{
  "index": "1940-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG",
    "ANA"
  ],
  "difficulty": "",
  "question": "12. Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface\n\\[\na x^{2}+b y^{2}+c z^{2}=1 \\quad(a b c \\neq 0)\n\\]\nis the sphere\n\\[\nx^{2}+y^{2}+z^{2}=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\n\\]",
  "solution": "We first determine the restrictions on the coefficients that ensure the plane  \n\\[\n\\alpha x+\\beta y+\\gamma z=\\delta \\tag{3}\n\\]\nis tangent to the quadric surface\n\\[\nQ:\\quad \\frac{x^{2}}{a}+\\frac{y^{2}}{b}+\\frac{z^{2}}{c}=1\\tag{1}\n\\]\n(the explicit form of~\\(Q\\) is used later).\n\n\\medskip\n\\noindent\nThe tangent plane to \\(Q\\) at the point \\((x_{1},y_{1},z_{1})\\) is  \n\\[\n\\alpha x_{1}x+\\beta y_{1}y+\\gamma z_{1}z=1. \\tag{4}\n\\]\nIf~(3) and~(4) coincide, then \\(\\delta\\neq0\\) and  \n\\[\nx_{1}=\\frac{\\alpha}{a\\delta},\\qquad \ny_{1}=\\frac{\\beta}{b\\delta},\\qquad \nz_{1}=\\frac{\\gamma}{c\\delta}. \\tag{5}\n\\]\nBecause the point \\((x_{1},y_{1},z_{1})\\) lies on \\(Q\\) we must have  \n\\[\n\\frac1{\\delta^{2}}\\Bigl(\\frac{\\alpha^{2}}{a}+\\frac{\\beta^{2}}{b}+\\frac{\\gamma^{2}}{c}\\Bigr)=1,\n\\]\nhence  \n\\[\n\\boxed{\\;\n\\frac{\\alpha^{2}}{a}+\\frac{\\beta^{2}}{b}+\\frac{\\gamma^{2}}{c}=\\delta^{2}\\;}. \\tag{6}\n\\]\nThus~(6) with \\(\\delta\\neq0\\) is \\emph{necessary and sufficient} for the plane~(3) to be tangent to \\(Q\\).\n\nIf~(6) holds with \\(\\delta=0\\) (assume \\(\\alpha,\\beta,\\gamma\\) not all zero) the plane~(3) is \\emph{asymptotic} to \\(Q\\).\nIn projective language,~(6) says that~(3) is \\emph{projectively} tangent to \\(Q\\);  \nthe point of tangency has homogeneous coordinates \\((\\alpha/a,\\beta/b,\\gamma/c,\\delta)\\).\nFor \\(\\delta=0\\) this is a point at infinity on \\(Q\\).\n\n\\bigskip\n\\noindent\nNow choose three mutually orthogonal unit vectors\n\\[\nu_{i}=(\\alpha_{i},\\beta_{i},\\gamma_{i}),\\qquad i=1,2,3,\n\\]\nso that the matrix  \n\\[\n\\begin{pmatrix}\n\\alpha_{1}&\\beta_{1}&\\gamma_{1}\\\\\n\\alpha_{2}&\\beta_{2}&\\gamma_{2}\\\\\n\\alpha_{3}&\\beta_{3}&\\gamma_{3}\n\\end{pmatrix}\\tag{7}\n\\]\nis orthogonal.  Consequently\n\\[\n\\sum_{i=1}^{3}\\alpha_{i}^{2}=\\sum_{i=1}^{3}\\beta_{i}^{2}=\\sum_{i=1}^{3}\\gamma_{i}^{2}=1.\n\\]\n\n\\medskip\\noindent\nIf planes having these vectors as normals are tangent to \\(Q\\) then they are pairwise perpendicular and have equations  \n\\[\n\\alpha_{i}x+\\beta_{i}y+\\gamma_{i}z=\\delta_{i},\\qquad i=1,2,3, \\tag{8}\n\\]\nwhere  \n\\[\n\\frac{\\alpha_{i}^{2}}{a}+\\frac{\\beta_{i}^{2}}{b}+\\frac{\\gamma_{i}^{2}}{c}=\\delta_{i}^{2}.\n\\]\nSince \\(|\\delta_{i}|\\) equals the distance from the origin \\(O\\) to the \\(i\\)-th plane, the\nPythagorean theorem yields  \n\\[\nOP^{2}=\\delta_{1}^{2}+\\delta_{2}^{2}+\\delta_{3}^{2}\n       =\\frac1a+\\frac1b+\\frac1c.\n\\]\nTherefore the intersection point \\(P\\) of the three mutually perpendicular tangent planes\nlies on the sphere\n\\[\n\\boxed{\\;x^{2}+y^{2}+z^{2}=\\frac1a+\\frac1b+\\frac1c\\;} \\tag{2}\n\\]\ncalled the \\emph{director sphere} of \\(Q\\).\n(The same argument, with only minor changes, works in any dimension, giving\na director circle for a conic and higher–dimensional ``director spheres'' for central quadrics.)\n\n\\bigskip\n\\noindent\\textbf{The converse problem.}\nDoes every point of the director sphere arise from a triple of\nperpendicular tangent planes?\nTo prepare for this we need an algebraic lemma.\n\n\\paragraph{Theorem.}\n\\emph{Let \\(M\\) be an \\(n\\times n\\) real matrix.\nThere exist \\(n\\) mutually orthogonal unit vectors \\(u_{1},\\dots,u_{n}\\in\\Bbb R^{n}\\)\nwith \\(u_{i}^{T}Mu_{i}=0\\;(i=1,\\dots,n)\\) if and only if \\(\\operatorname{tr}M=0\\).}\n\n\\emph{Proof.}\nInterpret \\(M\\) as the matrix of the quadratic form \\(F(x)=x^{T}Mx\\).\nThe quantity \\(\\operatorname{tr}F\\) (i.e.\\ the sum \\(F(v_{1})+\\cdots+F(v_{n})\\) for an orthonormal basis)  \nis basis–independent and additive on orthogonal direct sums.\nIf \\(\\operatorname{tr}F=0\\) and \\(\\dim V>0\\) we can pick a unit vector \\(u_{1}\\) with \\(F(u_{1})=0\\),\nthen iterate inside the orthogonal complement until the basis is complete.\nConversely, if such a basis exists, summing \\(F(u_{i})=0\\) gives \\(\\operatorname{tr}M=0\\). \\(\\square\\)\n\n\\bigskip\n\\noindent\nNow let \\(P=(r,s,t)\\) be any point on the director sphere.\nThrough \\(P\\) consider planes whose normals are three mutually orthogonal unit vectors\n\\(u_{i}=(\\alpha_{i},\\beta_{i},\\gamma_{i})\\):\n\\[\n\\alpha_{i}x+\\beta_{i}y+\\gamma_{i}z=\\alpha_{i}r+\\beta_{i}s+\\gamma_{i}t,\\qquad i=1,2,3.\n\\]\nThese planes are (projectively) tangent to \\(Q\\) precisely when  \n\\[\n\\frac{\\alpha_{i}^{2}}{a}+\\frac{\\beta_{i}^{2}}{b}+\\frac{\\gamma_{i}^{2}}{c}-\n(\\alpha_{i}r+\\beta_{i}s+\\gamma_{i}t)^{2}=0,\\qquad i=1,2,3.\\tag{9}\n\\]\nLet\n\\[\nM=\n\\begin{pmatrix}\n\\frac1a-r^{2} & -rs & -rt\\\\[4pt]\n-rs & \\frac1b-s^{2} & -st\\\\[4pt]\n-rt & -st & \\frac1c-t^{2}\n\\end{pmatrix},\n\\]\nso that~(9) is \\(u_{i}^{T}Mu_{i}=0\\).\nBy the theorem such an orthonormal triple exists \\emph{iff}\n\\[\n\\operatorname{tr}M=\\frac1a+\\frac1b+\\frac1c-r^{2}-s^{2}-t^{2}=0,\n\\]\nwhich is exactly the equation of the director sphere.\nHence \\emph{every} point of the sphere is the intersection of three mutually perpendicular planes projectively tangent to \\(Q\\).\n\n\\bigskip\n\\noindent\\textbf{Proper versus asymptotic tangency.}\nIf \\((1/a)+(1/b)+(1/c)=0\\) the planes can never be \\emph{properly} tangent; assume henceforth  \n\\((1/a)+(1/b)+(1/c)>0\\).\nThrough a given \\(P\\in S\\) we first take any plane \\(\\pi_{1}\\) projectively tangent to \\(Q\\).\nIts \\emph{polar conic} (intersection with \\(Q\\)) determines the remaining two planes \\(\\pi_{2},\\pi_{3}\\)\nperpendicular to \\(\\pi_{1}\\) and through \\(P\\); these form a unique mutually perpendicular triple.\nThere are infinitely many such triples, at most one of which involves an asymptotic plane,\nso in fact there are infinitely many \\emph{properly} tangent triples.\n\n\\bigskip\n\\noindent\\textbf{Exceptional points when \\(Q\\) is a hyperboloid of one sheet.}\nAssume \\(a>0>b>c\\).\nWriting out where the normal through \\(P\\) passes through the origin shows\nthat the exceptional set (where no proper triple exists) is\n\n\\[\n\\begin{cases}\n(\\pm a^{-1/2},0,0), & a\\neq b=-c,\\\\[4pt]\n(0,\\pm b^{-1/2},0), & b\\neq a=-c,\\\\[4pt]\n\\text{a circle in the $xy$--plane}, & a=b=-c,\\\\\n\\text{empty}, & \\text{otherwise}.\n\\end{cases}\n\\]\n\n\\bigskip\n\\noindent\\textbf{A related problem.}\nAny ellipsoid or hyperboloid can be put in the form~(1) by translation and rotation,\nso the problem above is solved for all of them.\nFor the paraboloid in canonical form  \n\\[\nz=ax^{2}+by^{2},\\qquad ab\\neq0,\n\\]\nshow that the locus of intersection of three mutually orthogonal tangent planes is the plane  \n\\[\n\\boxed{\\,z=-\\dfrac14\\Bigl(\\dfrac1a+\\dfrac1b\\Bigr).\\,}\n\\]\n\n\\bigskip\n\\noindent\\textbf{Remark.}\nIf we restrict to \\emph{rational} coefficients and points,\nthe first (``director sphere'') part remains valid but the converse can fail.\nFor instance, with \\(Q:x^{2}+y^{2}+\\tfrac12z^{2}=1\\) the rational point \\(P=(0,0,2)\\in S\\)\nadmits no triple of mutually perpendicular \\emph{rational} tangent planes.\nThe missing ingredient in the converse is a property of the real field not shared by the rationals.",
  "vars": [
    "x",
    "y",
    "z",
    "x_1",
    "y_1",
    "z_1",
    "r",
    "s",
    "t",
    "i",
    "n",
    "M",
    "F",
    "V",
    "P",
    "O",
    "Q",
    "S",
    "u_i",
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma",
    "\\\\delta",
    "\\\\alpha_i",
    "\\\\beta_i",
    "\\\\gamma_i",
    "\\\\delta_i",
    "\\\\pi",
    "\\\\pi_1",
    "\\\\pi_2",
    "\\\\pi_3"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "z": "applicate",
        "x_1": "abscissaone",
        "y_1": "ordinateone",
        "z_1": "applicateone",
        "r": "firstcoord",
        "s": "secondcoord",
        "t": "thirdcoord",
        "i": "indexvar",
        "n": "dimension",
        "M": "matrixm",
        "F": "quadform",
        "V": "vectorspace",
        "P": "pointp",
        "O": "originpt",
        "Q": "quadric",
        "S": "sphere",
        "u_i": "unitvectorindex",
        "\\alpha": "alphacoeff",
        "\\beta": "betacoeff",
        "\\gamma": "gammacoeff",
        "\\delta": "deltacoeff",
        "\\alpha_i": "alphacoeffi",
        "\\beta_i": "betacoeffi",
        "\\gamma_i": "gammacoeffi",
        "\\delta_i": "deltacoeffi",
        "\\pi": "planevar",
        "\\pi_1": "planeone",
        "\\pi_2": "planetwo",
        "\\pi_3": "planethree",
        "a": "coeffa",
        "b": "coeffb",
        "c": "coeffc"
      },
      "question": "12. Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface\n\\[\ncoeffa \\, abscissa^{2}+coeffb \\, ordinate^{2}+coeffc \\, applicate^{2}=1 \\quad(coeffa\\, coeffb\\, coeffc \\neq 0)\n\\]\nis the sphere\n\\[\nabscissa^{2}+ordinate^{2}+applicate^{2}=\\frac{1}{coeffa}+\\frac{1}{coeffb}+\\frac{1}{coeffc}\n\\]",
      "solution": "We first determine the restrictions on the coefficients that ensure the plane  \n\\[\nalphacoeff \\, abscissa + betacoeff \\, ordinate + gammacoeff \\, applicate = deltacoeff \\tag{3}\n\\]\nis tangent to the quadric surface\n\\[\nquadric:\\quad \\frac{abscissa^{2}}{coeffa}+\\frac{ordinate^{2}}{coeffb}+\\frac{applicate^{2}}{coeffc}=1\\tag{1}\n\\]\n(the explicit form of~\\(quadric\\) is used later).\n\n\\medskip\n\\noindent\nThe tangent plane to \\(quadric\\) at the point \\((abscissaone, ordinateone, applicateone)\\) is  \n\\[\nalphacoeff \\, abscissaone \\, abscissa + betacoeff \\, ordinateone \\, ordinate + gammacoeff \\, applicateone \\, applicate = 1. \\tag{4}\n\\]\nIf~(3) and~(4) coincide, then \\(deltacoeff\\neq0\\) and  \n\\[\nabscissaone=\\frac{alphacoeff}{coeffa \\, deltacoeff},\\qquad \nordinateone=\\frac{betacoeff}{coeffb \\, deltacoeff},\\qquad \napplicateone=\\frac{gammacoeff}{coeffc \\, deltacoeff}. \\tag{5}\n\\]\nBecause the point \\((abscissaone, ordinateone, applicateone)\\) lies on \\(quadric\\) we must have  \n\\[\n\\frac1{deltacoeff^{2}}\\Bigl(\\frac{alphacoeff^{2}}{coeffa}+\\frac{betacoeff^{2}}{coeffb}+\\frac{gammacoeff^{2}}{coeffc}\\Bigr)=1,\n\\]\nhence  \n\\[\n\\boxed{\\;\\frac{alphacoeff^{2}}{coeffa}+\\frac{betacoeff^{2}}{coeffb}+\\frac{gammacoeff^{2}}{coeffc}=deltacoeff^{2}\\;}\\tag{6}\n\\]\nThus~(6) with \\(deltacoeff\\neq0\\) is \\emph{necessary and sufficient} for the plane~(3) to be tangent to \\(quadric\\).\n\nIf~(6) holds with \\(deltacoeff=0\\) (assume \\(alphacoeff,betacoeff,gammacoeff\\) not all zero) the plane~(3) is \\emph{asymptotic} to \\(quadric\\).\nIn projective language,~(6) says that~(3) is \\emph{projectively} tangent to \\(quadric\\); the point of tangency has homogeneous coordinates \\((alphacoeff/coeffa,betacoeff/coeffb,gammacoeff/coeffc,deltacoeff)\\).\nFor \\(deltacoeff=0\\) this is a point at infinity on \\(quadric\\).\n\n\\bigskip\n\\noindent\nNow choose three mutually orthogonal unit vectors\n\\[\nunitvectorindex=(alphacoeffi,betacoeffi,gammacoeffi),\\qquad indexvar=1,2,3,\n\\]\nso that the matrix  \n\\[\n\\begin{pmatrix}\nalphacoeff_{1}&betacoeff_{1}&gammacoeff_{1}\\\\\nalphacoeff_{2}&betacoeff_{2}&gammacoeff_{2}\\\\\nalphacoeff_{3}&betacoeff_{3}&gammacoeff_{3}\n\\end{pmatrix}\\tag{7}\n\\]\nis orthogonal.  Consequently\n\\[\n\\sum_{indexvar=1}^{3}alphacoeff_{indexvar}^{2}=\\sum_{indexvar=1}^{3}betacoeff_{indexvar}^{2}=\\sum_{indexvar=1}^{3}gammacoeff_{indexvar}^{2}=1.\n\\]\n\n\\medskip\\noindent\nIf planes having these vectors as normals are tangent to \\(quadric\\) then they are pairwise perpendicular and have equations  \n\\[\nalphacoeffi \\, abscissa + betacoeffi \\, ordinate + gammacoeffi \\, applicate = deltacoeffi,\\qquad indexvar=1,2,3, \\tag{8}\n\\]\nwhere  \n\\[\n\\frac{alphacoeffi^{2}}{coeffa}+\\frac{betacoeffi^{2}}{coeffb}+\\frac{gammacoeffi^{2}}{coeffc}=deltacoeffi^{2}.\n\\]\nSince \\(|deltacoeffi|\\) equals the distance from the origin \\(originpt\\) to the \\(indexvar\\)-th plane, the Pythagorean theorem yields  \n\\[\noriginptpointp^{2}=deltacoeff_{1}^{2}+deltacoeff_{2}^{2}+deltacoeff_{3}^{2}=\\frac1{coeffa}+\\frac1{coeffb}+\\frac1{coeffc}.\n\\]\nTherefore the intersection point \\(pointp\\) of the three mutually perpendicular tangent planes lies on the sphere\n\\[\n\\boxed{\\;abscissa^{2}+ordinate^{2}+applicate^{2}=\\frac1{coeffa}+\\frac1{coeffb}+\\frac1{coeffc}\\;}\\tag{2}\n\\]\ncalled the \\emph{director sphere} of \\(quadric\\).\n(The same argument, with only minor changes, works in any dimension, giving a director circle for a conic and higher–dimensional \"director spheres\" for central quadrics.)\n\n\\bigskip\n\\noindent\\textbf{The converse problem.} Does every point of the director sphere arise from a triple of perpendicular tangent planes? To prepare for this we need an algebraic lemma.\n\n\\paragraph{Theorem.} \\emph{Let \\(matrixm\\) be a \\(dimension\\times dimension\\) real matrix. There exist \\(dimension\\) mutually orthogonal unit vectors \\(unitvectorindex_{1},\\dots,unitvectorindex_{dimension}\\in\\Bbb R^{dimension}\\) with \\(unitvectorindex_{indexvar}^{T}matrixm\\,unitvectorindex_{indexvar}=0\\;(indexvar=1,\\dots,dimension)\\) if and only if \\(\\operatorname{tr}matrixm=0\\).}\n\n\\emph{Proof.} Interpret \\(matrixm\\) as the matrix of the quadratic form \\(quadform(abscissa)=abscissa^{T}matrixm\\,abscissa\\). The quantity \\(\\operatorname{tr}quadform\\) (i.e. the sum \\(quadform(v_{1})+\\cdots+quadform(v_{dimension})\\) for an orthonormal basis) is basis–independent and additive on orthogonal direct sums. If \\(\\operatorname{tr}quadform=0\\) and \\(\\dim vectorspace>0\\) we can pick a unit vector \\(unitvectorindex_{1}\\) with \\(quadform(unitvectorindex_{1})=0\\), then iterate inside the orthogonal complement until the basis is complete. Conversely, if such a basis exists, summing \\(quadform(unitvectorindex_{indexvar})=0\\) gives \\(\\operatorname{tr}matrixm=0\\). \\(\\square\\)\n\n\\bigskip\n\\noindent\nNow let \\(pointp=(firstcoord,secondcoord,thirdcoord)\\) be any point on the director sphere. Through \\(pointp\\) consider planes whose normals are three mutually orthogonal unit vectors \\(unitvectorindex=(alphacoeffi,betacoeffi,gammacoeffi)\\):\n\\[\nalphacoeffi \\, abscissa + betacoeffi \\, ordinate + gammacoeffi \\, applicate = alphacoeffi firstcoord + betacoeffi secondcoord + gammacoeffi thirdcoord,\\qquad indexvar=1,2,3.\n\\]\nThese planes are (projectively) tangent to \\(quadric\\) precisely when  \n\\[\n\\frac{alphacoeffi^{2}}{coeffa}+\\frac{betacoeffi^{2}}{coeffb}+\\frac{gammacoeffi^{2}}{coeffc}-(alphacoeffi firstcoord + betacoeffi secondcoord + gammacoeffi thirdcoord)^{2}=0,\\qquad indexvar=1,2,3.\\tag{9}\n\\]\nLet\n\\[\nmatrixm=\\begin{pmatrix}\n\\frac1{coeffa}-firstcoord^{2} & -firstcoord\\,secondcoord & -firstcoord\\,thirdcoord\\\\[4pt]\n-firstcoord\\,secondcoord & \\frac1{coeffb}-secondcoord^{2} & -secondcoord\\,thirdcoord\\\\[4pt]\n-firstcoord\\,thirdcoord & -secondcoord\\,thirdcoord & \\frac1{coeffc}-thirdcoord^{2}\n\\end{pmatrix},\n\\]\nso that~(9) is \\(unitvectorindex^{T}matrixm\\,unitvectorindex=0\\). By the theorem such an orthonormal triple exists \\emph{iff}\n\\[\n\\operatorname{tr}matrixm=\\frac1{coeffa}+\\frac1{coeffb}+\\frac1{coeffc}-firstcoord^{2}-secondcoord^{2}-thirdcoord^{2}=0,\n\\]\nwhich is exactly the equation of the director sphere. Hence \\emph{every} point of the sphere is the intersection of three mutually perpendicular planes projectively tangent to \\(quadric\\).\n\n\\bigskip\n\\noindent\\textbf{Proper versus asymptotic tangency.} If \\((1/coeffa)+(1/coeffb)+(1/coeffc)=0\\) the planes can never be \\emph{properly} tangent; assume henceforth \\((1/coeffa)+(1/coeffb)+(1/coeffc)>0\\). Through a given \\(pointp\\in sphere\\) we first take any plane \\(planeone\\) projectively tangent to \\(quadric\\). Its \\emph{polar conic} (intersection with \\(quadric\\)) determines the remaining two planes \\(planetwo,planethree\\) perpendicular to \\(planeone\\) and through \\(pointp\\); these form a unique mutually perpendicular triple. There are infinitely many such triples, at most one of which involves an asymptotic plane, so in fact there are infinitely many \\emph{properly} tangent triples.\n\n\\bigskip\n\\noindent\\textbf{Exceptional points when \\(quadric\\) is a hyperboloid of one sheet.} Assume \\(coeffa>0>coeffb>coeffc\\). Writing out where the normal through \\(pointp\\) passes through the origin shows that the exceptional set (where no proper triple exists) is\n\\[\n\\begin{cases}\n(\\pm coeffa^{-1/2},0,0), & coeffa\\neq coeffb=-coeffc,\\\\[4pt]\n(0,\\pm coeffb^{-1/2},0), & coeffb\\neq coeffa=-coeffc,\\\\[4pt]\n\\text{a circle in the $xy$--plane}, & coeffa=coeffb=-coeffc,\\\\\n\\text{empty}, & \\text{otherwise}.\n\\end{cases}\n\\]\n\n\\bigskip\n\\noindent\\textbf{A related problem.} Any ellipsoid or hyperboloid can be put in the form~(1) by translation and rotation, so the problem above is solved for all of them. For the paraboloid in canonical form  \n\\[\napplicate=coeffa\\,abscissa^{2}+coeffb\\,ordinate^{2},\\qquad coeffa\\,coeffb\\neq0,\n\\]\nshow that the locus of intersection of three mutually orthogonal tangent planes is the plane  \n\\[\n\\boxed{\\,applicate=-\\dfrac14\\Bigl(\\dfrac1{coeffa}+\\dfrac1{coeffb}\\Bigr).\\,}\n\\]\n\n\\bigskip\n\\noindent\\textbf{Remark.} If we restrict to \\emph{rational} coefficients and points, the first (\"director sphere\") part remains valid but the converse can fail. For instance, with \\(quadric:abscissa^{2}+ordinate^{2}+\\tfrac12applicate^{2}=1\\) the rational point \\(pointp=(0,0,2)\\in sphere\\) admits no triple of mutually perpendicular \\emph{rational} tangent planes. The missing ingredient in the converse is a property of the real field not shared by the rationals."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "briefcase",
        "z": "compass",
        "x_1": "teacupset",
        "y_1": "lighthouse",
        "z_1": "rainstorm",
        "r": "avalanche",
        "s": "butterfly",
        "t": "marshland",
        "n": "bookshelf",
        "M": "blackboard",
        "F": "screwdriver",
        "V": "horseshoe",
        "P": "watermelon",
        "O": "sailboat",
        "Q": "newspaper",
        "S": "toothpaste",
        "u_i": "daydream",
        "\\alpha": "lemonade",
        "\\beta": "hairbrush",
        "\\gamma": "paintball",
        "\\delta": "gemstone",
        "\\alpha_i": "sailcloth",
        "\\beta_i": "autopilot",
        "\\gamma_i": "heartbreak",
        "\\delta_i": "flashlight",
        "\\pi_1": "courtship",
        "\\pi_2": "mothership",
        "\\pi_3": "stormfront",
        "a": "semaphore",
        "b": "playground",
        "c": "ventricle"
      },
      "question": "12. Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface\n\\[\nsemaphore\\, sunflower^{2}+playground\\, briefcase^{2}+ventricle\\, compass^{2}=1 \\quad(semaphore\\, playground\\, ventricle \\neq 0)\n\\]\nis the sphere\n\\[\nsunflower^{2}+briefcase^{2}+compass^{2}=\\frac{1}{semaphore}+\\frac{1}{playground}+\\frac{1}{ventricle}\n\\]\n",
      "solution": "We first determine the restrictions on the coefficients that ensure the plane  \n\\[\nlemonade\\, sunflower+hairbrush\\, briefcase+paintball\\, compass=gemstone \\tag{3}\n\\]\nis tangent to the quadric surface\n\\[\nnewspaper:\\quad \\frac{sunflower^{2}}{semaphore}+\\frac{briefcase^{2}}{playground}+\\frac{compass^{2}}{ventricle}=1\\tag{1}\n\\]\n(the explicit form of~\\(newspaper\\) is used later).\n\n\\medskip\n\\noindent\nThe tangent plane to \\(newspaper\\) at the point \\((teacupset,lighthouse,rainstorm)\\) is  \n\\[\nlemonade\\, teacupset\\, sunflower+hairbrush\\, lighthouse\\, briefcase+paintball\\, rainstorm\\, compass=1. \\tag{4}\n\\]\nIf~(3) and~(4) coincide, then \\(gemstone\\neq0\\) and  \n\\[\nteacupset=\\frac{lemonade}{semaphore gemstone},\\qquad \nlighthouse=\\frac{hairbrush}{playground gemstone},\\qquad \nrainstorm=\\frac{paintball}{ventricle gemstone}. \\tag{5}\n\\]\nBecause the point \\((teacupset,lighthouse,rainstorm)\\) lies on \\(newspaper\\) we must have  \n\\[\n\\frac1{gemstone^{2}}\\Bigl(\\frac{lemonade^{2}}{semaphore}+\\frac{hairbrush^{2}}{playground}+\\frac{paintball^{2}}{ventricle}\\Bigr)=1,\n\\]\nhence  \n\\[\n\\boxed{\\;\n\\frac{lemonade^{2}}{semaphore}+\\frac{hairbrush^{2}}{playground}+\\frac{paintball^{2}}{ventricle}=gemstone^{2}\\;} . \\tag{6}\n\\]\nThus~(6) with \\(gemstone\\neq0\\) is \\emph{necessary and sufficient} for the plane~(3) to be tangent to \\(newspaper\\).\n\nIf~(6) holds with \\(gemstone=0\\) (assume \\(lemonade,hairbrush,paintball\\) not all zero) the plane~(3) is \\emph{asymptotic} to \\(newspaper\\).\nIn projective language,~(6) says that~(3) is \\emph{projectively} tangent to \\(newspaper\\);  \nthe point of tangency has homogeneous coordinates \\((lemonade/semaphore, hairbrush/playground, paintball/ventricle, gemstone)\\).\nFor \\(gemstone=0\\) this is a point at infinity on \\(newspaper\\).\n\n\\bigskip\n\\noindent\nNow choose three mutually orthogonal unit vectors\n\\[\ndaydream_{i}=(sailcloth_{i},autopilot_{i},heartbreak_{i}),\\qquad i=1,2,3,\n\\]\nso that the matrix  \n\\[\n\\begin{pmatrix}\nsailcloth_{1}&autopilot_{1}&heartbreak_{1}\\\\\nsailcloth_{2}&autopilot_{2}&heartbreak_{2}\\\\\nsailcloth_{3}&autopilot_{3}&heartbreak_{3}\n\\end{pmatrix}\\tag{7}\n\\]\nis orthogonal.  Consequently\n\\[\n\\sum_{i=1}^{3}sailcloth_{i}^{2}=\\sum_{i=1}^{3}autopilot_{i}^{2}=\\sum_{i=1}^{3}heartbreak_{i}^{2}=1.\n\\]\n\n\\medskip\\noindent\nIf planes having these vectors as normals are tangent to \\(newspaper\\) then they are pairwise perpendicular and have equations  \n\\[\nsailcloth_{i}sunflower+autopilot_{i}briefcase+heartbreak_{i}compass=flashlight_{i},\\qquad i=1,2,3, \\tag{8}\n\\]\nwhere  \n\\[\n\\frac{sailcloth_{i}^{2}}{semaphore}+\\frac{autopilot_{i}^{2}}{playground}+\\frac{heartbreak_{i}^{2}}{ventricle}=flashlight_{i}^{2}.\n\\]\nSince \\(|flashlight_{i}|\\) equals the distance from the origin \\(sailboat\\) to the \\(i\\)-th plane, the\nPythagorean theorem yields  \n\\[\nsailboat\\, watermelon^{2}=flashlight_{1}^{2}+flashlight_{2}^{2}+flashlight_{3}^{2}\n       =\\frac1{semaphore}+\\frac1{playground}+\\frac1{ventricle}.\n\\]\nTherefore the intersection point \\(watermelon\\) of the three mutually perpendicular tangent planes\nlies on the sphere\n\\[\n\\boxed{\\;sunflower^{2}+briefcase^{2}+compass^{2}=\\frac1{semaphore}+\\frac1{playground}+\\frac1{ventricle}\\;} \\tag{2}\n\\]\ncalled the \\emph{director sphere} of \\(newspaper\\).\n(The same argument, with only minor changes, works in any dimension, giving\na director circle for a conic and higher–dimensional ``director spheres'' for central quadrics.)\n\n\\bigskip\n\\noindent\\textbf{The converse problem.}\nDoes every point of the director sphere arise from a triple of\nperpendicular tangent planes?\nTo prepare for this we need an algebraic lemma.\n\n\\paragraph{Theorem.}\n\\emph{Let \\(blackboard\\) be an \\(bookshelf\\times bookshelf\\) real matrix.\nThere exist \\(bookshelf\\) mutually orthogonal unit vectors \\(daydream_{1},\\dots,daydream_{bookshelf}\\in\\Bbb R^{bookshelf}\\)\nwith \\(daydream_{i}^{T}blackboard\\, daydream_{i}=0\\;(i=1,\\dots,bookshelf)\\) if and only if \\(\\operatorname{tr}blackboard=0\\).}\n\n\\emph{Proof.}\nInterpret \\(blackboard\\) as the matrix of the quadratic form \\(screwdriver(x)=x^{T}blackboard x\\).\nThe quantity \\(\\operatorname{tr}screwdriver\\) (i.e.\\ the sum \\(screwdriver(v_{1})+\\cdots+screwdriver(v_{bookshelf})\\) for an orthonormal basis)  \nis basis–independent and additive on orthogonal direct sums.\nIf \\(\\operatorname{tr}screwdriver=0\\) and \\(\\dim horseshoe>0\\) we can pick a unit vector \\(daydream_{1}\\) with \\(screwdriver(daydream_{1})=0\\),\nthen iterate inside the orthogonal complement until the basis is complete.\nConversely, if such a basis exists, summing \\(screwdriver(daydream_{i})=0\\) gives \\(\\operatorname{tr}blackboard=0\\). \\(\\square\\)\n\n\\bigskip\n\\noindent\nNow let \\(watermelon=(avalanche,butterfly,marshland)\\) be any point on the director sphere.\nThrough \\(watermelon\\) consider planes whose normals are three mutually orthogonal unit vectors\n\\(daydream_{i}=(sailcloth_{i},autopilot_{i},heartbreak_{i})\\):\n\\[\nsailcloth_{i}sunflower+autopilot_{i}briefcase+heartbreak_{i}compass=sailcloth_{i}avalanche+autopilot_{i}butterfly+heartbreak_{i}marshland,\\qquad i=1,2,3.\n\\]\nThese planes are (projectively) tangent to \\(newspaper\\) precisely when  \n\\[\n\\frac{sailcloth_{i}^{2}}{semaphore}+\\frac{autopilot_{i}^{2}}{playground}+\\frac{heartbreak_{i}^{2}}{ventricle}-\n(sailcloth_{i}avalanche+autopilot_{i}butterfly+heartbreak_{i}marshland)^{2}=0,\\qquad i=1,2,3.\\tag{9}\n\\]\nLet\n\\[\nblackboard=\n\\begin{pmatrix}\n\\frac1{semaphore}-avalanche^{2} & -avalanche butterfly & -avalanche marshland\\\\[4pt]\n-avalanche butterfly & \\frac1{playground}-butterfly^{2} & -butterfly marshland\\\\[4pt]\n-avalanche marshland & -butterfly marshland & \\frac1{ventricle}-marshland^{2}\n\\end{pmatrix},\n\\]\nso that~(9) is \\(daydream_{i}^{T}blackboard\\, daydream_{i}=0\\).\nBy the theorem such an orthonormal triple exists \\emph{iff}\n\\[\n\\operatorname{tr}blackboard=\\frac1{semaphore}+\\frac1{playground}+\\frac1{ventricle}-avalanche^{2}-butterfly^{2}-marshland^{2}=0,\n\\]\nwhich is exactly the equation of the director sphere.\nHence \\emph{every} point of the sphere is the intersection of three mutually perpendicular planes projectively tangent to \\(newspaper\\).\n\n\\bigskip\n\\noindent\\textbf{Proper versus asymptotic tangency.}\nIf \\((1/semaphore)+(1/playground)+(1/ventricle)=0\\) the planes can never be \\emph{properly} tangent; assume henceforth  \n\\((1/semaphore)+(1/playground)+(1/ventricle)>0\\).\nThrough a given \\(watermelon\\in toothpaste\\) we first take any plane \\(courtship\\) projectively tangent to \\(newspaper\\).\nIts \\emph{polar conic} (intersection with \\(newspaper\\)) determines the remaining two planes \\(mothership,stormfront\\)\nperpendicular to \\(courtship\\) and through \\(watermelon\\); these form a unique mutually perpendicular triple.\nThere are infinitely many such triples, at most one of which involves an asymptotic plane,\nso in fact there are infinitely many \\emph{properly} tangent triples.\n\n\\bigskip\n\\noindent\\textbf{Exceptional points when \\(newspaper\\) is a hyperboloid of one sheet.}\nAssume \\(semaphore>0>playground>ventricle\\).\nWriting out where the normal through \\(watermelon\\) passes through the origin shows\nthat the exceptional set (where no proper triple exists) is\n\n\\[\n\\begin{cases}\n(\\pm semaphore^{-1/2},0,0), & semaphore\\neq playground=-ventricle,\\\\[4pt]\n(0,\\pm playground^{-1/2},0), & playground\\neq semaphore=-ventricle,\\\\[4pt]\n\\text{a circle in the $sunflower briefcase$--plane}, & semaphore=playground=-ventricle,\\\\\n\\text{empty}, & \\text{otherwise}.\n\\end{cases}\n\\]\n\n\\bigskip\n\\noindent\\textbf{A related problem.}\nAny ellipsoid or hyperboloid can be put in the form~(1) by translation and rotation,\nso the problem above is solved for all of them.\nFor the paraboloid in canonical form  \n\\[\ncompass=semaphore\\,sunflower^{2}+playground\\,briefcase^{2},\\qquad semaphore\\,playground\\neq0,\n\\]\nshow that the locus of intersection of three mutually orthogonal tangent planes is the plane  \n\\[\n\\boxed{\\,compass=-\\dfrac14\\Bigl(\\dfrac1{semaphore}+\\dfrac1{playground}\\Bigr).\\,}\n\\]\n\n\\bigskip\n\\noindent\\textbf{Remark.}\nIf we restrict to \\emph{rational} coefficients and points,\nthe first (\"director sphere\") part remains valid but the converse can fail.\nFor instance, with \\(newspaper:sunflower^{2}+briefcase^{2}+\\tfrac12 compass^{2}=1\\) the rational point \\(watermelon=(0,0,2)\\in toothpaste\\)\nadmits no triple of mutually perpendicular \\emph{rational} tangent planes.\nThe missing ingredient in the converse is a property of the real field not shared by the rationals.\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "nonabscissa",
        "y": "nonordinate",
        "z": "nonaltitude",
        "x_1": "farabscissa",
        "y_1": "farordinate",
        "z_1": "faraltitude",
        "r": "nonradius",
        "s": "nonsegment",
        "t": "atemporal",
        "n": "antidimension",
        "M": "scalarform",
        "F": "linearform",
        "V": "scalarrealm",
        "P": "lineentity",
        "O": "terminus",
        "Q": "planarsheet",
        "S": "flatsurface",
        "u_i": "nonunitvector",
        "\\alpha": "counteralpha",
        "\\beta": "counterbeta",
        "\\gamma": "countergamma",
        "\\delta": "counterdelta",
        "\\alpha_i": "counteralphai",
        "\\beta_i": "counterbetai",
        "\\gamma_i": "countergammai",
        "\\delta_i": "counterdeltai",
        "\\pi_1": "curvatureone",
        "\\pi_2": "curvaturetwo",
        "\\pi_3": "curvaturethree",
        "a": "largesize",
        "b": "mediumsize",
        "c": "smallsized"
      },
      "question": "<<<\n12. Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface\n\\[\nlargesize nonabscissa^{2}+mediumsize nonordinate^{2}+smallsized nonaltitude^{2}=1 \\quad(largesize mediumsize smallsized \\neq 0)\n\\]\nis the sphere\n\\[\nnonabscissa^{2}+nonordinate^{2}+nonaltitude^{2}=\\frac{1}{largesize}+\\frac{1}{mediumsize}+\\frac{1}{smallsized}\n\\]\n>>>",
      "solution": "<<<\nWe first determine the restrictions on the coefficients that ensure the plane  \n\\[\ncounteralpha nonabscissa+counterbeta nonordinate+countergamma nonaltitude=counterdelta \\tag{3}\n\\]\nis tangent to the quadric surface\n\\[\nplanarsheet:\\quad \\frac{nonabscissa^{2}}{largesize}+\\frac{nonordinate^{2}}{mediumsize}+\\frac{nonaltitude^{2}}{smallsized}=1\\tag{1}\n\\]\n(the explicit form of~\\(planarsheet\\) is used later).\n\n\\medskip\n\\noindent\nThe tangent plane to \\(planarsheet\\) at the point \\((farabscissa,farordinate,faraltitude)\\) is  \n\\[\ncounteralpha farabscissa\\,nonabscissa+counterbeta farordinate\\,nonordinate+countergamma faraltitude\\,nonaltitude=1. \\tag{4}\n\\]\nIf~(3) and~(4) coincide, then \\(counterdelta\\neq0\\) and  \n\\[\nfarabscissa=\\frac{counteralpha}{largesize\\,counterdelta},\\qquad \nfarordinate=\\frac{counterbeta}{mediumsize\\,counterdelta},\\qquad \nfaraltitude=\\frac{countergamma}{smallsized\\,counterdelta}. \\tag{5}\n\\]\nBecause the point \\((farabscissa,farordinate,faraltitude)\\) lies on \\(planarsheet\\) we must have  \n\\[\n\\frac1{counterdelta^{2}}\\Bigl(\\frac{counteralpha^{2}}{largesize}+\\frac{counterbeta^{2}}{mediumsize}+\\frac{countergamma^{2}}{smallsized}\\Bigr)=1,\n\\]\nhence  \n\\[\n\\boxed{\\;\n\\frac{counteralpha^{2}}{largesize}+\\frac{counterbeta^{2}}{mediumsize}+\\frac{countergamma^{2}}{smallsized}=counterdelta^{2}\\;} . \\tag{6}\n\\]\nThus~(6) with \\(counterdelta\\neq0\\) is \\emph{necessary and sufficient} for the plane~(3) to be tangent to \\(planarsheet\\).\n\nIf~(6) holds with \\(counterdelta=0\\) (assume \\(counteralpha,counterbeta,countergamma\\) not all zero) the plane~(3) is \\emph{asymptotic} to \\(planarsheet\\).\nIn projective language,~(6) says that~(3) is \\emph{projectively} tangent to \\(planarsheet\\);  \nthe point of tangency has homogeneous coordinates \\((counteralpha/largesize,counterbeta/mediumsize,countergamma/smallsized,counterdelta)\\).\nFor \\(counterdelta=0\\) this is a point at infinity on \\(planarsheet\\).\n\n\\bigskip\n\\noindent\nNow choose three mutually orthogonal unit vectors\n\\[\nnonunitvector_{i}=(counteralpha_{i},counterbeta_{i},countergamma_{i}),\\qquad i=1,2,3,\n\\]\nso that the matrix  \n\\[\n\\begin{pmatrix}\ncounteralpha_{1}&counterbeta_{1}&countergamma_{1}\\\\\ncounteralpha_{2}&counterbeta_{2}&countergamma_{2}\\\\\ncounteralpha_{3}&counterbeta_{3}&countergamma_{3}\n\\end{pmatrix}\\tag{7}\n\\]\nis orthogonal.  Consequently\n\\[\n\\sum_{i=1}^{3}counteralpha_{i}^{2}=\\sum_{i=1}^{3}counterbeta_{i}^{2}=\\sum_{i=1}^{3}countergamma_{i}^{2}=1.\n\\]\n\n\\medskip\\noindent\nIf planes having these vectors as normals are tangent to \\(planarsheet\\) then they are pairwise perpendicular and have equations  \n\\[\ncounteralpha_{i}nonabscissa+counterbeta_{i}nonordinate+countergamma_{i}nonaltitude=counterdelta_{i},\\qquad i=1,2,3, \\tag{8}\n\\]\nwhere  \n\\[\n\\frac{counteralpha_{i}^{2}}{largesize}+\\frac{counterbeta_{i}^{2}}{mediumsize}+\\frac{countergamma_{i}^{2}}{smallsized}=counterdelta_{i}^{2}.\n\\]\nSince \\(|counterdelta_{i}|\\) equals the distance from the terminus \\(terminus\\) to the \\(i\\)-th plane, the\nPythagorean theorem yields  \n\\[\nterminus lineentity^{2}=counterdelta_{1}^{2}+counterdelta_{2}^{2}+counterdelta_{3}^{2}\n       =\\frac1{largesize}+\\frac1{mediumsize}+\\frac1{smallsized}.\n\\]\nTherefore the intersection point \\(lineentity\\) of the three mutually perpendicular tangent planes\nlies on the sphere\n\\[\n\\boxed{\\;nonabscissa^{2}+nonordinate^{2}+nonaltitude^{2}=\\frac1{largesize}+\\frac1{mediumsize}+\\frac1{smallsized}\\;} \\tag{2}\n\\]\ncalled the \\emph{director sphere} of \\(planarsheet\\).\n(The same argument, with only minor changes, works in any dimension, giving\na director circle for a conic and higher–dimensional ``director spheres'' for central quadrics.)\n\n\\bigskip\n\\noindent\\textbf{The converse problem.}\nDoes every point of the director sphere arise from a triple of\nperpendicular tangent planes?\nTo prepare for this we need an algebraic lemma.\n\n\\paragraph{Theorem.}\n\\emph{Let \\(scalarform\\) be an \\(antidimension\\times antidimension\\) real matrix.\nThere exist \\(antidimension\\) mutually orthogonal unit vectors \\(nonunitvector_{1},\\dots,nonunitvector_{antidimension}\\in\\Bbb R^{antidimension}\\)\nwith \\(nonunitvector_{i}^{T}scalarform\\,nonunitvector_{i}=0\\;(i=1,\\dots,antidimension)\\) if and only if \\(\\operatorname{tr}scalarform=0\\).}\n\n\\emph{Proof.}\nInterpret \\(scalarform\\) as the matrix of the quadratic form \\(linearform(nonabscissa)=nonabscissa^{T}scalarform nonabscissa\\).\nThe quantity \\(\\operatorname{tr}linearform\\) (i.e.\\ the sum \\(linearform(v_{1})+\\cdots+linearform(v_{antidimension})\\) for an orthonormal basis)  \nis basis–independent and additive on orthogonal direct sums.\nIf \\(\\operatorname{tr}linearform=0\\) and \\(\\dim scalarrealm>0\\) we can pick a unit vector \\(nonunitvector_{1}\\) with \\(linearform(nonunitvector_{1})=0\\),\nthen iterate inside the orthogonal complement until the basis is complete.\nConversely, if such a basis exists, summing \\(linearform(nonunitvector_{i})=0\\) gives \\(\\operatorname{tr}scalarform=0\\). \\(\\square\\)\n\n\\bigskip\n\\noindent\nNow let \\(lineentity=(nonradius,nonsegment,atemporal)\\) be any point on the director sphere.\nThrough \\(lineentity\\) consider planes whose normals are three mutually orthogonal unit vectors\n\\(nonunitvector_{i}=(counteralpha_{i},counterbeta_{i},countergamma_{i})\\):\n\\[\ncounteralpha_{i}nonabscissa+counterbeta_{i}nonordinate+countergamma_{i}nonaltitude=counteralpha_{i}nonradius+counterbeta_{i}nonsegment+countergamma_{i}atemporal,\\qquad i=1,2,3.\n\\]\nThese planes are (projectively) tangent to \\(planarsheet\\) precisely when  \n\\[\n\\frac{counteralpha_{i}^{2}}{largesize}+\\frac{counterbeta_{i}^{2}}{mediumsize}+\\frac{countergamma_{i}^{2}}{smallsized}-\n(counteralpha_{i}nonradius+counterbeta_{i}nonsegment+countergamma_{i}atemporal)^{2}=0,\\qquad i=1,2,3.\\tag{9}\n\\]\nLet\n\\[\nscalarform=\n\\begin{pmatrix}\n\\frac1{largesize}-nonradius^{2} & -nonradius\\,nonsegment & -nonradius\\,atemporal\\\\[4pt]\n-nonradius\\,nonsegment & \\frac1{mediumsize}-nonsegment^{2} & -nonsegment\\,atemporal\\\\[4pt]\n-nonradius\\,atemporal & -nonsegment\\,atemporal & \\frac1{smallsized}-atemporal^{2}\n\\end{pmatrix},\n\\]\nso that~(9) is \\(nonunitvector_{i}^{T}scalarform\\,nonunitvector_{i}=0\\).\nBy the theorem such an orthonormal triple exists \\emph{iff}\n\\[\n\\operatorname{tr}scalarform=\\frac1{largesize}+\\frac1{mediumsize}+\\frac1{smallsized}-nonradius^{2}-nonsegment^{2}-atemporal^{2}=0,\n\\]\nwhich is exactly the equation of the director sphere.\nHence \\emph{every} point of the sphere is the intersection of three mutually perpendicular planes projectively tangent to \\(planarsheet\\).\n\n\\bigskip\n\\noindent\\textbf{Proper versus asymptotic tangency.}\nIf \\((1/largesize)+(1/mediumsize)+(1/smallsized)=0\\) the planes can never be \\emph{properly} tangent; assume henceforth  \n\\((1/largesize)+(1/mediumsize)+(1/smallsized)>0\\).\nThrough a given \\(lineentity\\in flatsurface\\) we first take any plane \\(curvatureone\\) projectively tangent to \\(planarsheet\\).\nIts \\emph{polar conic} (intersection with \\(planarsheet\\)) determines the remaining two planes \\(curvaturetwo,curvaturethree\\)\nperpendicular to \\(curvatureone\\) and through \\(lineentity\\); these form a unique mutually perpendicular triple.\nThere are infinitely many such triples, at most one of which involves an asymptotic plane,\nso in fact there are infinitely many \\emph{properly} tangent triples.\n\n\\bigskip\n\\noindent\\textbf{Exceptional points when \\(planarsheet\\) is a hyperboloid of one sheet.}\nAssume \\(largesize>0>mediumsize>smallsized\\).\nWriting out where the normal through \\(lineentity\\) passes through the terminus shows\nthat the exceptional set (where no proper triple exists) is\n\n\\[\n\\begin{cases}\n(\\pm largesize^{-1/2},0,0), & largesize\\neq mediumsize=-smallsized,\\\\[4pt]\n(0,\\pm mediumsize^{-1/2},0), & mediumsize\\neq largesize=-smallsized,\\\\[4pt]\n\\text{a circle in the $nonabscissa nonordinate$--plane}, & largesize=mediumsize=-smallsized,\\\\\n\\text{empty}, & \\text{otherwise}.\n\\end{cases}\n\\]\n\n\\bigskip\n\\noindent\\textbf{A related problem.}\nAny ellipsoid or hyperboloid can be put in the form~(1) by translation and rotation,\nso the problem above is solved for all of them.\nFor the paraboloid in canonical form  \n\\[\nnonaltitude=largesize nonabscissa^{2}+mediumsize nonordinate^{2},\\qquad largesize mediumsize\\neq0,\n\\]\nshow that the locus of intersection of three mutually orthogonal tangent planes is the plane  \n\\[\n\\boxed{\\,nonaltitude=-\\dfrac14\\Bigl(\\dfrac1{largesize}+\\dfrac1{mediumsize}\\Bigr).\\,}\n\\]\n\n\\bigskip\n\\noindent\\textbf{Remark.}\nIf we restrict to \\emph{rational} coefficients and points,\nthe first (``director sphere'') part remains valid but the converse can fail.\nFor instance, with \\(planarsheet:nonabscissa^{2}+nonordinate^{2}+\\tfrac12 nonaltitude^{2}=1\\) the rational point \\(lineentity=(0,0,2)\\in flatsurface\\)\nadmits no triple of mutually perpendicular \\emph{rational} tangent planes.\nThe missing ingredient in the converse is a property of the real field not shared by the rationals.\n>>>"
    },
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        "y_1": "hslwqxzn",
        "z_1": "mvplqser",
        "r": "kldqmsnv",
        "s": "plkswnez",
        "t": "vbnwmqse",
        "i": "gzcxptnh",
        "n": "fkrmzqws",
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        "S": "jkwmrtzu",
        "u_i": "xnvqpsrz_{gzcxptnh}",
        "\\alpha": "kzsomnfa",
        "\\beta": "pldfqzew",
        "\\gamma": "nxmpqrsl",
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        "\\alpha_i": "kzsomnfa_{gzcxptnh}",
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        "\\gamma_i": "nxmpqrsl_{gzcxptnh}",
        "\\delta_i": "rqplkzmn_{gzcxptnh}",
        "\\pi_1": "cvbnqwer_{1}",
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        "a": "smlqtnvz",
        "b": "dgfmrpqw",
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      },
      "question": "12. Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface\n\\[\nsmlqtnvz qzxwvtnp^{2}+dgfmrpqw hjgrksla^{2}+wqscznab mvbqplse^{2}=1 \\quad(smlqtnvz dgfmrpqw wqscznab \\neq 0)\n\\]\nis the sphere\n\\[\nqzxwvtnp^{2}+hjgrksla^{2}+mvbqplse^{2}=\\frac{1}{smlqtnvz}+\\frac{1}{dgfmrpqw}+\\frac{1}{wqscznab}\n\\]",
      "solution": "We first determine the restrictions on the coefficients that ensure the plane  \n\\[\nkzsomnfa qzxwvtnp+pldfqzew hjgrksla+nxmpqrsl mvbqplse=rqplkzmn \\tag{3}\n\\]\nis tangent to the quadric surface\n\\[\nabdjcefh:\\quad \\frac{qzxwvtnp^{2}}{smlqtnvz}+\\frac{hjgrksla^{2}}{dgfmrpqw}+\\frac{mvbqplse^{2}}{wqscznab}=1\\tag{1}\n\\]\n(the explicit form of~\\(abdjcefh\\) is used later).\n\n\\medskip\n\\noindent\nThe tangent plane to \\(abdjcefh\\) at the point \\((qzmcprlw,hslwqxzn,mvplqser)\\) is  \n\\[\nkzsomnfa qzmcprlw qzxwvtnp+pldfqzew hslwqxzn hjgrksla+nxmpqrsl mvplqser mvbqplse=1. \\tag{4}\n\\]\nIf~(3) and~(4) coincide, then \\(rqplkzmn\\ne 0\\) and  \n\\[\nqzmcprlw=\\frac{kzsomnfa}{smlqtnvz rqplkzmn},\\qquad \nhslwqxzn=\\frac{pldfqzew}{dgfmrpqw rqplkzmn},\\qquad \nmvplqser=\\frac{nxmpqrsl}{wqscznab rqplkzmn}. \\tag{5}\n\\]\nBecause the point \\((qzmcprlw,hslwqxzn,mvplqser)\\) lies on \\(abdjcefh\\) we must have  \n\\[\n\\frac1{rqplkzmn^{2}}\\Bigl(\\frac{kzsomnfa^{2}}{smlqtnvz}+\\frac{pldfqzew^{2}}{dgfmrpqw}+\\frac{nxmpqrsl^{2}}{wqscznab}\\Bigr)=1,\n\\]\nhence  \n\\[\n\\boxed{\\;\\frac{kzsomnfa^{2}}{smlqtnvz}+\\frac{pldfqzew^{2}}{dgfmrpqw}+\\frac{nxmpqrsl^{2}}{wqscznab}=rqplkzmn^{2}\\;}. \\tag{6}\n\\]\nThus~(6) with \\(rqplkzmn\\ne 0\\) is \\emph{necessary and sufficient} for the plane~(3) to be tangent to \\(abdjcefh\\).\n\nIf~(6) holds with \\(rqplkzmn=0\\) (assume \\(kzsomnfa,pldfqzew,nxmpqrsl\\) not all zero) the plane~(3) is \\emph{asymptotic} to \\(abdjcefh\\).\nIn projective language,~(6) says that~(3) is \\emph{projectively} tangent to \\(abdjcefh\\);  \nthe point of tangency has homogeneous coordinates \\((kzsomnfa/smlqtnvz,pldfqzew/dgfmrpqw,nxmpqrsl/wqscznab,rqplkzmn)\\).\nFor \\(rqplkzmn=0\\) this is a point at infinity on \\(abdjcefh\\).\n\n\\bigskip\n\\noindent\nNow choose three mutually orthogonal unit vectors\n\\[\nxnvqpsrz_{gzcxptnh}=(kzsomnfa_{gzcxptnh},pldfqzew_{gzcxptnh},nxmpqrsl_{gzcxptnh}),\\qquad gzcxptnh=1,2,3,\n\\]\nso that the matrix  \n\\[\n\\begin{pmatrix}\nkzsomnfa_{1}&pldfqzew_{1}&nxmpqrsl_{1}\\\\\nkzsomnfa_{2}&pldfqzew_{2}&nxmpqrsl_{2}\\\\\nkzsomnfa_{3}&pldfqzew_{3}&nxmpqrsl_{3}\n\\end{pmatrix}\\tag{7}\n\\]\nis orthogonal.  Consequently\n\\[\n\\sum_{gzcxptnh=1}^{3}kzsomnfa_{gzcxptnh}^{2}=\\sum_{gzcxptnh=1}^{3}pldfqzew_{gzcxptnh}^{2}=\\sum_{gzcxptnh=1}^{3}nxmpqrsl_{gzcxptnh}^{2}=1.\n\\]\n\n\\medskip\\noindent\nIf planes having these vectors as normals are tangent to \\(abdjcefh\\) then they are pairwise perpendicular and have equations  \n\\[\nkzsomnfa_{gzcxptnh} qzxwvtnp+pldfqzew_{gzcxptnh} hjgrksla+nxmpqrsl_{gzcxptnh} mvbqplse=rqplkzmn_{gzcxptnh},\\qquad gzcxptnh=1,2,3, \\tag{8}\n\\]\nwhere  \n\\[\n\\frac{kzsomnfa_{gzcxptnh}^{2}}{smlqtnvz}+\\frac{pldfqzew_{gzcxptnh}^{2}}{dgfmrpqw}+\\frac{nxmpqrsl_{gzcxptnh}^{2}}{wqscznab}=rqplkzmn_{gzcxptnh}^{2}.\n\\]\nSince \\(|rqplkzmn_{gzcxptnh}|\\) equals the distance from the origin \\(tuvsckwm\\) to the \\(gzcxptnh\\)-th plane, the\nPythagorean theorem yields  \n\\[\ntuvsckwmsdqplwer^{2}=rqplkzmn_{1}^{2}+rqplkzmn_{2}^{2}+rqplkzmn_{3}^{2}\n       =\\frac1{smlqtnvz}+\\frac1{dgfmrpqw}+\\frac1{wqscznab}.\n\\]\nTherefore the intersection point \\(sdqplwer\\) of the three mutually perpendicular tangent planes\nlies on the sphere\n\\[\n\\boxed{\\;qzxwvtnp^{2}+hjgrksla^{2}+mvbqplse^{2}=\\frac1{smlqtnvz}+\\frac1{dgfmrpqw}+\\frac1{wqscznab}\\;} \\tag{2}\n\\]\ncalled the \\emph{director sphere} of \\(abdjcefh\\).\n(The same argument, with only minor changes, works in any dimension, giving\na director circle for a conic and higher–dimensional ``director spheres'' for central quadrics.)\n\n\\bigskip\n\\noindent\\textbf{The converse problem.}\nDoes every point of the director sphere arise from a triple of\nperpendicular tangent planes?\nTo prepare for this we need an algebraic lemma.\n\n\\paragraph{Theorem.}\n\\emph{Let \\(lqwerxmn\\) be an \\(fkrmzqws\\times fkrmzqws\\) real matrix.\nThere exist \\(fkrmzqws\\) mutually orthogonal unit vectors \\(xnvqpsrz_{1},\\dots,xnvqpsrz_{fkrmzqws}\\in\\Bbb R^{fkrmzqws}\\)\nwith \\(xnvqpsrz_{gzcxptnh}^{T}lqwerxmn xnvqpsrz_{gzcxptnh}=0\\;(gzcxptnh=1,\\dots,fkrmzqws)\\) if and only if \\(\\operatorname{tr}lqwerxmn=0\\).}\n\n\\emph{Proof.}\nInterpret \\(lqwerxmn\\) as the matrix of the quadratic form \\(bhlptuxq(x)=x^{T}lqwerxmn x\\).\nThe quantity \\(\\operatorname{tr}bhlptuxq\\) (i.e.\\ the sum \\(bhlptuxq(v_{1})+\\cdots+bhlptuxq(v_{fkrmzqws})\\) for an orthonormal basis)  \nis basis–independent and additive on orthogonal direct sums.\nIf \\(\\operatorname{tr}bhlptuxq=0\\) and \\(\\dim zkrmuvcn>0\\) we can pick a unit vector \\(xnvqpsrz_{1}\\) with \\(bhlptuxq(xnvqpsrz_{1})=0\\),\nthen iterate inside the orthogonal complement until the basis is complete.\nConversely, if such a basis exists, summing \\(bhlptuxq(xnvqpsrz_{gzcxptnh})=0\\) gives \\(\\operatorname{tr}lqwerxmn=0\\). \\(\\square\\)\n\n\\bigskip\n\\noindent\nNow let \\(sdqplwer=(kldqmsnv,plkswnez,vbnwmqse)\\) be any point on the director sphere.\nThrough \\(sdqplwer\\) consider planes whose normals are three mutually orthogonal unit vectors\n\\(xnvqpsrz_{gzcxptnh}=(kzsomnfa_{gzcxptnh},pldfqzew_{gzcxptnh},nxmpqrsl_{gzcxptnh})\\):\n\\[\nkzsomnfa_{gzcxptnh}qzxwvtnp+pldfqzew_{gzcxptnh}hjgrksla+nxmpqrsl_{gzcxptnh}mvbqplse=kzsomnfa_{gzcxptnh}kldqmsnv+pldfqzew_{gzcxptnh}plkswnez+nxmpqrsl_{gzcxptnh}vbnwmqse,\\qquad gzcxptnh=1,2,3.\n\\]\nThese planes are (projectively) tangent to \\(abdjcefh\\) precisely when  \n\\[\n\\frac{kzsomnfa_{gzcxptnh}^{2}}{smlqtnvz}+\\frac{pldfqzew_{gzcxptnh}^{2}}{dgfmrpqw}+\\frac{nxmpqrsl_{gzcxptnh}^{2}}{wqscznab}-\n(kzsomnfa_{gzcxptnh}kldqmsnv+pldfqzew_{gzcxptnh}plkswnez+nxmpqrsl_{gzcxptnh}vbnwmqse)^{2}=0,\\qquad gzcxptnh=1,2,3.\\tag{9}\n\\]\nLet\n\\[\nlqwerxmn=\n\\begin{pmatrix}\n\\frac1{smlqtnvz}-kldqmsnv^{2} & -kldqmsnv\\,plkswnez & -kldqmsnv\\,vbnwmqse\\\\[4pt]\n-kldqmsnv\\,plkswnez & \\frac1{dgfmrpqw}-plkswnez^{2} & -plkswnez\\,vbnwmqse\\\\[4pt]\n-kldqmsnv\\,vbnwmqse & -plkswnez\\,vbnwmqse & \\frac1{wqscznab}-vbnwmqse^{2}\n\\end{pmatrix},\n\\]\nso that~(9) is \\(xnvqpsrz_{gzcxptnh}^{T}lqwerxmn xnvqpsrz_{gzcxptnh}=0\\).\nBy the theorem such an orthonormal triple exists \\emph{iff}\n\\[\n\\operatorname{tr}lqwerxmn=\\frac1{smlqtnvz}+\\frac1{dgfmrpqw}+\\frac1{wqscznab}-kldqmsnv^{2}-plkswnez^{2}-vbnwmqse^{2}=0,\n\\]\nwhich is exactly the equation of the director sphere.\nHence \\emph{every} point of the sphere is the intersection of three mutually perpendicular planes projectively tangent to \\(abdjcefh\\).\n\n\\bigskip\n\\noindent\\textbf{Proper versus asymptotic tangency.}\nIf \\((1/smlqtnvz)+(1/dgfmrpqw)+(1/wqscznab)=0\\) the planes can never be \\emph{properly} tangent; assume henceforth  \n\\((1/smlqtnvz)+(1/dgfmrpqw)+(1/wqscznab)>0\\).\nThrough a given \\(sdqplwer\\in jkwmrtzu\\) we first take any plane \\(cvbnqwer_{1}\\) projectively tangent to \\(abdjcefh\\).\nIts \\emph{polar conic} (intersection with \\(abdjcefh\\)) determines the remaining two planes \\(cvbnqwer_{2},cvbnqwer_{3}\\)\nperpendicular to \\(cvbnqwer_{1}\\) and through \\(sdqplwer\\); these form a unique mutually perpendicular triple.\nThere are infinitely many such triples, at most one of which involves an asymptotic plane,\nso in fact there are infinitely many \\emph{properly} tangent triples.\n\n\\bigskip\n\\noindent\\textbf{Exceptional points when \\(abdjcefh\\) is a hyperboloid of one sheet.}\nAssume \\(smlqtnvz>0>dgfmrpqw>wqscznab\\).\nWriting out where the normal through \\(sdqplwer\\) passes through the origin shows\nthat the exceptional set (where no proper triple exists) is\n\n\\[\n\\begin{cases}\n(\\pm smlqtnvz^{-1/2},0,0), & smlqtnvz\\neq dgfmrpqw=-wqscznab,\\\\[4pt]\n(0,\\pm dgfmrpqw^{-1/2},0), & dgfmrpqw\\neq smlqtnvz=-wqscznab,\\\\[4pt]\n\\text{a circle in the $xy$--plane}, & smlqtnvz=dgfmrpqw=-wqscznab,\\\\\n\\text{empty}, & \\text{otherwise}.\n\\end{cases}\n\\]\n\n\\bigskip\n\\noindent\\textbf{A related problem.}\nAny ellipsoid or hyperboloid can be put in the form~(1) by translation and rotation,\nso the problem above is solved for all of them.\nFor the paraboloid in canonical form  \n\\[\nmvbqplse=smlqtnvz qzxwvtnp^{2}+dgfmrpqw hjgrksla^{2},\\qquad smlqtnvz dgfmrpqw\\neq0,\n\\]\nshow that the locus of intersection of three mutually orthogonal tangent planes is the plane  \n\\[\n\\boxed{\\,mvbqplse=-\\dfrac14\\Bigl(\\dfrac1{smlqtnvz}+\\dfrac1{dgfmrpqw}\\Bigr).\\,}\n\\]\n\n\\bigskip\n\\noindent\\textbf{Remark.}\nIf we restrict to \\emph{rational} coefficients and points,\nthe first (\"director sphere\") part remains valid but the converse can fail.\nFor instance, with \\(abdjcefh:qzxwvtnp^{2}+hjgrksla^{2}+\\tfrac12 mvbqplse^{2}=1\\) the rational point \\((0,0,2)\\in jkwmrtzu\\)\nadmits no triple of mutually perpendicular \\emph{rational} tangent planes.\nThe missing ingredient in the converse is a property of the real field not shared by the rationals."
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 be an integer and let d , d₁ , … , dₙ be non–zero real numbers.\n\nCentral quadric\n                    d₁x₁² + d₂x₂² + ⋯ + dₙxₙ² = d.                                    (1)\n\nA (real) hyperplane is called tangent to (1) if it meets the quadric in exactly one point (counted with multiplicity).  Two hyperplanes are called orthogonal when their Euclidean normal vectors are orthogonal.\n\nProve that the set of all points that can be written as the common intersection of n mutually orthogonal hyperplanes tangent to (1) is precisely the (possibly empty) sphere\n                    x₁² + x₂² + ⋯ + xₙ² = d  ( 1/d₁ + 1/d₂ + ⋯ + 1/dₙ ).                  (2)\n\nThe right–hand side of (2) is positive exactly when\n                    d ( 1/d₁ + 1/d₂ + ⋯ + 1/dₙ ) > 0.                                     (*)\nIf (*) fails, both members of (2) are non–positive; the “sphere’’ is then empty and the statement is vacuous.",
      "solution": "Throughout d , d₁ , … , dₙ are non–zero reals.  The standard Euclidean inner product and norm are denoted by ⟨·,·⟩ and |·|.\n\n1.  A tangency criterion.\n   Fix a non–zero vector a ∈ ℝⁿ and a real number δ.  The hyperplane\n                          H(a,δ):   ⟨a,x⟩ = δ                                    (3)\n   meets (1) in finitely many points unless δ = 0.  Because the origin does not belong to (1) (d ≠ 0), a necessary condition for tangency is δ ≠ 0, and we impose it from now on.\n\n   Put f(x)=∑_{i=1}^{n} d_i x_i².  The hyperplane (3) is tangent to the level surface f(x)=d at the point P if and only if ∇f(P) is parallel to a.  Writing P=(p₁,…,pₙ) this reads\n                          a_i = 2λ d_i p_i    (i=1,…,n)                            (4)\n   for some scalar λ.  Taking inner product with P gives λ = δ/(2d), whence\n                          p_i = a_i d /(d_i δ).                                   (5a)\n   Substituting (5a) in (1) we obtain a single relation between a and δ:\n                          ∑_{i=1}^{n} a_i² / d_i = δ² / d.                         (5b)\n   Conversely, if δ ≠ 0 and (5b) holds, (5a) produces the unique point of contact.  Thus (5b) with δ ≠ 0 is necessary and sufficient for H(a,δ) to be tangent to (1).\n\n2.  From n orthogonal tangent hyperplanes to the sphere.\n   Let n mutually orthogonal tangent hyperplanes be\n                          ⟨a^{(k)},x⟩ = δ_k        (k = 1,…,n).                      (6)\n   Replacing each normal by a^{(k)}/|a^{(k)}| we may assume |a^{(k)}| = 1; then (a^{(1)},…,a^{(n)}) is an orthonormal basis of ℝⁿ.  Applying (5b) to every k and adding, we get\n            (1/d) ∑_{k=1}^{n} δ_k² = ∑_{k=1}^{n} ∑_{i=1}^{n} (a^{(k)}_i)² / d_i.       (7)\n   Because the matrix (a^{(k)}_i) is orthogonal,  ∑_{k} (a^{(k)}_i)² = 1 for every i; hence the right–hand side equals ∑_{i=1}^{n} 1/d_i.  Multiplying by d gives\n                          ∑_{k=1}^{n} δ_k² = d ∑_{i=1}^{n} 1/d_i.                  (8)\n   But δ_k = ⟨a^{(k)},P⟩, so orthonormality implies ∑_{k} δ_k² = |P|².  Therefore |P|² satisfies (2): every point obtained from n orthogonal tangent hyperplanes lies on the sphere (2).\n\n3.  Converse: from a point of the sphere to n perpendicular tangent hyperplanes.\n   Let P=(p₁,…,pₙ) satisfy (2).  Introduce the quadratic form\n        q(v)=∑_{i=1}^{n} v_i² / d_i − (⟨v,P⟩)² / d    (v∈ℝⁿ).                      (9)\n   Write its symmetric matrix as M.  Two elementary facts will be used:\n     • tr M = 0; indeed the diagonal entries add up to ∑ 1/d_i − |P|²/d = 0 by (2).\n     • q is not the zero form, hence is indefinite (because tr q = 0).  One sees this by evaluating q on P and on vectors orthogonal to P.\n\n   Lemma.  Let V be an n–dimensional Euclidean space (n ≥ 2) and let q be a symmetric bilinear form on V with tr q = 0 that is not identically zero.  For any non–zero vector w ∈ V there exists an orthonormal basis u₁,…,uₙ of V such that\n                          q(u_k)=0  and  ⟨u_k,w⟩≠0 (k=1,…,n).                 (10)\n\n   Proof.\n   We build the basis inductively.\n   Initialise V₀ = V and w₀ = w.\n\n   Inductive step.  Assume 0 ≤ k < n and we have already chosen orthonormal isotropic vectors u₁,…,u_k with ⟨u_j,w⟩ ≠ 0, and that w_k (the orthogonal projection of w_{k-1} onto V_k:=span{u₁,…,u_k}^⊥) is non–zero (for k=0 put w₀=w).\n\n   • The restriction q|_{V_k} still has trace 0; if dim V_k ≥ 2 it is again indefinite, hence its isotropic cone Z_k := {v∈V_k : |v|=1, q(v)=0} is a smooth codimension-1 submanifold of the unit sphere S(V_k) and in particular non–empty.\n   • The function v ↦ ⟨v,w_k⟩ does not vanish identically on Z_k, so the open dense set Z_k \\ {v : ⟨v,w_k⟩ = 0} is non–empty.  Choose u_{k+1} in this set with u_{k+1} not parallel to w_k (possible because dim V_k ≥ 2).\n   • Then u_{k+1} is unit, orthogonal to the previous u_j’s, isotropic, and satisfies ⟨u_{k+1},w⟩ = ⟨u_{k+1},w_k⟩ ≠ 0.\n   • Because u_{k+1} not parallel to w_k the new projection w_{k+1} = w_k − ⟨w_k,u_{k+1}⟩u_{k+1} is still non–zero, so the construction continues.\n\n   When dim V_k = 1, V_k = span{v} with |v|=1 and trace zero forces q(v)=0; moreover w_k ≠ 0 lies in V_k, so v is not orthogonal to w_k.  Setting u_{k+1}=±v (sign chosen so ⟨u_{k+1},w_k⟩>0) completes the basis.\n\n   By induction we eventually obtain an orthonormal basis satisfying (10). ∎\n\n   Apply the lemma with V = ℝⁿ, q as in (9) and w = P.  We obtain an orthonormal basis u₁,…,uₙ with q(u_k)=0 and δ_k := ⟨u_k,P⟩ ≠ 0.  Because q(u_k)=0, relation (5b) holds with a = u_k and δ = δ_k, so the hyperplanes\n                          H_k : ⟨u_k,x⟩ = δ_k  (k = 1,…,n)                  (11)\n   are mutually orthogonal, tangent to (1) and meet at P.\n\n4.  Conclusion.\n   When (*) holds the locus of intersection points of n mutually orthogonal hyperplanes tangent to (1) is exactly the sphere (2).  If (*) fails no such hyperplane system exists, hence the locus is empty.  This completes the proof.",
      "_meta": {
        "core_steps": [
          "Compute tangency condition: a plane αx+βy+γz=δ is tangent to ax²+by²+cz²=1 iff α²/a+β²/b+γ²/c=δ².",
          "Choose three orthonormal normal vectors u₁,u₂,u₃ for the mutually perpendicular planes and apply the tangency condition to each (giving δ₁,δ₂,δ₃).",
          "Use the Pythagorean theorem together with Σαᵢ²=Σβᵢ²=Σγᵢ²=1 to obtain OP² = δ₁²+δ₂²+δ₃² = 1/a+1/b+1/c, so the intersection point lies on the sphere.",
          "Invoke the trace-zero lemma (existence of an orthonormal set with uᵢᵀMuᵢ=0 ⇔ tr M=0) to show that every point on this sphere arises from such a triple, completing the converse."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Diagonal coefficients of the quadric; any non-zero real numbers work.",
            "original": [
              "a",
              "b",
              "c"
            ]
          },
          "slot2": {
            "description": "Right-hand-side constant of the quadric equation (currently 1); replacing it with any non-zero constant rescales the final sphere accordingly.",
            "original": 1
          },
          "slot3": {
            "description": "Ambient dimension / number of mutually orthogonal tangent hyperplanes; 3 can be replaced by n with identical reasoning.",
            "original": 3
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}