb{R}^2$. Prove that if a list of $\\binom{krtuzsba}{3}$ numbers is area definite for $\\mathbb{R}^2$, then it is area definite for $\\mathbb{R}^3$.",
      "solution": "Let eumlscvo_1,\\ldots,eumlscvo_{krtuzsba} be points in $\\mathbb{R}^3$, and let $hvnesylq_{ijk}$ denote a unit vector normal to $\\Delta eumlscvo_{lxadrope}eumlscvo_{fsygmkzn}eumlscvo_{qbvrejhi}$ (unless $eumlscvo_{lxadrope},eumlscvo_{fsygmkzn},eumlscvo_{qbvrejhi}$ are collinear, there are two possible choices for $hvnesylq_{ijk}$). If $jpawrcgy$ is a unit vector in $\\mathbb{R}^3$, and $xgodflkb$ is a plane perpendicular to $jpawrcgy$, then the area of the orthogonal projection of $\\Delta eumlscvo_{lxadrope}eumlscvo_{fsygmkzn}eumlscvo_{qbvrejhi}$ onto $xgodflkb$ is $\\text{Area}(\\Delta eumlscvo_{lxadrope}eumlscvo_{fsygmkzn}eumlscvo_{qbvrejhi}) |hvnesylq_{ijk} \\cdot jpawrcgy|$. Thus if $\\{phqvzctu_{ijk}\\}$ is area definite for $\\mathbb{R}^2$, then for any $jpawrcgy$,\n\\[\n\\sum phqvzctu_{ijk} \\text{Area}(\\Delta eumlscvo_{lxadrope}eumlscvo_{fsygmkzn}eumlscvo_{qbvrejhi}) |hvnesylq_{ijk} \\cdot jpawrcgy| \\geq 0.\n\\]\nNote that integrating $|hvnesylq_{ijk} \\cdot jpawrcgy|$ over $jpawrcgy \\in tqsprnzl^2$, the unit sphere in $\\mathbb{R}^3$, with respect to the natural measure on $tqsprnzl^2$ gives a positive number $mldoynve$, which is independent of $hvnesylq_{ijk}$ since the measure on $tqsprnzl^2$ is rotation-independent. Thus integrating the above inequality over $jpawrcgy$ gives\n$mldoynve \\sum phqvzctu_{ijk} \\text{Area}(\\Delta eumlscvo_{lxadrope}eumlscvo_{fsygmkzn}eumlscvo_{qbvrejhi}) \\geq 0$. It follows that $\\{phqvzctu_{ijk}\\}$ is area definite for $\\mathbb{R}^3$, as desired. \n\n\\noindent\n\\textbf{Remark:}\nIt is not hard to check (e.g., by integration in spherical coordinates) that the constant $mldoynve$ occurring above is equal to $2\\pi$. It follows that for any convex body $rozybten$ in $\\mathbb{R}^3$, the average over $jpawrcgy$ of the area of the projection of $rozybten$ onto $xgodflkb$ equals $1/4$ of the surface area of $rozybten$. \n\nMore generally, let $rozybten$ be a convex body in $\\mathbb{R}^{wqovipher}$. For $jpawrcgy$ a unit vector, let $xgodflkb$ denote the hyperplane through the origin perpendicular to $jpawrcgy$. Then the average over $jpawrcgy$ of the volume of the projection of $rozybten$ onto $xgodflkb$ equals a constant (depending only on $wqovipher$) times the $(wqovipher-1)$-dimensional surface area of $rozybten$. \n\nStatements of this form inhabit the field of \\emph{inverse problems}, in which one attempts to reconstruct information about a geometric object from low-dimensional samples. This field has important applications in imaging and tomography."
    },
    "kernel_variant": {
      "question": "Fix integers m,r with \n • r ≥ 2  and  \n • m ≥ r+2 .  \n\nFor every (r+1)–subset I={i₀,…,i_r}⊂{1,…,m} with i₀<⋯<i_r let a_I∈ℝ be given.  \n\nThe family (a_I) is called r–volume–definite for ℝ^n if, for every choice of points  \nA₁,…,A_m∈ℝ^n, the inequality  \n        Σ_{|I|=r+1} a_I · Vol_r(Δ_I) ≥ 0                        (★_n)  \n\nholds, where Vol_r(Δ_I) denotes the (unsigned) r–dimensional volume of the simplex  \nΔ_I = conv{A_i : i∈I}.  \n\nProve the following general transfer theorem.\n\nGENERAL  TRANSFER  THEOREM.  \nIf (a_I) is r–volume–definite for ℝ^{r+1}, then it is r–volume–definite for every Euclidean space ℝ^n with n ≥ r+1; i.e.  \n                (★_{r+1})  ⇒  (★_n) for all n ≥ r+1.",
      "solution": "We keep the original four–step strategy; only the quantitative part of Step 3 needed repair.\n\nPreliminaries and notation.  \n•  G_{n,r+1}: Grassmannian of (r+1)–dimensional linear subspaces of ℝ^n, equipped with the O(n)–invariant probability measure dL.  \n•  π_L : ℝ^n → L denotes orthogonal projection.  \n•  For an r–simplex Δ write Vol_r(Δ) for its (unsigned) r–volume.  \n•  For vectors v₁,…,v_r set Vol_r(v₁,…,v_r)=1/r!·√det(Gram(v₁,…,v_r)).\n\nStep 1 (projection).  \nGiven an arbitrary configuration A₁,…,A_m∈ℝ^n (n ≥ r+1), fix L∈G_{n,r+1}.  \nThe m images π_L(A_i) lie in the (r+1)–space L ≅ ℝ^{r+1}; therefore (★_{r+1}) yields  \n        Σ_I a_I·Vol_r(π_L(Δ_I))  ≥  0.                                (1)\n\nStep 2 (averaging).  \nIntegrate (1) over G_{n,r+1}:\n\n        0 ≤ ∫_{G_{n,r+1}} Σ_I a_I·Vol_r(π_L(Δ_I)) dL  \n           = Σ_I a_I·∫_{G_{n,r+1}} Vol_r(π_L(Δ_I)) dL .                (2)\n\nFor each fixed r–simplex Δ we abbreviate  \n\n        E_{n,r}[Δ] := ∫_{G_{n,r+1}} Vol_r(π_L(Δ)) dL .                (3)\n\nStep 3 (the kinematic factor).  \nWe show that E_{n,r}[Δ] is a positive constant multiple of Vol_r(Δ), and that the\nmultiplicative constant depends only on r and n.\n\nLet Δ=conv{P₀,…,P_r}.  Translate so that P₀=0 and write v_j=P_j (j=1,…,r).\nSet Λ:=v₁∧⋯∧v_r∈∧^rℝ^n and denote |Λ|:=r!·Vol_r(Δ).\nFor L∈G_{n,r+1}, orthogonal projection induces ∧^rπ_L : ∧^rℝ^n→∧^rL; hence\n\n        Vol_r(π_L(Δ)) = 1/r!· |∧^rπ_L(Λ)| .                          (4)\n\nBecause the probability measure dL is O(n)–invariant, the random variable\n|∧^rπ_L(Λ)| depends on Λ only through its length.  Consequently there is a\nconstant κ_{n,r}>0 such that\n\n        E_{n,r}[Δ] = κ_{n,r} · Vol_r(Δ)               for every r–simplex Δ.     (5)\n\nExistence and positivity of κ_{n,r}.  \nChoose Λ with |Λ|=1.  Then |∧^rπ_L(Λ)| is continuous and strictly positive on a\nset of positive measure (indeed, whenever Λ is not contained in L it projects to\na non–zero r–vector); hence its integral (3) is strictly positive.  Finally,\nby homogeneity (4) we obtain (5), where κ_{n,r}:=E_{n,r}[Δ₀] with any unit–volume\nreference simplex Δ₀.  For completeness we record two easy checks:  \n• κ_{r+1,r}=1 (because G_{r+1,r+1}={ℝ^{r+1}} consists of a single subspace);  \n• κ_{n,r} decreases strictly with n and satisfies 0<κ_{n,r}<1 for n>r+1.  \nAn explicit closed formula can be expressed through Beta–functions, e.g.  \n\n        κ_{n,r}= \\frac{1}{B(\\frac{1}{2},\\frac{n-r}{2})}\n                 \\int_{0}^{1} t^{r/2}(1-t)^{(n-r-3)/2}\\,dt,             (6)\n\nbut the present argument needs only (5) and κ_{n,r}>0.\n\nStep 4 (conclusion).  \nInsert (5) into (2):\n\n        0 ≤ κ_{n,r} · Σ_I a_I·Vol_r(Δ_I).                             (7)\n\nSince κ_{n,r}>0, the inequality (7) is exactly (★_n).  \nBecause the initial configuration of points in ℝ^n was arbitrary, (a_I) is\nr–volume–definite for ℝ^n.  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.827113",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher–Dimensional Generalisation.  \n   • The original problem dealt with 2–dimensional areas; the current variant treats arbitrary r–dimensional volumes with r ≥ 2.  \n   • Instead of a single step (ℝ²→ℝ³) we must prove a universally quantified transfer (ℝ^{r+1}→ℝⁿ for all n≥r+1).\n\n2. Advanced Geometric Measure Theory.  \n   • The solution uses integration over the Grassmannian G_{n,r+1}, a manifold whose measure theory is far subtler than the unit sphere exploited in the original.  \n   • One must invoke exterior algebra and the determinant/Gram–matrix description of simplex volume.\n\n3. Kinematic/Crofton-type Identity.  \n   • Establishing formula (4) requires knowledge of classical integral–geometric results or a careful derivation via rotational invariance; neither is elementary.\n\n4. Abstraction and Notation.  \n   • Multiple indices (|I|=r+1) and wedge products demand fluency with multilinear algebra.  \n   • The argument isolates an invariant κ_{n,r} by group–theoretic reasoning, unlike the elementary constant 2π in the original solution.\n\n5. Non-trivial Iteration.  \n   • Because κ_{n,r}>0 for every n, the theorem folds an infinite cascade of dimensions into a single proof, establishing much more than a one-step lift.\n\nCollectively these enhancements force competitors to blend combinatorial simplex bookkeeping, advanced linear/​multilinear algebra, and integral geometry—considerably deeper and more technical than the original setting."
      }
    },
    "original_kernel_variant": {
      "question": "Fix integers m,r with \n • r ≥ 2  and  \n • m ≥ r+2 .  \n\nFor every (r+1)–subset I={i₀,…,i_r}⊂{1,…,m} with i₀<⋯<i_r let a_I∈ℝ be given.  \n\nThe family (a_I) is called r–volume–definite for ℝ^n if, for every choice of points  \nA₁,…,A_m∈ℝ^n, the inequality  \n        Σ_{|I|=r+1} a_I · Vol_r(Δ_I) ≥ 0                        (★_n)  \n\nholds, where Vol_r(Δ_I) denotes the (unsigned) r–dimensional volume of the simplex  \nΔ_I = conv{A_i : i∈I}.  \n\nProve the following general transfer theorem.\n\nGENERAL  TRANSFER  THEOREM.  \nIf (a_I) is r–volume–definite for ℝ^{r+1}, then it is r–volume–definite for every Euclidean space ℝ^n with n ≥ r+1; i.e.  \n                (★_{r+1})  ⇒  (★_n) for all n ≥ r+1.",
      "solution": "We keep the original four–step strategy; only the quantitative part of Step 3 needed repair.\n\nPreliminaries and notation.  \n•  G_{n,r+1}: Grassmannian of (r+1)–dimensional linear subspaces of ℝ^n, equipped with the O(n)–invariant probability measure dL.  \n•  π_L : ℝ^n → L denotes orthogonal projection.  \n•  For an r–simplex Δ write Vol_r(Δ) for its (unsigned) r–volume.  \n•  For vectors v₁,…,v_r set Vol_r(v₁,…,v_r)=1/r!·√det(Gram(v₁,…,v_r)).\n\nStep 1 (projection).  \nGiven an arbitrary configuration A₁,…,A_m∈ℝ^n (n ≥ r+1), fix L∈G_{n,r+1}.  \nThe m images π_L(A_i) lie in the (r+1)–space L ≅ ℝ^{r+1}; therefore (★_{r+1}) yields  \n        Σ_I a_I·Vol_r(π_L(Δ_I))  ≥  0.                                (1)\n\nStep 2 (averaging).  \nIntegrate (1) over G_{n,r+1}:\n\n        0 ≤ ∫_{G_{n,r+1}} Σ_I a_I·Vol_r(π_L(Δ_I)) dL  \n           = Σ_I a_I·∫_{G_{n,r+1}} Vol_r(π_L(Δ_I)) dL .                (2)\n\nFor each fixed r–simplex Δ we abbreviate  \n\n        E_{n,r}[Δ] := ∫_{G_{n,r+1}} Vol_r(π_L(Δ)) dL .                (3)\n\nStep 3 (the kinematic factor).  \nWe show that E_{n,r}[Δ] is a positive constant multiple of Vol_r(Δ), and that the\nmultiplicative constant depends only on r and n.\n\nLet Δ=conv{P₀,…,P_r}.  Translate so that P₀=0 and write v_j=P_j (j=1,…,r).\nSet Λ:=v₁∧⋯∧v_r∈∧^rℝ^n and denote |Λ|:=r!·Vol_r(Δ).\nFor L∈G_{n,r+1}, orthogonal projection induces ∧^rπ_L : ∧^rℝ^n→∧^rL; hence\n\n        Vol_r(π_L(Δ)) = 1/r!· |∧^rπ_L(Λ)| .                          (4)\n\nBecause the probability measure dL is O(n)–invariant, the random variable\n|∧^rπ_L(Λ)| depends on Λ only through its length.  Consequently there is a\nconstant κ_{n,r}>0 such that\n\n        E_{n,r}[Δ] = κ_{n,r} · Vol_r(Δ)               for every r–simplex Δ.     (5)\n\nExistence and positivity of κ_{n,r}.  \nChoose Λ with |Λ|=1.  Then |∧^rπ_L(Λ)| is continuous and strictly positive on a\nset of positive measure (indeed, whenever Λ is not contained in L it projects to\na non–zero r–vector); hence its integral (3) is strictly positive.  Finally,\nby homogeneity (4) we obtain (5), where κ_{n,r}:=E_{n,r}[Δ₀] with any unit–volume\nreference simplex Δ₀.  For completeness we record two easy checks:  \n• κ_{r+1,r}=1 (because G_{r+1,r+1}={ℝ^{r+1}} consists of a single subspace);  \n• κ_{n,r} decreases strictly with n and satisfies 0<κ_{n,r}<1 for n>r+1.  \nAn explicit closed formula can be expressed through Beta–functions, e.g.  \n\n        κ_{n,r}= \\frac{1}{B(\\frac{1}{2},\\frac{n-r}{2})}\n                 \\int_{0}^{1} t^{r/2}(1-t)^{(n-r-3)/2}\\,dt,             (6)\n\nbut the present argument needs only (5) and κ_{n,r}>0.\n\nStep 4 (conclusion).  \nInsert (5) into (2):\n\n        0 ≤ κ_{n,r} · Σ_I a_I·Vol_r(Δ_I).                             (7)\n\nSince κ_{n,r}>0, the inequality (7) is exactly (★_n).  \nBecause the initial configuration of points in ℝ^n was arbitrary, (a_I) is\nr–volume–definite for ℝ^n.  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.632373",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher–Dimensional Generalisation.  \n   • The original problem dealt with 2–dimensional areas; the current variant treats arbitrary r–dimensional volumes with r ≥ 2.  \n   • Instead of a single step (ℝ²→ℝ³) we must prove a universally quantified transfer (ℝ^{r+1}→ℝⁿ for all n≥r+1).\n\n2. Advanced Geometric Measure Theory.  \n   • The solution uses integration over the Grassmannian G_{n,r+1}, a manifold whose measure theory is far subtler than the unit sphere exploited in the original.  \n   • One must invoke exterior algebra and the determinant/Gram–matrix description of simplex volume.\n\n3. Kinematic/Crofton-type Identity.  \n   • Establishing formula (4) requires knowledge of classical integral–geometric results or a careful derivation via rotational invariance; neither is elementary.\n\n4. Abstraction and Notation.  \n   • Multiple indices (|I|=r+1) and wedge products demand fluency with multilinear algebra.  \n   • The argument isolates an invariant κ_{n,r} by group–theoretic reasoning, unlike the elementary constant 2π in the original solution.\n\n5. Non-trivial Iteration.  \n   • Because κ_{n,r}>0 for every n, the theorem folds an infinite cascade of dimensions into a single proof, establishing much more than a one-step lift.\n\nCollectively these enhancements force competitors to blend combinatorial simplex bookkeeping, advanced linear/​multilinear algebra, and integral geometry—considerably deeper and more technical than the original setting."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}