{
  "index": "1946-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "5. Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid\n\\[\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}+\\frac{z^{2}}{c^{2}}=1\n\\]",
  "solution": "Solution. The tangent plane to the given ellipsoid at the point \\( \\left(x_{1}, y_{1}, z_{1}\\right) \\) has the equation\n\\[\n\\frac{x x_{1}}{a^{2}}+\\frac{y y_{1}}{b^{2}}+\\frac{z z_{1}}{c^{2}}=1 .\n\\]\n\nIts intercepts on the \\( x, y \\), and \\( z \\)-axes, respectively, are\n\\[\n\\frac{a^{2}}{x_{1}}, \\frac{b^{2}}{y_{1}}, \\frac{c^{2}}{z_{1}}\n\\]\n\nThe volume of the solid cut off by the tangent plane and the three coordinate planes is\n\\[\nV=\\frac{1}{6}\\left|\\frac{a^{2} b^{2} c^{2}}{x_{1} y_{1} z_{1}}\\right| .\n\\]\n(If \\( x_{1} y_{1} z_{1}=0 \\), then the four planes do not bound a finite region.) Hence\n\\[\nV^{2}=\\frac{1}{36} a^{2} b^{2} c^{2}\\left(\\frac{x_{1}{ }^{2}}{a^{2}} \\frac{y_{1}{ }^{2}}{b^{2}} \\frac{z_{1}{ }^{2}}{c^{2}}\\right)^{-1}\n\\]\n\nBut\n\\[\n\\left(\\frac{x_{1}{ }^{2}}{a^{2}} \\cdot \\frac{y_{1}{ }^{2}}{b^{2}} \\cdot \\frac{z_{1}{ }^{2}}{c^{2}}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{x_{1}{ }^{2}}{a^{2}}+\\frac{y_{1}{ }^{2}}{b^{2}}+\\frac{z_{1}{ }^{2}}{c^{2}}\\right)=\\frac{1}{3}\n\\]\nwith equality if and only if\n\\[\n\\frac{x_{1}{ }^{2}}{a^{2}}=\\frac{y_{1}{ }^{2}}{b^{2}}=\\frac{z_{1}{ }^{2}}{c^{2}}=\\frac{1}{3}\n\\]\n(the arithmetic-geometric mean inequality). Hence\n\\[\nV^{2} \\geq \\frac{27}{36} a^{2} b^{2} c^{2} \\quad \\text { and } \\quad V \\geq \\frac{1}{2} \\sqrt{3} a b c\n\\]\nwith equality if and only if \\( \\left(x_{1}, y_{1}, z_{1}\\right) \\) is one of the eight points for which (3) holds, namely\n\\[\n( \\pm a / \\sqrt{3}, \\quad \\pm b / \\sqrt{3}, \\quad \\pm c / \\sqrt{3}) .\n\\]\n\nIt is also possible, of course, to minimize \\( V \\) straightforwardly by maximizing the product \\( x_{1} y_{1} z_{1} \\) under the constraint that \\( \\left(x_{1}, y_{1}, z_{1}\\right) \\) is a point",
  "vars": [
    "x",
    "y",
    "z",
    "x_1",
    "y_1",
    "z_1",
    "V"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "z": "altitude",
        "x_1": "tangentx",
        "y_1": "tangenty",
        "z_1": "tangentz",
        "V": "volumevar",
        "a": "semiaxisx",
        "b": "semiaxisy",
        "c": "semiaxisz"
      },
      "question": "5. Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid\n\\[\n\\frac{\\abscissa^{2}}{\\semiaxisx^{2}}+\\frac{\\ordinate^{2}}{\\semiaxisy^{2}}+\\frac{\\altitude^{2}}{\\semiaxisz^{2}}=1\n\\]",
      "solution": "Solution. The tangent plane to the given ellipsoid at the point \\( \\left(\\tangentx, \\tangenty, \\tangentz\\right) \\) has the equation\n\\[\n\\frac{\\abscissa \\tangentx}{\\semiaxisx^{2}}+\\frac{\\ordinate \\tangenty}{\\semiaxisy^{2}}+\\frac{\\altitude \\tangentz}{\\semiaxisz^{2}}=1 .\n\\]\n\nIts intercepts on the \\( \\abscissa, \\ordinate \\), and \\( \\altitude \\)-axes, respectively, are\n\\[\n\\frac{\\semiaxisx^{2}}{\\tangentx}, \\frac{\\semiaxisy^{2}}{\\tangenty}, \\frac{\\semiaxisz^{2}}{\\tangentz}\n\\]\n\nThe volume of the solid cut off by the tangent plane and the three coordinate planes is\n\\[\n\\volumevar=\\frac{1}{6}\\left|\\frac{\\semiaxisx^{2} \\semiaxisy^{2} \\semiaxisz^{2}}{\\tangentx \\tangenty \\tangentz}\\right| .\n\\]\n(If \\( \\tangentx \\tangenty \\tangentz=0 \\), then the four planes do not bound a finite region.) Hence\n\\[\n\\volumevar^{2}=\\frac{1}{36} \\semiaxisx^{2} \\semiaxisy^{2} \\semiaxisz^{2}\\left(\\frac{\\tangentx^{2}}{\\semiaxisx^{2}} \\frac{\\tangenty^{2}}{\\semiaxisy^{2}} \\frac{\\tangentz^{2}}{\\semiaxisz^{2}}\\right)^{-1}\n\\]\n\nBut\n\\[\n\\left(\\frac{\\tangentx^{2}}{\\semiaxisx^{2}} \\cdot \\frac{\\tangenty^{2}}{\\semiaxisy^{2}} \\cdot \\frac{\\tangentz^{2}}{\\semiaxisz^{2}}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{\\tangentx^{2}}{\\semiaxisx^{2}}+\\frac{\\tangenty^{2}}{\\semiaxisy^{2}}+\\frac{\\tangentz^{2}}{\\semiaxisz^{2}}\\right)=\\frac{1}{3}\n\\]\nwith equality if and only if\n\\[\n\\frac{\\tangentx^{2}}{\\semiaxisx^{2}}=\\frac{\\tangenty^{2}}{\\semiaxisy^{2}}=\\frac{\\tangentz^{2}}{\\semiaxisz^{2}}=\\frac{1}{3}\n\\]\n(the arithmetic-geometric mean inequality). Hence\n\\[\n\\volumevar^{2} \\geq \\frac{27}{36} \\semiaxisx^{2} \\semiaxisy^{2} \\semiaxisz^{2} \\quad \\text { and } \\quad \\volumevar \\geq \\frac{1}{2} \\sqrt{3} \\semiaxisx \\semiaxisy \\semiaxisz\n\\]\nwith equality if and only if \\( \\left(\\tangentx, \\tangenty, \\tangentz\\right) \\) is one of the eight points for which (3) holds, namely\n\\[\n( \\pm \\semiaxisx / \\sqrt{3}, \\quad \\pm \\semiaxisy / \\sqrt{3}, \\quad \\pm \\semiaxisz / \\sqrt{3}) .\n\\]\n\nIt is also possible, of course, to minimize \\( \\volumevar \\) straightforwardly by maximizing the product \\( \\tangentx \\tangenty \\tangentz \\) under the constraint that \\( \\left(\\tangentx, \\tangenty, \\tangentz\\right) \\) is a point"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orangetree",
        "y": "blueberry",
        "z": "watermelon",
        "x_1": "pineapple",
        "y_1": "strawberry",
        "z_1": "pomegranate",
        "V": "raspberry",
        "a": "sandalwood",
        "b": "elderberry",
        "c": "gooseberry"
      },
      "question": "5. Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid\n\\[\n\\frac{orangetree^{2}}{sandalwood^{2}}+\\frac{blueberry^{2}}{elderberry^{2}}+\\frac{watermelon^{2}}{gooseberry^{2}}=1\n\\]\n",
      "solution": "Solution. The tangent plane to the given ellipsoid at the point \\( \\left(pineapple, strawberry, pomegranate\\right) \\) has the equation\n\\[\n\\frac{orangetree\\,pineapple}{sandalwood^{2}}+\\frac{blueberry\\,strawberry}{elderberry^{2}}+\\frac{watermelon\\,pomegranate}{gooseberry^{2}}=1 .\n\\]\n\nIts intercepts on the \\( x, y \\), and \\( z \\)-axes, respectively, are\n\\[\n\\frac{sandalwood^{2}}{pineapple}, \\frac{elderberry^{2}}{strawberry}, \\frac{gooseberry^{2}}{pomegranate}\n\\]\n\nThe volume of the solid cut off by the tangent plane and the three coordinate planes is\n\\[\nraspberry=\\frac{1}{6}\\left|\\frac{sandalwood^{2}\\,elderberry^{2}\\,gooseberry^{2}}{pineapple\\,strawberry\\,pomegranate}\\right| .\n\\]\n(If \\( pineapple\\,strawberry\\,pomegranate=0 \\), then the four planes do not bound a finite region.) Hence\n\\[\nraspberry^{2}=\\frac{1}{36}\\,sandalwood^{2}\\,elderberry^{2}\\,gooseberry^{2}\\left(\\frac{pineapple{ }^{2}}{sandalwood^{2}} \\frac{strawberry{ }^{2}}{elderberry^{2}} \\frac{pomegranate{ }^{2}}{gooseberry^{2}}\\right)^{-1}\n\\]\n\nBut\n\\[\n\\left(\\frac{pineapple{ }^{2}}{sandalwood^{2}} \\cdot \\frac{strawberry{ }^{2}}{elderberry^{2}} \\cdot \\frac{pomegranate{ }^{2}}{gooseberry^{2}}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{pineapple{ }^{2}}{sandalwood^{2}}+\\frac{strawberry{ }^{2}}{elderberry^{2}}+\\frac{pomegranate{ }^{2}}{gooseberry^{2}}\\right)=\\frac{1}{3}\n\\]\nwith equality if and only if\n\\[\n\\frac{pineapple{ }^{2}}{sandalwood^{2}}=\\frac{strawberry{ }^{2}}{elderberry^{2}}=\\frac{pomegranate{ }^{2}}{gooseberry^{2}}=\\frac{1}{3}\n\\]\n(the arithmetic-geometric mean inequality). Hence\n\\[\nraspberry^{2} \\geq \\frac{27}{36} sandalwood^{2} elderberry^{2} gooseberry^{2} \\quad \\text { and } \\quad raspberry \\geq \\frac{1}{2} \\sqrt{3} sandalwood\\,elderberry\\,gooseberry\n\\]\nwith equality if and only if \\( \\left(pineapple, strawberry, pomegranate\\right) \\) is one of the eight points for which (3) holds, namely\n\\[\n( \\pm sandalwood / \\sqrt{3}, \\quad \\pm elderberry / \\sqrt{3}, \\quad \\pm gooseberry / \\sqrt{3}) .\n\\]\n\nIt is also possible, of course, to minimize \\( raspberry \\) straightforwardly by maximizing the product \\( pineapple\\,strawberry\\,pomegranate \\) under the constraint that \\( \\left(pineapple, strawberry, pomegranate\\right) \\) is a point"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "longitudinalaxis",
        "z": "planaraxis",
        "x_1": "verticalpoint",
        "y_1": "longitudinalpoint",
        "z_1": "planarpoint",
        "V": "voidvalue",
        "a": "contractaxis",
        "b": "narrowaxis",
        "c": "thinaxis"
      },
      "question": "5. Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid\n\\[\n\\frac{verticalaxis^{2}}{contractaxis^{2}}+\\frac{longitudinalaxis^{2}}{narrowaxis^{2}}+\\frac{planaraxis^{2}}{thinaxis^{2}}=1\n\\]",
      "solution": "Solution. The tangent plane to the given ellipsoid at the point \\( \\left(verticalpoint, longitudinalpoint, planarpoint\\right) \\) has the equation\n\\[\n\\frac{verticalaxis\\, verticalpoint}{contractaxis^{2}}+\\frac{longitudinalaxis\\, longitudinalpoint}{narrowaxis^{2}}+\\frac{planaraxis\\, planarpoint}{thinaxis^{2}}=1 .\n\\]\n\nIts intercepts on the \\( verticalaxis, longitudinalaxis \\), and \\( planaraxis \\)-axes, respectively, are\n\\[\n\\frac{contractaxis^{2}}{verticalpoint}, \\frac{narrowaxis^{2}}{longitudinalpoint}, \\frac{thinaxis^{2}}{planarpoint}\n\\]\n\nThe volume of the solid cut off by the tangent plane and the three coordinate planes is\n\\[\nvoidvalue=\\frac{1}{6}\\left|\\frac{contractaxis^{2} narrowaxis^{2} thinaxis^{2}}{verticalpoint longitudinalpoint planarpoint}\\right| .\n\\]\n(If \\( verticalpoint longitudinalpoint planarpoint=0 \\), then the four planes do not bound a finite region.) Hence\n\\[\nvoidvalue^{2}=\\frac{1}{36} contractaxis^{2} narrowaxis^{2} thinaxis^{2}\\left(\\frac{verticalpoint{ }^{2}}{contractaxis^{2}} \\frac{longitudinalpoint{ }^{2}}{narrowaxis^{2}} \\frac{planarpoint{ }^{2}}{thinaxis^{2}}\\right)^{-1}\n\\]\n\nBut\n\\[\n\\left(\\frac{verticalpoint{ }^{2}}{contractaxis^{2}} \\cdot \\frac{longitudinalpoint{ }^{2}}{narrowaxis^{2}} \\cdot \\frac{planarpoint{ }^{2}}{thinaxis^{2}}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{verticalpoint{ }^{2}}{contractaxis^{2}}+\\frac{longitudinalpoint{ }^{2}}{narrowaxis^{2}}+\\frac{planarpoint{ }^{2}}{thinaxis^{2}}\\right)=\\frac{1}{3}\n\\]\nwith equality if and only if\n\\[\n\\frac{verticalpoint{ }^{2}}{contractaxis^{2}}=\\frac{longitudinalpoint{ }^{2}}{narrowaxis^{2}}=\\frac{planarpoint{ }^{2}}{thinaxis^{2}}=\\frac{1}{3}\n\\]\n(the arithmetic-geometric mean inequality). Hence\n\\[\nvoidvalue^{2} \\geq \\frac{27}{36} contractaxis^{2} narrowaxis^{2} thinaxis^{2} \\quad \\text { and } \\quad voidvalue \\geq \\frac{1}{2} \\sqrt{3} contractaxis narrowaxis thinaxis\n\\]\nwith equality if and only if \\( \\left(verticalpoint, longitudinalpoint, planarpoint\\right) \\) is one of the eight points for which (3) holds, namely\n\\[\n( \\pm contractaxis / \\sqrt{3}, \\quad \\pm narrowaxis / \\sqrt{3}, \\quad \\pm thinaxis / \\sqrt{3}) .\n\\]\n\nIt is also possible, of course, to minimize \\( voidvalue \\) straightforwardly by maximizing the product \\( verticalpoint longitudinalpoint planarpoint \\) under the constraint that \\( \\left(verticalpoint, longitudinalpoint, planarpoint\\right) \\) is a point"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mndfqtly",
        "x_1": "vchplrwo",
        "y_1": "sgnkfduc",
        "z_1": "btqrwexy",
        "V": "ksjmztrp",
        "a": "pfwekrql",
        "b": "zfqtbnma",
        "c": "dxrglsvo"
      },
      "question": "5. Find the smallest volume bounded by the coordinate planes and by a tangent plane to the ellipsoid\n\\[\n\\frac{qzxwvtnp^{2}}{pfwekrql^{2}}+\\frac{hjgrksla^{2}}{zfqtbnma^{2}}+\\frac{mndfqtly^{2}}{dxrglsvo^{2}}=1\n\\]\n",
      "solution": "Solution. The tangent plane to the given ellipsoid at the point \\( \\left(vchplrwo, sgnkfduc, btqrwexy\\right) \\) has the equation\n\\[\n\\frac{qzxwvtnp vchplrwo}{pfwekrql^{2}}+\\frac{hjgrksla sgnkfduc}{zfqtbnma^{2}}+\\frac{mndfqtly btqrwexy}{dxrglsvo^{2}}=1 .\n\\]\n\nIts intercepts on the \\( qzxwvtnp, hjgrksla \\), and \\( mndfqtly \\)-axes, respectively, are\n\\[\n\\frac{pfwekrql^{2}}{vchplrwo}, \\frac{zfqtbnma^{2}}{sgnkfduc}, \\frac{dxrglsvo^{2}}{btqrwexy}\n\\]\n\nThe volume of the solid cut off by the tangent plane and the three coordinate planes is\n\\[\nksjmztrp=\\frac{1}{6}\\left|\\frac{pfwekrql^{2} zfqtbnma^{2} dxrglsvo^{2}}{vchplrwo sgnkfduc btqrwexy}\\right| .\n\\]\n(If \\( vchplrwo sgnkfduc btqrwexy=0 \\), then the four planes do not bound a finite region.) Hence\n\\[\nksjmztrp^{2}=\\frac{1}{36} pfwekrql^{2} zfqtbnma^{2} dxrglsvo^{2}\\left(\\frac{vchplrwo^{2}}{pfwekrql^{2}} \\frac{sgnkfduc^{2}}{zfqtbnma^{2}} \\frac{btqrwexy^{2}}{dxrglsvo^{2}}\\right)^{-1}\n\\]\n\nBut\n\\[\n\\left(\\frac{vchplrwo^{2}}{pfwekrql^{2}} \\cdot \\frac{sgnkfduc^{2}}{zfqtbnma^{2}} \\cdot \\frac{btqrwexy^{2}}{dxrglsvo^{2}}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{vchplrwo^{2}}{pfwekrql^{2}}+\\frac{sgnkfduc^{2}}{zfqtbnma^{2}}+\\frac{btqrwexy^{2}}{dxrglsvo^{2}}\\right)=\\frac{1}{3}\n\\]\nwith equality if and only if\n\\[\n\\frac{vchplrwo^{2}}{pfwekrql^{2}}=\\frac{sgnkfduc^{2}}{zfqtbnma^{2}}=\\frac{btqrwexy^{2}}{dxrglsvo^{2}}=\\frac{1}{3}\n\\]\n(the arithmetic-geometric mean inequality). Hence\n\\[\nksjmztrp^{2} \\geq \\frac{27}{36} pfwekrql^{2} zfqtbnma^{2} dxrglsvo^{2} \\quad \\text { and } \\quad ksjmztrp \\geq \\frac{1}{2} \\sqrt{3} pfwekrql zfqtbnma dxrglsvo\n\\]\nwith equality if and only if \\( \\left(vchplrwo, sgnkfduc, btqrwexy\\right) \\) is one of the eight points for which (3) holds, namely\n\\[\n( \\pm pfwekrql / \\sqrt{3}, \\quad \\pm zfqtbnma / \\sqrt{3}, \\quad \\pm dxrglsvo / \\sqrt{3}) .\n\\]\n\nIt is also possible, of course, to minimize \\( ksjmztrp \\) straightforwardly by maximizing the product \\( vchplrwo sgnkfduc btqrwexy \\) under the constraint that \\( \\left(vchplrwo, sgnkfduc, btqrwexy\\right) \\) is a point\n"
    },
    "kernel_variant": {
      "question": "Let $\\lambda$ be a real number with $0<\\lambda<1$ and consider in $\\mathbb R^{6}$ the centred “correlated’’ ellipsoid  \n\\[\nE_{\\lambda}\\;:\\qquad \\sum_{i=1}^{6}x_{i}^{2}+2\\lambda\\!\\!\\sum_{1\\le i<j\\le 6}\\!x_{i}x_{j}=1. \\tag{$\\star$}\n\\]\nEquivalently $E_{\\lambda}$ is the quadric $x^{\\mathsf T}Ax=1$, where\n\\[\nA=(1-\\lambda)I_{6}+\\lambda J_{6},\\qquad\n0<\\lambda<1 ,\n\\]\n$I_{6}$ is the $6\\times 6$ identity matrix and $J_{6}$ the $6\\times 6$ all–ones matrix.\n\nFor a point $P=(p_{1},\\dots ,p_{6})\\in E_{\\lambda}$ let $\\Pi_{P}$ be the hyper-plane tangent to $E_{\\lambda}$ at $P$.  \nTogether with the six coordinate hyper-planes $x_{1}=0,\\dots ,x_{6}=0$, the plane $\\Pi_{P}$ encloses a $6$-simplex $S(P)$ contained in the first orthant whenever that simplex is bounded.  \nDenote its (six–dimensional) hyper-volume by $V(P)$.\n\n(a)  Prove that $S(P)$ is bounded if and only if  \n\\[\n(AP)_{i}>0\\qquad\\text{for all }i=1,\\dots ,6 , \\tag{1}\n\\]\ni.e. iff every axis–intercept $t_{i}=1/(AP)_{i}$ of $\\Pi_{P}$ is positive.  (No explicit sign condition on the $p_{i}$ is required.)\n\n(b)  Determine, in closed form,\n\\[\nV_{\\min}= \\min_{\\substack{P\\in E_{\\lambda}\\\\ (AP)_{i}>0}} V(P), \\tag{2}\n\\]\nand describe precisely the set of all points $P$ at which that minimum is attained.  Express your answer solely in terms of $\\lambda$.",
      "solution": "Throughout let $\\boldsymbol 1=(1,\\dots ,1)^{\\mathsf T}\\in\\mathbb R^{6}$ and set $\\Sigma:=\\sum_{j=1}^{6}p_{j}$ for brevity.\n\n--------------------------------------------------------------------\nPart (a)  (boundedness criterion)\n\nThe tangent hyper-plane at $P\\in E_{\\lambda}$ is obtained from the gradient of the quadratic form $F(x)=x^{\\mathsf T}Ax$:\n\\[\n\\nabla F(P)=2AP\\quad\\Longrightarrow\\quad (AP)\\cdot(x-P)=0.\n\\]\nBecause $P^{\\mathsf T}AP=1$, this rewrites as\n\\[\n(AP)\\cdot x=1. \\tag{3}\n\\]\n\nIntercept with the $x_{i}$-axis: insert $x=t\\,e_{i}$ in (3) to get $t_{i}=1/(AP)_{i}$.  \nIf some $(AP)_{i}\\le 0$, then either $t_{i}\\le 0$ or $t_{i}=\\infty$; in both cases $\\Pi_{P}$ fails to intersect the positive $x_{i}$-axis at a positive finite point, whence the coordinate hyper-planes cannot cut off a bounded region in the first orthant.  \nConversely, if every $(AP)_{i}>0$ then each $t_{i}$ is positive and finite, and the six coordinate hyper-planes together with $\\Pi_{P}$ bound the right $6$-simplex\n\\[\nS(P)=\\bigl\\{(x_{1},\\dots ,x_{6})\\;|\\;x_{i}\\ge 0,\\;\\; (AP)\\cdot x\\le 1\\bigr\\}\n\\]\nwithin the first orthant.  Hence (1) is both necessary and sufficient for $S(P)$ to be bounded. $\\square$\n\n--------------------------------------------------------------------\nPart (b)  (minimisation of the volume)\n\nStep 1.  Volume in terms of $AP$  \nBecause $S(P)$ is a right $6$-simplex whose edges along the positive coordinate axes have lengths $t_{i}=1/(AP)_{i}$, its hyper-volume is\n\\[\nV(P)=\\frac{1}{6!}\\prod_{i=1}^{6}\\frac{1}{(AP)_{i}}=\\frac{1}{720}\\prod_{i=1}^{6}(AP)_{i}^{\\,-1}. \\tag{4}\n\\]\nThus minimising $V(P)$ is equivalent to maximising\n\\[\n\\Phi(P):=\\prod_{i=1}^{6}(AP)_{i} \\tag{5}\n\\]\nunder the constraints $P\\in E_{\\lambda}$ and $(AP)_{i}>0$.\n\nStep 2.  A convenient expression for $(AP)_{i}$  \nSince $A$ has $1$ on the diagonal and $\\lambda$ off the diagonal,\n\\[\n(AP)_{i}=p_{i}+\\lambda\\!\\!\\sum_{j\\ne i}\\!p_{j}=(1-\\lambda)p_{i}+\\lambda\\Sigma. \\tag{6}\n\\]\n\nStep 3.  Parameterisation by mean and deviation  \nWrite $P=t\\boldsymbol 1+u$, where $u=(u_{1},\\dots ,u_{6})$ satisfies $\\boldsymbol 1^{\\mathsf T}u=0$.  \nThen $\\Sigma=6t$ and (6) becomes\n\\[\n(AP)_{i}=t(1+5\\lambda)+(1-\\lambda)u_{i}. \\tag{7}\n\\]\nSet\n\\[\n\\alpha:=t(1+5\\lambda),\\qquad \\beta:=1-\\lambda>0 .\n\\]\nHence\n\\[\n(AP)_{i}= \\alpha+\\beta u_{i}. \\tag{8}\n\\]\n\nStep 4.  The ellipsoid constraint in the new variables  \nBecause $A=(1-\\lambda)I_{6}+\\lambda J_{6}$ is positive definite,\n\\[\n1=P^{\\mathsf T}AP=(t\\boldsymbol 1+u)^{\\mathsf T}A(t\\boldsymbol 1+u)\n        =6(1+5\\lambda)t^{2}+\\beta\\,\\lVert u\\rVert^{2}. \\tag{9}\n\\]\nSolving (9) for $t^{2}$ gives\n\\[\nt^{2}=\\frac{1-\\beta\\lVert u\\rVert^{2}}{6(1+5\\lambda)}\\quad\\text{with}\\quad\n0\\le\\beta\\lVert u\\rVert^{2}\\le 1. \\tag{10}\n\\]\nConsequently\n\\[\n\\alpha=\\sqrt{\\frac{1+5\\lambda}{6}}\\,\n       \\sqrt{1-\\beta\\lVert u\\rVert^{2}}, \\quad 0\\le\\beta\\lVert u\\rVert^{2}\\le 1. \\tag{11}\n\\]\nObserve that $\\alpha$ is a strictly decreasing function of $\\lVert u\\rVert^{2}$.\n\nStep 5.  Maximising $\\Phi$  \n\nWe have\n\\[\n\\Phi(P)=\\prod_{i=1}^{6}\\bigl(\\alpha+\\beta u_{i}\\bigr). \\tag{12}\n\\]\nBecause $\\sum_{i=1}^{6}u_{i}=0$, the numbers $\\alpha+\\beta u_{i}$ share the arithmetic mean $\\alpha$.  The AM–GM inequality yields\n\\[\n\\Phi(P)\\le \\alpha^{6}; \\tag{13}\n\\]\nequality in (13) occurs if and only if every $u_{i}=0$.  \nBut by (11) $\\alpha$ itself is maximised when $\\lVert u\\rVert^{2}=0$.  \nCombining both facts shows\n\\[\n\\Phi(P)\\le \\alpha^{6}\\le \\alpha_{0}^{6},\\qquad\n\\alpha_{0}:=\\sqrt{\\frac{1+5\\lambda}{6}}, \\tag{14}\n\\]\nwith equality throughout if and only if $u=0$.  Hence every maximiser of $\\Phi$ (and therefore every minimiser of $V$) must satisfy\n\\[\nP=t\\boldsymbol 1\\quad\\text{with}\\quad u=0. \\tag{15}\n\\]\n\nStep 6.  Determining $t$  \nInsert $P=t\\boldsymbol 1$ into the constraint $P^{\\mathsf T}AP=1$:\n\\[\n1=6t^{2}+2\\lambda\\cdot 15t^{2}=6(1+5\\lambda)t^{2}\n\\quad\\Longrightarrow\\quad\nt=\\pm\\frac{1}{\\sqrt{6(1+5\\lambda)}}. \\tag{16}\n\\]\nStep (a) demands $(AP)_{i}>0$; by (7) this reads $(1+5\\lambda)t>0$, hence $t>0$.  Therefore the unique feasible optimiser is\n\\[\nP_{*}=\\frac{\\boldsymbol 1}{\\sqrt{6(1+5\\lambda)}}. \\tag{17}\n\\]\n\nStep 7.  Minimal volume  \nUsing (4) at $P_{*}$ and $(AP_{*})_{i}=\\alpha_{0}$,\n\\[\nV_{\\min}= \\frac{1}{720}\\,\\alpha_{0}^{-6}\n        =\\frac{1}{720}\\left(\\frac{6}{1+5\\lambda}\\right)^{3}\n        =\\frac{216}{720}(1+5\\lambda)^{-3}\n        =\\frac{3}{10}\\,(1+5\\lambda)^{-3}. \\tag{18}\n\\]\n\n--------------------------------------------------------------------\nResult  \n\n\\[\n\\boxed{\\,V_{\\min}=\\dfrac{3}{10}\\,(1+5\\lambda)^{-3}\\,},\\qquad\n\\boxed{\\,P_{*}=\\dfrac{(1,\\dots ,1)}{\\sqrt{6(1+5\\lambda)}}\\,}.\n\\]\nThe volume $V(P)$ attains its minimum if and only if $P=P_{*}$; for all other feasible $P$ one has $V(P)>V_{\\min}$.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.404181",
        "was_fixed": false,
        "difficulty_analysis": "Compared with both the original three–dimensional problem and the current four–dimensional kernel variant, this enhancement is markedly tougher for several reasons.\n\n• Higher dimension: the problem is lifted to six dimensions, inflating the combinatorial and algebraic workload (e.g. 6! = 720 factors into the volume formula).\n\n• Non-diagonal quadric: the ellipsoid contains equal cross terms controlled by λ, so the gradient and tangent-plane intercepts are no longer simple coordinate scalings.  One must manipulate the full matrix A with off-diagonal entries, which forces the introduction of linear‐form products (A P)_i rather than single coordinates.\n\n• Symmetry yet non-orthogonality: while A is permutation-invariant, its non-orthogonal nature obliges a careful symmetry argument (or an appeal to convex optimisation theory) to justify that the extremum occurs at equal coordinates.  A naive arithmetic–geometric mean inequality does not suffice.\n\n• Advanced optimisation: maximising the product of six affine forms under a quadratic constraint demands either Lagrange multipliers in matrix language or a convexity/majorisation argument, both substantially more sophisticated than the scalar AM–GM step used in the original.\n\n• Final expression: obtaining an explicit closed-form minimum volume, (3/10)(1+5λ)^{−3}, requires tracking the dependence on λ through each stage, including the non-trivial evaluation of the quadratic constraint and the intercepts.\n\nThese layers of additional structure and proof complexity raise the problem well beyond “plug-and-play” techniques, satisfying all the stated guidelines for a significantly more difficult kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\lambda$ be a real number with $0<\\lambda<1$ and consider in $\\mathbb R^{6}$ the centred “correlated’’ ellipsoid  \n\\[\nE_{\\lambda}\\;:\\qquad \\sum_{i=1}^{6}x_{i}^{2}+2\\lambda\\!\\!\\sum_{1\\le i<j\\le 6}\\!x_{i}x_{j}=1. \\tag{$\\star$}\n\\]\nEquivalently $E_{\\lambda}$ is the quadric $x^{\\mathsf T}Ax=1$, where\n\\[\nA=(1-\\lambda)I_{6}+\\lambda J_{6},\\qquad\n0<\\lambda<1 ,\n\\]\n$I_{6}$ is the $6\\times 6$ identity matrix and $J_{6}$ the $6\\times 6$ all–ones matrix.\n\nFor a point $P=(p_{1},\\dots ,p_{6})\\in E_{\\lambda}$ let $\\Pi_{P}$ be the hyper-plane tangent to $E_{\\lambda}$ at $P$.  \nTogether with the six coordinate hyper-planes $x_{1}=0,\\dots ,x_{6}=0$, the plane $\\Pi_{P}$ encloses a $6$-simplex $S(P)$ contained in the first orthant whenever that simplex is bounded.  \nDenote its (six–dimensional) hyper-volume by $V(P)$.\n\n(a)  Prove that $S(P)$ is bounded if and only if  \n\\[\n(AP)_{i}>0\\qquad\\text{for all }i=1,\\dots ,6 , \\tag{1}\n\\]\ni.e. iff every axis–intercept $t_{i}=1/(AP)_{i}$ of $\\Pi_{P}$ is positive.  (No explicit sign condition on the $p_{i}$ is required.)\n\n(b)  Determine, in closed form,\n\\[\nV_{\\min}= \\min_{\\substack{P\\in E_{\\lambda}\\\\ (AP)_{i}>0}} V(P), \\tag{2}\n\\]\nand describe precisely the set of all points $P$ at which that minimum is attained.  Express your answer solely in terms of $\\lambda$.",
      "solution": "Throughout let $\\boldsymbol 1=(1,\\dots ,1)^{\\mathsf T}\\in\\mathbb R^{6}$ and set $\\Sigma:=\\sum_{j=1}^{6}p_{j}$ for brevity.\n\n--------------------------------------------------------------------\nPart (a)  (boundedness criterion)\n\nThe tangent hyper-plane at $P\\in E_{\\lambda}$ is obtained from the gradient of the quadratic form $F(x)=x^{\\mathsf T}Ax$:\n\\[\n\\nabla F(P)=2AP\\quad\\Longrightarrow\\quad (AP)\\cdot(x-P)=0.\n\\]\nBecause $P^{\\mathsf T}AP=1$, this rewrites as\n\\[\n(AP)\\cdot x=1. \\tag{3}\n\\]\n\nIntercept with the $x_{i}$-axis: insert $x=t\\,e_{i}$ in (3) to get $t_{i}=1/(AP)_{i}$.  \nIf some $(AP)_{i}\\le 0$, then either $t_{i}\\le 0$ or $t_{i}=\\infty$; in both cases $\\Pi_{P}$ fails to intersect the positive $x_{i}$-axis at a positive finite point, whence the coordinate hyper-planes cannot cut off a bounded region in the first orthant.  \nConversely, if every $(AP)_{i}>0$ then each $t_{i}$ is positive and finite, and the six coordinate hyper-planes together with $\\Pi_{P}$ bound the right $6$-simplex\n\\[\nS(P)=\\bigl\\{(x_{1},\\dots ,x_{6})\\;|\\;x_{i}\\ge 0,\\;\\; (AP)\\cdot x\\le 1\\bigr\\}\n\\]\nwithin the first orthant.  Hence (1) is both necessary and sufficient for $S(P)$ to be bounded. $\\square$\n\n--------------------------------------------------------------------\nPart (b)  (minimisation of the volume)\n\nStep 1.  Volume in terms of $AP$  \nBecause $S(P)$ is a right $6$-simplex whose edges along the positive coordinate axes have lengths $t_{i}=1/(AP)_{i}$, its hyper-volume is\n\\[\nV(P)=\\frac{1}{6!}\\prod_{i=1}^{6}\\frac{1}{(AP)_{i}}=\\frac{1}{720}\\prod_{i=1}^{6}(AP)_{i}^{\\,-1}. \\tag{4}\n\\]\nThus minimising $V(P)$ is equivalent to maximising\n\\[\n\\Phi(P):=\\prod_{i=1}^{6}(AP)_{i} \\tag{5}\n\\]\nunder the constraints $P\\in E_{\\lambda}$ and $(AP)_{i}>0$.\n\nStep 2.  A convenient expression for $(AP)_{i}$  \nSince $A$ has $1$ on the diagonal and $\\lambda$ off the diagonal,\n\\[\n(AP)_{i}=p_{i}+\\lambda\\!\\!\\sum_{j\\ne i}\\!p_{j}=(1-\\lambda)p_{i}+\\lambda\\Sigma. \\tag{6}\n\\]\n\nStep 3.  Parameterisation by mean and deviation  \nWrite $P=t\\boldsymbol 1+u$, where $u=(u_{1},\\dots ,u_{6})$ satisfies $\\boldsymbol 1^{\\mathsf T}u=0$.  \nThen $\\Sigma=6t$ and (6) becomes\n\\[\n(AP)_{i}=t(1+5\\lambda)+(1-\\lambda)u_{i}. \\tag{7}\n\\]\nSet\n\\[\n\\alpha:=t(1+5\\lambda),\\qquad \\beta:=1-\\lambda>0 .\n\\]\nHence\n\\[\n(AP)_{i}= \\alpha+\\beta u_{i}. \\tag{8}\n\\]\n\nStep 4.  The ellipsoid constraint in the new variables  \nBecause $A=(1-\\lambda)I_{6}+\\lambda J_{6}$ is positive definite,\n\\[\n1=P^{\\mathsf T}AP=(t\\boldsymbol 1+u)^{\\mathsf T}A(t\\boldsymbol 1+u)\n        =6(1+5\\lambda)t^{2}+\\beta\\,\\lVert u\\rVert^{2}. \\tag{9}\n\\]\nSolving (9) for $t^{2}$ gives\n\\[\nt^{2}=\\frac{1-\\beta\\lVert u\\rVert^{2}}{6(1+5\\lambda)}\\quad\\text{with}\\quad\n0\\le\\beta\\lVert u\\rVert^{2}\\le 1. \\tag{10}\n\\]\nConsequently\n\\[\n\\alpha=\\sqrt{\\frac{1+5\\lambda}{6}}\\,\n       \\sqrt{1-\\beta\\lVert u\\rVert^{2}}, \\quad 0\\le\\beta\\lVert u\\rVert^{2}\\le 1. \\tag{11}\n\\]\nObserve that $\\alpha$ is a strictly decreasing function of $\\lVert u\\rVert^{2}$.\n\nStep 5.  Maximising $\\Phi$  \n\nWe have\n\\[\n\\Phi(P)=\\prod_{i=1}^{6}\\bigl(\\alpha+\\beta u_{i}\\bigr). \\tag{12}\n\\]\nBecause $\\sum_{i=1}^{6}u_{i}=0$, the numbers $\\alpha+\\beta u_{i}$ share the arithmetic mean $\\alpha$.  The AM–GM inequality yields\n\\[\n\\Phi(P)\\le \\alpha^{6}; \\tag{13}\n\\]\nequality in (13) occurs if and only if every $u_{i}=0$.  \nBut by (11) $\\alpha$ itself is maximised when $\\lVert u\\rVert^{2}=0$.  \nCombining both facts shows\n\\[\n\\Phi(P)\\le \\alpha^{6}\\le \\alpha_{0}^{6},\\qquad\n\\alpha_{0}:=\\sqrt{\\frac{1+5\\lambda}{6}}, \\tag{14}\n\\]\nwith equality throughout if and only if $u=0$.  Hence every maximiser of $\\Phi$ (and therefore every minimiser of $V$) must satisfy\n\\[\nP=t\\boldsymbol 1\\quad\\text{with}\\quad u=0. \\tag{15}\n\\]\n\nStep 6.  Determining $t$  \nInsert $P=t\\boldsymbol 1$ into the constraint $P^{\\mathsf T}AP=1$:\n\\[\n1=6t^{2}+2\\lambda\\cdot 15t^{2}=6(1+5\\lambda)t^{2}\n\\quad\\Longrightarrow\\quad\nt=\\pm\\frac{1}{\\sqrt{6(1+5\\lambda)}}. \\tag{16}\n\\]\nStep (a) demands $(AP)_{i}>0$; by (7) this reads $(1+5\\lambda)t>0$, hence $t>0$.  Therefore the unique feasible optimiser is\n\\[\nP_{*}=\\frac{\\boldsymbol 1}{\\sqrt{6(1+5\\lambda)}}. \\tag{17}\n\\]\n\nStep 7.  Minimal volume  \nUsing (4) at $P_{*}$ and $(AP_{*})_{i}=\\alpha_{0}$,\n\\[\nV_{\\min}= \\frac{1}{720}\\,\\alpha_{0}^{-6}\n        =\\frac{1}{720}\\left(\\frac{6}{1+5\\lambda}\\right)^{3}\n        =\\frac{216}{720}(1+5\\lambda)^{-3}\n        =\\frac{3}{10}\\,(1+5\\lambda)^{-3}. \\tag{18}\n\\]\n\n--------------------------------------------------------------------\nResult  \n\n\\[\n\\boxed{\\,V_{\\min}=\\dfrac{3}{10}\\,(1+5\\lambda)^{-3}\\,},\\qquad\n\\boxed{\\,P_{*}=\\dfrac{(1,\\dots ,1)}{\\sqrt{6(1+5\\lambda)}}\\,}.\n\\]\nThe volume $V(P)$ attains its minimum if and only if $P=P_{*}$; for all other feasible $P$ one has $V(P)>V_{\\min}$.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.345804",
        "was_fixed": false,
        "difficulty_analysis": "Compared with both the original three–dimensional problem and the current four–dimensional kernel variant, this enhancement is markedly tougher for several reasons.\n\n• Higher dimension: the problem is lifted to six dimensions, inflating the combinatorial and algebraic workload (e.g. 6! = 720 factors into the volume formula).\n\n• Non-diagonal quadric: the ellipsoid contains equal cross terms controlled by λ, so the gradient and tangent-plane intercepts are no longer simple coordinate scalings.  One must manipulate the full matrix A with off-diagonal entries, which forces the introduction of linear‐form products (A P)_i rather than single coordinates.\n\n• Symmetry yet non-orthogonality: while A is permutation-invariant, its non-orthogonal nature obliges a careful symmetry argument (or an appeal to convex optimisation theory) to justify that the extremum occurs at equal coordinates.  A naive arithmetic–geometric mean inequality does not suffice.\n\n• Advanced optimisation: maximising the product of six affine forms under a quadratic constraint demands either Lagrange multipliers in matrix language or a convexity/majorisation argument, both substantially more sophisticated than the scalar AM–GM step used in the original.\n\n• Final expression: obtaining an explicit closed-form minimum volume, (3/10)(1+5λ)^{−3}, requires tracking the dependence on λ through each stage, including the non-trivial evaluation of the quadratic constraint and the intercepts.\n\nThese layers of additional structure and proof complexity raise the problem well beyond “plug-and-play” techniques, satisfying all the stated guidelines for a significantly more difficult kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}