{
  "index": "1970-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { A-5. Determine the radius of the largest circle which can lie on the ellipsoid }\\\\\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}+\\frac{z^{2}}{c^{2}}=1 \\quad(a>b>c)\n\\end{array}",
  "solution": "A-5 Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through ( \\( 0,0,0 \\) ) which makes a circular cross section must intersect the \\( y-z \\) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \\( x=0, y^{2} / b^{2}+z^{2} / c^{2}=1 \\). Hence the radius of the circle is at most \\( b \\). Similar reasoning with the \\( x-y \\) plane shows that the radius of the circle is at least \\( b \\), so that any circular cross section formed by a plane through ( \\( 0,0,0 \\) ) must have radius \\( b \\), and this will be the required maximum radius. To show that circular cross sections of radius \\( b \\) actually exist, consider all planes through the \\( y \\)-axis. It can be verified that the two planes given by \\( a^{2}\\left(b^{2}-c^{2}\\right) z^{2}=c^{2}\\left(a^{2}-b^{2}\\right) x^{2} \\) give circular cross sections of radius \\( b \\).",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "coordx",
        "y": "coordy",
        "z": "coordz",
        "a": "semiaxisa",
        "b": "semiaaxisb",
        "c": "semiaxisc"
      },
      "question": "\\begin{array}{l}\n\\text { A-5. Determine the radius of the largest circle which can lie on the ellipsoid }\\\\\n\\frac{coordx^{2}}{semiaxisa^{2}}+\\frac{coordy^{2}}{semiaaxisb^{2}}+\\frac{coordz^{2}}{semiaxisc^{2}}=1 \\quad(semiaxisa>semiaaxisb>semiaxisc)\n\\end{array}",
      "solution": "A-5 Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through ( \\( 0,0,0 \\) ) which makes a circular cross section must intersect the \\( coordy-coordz \\) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \\( coordx=0,\\; coordy^{2} / semiaaxisb^{2}+coordz^{2} / semiaxisc^{2}=1 \\). Hence the radius of the circle is at most \\( semiaaxisb \\). Similar reasoning with the \\( coordx-coordy \\) plane shows that the radius of the circle is at least \\( semiaaxisb \\), so that any circular cross section formed by a plane through ( \\( 0,0,0 \\) ) must have radius \\( semiaaxisb \\), and this will be the required maximum radius. To show that circular cross sections of radius \\( semiaaxisb \\) actually exist, consider all planes through the \\( coordy \\)-axis. It can be verified that the two planes given by \\( semiaxisa^{2}\\left(semiaaxisb^{2}-semiaxisc^{2}\\right) coordz^{2}=semiaxisc^{2}\\left(semiaxisa^{2}-semiaaxisb^{2}\\right) coordx^{2} \\) give circular cross sections of radius \\( semiaaxisb \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanternfly",
        "y": "dragonfruit",
        "z": "merrymaker",
        "a": "butterscotch",
        "b": "parchment",
        "c": "lemonade"
      },
      "question": "<<<\n\\begin{array}{l}\n\\text { A-5. Determine the radius of the largest circle which can lie on the ellipsoid }\\\\\n\\frac{lanternfly^{2}}{butterscotch^{2}}+\\frac{dragonfruit^{2}}{parchment^{2}}+\\frac{merrymaker^{2}}{lemonade^{2}}=1 \\quad(butterscotch>parchment>lemonade)\n\\end{array}\n>>>",
      "solution": "<<<\nA-5 Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through ( \\( 0,0,0 \\) ) which makes a circular cross section must intersect the \\( dragonfruit-merrymaker \\) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \\( lanternfly=0,\\; dragonfruit^{2} / parchment^{2}+merrymaker^{2} / lemonade^{2}=1 \\). Hence the radius of the circle is at most \\( parchment \\). Similar reasoning with the \\( lanternfly-dragonfruit \\) plane shows that the radius of the circle is at least \\( parchment \\), so that any circular cross section formed by a plane through ( \\( 0,0,0 \\) ) must have radius \\( parchment \\), and this will be the required maximum radius. To show that circular cross sections of radius \\( parchment \\) actually exist, consider all planes through the \\( dragonfruit \\)-axis. It can be verified that the two planes given by \\( butterscotch^{2}\\left(parchment^{2}-lemonade^{2}\\right) merrymaker^{2}=lemonade^{2}\\left(butterscotch^{2}-parchment^{2}\\right) lanternfly^{2} \\) give circular cross sections of radius \\( parchment \\).\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "staticline",
        "z": "planepoint",
        "a": "zerolength",
        "b": "minorbend",
        "c": "widescope"
      },
      "question": "\\begin{array}{l}\n\\text { A-5. Determine the radius of the largest circle which can lie on the ellipsoid }\\\\\n\\frac{verticalaxis^{2}}{zerolength^{2}}+\\frac{staticline^{2}}{minorbend^{2}}+\\frac{planepoint^{2}}{widescope^{2}}=1 \\quad(zerolength>minorbend>widescope)\n\\end{array}",
      "solution": "A-5 Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through ( \\( 0,0,0 \\) ) which makes a circular cross section must intersect the \\( staticline-planepoint \\) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \\( verticalaxis=0, staticline^{2} / minorbend^{2}+planepoint^{2} / widescope^{2}=1 \\). Hence the radius of the circle is at most \\( minorbend \\). Similar reasoning with the \\( verticalaxis-staticline \\) plane shows that the radius of the circle is at least \\( minorbend \\), so that any circular cross section formed by a plane through ( \\( 0,0,0 \\) ) must have radius \\( minorbend \\), and this will be the required maximum radius. To show that circular cross sections of radius \\( minorbend \\) actually exist, consider all planes through the \\( staticline \\)-axis. It can be verified that the two planes given by \\( zerolength^{2}\\left(minorbend^{2}-widescope^{2}\\right) planepoint^{2}=widescope^{2}\\left(zerolength^{2}-minorbend^{2}\\right) verticalaxis^{2} \\) give circular cross sections of radius \\( minorbend \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mbdqlfve",
        "a": "tsroiyeg",
        "b": "wkzcapud",
        "c": "vnglmefa"
      },
      "question": "\\begin{array}{l}\n\\text { A-5. Determine the radius of the largest circle which can lie on the ellipsoid }\\\\\n\\frac{qzxwvtnp^{2}}{tsroiyeg^{2}}+\\frac{hjgrksla^{2}}{wkzcapud^{2}}+\\frac{mbdqlfve^{2}}{vnglmefa^{2}}=1 \\quad(tsroiyeg>wkzcapud>vnglmefa)\n\\end{array}",
      "solution": "A-5 Since parallel cross sections of the ellipsoid are always similar ellipses, any circular cross section can be increased in size by taking a parallel cutting plane passing through the center. Every plane through ( \\( 0,0,0 \\) ) which makes a circular cross section must intersect the \\( hjgrksla-mbdqlfve \\) plane. But this means that a diameter of the circular cross section must be a diameter of the ellipse \\( qzxwvtnp=0, hjgrksla^{2} / wkzcapud^{2}+mbdqlfve^{2} / vnglmefa^{2}=1 \\). Hence the radius of the circle is at most \\( wkzcapud \\). Similar reasoning with the \\( qzxwvtnp-hjgrksla \\) plane shows that the radius of the circle is at least \\( wkzcapud \\), so that any circular cross section formed by a plane through ( \\( 0,0,0 \\) ) must have radius \\( wkzcapud \\), and this will be the required maximum radius. To show that circular cross sections of radius \\( wkzcapud \\) actually exist, consider all planes through the \\( hjgrksla \\)-axis. It can be verified that the two planes given by \\( tsroiyeg^{2}\\left(wkzcapud^{2}-vnglmefa^{2}\\right) mbdqlfve^{2}=vnglmefa^{2}\\left(tsroiyeg^{2}-wkzcapud^{2}\\right) qzxwvtnp^{2} \\) give circular cross sections of radius \\( wkzcapud \\ )."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 3$ and let  \n\\[\na_{1}>a_{2}>\\dots>a_{n}>0\n\\]\nbe pairwise–distinct semi–axes.  Put $d_{i}:=1/a_{i}^{2}$, so  \n\\[\nd_{1}<d_{2}<\\dots<d_{n}.\n\\]\nConsider the $n$–dimensional ellipsoid  \n\\[\nE:=\\Bigl\\{\\,x\\in\\mathbb R^{n}\\;\\Bigm|\\;\nd_{1}x_{1}^{2}+d_{2}x_{2}^{2}+\\dots+d_{n}x_{n}^{2}=1\\Bigr\\}.\n\\]\n\nFor a $2$–dimensional linear subspace (plane through the origin) $P\\subset\\mathbb R^{n}$ define  \n\\[\nC(P):=E\\cap P.\n\\]\nWe call $C(P)$ a \\emph{circle} if, with the Euclidean metric inherited from $\\mathbb R^{n}$, it is a $1$–sphere; we write its radius as $R(P)$.\n\n(a) Determine  \n\\[\nR_{\\max}:=\\sup\\bigl\\{\\,R(P)\\,\\bigm|\\;P\\text{ is a $2$–plane and }C(P)\\text{ is a circle}\\bigr\\}.\n\\]\n\n(b) Describe explicitly all $2$–planes $P$ for which $C(P)$ is a circle of radius $R_{\\max}$.  \n(Two planes that differ only by \\emph{independent sign changes} of the coordinate axes are regarded as identical; those are precisely the orthogonal maps that leave the diagonal ellipsoid $E$ invariant.)\n\n\\vspace{1.5em}",
      "solution": "Throughout write $D:=\\operatorname{diag}(d_{1},\\dots,d_{n})$.\n\n\\textbf{Step 0.  Circle criterion inside a plane.}  \nLet $P$ be a $2$–plane with an orthonormal basis $\\{u,v\\}$.  Restricted to $P$, the quadratic form $x\\mapsto x^{\\mathrm T}Dx$ is represented by  \n\\[\nM_{P}:=\n\\begin{bmatrix}\nu^{\\mathrm T}Du & u^{\\mathrm T}Dv\\\\\nv^{\\mathrm T}Du & v^{\\mathrm T}Dv\n\\end{bmatrix}.\n\\tag{1}\n\\]\nIn coordinates $(s,t)\\in\\mathbb R^{2}$ the section is  \n\\[\nC(P)=\\{(s,t)\\mid (s,t)M_{P}(s,t)^{\\mathrm T}=1\\}.\n\\]\nHence $C(P)$ is a Euclidean circle of radius $R(P)$\niff  \n\\[\nM_{P}=\\frac1{R(P)^{2}}I_{2}.\n\\tag{2}\n\\]\n\n\\textbf{Step 1.  Universal upper bound $R(P)\\le a_{2}$.}  \nEvery $2$–plane $P$ intersects the hyperplane $H:=\\{x\\mid x_{1}=0\\}$ in a line.  \nChoose a unit vector $w\\in P\\cap H$.  Because $w_{1}=0$,\n\\[\nw^{\\mathrm T}Dw=\\sum_{j=2}^{n}d_{j}w_{j}^{2}\\ge d_{2}.\n\\tag{3}\n\\]\nIf $C(P)$ is a circle, (2) gives $w^{\\mathrm T}Dw=1/R(P)^{2}$; therefore\n\\[\n\\frac1{R(P)^{2}}\\;\\ge\\;d_{2}\\quad\\Longleftrightarrow\\quad\nR(P)\\le a_{2}.\n\\tag{4}\n\\]\nConsequently $R_{\\max}\\le a_{2}$.\n\n\\textbf{Step 2.  Construction of a plane with radius $a_{2}$.}  \nFix a non–empty subset $K\\subset\\{3,\\dots,n\\}$ and numbers $\\alpha_{k}$ $(k\\in K)$ with  \n\\[\n\\sum_{k\\in K}\\alpha_{k}^{2}=1.\n\\tag{5}\n\\]\nInside the subspace $\\operatorname{span}\\{e_{1}\\}\\cup\\{e_{k}\\mid k\\in K\\}$ set  \n\\[\nu(\\theta):=\\cos\\theta\\,e_{1}+\\sin\\theta\\sum_{k\\in K}\\alpha_{k}e_{k},\n\\qquad 0\\le\\theta\\le\\frac\\pi2.\n\\tag{6}\n\\]\nBecause $d_{1}<d_{2}<d_{k}$ $(k\\ge 3)$ there is a unique $\\theta$ satisfying  \n\\[\nu(\\theta)^{\\mathrm T}Du(\\theta)=d_{2},\n\\tag{7}\n\\]\nnamely\n\\[\n\\cos^{2}\\theta=\\frac{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{2})}\n{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{1})},\n\\qquad\n\\sin^{2}\\theta=\\frac{d_{2}-d_{1}}\n{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{1})}.\n\\tag{8}\n\\]\nPut $\\theta=\\theta(K,\\alpha)$ from (8) and define  \n\\[\nu:=u(\\theta),\\qquad v:=e_{2},\\qquad P:=\\operatorname{span}\\{u,v\\}.\n\\tag{9}\n\\]\nThen $\\lVert u\\rVert=\\lVert v\\rVert=1$, $u\\perp v$, and by (8)  \n\\[\nu^{\\mathrm T}Du=v^{\\mathrm T}Dv=d_{2},\n\\qquad\nu^{\\mathrm T}Dv=0.\n\\]\nThus (2) holds with $1/R(P)^{2}=d_{2}$, i.e.\\ $R(P)=a_{2}$.  Therefore  \n\\[\nR_{\\max}\\ge a_{2}.\n\\tag{10}\n\\]\nCombining (4) and (10) we get  \n\\[\nR_{\\max}=a_{2}.\n\\tag{11}\n\\]\n\n\\textbf{Step 3.  Description of all maximising planes.}  \nLet $P$ be any $2$–plane with $R(P)=a_{2}$; so (2) holds with $1/R^{2}=d_{2}$, i.e.  \n\\[\nx^{\\mathrm T}Dx=d_{2}\\lVert x\\rVert^{2}\\quad\\forall\\,x\\in P.\n\\tag{12}\n\\]\n\n\\emph{(3A)  The vector $e_{2}$ lies in $P$.}  \nAs in Step 1, $P\\cap H$ contains a unit vector $w$ with $w_{1}=0$.  \nBy (3) and (12) we must have $w^{\\mathrm T}Dw=d_{2}$, and the quadratic form  \n$x\\mapsto x^{\\mathrm T}Dx$ attains its strict minimum $d_{2}$ on the unit sphere\ninside $H$ precisely at $\\pm e_{2}$; hence $w=\\pm e_{2}\\in P$.\n\n\\emph{(3B)  Vectors orthogonal to $e_{2}$ inside $P$.}  \nChoose a unit vector $u\\in P$ with $u\\perp e_{2}$.  Then\n\\[\nu=(u_{1},0,u_{3},\\dots,u_{n}),\\qquad\nd_{1}u_{1}^{2}+\\sum_{k\\ge 3}d_{k}u_{k}^{2}=d_{2},\n\\qquad\nu_{1}^{2}+\\sum_{k\\ge 3}u_{k}^{2}=1.\n\\tag{13}\n\\]\nConversely, every unit vector fulfilling (13) together with $e_{2}$ generates a maximising plane, because (2) is then immediate.\n\nDefine  \n\\[\n\\mathcal U:=\\bigl\\{\\,u\\in S^{\\,n-1}\\bigm|\\;u_{2}=0,\\;u^{\\mathrm T}Du=d_{2}\\bigr\\}.\n\\tag{14}\n\\]\nEvery maximising plane is therefore\n\\[\nP(u):=\\operatorname{span}\\{e_{2},u\\}\\quad\\text{for some }u\\in\\mathcal U.\n\\tag{15}\n\\]\n\n\\emph{Dimension count.}  \n\\begin{itemize}\n\\item If $n\\ge 4$, the two independent equations in (13) cut the $(n-2)$–sphere given by $u_{2}=0$ down to a smooth manifold of dimension $n-3$.  Hence $\\mathcal U$ and the corresponding family of planes each have dimension $n-3$.\n\\item If $n=3$, (13) becomes  \n\\[\nd_{1}u_{1}^{2}+d_{3}u_{3}^{2}=d_{2},\n\\qquad\nu_{1}^{2}+u_{3}^{2}=1,\n\\tag{16}\n\\]\nwhose solutions are  \n\\[\nu=\\Bigl(\\pm\\sqrt{\\tfrac{d_{3}-d_{2}}{d_{3}-d_{1}}},\\;0,\\;\n\\pm\\sqrt{\\tfrac{d_{2}-d_{1}}{d_{3}-d_{1}}}\\Bigr).\n\\tag{17}\n\\]\nThere are four such unit vectors; replacing $u$ by $-u$ does not change the span, so we obtain two distinct $2$–planes  \n\\[\nP_{\\pm}:=\\operatorname{span}\\Bigl\\{e_{2},\n\\bigl(\\sqrt{\\tfrac{d_{3}-d_{2}}{d_{3}-d_{1}}},\\;0,\\;\n\\pm\\sqrt{\\tfrac{d_{2}-d_{1}}{d_{3}-d_{1}}}\\bigr)\\Bigr\\}.\n\\tag{18}\n\\]\nUnder independent sign changes of the $x$– and $z$–axes these two coincide, furnishing a single equivalence class.\n\\end{itemize}\n\n\\textbf{Step 4.  Action of coordinate sign changes.}  \nLet $G:=\\{\\operatorname{diag}(\\varepsilon_{1},\\dots,\\varepsilon_{n})\\mid\n\\varepsilon_{j}\\in\\{\\pm1\\}\\}\\cong(\\mathbb Z/2\\mathbb Z)^{n}$ be the group of coordinate sign changes; $G$ acts orthogonally on $\\mathbb R^{n}$ and leaves $E$ invariant.\n\nFix $u\\in\\mathcal U$ and write  \n\\[\nS(u):=\\{\\,j\\neq 2\\mid u_{j}=0\\}.\n\\]\nFor $\\sigma=\\operatorname{diag}(\\varepsilon_{1},\\dots,\\varepsilon_{n})\\in G$ we have  \n\\[\n\\sigma\\bigl(P(u)\\bigr)=P(u)\\quad\\Longleftrightarrow\\quad\n\\varepsilon_{2}=1\\text{ and }\\varepsilon_{j}=1\\text{ for every }j\\notin S(u)\\cup\\{2\\}.\n\\]\nHence the stabiliser of $P(u)$ is isomorphic to $(\\mathbb Z/2\\mathbb Z)^{\\,|S(u)|}$ and the $G$–orbit of $P(u)$ has cardinality $2^{\\,n-2-|S(u)|}$.  In particular:\n\\begin{itemize}\n\\item If \\emph{no} coordinate of $u$ (other than $u_{2}=0$) vanishes, then $|S(u)|=0$ and the orbit has the maximal size $2^{\\,n-2}$.\n\\item Should $u$ possess additional zero coordinates, the orbit size drops accordingly.  (This occurs, for example, in the explicit construction of Step 2 whenever $|K|\\le n-2$.)\n\\end{itemize}\nThus the earlier uniform count $2^{\\,n-2}$ is incorrect except when $|S(u)|=0$; the corrected formula above gives the precise size of each orbit.\n\n\\bigskip\n\\textbf{Answer.}\n\n(a) $\\displaystyle R_{\\max}=a_{2}$.\n\n(b) All circles of maximal radius $a_{2}$ occur precisely in the $2$–planes  \n\\[\nP(u)=\\operatorname{span}\\{e_{2},u\\},\\qquad\nu\\in\\mathcal U,\n\\]\nwhere\n\\[\n\\mathcal U=\\Bigl\\{u=(u_{1},0,u_{3},\\dots,u_{n})\\in S^{\\,n-1}\\;\\Bigm|\\;\nd_{1}u_{1}^{2}+\\sum_{k=3}^{n}d_{k}u_{k}^{2}=d_{2}\\Bigr\\}.\n\\]\nFor $n\\ge 4$ these form an $(n-3)$–parameter family of mutually non–congruent planes; for $n=3$ they reduce to the single equivalence class represented by the two planes $P_{\\pm}$ in (18).  The description is complete up to the natural action of independent coordinate sign changes.\n\n\\vspace{1em}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.590967",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem is lifted from ℝ³ to arbitrary ℝⁿ, forcing the solver to juggle Grassmann dimension counts, subspace intersections and the restriction of quadratic forms.  \n• Additional structure: one must translate “circle on an ellipsoid” into the matrix condition (1) and handle general positive–definite diagonal forms.  \n• Deeper theory: the proof merges linear-algebraic spectral arguments (simultaneous diagonalisation), projective dimension counting, and the classical 3-dimensional result as a sub-case.  \n• Multiple interacting concepts: eigenvalue bounds yield the universal inequality R ≤ a₂, while an explicit geometric construction in a chosen 3-subspace is needed to prove sharpness.  \nThese layers of abstraction and the need to manage n-dimensional geometry make the enhanced variant decisively more challenging than both the original and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 3$ and let  \n\\[\na_{1}>a_{2}>\\dots>a_{n}>0\n\\]\nbe pairwise–distinct semi–axes.  Put $d_{i}:=1/a_{i}^{2}$, so  \n\\[\nd_{1}<d_{2}<\\dots<d_{n}.\n\\]\nConsider the $n$–dimensional ellipsoid  \n\\[\nE:=\\Bigl\\{\\,x\\in\\mathbb R^{n}\\;\\Bigm|\\;\nd_{1}x_{1}^{2}+d_{2}x_{2}^{2}+\\dots+d_{n}x_{n}^{2}=1\\Bigr\\}.\n\\]\n\nFor a $2$–dimensional linear subspace (plane through the origin) $P\\subset\\mathbb R^{n}$ define  \n\\[\nC(P):=E\\cap P.\n\\]\nWe call $C(P)$ a \\emph{circle} if, with the Euclidean metric inherited from $\\mathbb R^{n}$, it is a $1$–sphere; we write its radius as $R(P)$.\n\n(a) Determine  \n\\[\nR_{\\max}:=\\sup\\bigl\\{\\,R(P)\\,\\bigm|\\;P\\text{ is a $2$–plane and }C(P)\\text{ is a circle}\\bigr\\}.\n\\]\n\n(b) Describe explicitly all $2$–planes $P$ for which $C(P)$ is a circle of radius $R_{\\max}$.  \n(Two planes that differ only by \\emph{independent sign changes} of the coordinate axes are regarded as identical; those are precisely the orthogonal maps that leave the diagonal ellipsoid $E$ invariant.)\n\n\\vspace{1.5em}",
      "solution": "Throughout write $D:=\\operatorname{diag}(d_{1},\\dots,d_{n})$.\n\n\\textbf{Step 0.  Circle criterion inside a plane.}  \nLet $P$ be a $2$–plane with an orthonormal basis $\\{u,v\\}$.  Restricted to $P$, the quadratic form $x\\mapsto x^{\\mathrm T}Dx$ is represented by  \n\\[\nM_{P}:=\n\\begin{bmatrix}\nu^{\\mathrm T}Du & u^{\\mathrm T}Dv\\\\\nv^{\\mathrm T}Du & v^{\\mathrm T}Dv\n\\end{bmatrix}.\n\\tag{1}\n\\]\nIn coordinates $(s,t)\\in\\mathbb R^{2}$ the section is  \n\\[\nC(P)=\\{(s,t)\\mid (s,t)M_{P}(s,t)^{\\mathrm T}=1\\}.\n\\]\nHence $C(P)$ is a Euclidean circle of radius $R(P)$\niff  \n\\[\nM_{P}=\\frac1{R(P)^{2}}I_{2}.\n\\tag{2}\n\\]\n\n\\textbf{Step 1.  Universal upper bound $R(P)\\le a_{2}$.}  \nEvery $2$–plane $P$ intersects the hyperplane $H:=\\{x\\mid x_{1}=0\\}$ in a line.  \nChoose a unit vector $w\\in P\\cap H$.  Because $w_{1}=0$,\n\\[\nw^{\\mathrm T}Dw=\\sum_{j=2}^{n}d_{j}w_{j}^{2}\\ge d_{2}.\n\\tag{3}\n\\]\nIf $C(P)$ is a circle, (2) gives $w^{\\mathrm T}Dw=1/R(P)^{2}$; therefore\n\\[\n\\frac1{R(P)^{2}}\\;\\ge\\;d_{2}\\quad\\Longleftrightarrow\\quad\nR(P)\\le a_{2}.\n\\tag{4}\n\\]\nConsequently $R_{\\max}\\le a_{2}$.\n\n\\textbf{Step 2.  Construction of a plane with radius $a_{2}$.}  \nFix a non–empty subset $K\\subset\\{3,\\dots,n\\}$ and numbers $\\alpha_{k}$ $(k\\in K)$ with  \n\\[\n\\sum_{k\\in K}\\alpha_{k}^{2}=1.\n\\tag{5}\n\\]\nInside the subspace $\\operatorname{span}\\{e_{1}\\}\\cup\\{e_{k}\\mid k\\in K\\}$ set  \n\\[\nu(\\theta):=\\cos\\theta\\,e_{1}+\\sin\\theta\\sum_{k\\in K}\\alpha_{k}e_{k},\n\\qquad 0\\le\\theta\\le\\frac\\pi2.\n\\tag{6}\n\\]\nBecause $d_{1}<d_{2}<d_{k}$ $(k\\ge 3)$ there is a unique $\\theta$ satisfying  \n\\[\nu(\\theta)^{\\mathrm T}Du(\\theta)=d_{2},\n\\tag{7}\n\\]\nnamely\n\\[\n\\cos^{2}\\theta=\\frac{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{2})}\n{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{1})},\n\\qquad\n\\sin^{2}\\theta=\\frac{d_{2}-d_{1}}\n{\\displaystyle\\sum_{k\\in K}\\alpha_{k}^{2}(d_{k}-d_{1})}.\n\\tag{8}\n\\]\nPut $\\theta=\\theta(K,\\alpha)$ from (8) and define  \n\\[\nu:=u(\\theta),\\qquad v:=e_{2},\\qquad P:=\\operatorname{span}\\{u,v\\}.\n\\tag{9}\n\\]\nThen $\\lVert u\\rVert=\\lVert v\\rVert=1$, $u\\perp v$, and by (8)  \n\\[\nu^{\\mathrm T}Du=v^{\\mathrm T}Dv=d_{2},\n\\qquad\nu^{\\mathrm T}Dv=0.\n\\]\nThus (2) holds with $1/R(P)^{2}=d_{2}$, i.e.\\ $R(P)=a_{2}$.  Therefore  \n\\[\nR_{\\max}\\ge a_{2}.\n\\tag{10}\n\\]\nCombining (4) and (10) we get  \n\\[\nR_{\\max}=a_{2}.\n\\tag{11}\n\\]\n\n\\textbf{Step 3.  Description of all maximising planes.}  \nLet $P$ be any $2$–plane with $R(P)=a_{2}$; so (2) holds with $1/R^{2}=d_{2}$, i.e.  \n\\[\nx^{\\mathrm T}Dx=d_{2}\\lVert x\\rVert^{2}\\quad\\forall\\,x\\in P.\n\\tag{12}\n\\]\n\n\\emph{(3A)  The vector $e_{2}$ lies in $P$.}  \nAs in Step 1, $P\\cap H$ contains a unit vector $w$ with $w_{1}=0$.  \nBy (3) and (12) we must have $w^{\\mathrm T}Dw=d_{2}$, and the quadratic form  \n$x\\mapsto x^{\\mathrm T}Dx$ attains its strict minimum $d_{2}$ on the unit sphere\ninside $H$ precisely at $\\pm e_{2}$; hence $w=\\pm e_{2}\\in P$.\n\n\\emph{(3B)  Vectors orthogonal to $e_{2}$ inside $P$.}  \nChoose a unit vector $u\\in P$ with $u\\perp e_{2}$.  Then\n\\[\nu=(u_{1},0,u_{3},\\dots,u_{n}),\\qquad\nd_{1}u_{1}^{2}+\\sum_{k\\ge 3}d_{k}u_{k}^{2}=d_{2},\n\\qquad\nu_{1}^{2}+\\sum_{k\\ge 3}u_{k}^{2}=1.\n\\tag{13}\n\\]\nConversely, every unit vector fulfilling (13) together with $e_{2}$ generates a maximising plane, because (2) is then immediate.\n\nDefine  \n\\[\n\\mathcal U:=\\bigl\\{\\,u\\in S^{\\,n-1}\\bigm|\\;u_{2}=0,\\;u^{\\mathrm T}Du=d_{2}\\bigr\\}.\n\\tag{14}\n\\]\nEvery maximising plane is therefore\n\\[\nP(u):=\\operatorname{span}\\{e_{2},u\\}\\quad\\text{for some }u\\in\\mathcal U.\n\\tag{15}\n\\]\n\n\\emph{Dimension count.}  \n\\begin{itemize}\n\\item If $n\\ge 4$, the two independent equations in (13) cut the $(n-2)$–sphere given by $u_{2}=0$ down to a smooth manifold of dimension $n-3$.  Hence $\\mathcal U$ and the corresponding family of planes each have dimension $n-3$.\n\\item If $n=3$, (13) becomes  \n\\[\nd_{1}u_{1}^{2}+d_{3}u_{3}^{2}=d_{2},\n\\qquad\nu_{1}^{2}+u_{3}^{2}=1,\n\\tag{16}\n\\]\nwhose solutions are  \n\\[\nu=\\Bigl(\\pm\\sqrt{\\tfrac{d_{3}-d_{2}}{d_{3}-d_{1}}},\\;0,\\;\n\\pm\\sqrt{\\tfrac{d_{2}-d_{1}}{d_{3}-d_{1}}}\\Bigr).\n\\tag{17}\n\\]\nThere are four such unit vectors; replacing $u$ by $-u$ does not change the span, so we obtain two distinct $2$–planes  \n\\[\nP_{\\pm}:=\\operatorname{span}\\Bigl\\{e_{2},\n\\bigl(\\sqrt{\\tfrac{d_{3}-d_{2}}{d_{3}-d_{1}}},\\;0,\\;\n\\pm\\sqrt{\\tfrac{d_{2}-d_{1}}{d_{3}-d_{1}}}\\bigr)\\Bigr\\}.\n\\tag{18}\n\\]\nUnder independent sign changes of the $x$– and $z$–axes these two coincide, furnishing a single equivalence class.\n\\end{itemize}\n\n\\textbf{Step 4.  Action of coordinate sign changes.}  \nLet $G:=\\{\\operatorname{diag}(\\varepsilon_{1},\\dots,\\varepsilon_{n})\\mid\n\\varepsilon_{j}\\in\\{\\pm1\\}\\}\\cong(\\mathbb Z/2\\mathbb Z)^{n}$ be the group of coordinate sign changes; $G$ acts orthogonally on $\\mathbb R^{n}$ and leaves $E$ invariant.\n\nFix $u\\in\\mathcal U$ and write  \n\\[\nS(u):=\\{\\,j\\neq 2\\mid u_{j}=0\\}.\n\\]\nFor $\\sigma=\\operatorname{diag}(\\varepsilon_{1},\\dots,\\varepsilon_{n})\\in G$ we have  \n\\[\n\\sigma\\bigl(P(u)\\bigr)=P(u)\\quad\\Longleftrightarrow\\quad\n\\varepsilon_{2}=1\\text{ and }\\varepsilon_{j}=1\\text{ for every }j\\notin S(u)\\cup\\{2\\}.\n\\]\nHence the stabiliser of $P(u)$ is isomorphic to $(\\mathbb Z/2\\mathbb Z)^{\\,|S(u)|}$ and the $G$–orbit of $P(u)$ has cardinality $2^{\\,n-2-|S(u)|}$.  In particular:\n\\begin{itemize}\n\\item If \\emph{no} coordinate of $u$ (other than $u_{2}=0$) vanishes, then $|S(u)|=0$ and the orbit has the maximal size $2^{\\,n-2}$.\n\\item Should $u$ possess additional zero coordinates, the orbit size drops accordingly.  (This occurs, for example, in the explicit construction of Step 2 whenever $|K|\\le n-2$.)\n\\end{itemize}\nThus the earlier uniform count $2^{\\,n-2}$ is incorrect except when $|S(u)|=0$; the corrected formula above gives the precise size of each orbit.\n\n\\bigskip\n\\textbf{Answer.}\n\n(a) $\\displaystyle R_{\\max}=a_{2}$.\n\n(b) All circles of maximal radius $a_{2}$ occur precisely in the $2$–planes  \n\\[\nP(u)=\\operatorname{span}\\{e_{2},u\\},\\qquad\nu\\in\\mathcal U,\n\\]\nwhere\n\\[\n\\mathcal U=\\Bigl\\{u=(u_{1},0,u_{3},\\dots,u_{n})\\in S^{\\,n-1}\\;\\Bigm|\\;\nd_{1}u_{1}^{2}+\\sum_{k=3}^{n}d_{k}u_{k}^{2}=d_{2}\\Bigr\\}.\n\\]\nFor $n\\ge 4$ these form an $(n-3)$–parameter family of mutually non–congruent planes; for $n=3$ they reduce to the single equivalence class represented by the two planes $P_{\\pm}$ in (18).  The description is complete up to the natural action of independent coordinate sign changes.\n\n\\vspace{1em}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.474011",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem is lifted from ℝ³ to arbitrary ℝⁿ, forcing the solver to juggle Grassmann dimension counts, subspace intersections and the restriction of quadratic forms.  \n• Additional structure: one must translate “circle on an ellipsoid” into the matrix condition (1) and handle general positive–definite diagonal forms.  \n• Deeper theory: the proof merges linear-algebraic spectral arguments (simultaneous diagonalisation), projective dimension counting, and the classical 3-dimensional result as a sub-case.  \n• Multiple interacting concepts: eigenvalue bounds yield the universal inequality R ≤ a₂, while an explicit geometric construction in a chosen 3-subspace is needed to prove sharpness.  \nThese layers of abstraction and the need to manage n-dimensional geometry make the enhanced variant decisively more challenging than both the original and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}