{
  "index": "1953-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. Determine the equation of a surface in three dimensional cartesian space which has the following properties: (a) it passes through the point \\( (1,1 \\), 1); and (b) if the tangent plane be drawn at any point \\( P \\), and \\( A, B \\) and \\( C \\) are the intersections of this plane with the \\( x, y \\) and \\( z \\) axes respectively, then \\( P \\) is the orthocenter (intersection of the altitudes) of the triangle \\( A B C \\).",
  "solution": "Solution. Consider any plane \\( \\Pi \\) that crosses the coordinate axes at three distinct points \\( A, B \\), and \\( C \\), and let \\( P \\) be the orthocenter of the triangle \\( A B C \\). Treat the points as vectors from the origin, use \"( , )\" to denote the inner product of two vectors, and recall that \\( (A, B)=(B, C)= \\) \\( (C, A)=0 \\) since the axes are orthogonal. Then the orthocenter property becomes\n\\[\n(P-A, B-C)=(P-B, C-A)=(P-C, A-B)=0\n\\]\n\nUsing the linearity properties of the inner product, we get\n\\[\n(P, B-C)=(P, C-A)=(P, A-B)=0\n\\]\n\nThus \\( P \\), as a vector, is orthogonal to the plane \\( \\Pi \\). This means that \\( P \\) is the foot of the perpendicular on \\( \\Pi \\) from the origin \\( O \\).\n\nThe condition on the surface is, then, that every normal passes through the origin. Obviously the sphere\n\\[\nx^{2}+y^{2}+z^{2}=3\n\\]\nhas this property and passes through the required point \\( (1,1,1) \\).\nA surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere (1) satisfying \\( x>0 \\), \\( y>0, z>0 \\). [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form\n\\[\nx d x+y d y+z d z=\\frac{1}{2} d\\left(x^{2}+y^{2}+z^{2}\\right)\n\\]\nvanish on the surface. We conclude that \\( x^{2}+y^{2}+z^{2} \\) is constant.\nSince the surface is required to contain \\( (1,1,1) \\) it must be a part of the sphere (1). Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant.",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "A",
    "B",
    "C",
    "P",
    "O",
    "\\\\Pi"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "z": "applicate",
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "P": "orthopnt",
        "O": "originpt",
        "\\Pi": "planeone"
      },
      "question": "4. Determine the equation of a surface in three dimensional cartesian space which has the following properties: (a) it passes through the point \\( (1,1 \\), 1); and (b) if the tangent plane be drawn at any point \\( orthopnt \\), and \\( vertexa, vertexb \\) and \\( vertexc \\) are the intersections of this plane with the \\( abscissa, ordinate \\) and \\( applicate \\) axes respectively, then \\( orthopnt \\) is the orthocenter (intersection of the altitudes) of the triangle \\( vertexa vertexb vertexc \\).",
      "solution": "Solution. Consider any plane \\( planeone \\) that crosses the coordinate axes at three distinct points \\( vertexa, vertexb \\), and \\( vertexc \\), and let \\( orthopnt \\) be the orthocenter of the triangle \\( vertexa vertexb vertexc \\). Treat the points as vectors from the origin, use \"( , )\" to denote the inner product of two vectors, and recall that \\( (vertexa, vertexb)=(vertexb, vertexc)=(vertexc, vertexa)=0 \\) since the axes are orthogonal. Then the orthocenter property becomes\n\\[\n(orthopnt-vertexa, vertexb-vertexc)=(orthopnt-vertexb, vertexc-vertexa)=(orthopnt-vertexc, vertexa-vertexb)=0\n\\]\n\nUsing the linearity properties of the inner product, we get\n\\[\n(orthopnt, vertexb-vertexc)=(orthopnt, vertexc-vertexa)=(orthopnt, vertexa-vertexb)=0\n\\]\n\nThus \\( orthopnt \\), as a vector, is orthogonal to the plane \\( planeone \\). This means that \\( orthopnt \\) is the foot of the perpendicular on \\( planeone \\) from the origin \\( originpt \\).\n\nThe condition on the surface is, then, that every normal passes through the origin. Obviously the sphere\n\\[\nabscissa^{2}+ordinate^{2}+applicate^{2}=3\n\\]\nhas this property and passes through the required point \\( (1,1,1) \\).\nA surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere (1) satisfying \\( abscissa>0 \\), \\( ordinate>0, applicate>0 \\). [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form\n\\[\nabscissa\\, d\\, abscissa+ordinate\\, d\\, ordinate+applicate\\, d\\, applicate=\\frac{1}{2} d\\left(abscissa^{2}+ordinate^{2}+applicate^{2}\\right)\n\\]\nvanish on the surface. We conclude that \\( abscissa^{2}+ordinate^{2}+applicate^{2} \\) is constant.\nSince the surface is required to contain \\( (1,1,1) \\) it must be a part of the sphere (1). Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "bookshelf",
        "z": "tapestry",
        "A": "marigold",
        "B": "blackbird",
        "C": "cinnamon",
        "P": "goldfinch",
        "O": "rainstorm",
        "\\Pi": "silhouette"
      },
      "question": "Determine the equation of a surface in three dimensional cartesian space which has the following properties: (a) it passes through the point \\( (1,1 ), 1); and (b) if the tangent plane be drawn at any point \\( goldfinch \\), and \\( marigold, blackbird \\) and \\( cinnamon \\) are the intersections of this plane with the \\( sunflower, bookshelf \\) and \\( tapestry \\) axes respectively, then \\( goldfinch \\) is the orthocenter (intersection of the altitudes) of the triangle \\( marigold blackbird cinnamon \\).",
      "solution": "Solution. Consider any plane \\( silhouette \\) that crosses the coordinate axes at three distinct points \\( marigold, blackbird \\), and \\( cinnamon \\), and let \\( goldfinch \\) be the orthocenter of the triangle \\( marigold blackbird cinnamon \\). Treat the points as vectors from the origin, use \"( , )\" to denote the inner product of two vectors, and recall that \\( (marigold, blackbird)=(blackbird, cinnamon)=(cinnamon, marigold)=0 \\) since the axes are orthogonal. Then the orthocenter property becomes\n\\[\n(goldfinch-marigold, blackbird-cinnamon)=(goldfinch-blackbird, cinnamon-marigold)=(goldfinch-cinnamon, marigold-blackbird)=0\n\\]\n\nUsing the linearity properties of the inner product, we get\n\\[\n(goldfinch, blackbird-cinnamon)=(goldfinch, cinnamon-marigold)=(goldfinch, marigold-blackbird)=0\n\\]\n\nThus \\( goldfinch \\), as a vector, is orthogonal to the plane \\( silhouette \\). This means that \\( goldfinch \\) is the foot of the perpendicular on \\( silhouette \\) from the origin \\( rainstorm \\).\n\nThe condition on the surface is, then, that every normal passes through the origin. Obviously the sphere\n\\[\nsunflower^{2}+bookshelf^{2}+tapestry^{2}=3\n\\]\nhas this property and passes through the required point \\( (1,1,1) \\).\nA surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere (1) satisfying \\( sunflower>0 \\), \\( bookshelf>0, tapestry>0 \\). [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form\n\\[\nsunflower d sunflower+bookshelf d bookshelf+tapestry d tapestry=\\frac{1}{2} d\\left(sunflower^{2}+bookshelf^{2}+tapestry^{2}\\right)\n\\]\nvanish on the surface. We conclude that \\( sunflower^{2}+bookshelf^{2}+tapestry^{2} \\) is constant.\nSince the surface is required to contain \\( (1,1,1) \\) it must be a part of the sphere (1). Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "unvarying",
        "y": "timeless",
        "z": "placidval",
        "A": "emptiness",
        "B": "nothingness",
        "C": "nullspace",
        "P": "circumspot",
        "O": "infinitypt",
        "\\\\Pi": "curvedsurf"
      },
      "question": "4. Determine the equation of a surface in three dimensional cartesian space which has the following properties: (a) it passes through the point \\( (1,1 \\), 1); and (b) if the tangent plane be drawn at any point \\( circumspot \\), and \\( emptiness, nothingness \\) and \\( nullspace \\) are the intersections of this plane with the \\( unvarying, timeless \\) and \\( placidval \\) axes respectively, then \\( circumspot \\) is the orthocenter (intersection of the altitudes) of the triangle \\( emptiness nothingness nullspace \\).",
      "solution": "Solution. Consider any plane \\( curvedsurf \\) that crosses the coordinate axes at three distinct points \\( emptiness, nothingness \\), and \\( nullspace \\), and let \\( circumspot \\) be the orthocenter of the triangle \\( emptiness nothingness nullspace \\). Treat the points as vectors from the origin, use \"( , )\" to denote the inner product of two vectors, and recall that \\( (emptiness, nothingness)=(nothingness, nullspace)= \\) \\( (nullspace, emptiness)=0 \\) since the axes are orthogonal. Then the orthocenter property becomes\n\\[\n(circumspot-emptiness, nothingness-nullspace)=(circumspot-nothingness, nullspace-emptiness)=(circumspot-nullspace, emptiness-nothingness)=0\n\\]\n\nUsing the linearity properties of the inner product, we get\n\\[\n(circumspot, nothingness-nullspace)=(circumspot, nullspace-emptiness)=(circumspot, emptiness-nothingness)=0\n\\]\n\nThus \\( circumspot \\), as a vector, is orthogonal to the plane \\( curvedsurf \\). This means that \\( circumspot \\) is the foot of the perpendicular on \\( curvedsurf \\) from the origin \\( infinitypt \\).\n\nThe condition on the surface is, then, that every normal passes through the origin. Obviously the sphere\n\\[\nunvarying^{2}+timeless^{2}+placidval^{2}=3\n\\]\nhas this property and passes through the required point \\( (1,1,1) \\).\nA surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere (1) satisfying \\( unvarying>0 \\), \\( timeless>0, placidval>0 \\). [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form\n\\[\nunvarying d unvarying+timeless d timeless+placidval d placidval=\\frac{1}{2} d\\left(unvarying^{2}+timeless^{2}+placidval^{2}\\right)\n\\]\nvanish on the surface. We conclude that \\( unvarying^{2}+timeless^{2}+placidval^{2} \\) is constant.\nSince the surface is required to contain \\( (1,1,1) \\) it must be a part of the sphere (1). Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mfldpqrs",
        "A": "kjsdoqwe",
        "B": "flmqivnr",
        "C": "tndrhsap",
        "P": "ghazuepl",
        "O": "weuirtcv",
        "\\Pi": "zxclmnvb"
      },
      "question": "4. Determine the equation of a surface in three dimensional cartesian space which has the following properties: (a) it passes through the point \\( (1,1 , 1)\\); and (b) if the tangent plane be drawn at any point \\( ghazuepl \\), and \\( kjsdoqwe, flmqivnr \\) and \\( tndrhsap \\) are the intersections of this plane with the \\( qzxwvtnp, hjgrksla \\) and \\( mfldpqrs \\) axes respectively, then \\( ghazuepl \\) is the orthocenter (intersection of the altitudes) of the triangle \\( kjsdoqwe flmqivnr tndrhsap \\).",
      "solution": "Solution. Consider any plane \\( zxclmnvb \\) that crosses the coordinate axes at three distinct points \\( kjsdoqwe, flmqivnr \\), and \\( tndrhsap \\), and let \\( ghazuepl \\) be the orthocenter of the triangle \\( kjsdoqwe flmqivnr tndrhsap \\). Treat the points as vectors from the origin, use \"( , )\" to denote the inner product of two vectors, and recall that \\( (kjsdoqwe, flmqivnr)=(flmqivnr, tndrhsap)= (tndrhsap, kjsdoqwe)=0 \\) since the axes are orthogonal. Then the orthocenter property becomes\n\\[\n(ghazuepl-kjsdoqwe, flmqivnr-tndrhsap)=(ghazuepl-flmqivnr, tndrhsap-kjsdoqwe)=(ghazuepl-tndrhsap, kjsdoqwe-flmqivnr)=0\n\\]\n\nUsing the linearity properties of the inner product, we get\n\\[\n(ghazuepl, flmqivnr-tndrhsap)=(ghazuepl, tndrhsap-kjsdoqwe)=(ghazuepl, kjsdoqwe-flmqivnr)=0\n\\]\n\nThus \\( ghazuepl \\), as a vector, is orthogonal to the plane \\( zxclmnvb \\). This means that \\( ghazuepl \\) is the foot of the perpendicular on \\( zxclmnvb \\) from the origin \\( weuirtcv \\).\n\nThe condition on the surface is, then, that every normal passes through the origin. Obviously the sphere\n\\[\nqzxwvtnp^{2}+hjgrksla^{2}+mfldpqrs^{2}=3\n\\]\nhas this property and passes through the required point \\( (1,1,1) \\).\nA surface that satisfies the original conditions, however, cannot contain a point of a coordinate plane, since it is impossible (by the above analysis) that such a point be the orthocenter of a triangle whose vertices are on the coordinate axes. We shall show that the largest connected surface satisfying the conditions is that portion of the sphere (1) satisfying \\( qzxwvtnp>0 \\), \\( hjgrksla>0, mfldpqrs>0 \\). [We consider only connected surfaces, since without this restriction we can adjoin portions of other spheres at will.] The condition on the surface is that the differential form\n\\[\nqzxwvtnp d qzxwvtnp+hjgrksla d hjgrksla+mfldpqrs d mfldpqrs=\\frac{1}{2} d\\left(qzxwvtnp^{2}+hjgrksla^{2}+mfldpqrs^{2}\\right)\n\\]\nvanish on the surface. We conclude that \\( qzxwvtnp^{2}+hjgrksla^{2}+mfldpqrs^{2} \\) is constant.\nSince the surface is required to contain \\( (1,1,1) \\) it must be a part of the sphere (1). Since, as we have seen, it can contain no point of the coordinate planes, it must lie entirely in the first octant."
    },
    "kernel_variant": {
      "question": "Let ℝ⁴ be endowed with its standard Euclidean inner product ⟨·,·⟩.  \nConsider the four pair-wise orthogonal non-zero vectors  \n\n  v₁ =(1, 1, 1, 1), v₂ =(1, −1, 1, −1), v₃ =(1, 1, −1, −1), v₄ =(1, −1, −1, 1)\n\nand the four lines through the origin  \n\n  ℓᵢ = { t vᵢ : t ∈ ℝ }, i = 1,2,3,4.\n\nA smooth connected three-dimensional hypersurface Σ ⊂ ℝ⁴ satisfies\n\n(a)  Σ passes through the point P₀ =(4, 1, 1, 1);  \n\n(b)  for every P ∈ Σ the tangent hyperplane Πₚ to Σ at P meets each line ℓᵢ in exactly one finite point Aᵢ (i = 1,2,3,4); the four points A₁, A₂, A₃, A₄ are pair-wise distinct, and P is the orthocentre of the tetrahedron A₁A₂A₃A₄ (i.e. the four segments PAᵢ are its altitudes);  \n\n(c)  Σ is maximal with respect to (a) and (b): if Σ ⊂ Σ′ and the connected hypersurface Σ′ also fulfils (a) and (b), then Σ′ = Σ.\n\nDetermine explicitly all points of Σ; that is, give a system of equalities and strict inequalities that a point  \nx = (x₁,x₂,x₃,x₄) must satisfy in order to belong to Σ.",
      "solution": "Notation.  Orthogonality is always understood with respect to the standard inner product.\n\nStep 1.  The intersection points Aᵢ.  \nBecause Πₚ meets ℓᵢ in exactly one finite point, there exists a non-zero scalar αᵢ such that\n\n  Aᵢ = αᵢ vᵢ  (αᵢ ≠ 0, i = 1,…,4). (1)\n\nIf n is any non-zero normal of Πₚ we have the affine equation\n\n  Πₚ : ⟨n,x⟩ = ⟨n,P⟩. (2)\n\nConsequently αᵢ = ⟨n,P⟩ / ⟨n,vᵢ⟩.  In particular αᵢ ≠ 0 implies ⟨n,vᵢ⟩ ≠ 0.\n\nStep 2.  Vector form of the orthocentre condition.  \nLet H be the orthocentre of the tetrahedron A₁A₂A₃A₄ that lies in Πₚ.  \nIn an affine 3-space, H is the common intersection of the four altitudes; equivalently\n\n  ⟨H − Aᵢ , Aⱼ − A_k⟩ = 0 for every choice of pairwise distinct indices i,j,k. (3)\n\nHypothesis (b) states H = P, so we substitute H = P and Aᵢ = αᵢ vᵢ in (3).\n\nStep 3.  Using the orthogonality of the vᵢ.  \nFix i and two distinct indices j,k different from i.  Because vᵢ ⟂ vⱼ and vᵢ ⟂ v_k we have\n\n  ⟨vᵢ , Aⱼ − A_k⟩ = 0. (4)\n\nInsert (4) and (1) into (3) to obtain\n\n  ⟨P , Aⱼ − A_k⟩ = 0 for every unordered pair {j,k}. (5)\n\nStep 4.  The tangent hyperplane is radial.  \nFor a non-degenerate tetrahedron the three vectors A₂−A₁, A₃−A₁, A₄−A₁ span the direction space of Πₚ; hence (5) means precisely\n\n  P ⟂ Πₚ. (6)\n\nThus P itself can be taken as a normal of Πₚ.  Choosing n = P in (2) gives\n\n  Πₚ : ⟨P,x⟩ = ⟨P,P⟩ = ‖P‖². (7)\n\nConsequently, the normal line of Σ at every point is its radial line through the origin.\n\nStep 5.  Σ lies on a sphere centred at the origin.  \nA smooth hypersurface Φ whose position vector is normal to every tangent space is necessarily contained in a level set of the function f(x)=‖x‖², because the 1-form\n\n  ω := x₁ dx₁ + x₂ dx₂ + x₃ dx₃ + x₄ dx₄ = ½ d‖x‖²\n\nvanishes on every tangent space of Φ.  On any connected component, ‖x‖² is therefore constant.\n\nSince P₀ ∈ Σ, we conclude that every x ∈ Σ satisfies\n\n  ‖x‖² = ‖P₀‖² = 4² + 1² + 1² + 1² = 19. (8)\n\nThus  \n\n  Σ ⊂ 𝕊³(√19) := { x ∈ ℝ⁴ : x₁² + x₂² + x₃² + x₄² = 19 }. (9)\n\nStep 6.  The hyperplane is never parallel to any ℓᵢ.  \nCondition (b) also requires that Πₚ meet every ℓᵢ in a finite point, i.e. Πₚ is not parallel to vᵢ.  By (7) this is equivalent to\n\n  ⟨P,vᵢ⟩ ≠ 0 (i = 1,2,3,4). (10)\n\nAt P₀ we compute\n\n  ⟨v₁,P₀⟩ = 7, ⟨v₂,P₀⟩ = 3, ⟨v₃,P₀⟩ = 3, ⟨v₄,P₀⟩ = 3,\n\nall of which are positive.  Because Σ is connected and each function x ↦ ⟨vᵢ,x⟩ has no zero on Σ by (10), every ⟨vᵢ,x⟩ keeps its sign throughout Σ; hence\n\n  ⟨vᵢ,x⟩ > 0 for i = 1,2,3,4 and all x ∈ Σ. (11)\n\nStep 7.  Re-establishing the orthocentre property for every point.  \nLet P be an arbitrary point of the set\n\n  𝕊³₊ := { x ∈ ℝ⁴ : ‖x‖² = 19 and ⟨vᵢ,x⟩ > 0 for i=1,2,3,4 }. (12)\n\nDefine\n\n  αᵢ := ‖P‖² / ⟨P,vᵢ⟩ = 19 / ⟨P,vᵢ⟩  (positive by (11)), and Aᵢ := αᵢ vᵢ. (13)\n\nBecause Πₚ is given by (7), each Aᵢ lies on Πₚ (indeed ⟨P,Aᵢ⟩ = ‖P‖²), so Aᵢ is the unique intersection point Πₚ ∩ ℓᵢ.\n\nNow fix distinct indices i,j,k.  Using (13) and the orthogonality of the vᵢ we compute\n\n  ⟨P , Aⱼ − A_k⟩  \n   = αⱼ⟨P , vⱼ⟩ − α_k⟨P , v_k⟩  \n   = ‖P‖² − ‖P‖² = 0. (14)\n\nBecause ⟨vᵢ , Aⱼ − A_k⟩ = 0, we finally get\n\n  ⟨P − Aᵢ , Aⱼ − A_k⟩ = ⟨P , Aⱼ − A_k⟩ − αᵢ⟨vᵢ , Aⱼ − A_k⟩ = 0 − 0 = 0. (15)\n\nHence P is indeed the orthocentre of the tetrahedron A₁A₂A₃A₄.  This verifies property (b) for every point of 𝕊³₊.\n\nStep 8.  Maximality.  \nThe subset 𝕊³₊ of the sphere (9) is connected (it is the intersection of the sphere with four open hemispheres).  We have shown that every point of 𝕊³₊ satisfies (a) and (b), and Σ itself is contained in 𝕊³₊ by Steps 5–6.  Property (c) (maximality) therefore forces\n\n  Σ = 𝕊³₊. (16)\n\nFinal description.  \nA point x = (x₁,x₂,x₃,x₄) lies on Σ iff\n\n  x₁² + x₂² + x₃² + x₄² = 19           (equal­ity)  \n  x₁ + x₂ + x₃ + x₄      > 0  \n  x₁ − x₂ + x₃ − x₄     > 0  \n  x₁ + x₂ − x₃ − x₄     > 0  \n  x₁ − x₂ − x₃ + x₄     > 0.       (in­e­qual­ities)\n\nConsequently Σ is the “positive vᵢ-orthant’’ of the 3-sphere of radius √19.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.459899",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The problem moves from ℝ³ (surface of codimension 1) to ℝ⁴ (hypersurface), obliging the solver to juggle four coordinates, a three-dimensional tangent space, and a tetrahedral rather than triangular configuration.  \n\n• More variables and objects.  Instead of three axis-parallel lines we have four lines in arbitrary (yet orthogonal) directions, producing a tetrahedron inside the tangent hyperplane.  \n\n• Sophisticated geometry.  Understanding concurrency of altitudes in a tetrahedron and translating it into inner-product identities (step 3) is markedly subtler than the planar orthocentre of a triangle.  \n\n• Linear-algebraic subtlety.  The solver must recognise that radial normal vectors imply constancy of the squared norm via the differential 1-form ω = ½ d(‖x‖²), a step absent from the original exercise.  \n\n• Sign analysis.  Additional inequalities (13) are required to guarantee that the intersections with the lines are finite and distinct, something the original problem could avoid by working in the positive octant only.  \n\n• Maximality argument.  After finding the sphere the solver must still determine which connected component is selected by the sign constraints and justify that no further enlargement is possible, integrating topology (connected components) with geometry.  \n\nThese layers demand substantially deeper geometric insight, broader linear-algebraic competence, and more elaborate reasoning than the original and the first kernel variant, fulfilling the brief of a significantly harder problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let ℝ⁴ be endowed with its standard Euclidean inner product ⟨·,·⟩.  \nConsider the four pair-wise orthogonal non-zero vectors  \n\n  v₁ =(1, 1, 1, 1), v₂ =(1, −1, 1, −1), v₃ =(1, 1, −1, −1), v₄ =(1, −1, −1, 1)\n\nand the four lines through the origin  \n\n  ℓᵢ = { t vᵢ : t ∈ ℝ }, i = 1,2,3,4.\n\nA smooth connected three-dimensional hypersurface Σ ⊂ ℝ⁴ satisfies\n\n(a)  Σ passes through the point P₀ =(4, 1, 1, 1);  \n\n(b)  for every P ∈ Σ the tangent hyperplane Πₚ to Σ at P meets each line ℓᵢ in exactly one finite point Aᵢ (i = 1,2,3,4); the four points A₁, A₂, A₃, A₄ are pair-wise distinct, and P is the orthocentre of the tetrahedron A₁A₂A₃A₄ (i.e. the four segments PAᵢ are its altitudes);  \n\n(c)  Σ is maximal with respect to (a) and (b): if Σ ⊂ Σ′ and the connected hypersurface Σ′ also fulfils (a) and (b), then Σ′ = Σ.\n\nDetermine explicitly all points of Σ; that is, give a system of equalities and strict inequalities that a point  \nx = (x₁,x₂,x₃,x₄) must satisfy in order to belong to Σ.",
      "solution": "Notation.  Orthogonality is always understood with respect to the standard inner product.\n\nStep 1.  The intersection points Aᵢ.  \nBecause Πₚ meets ℓᵢ in exactly one finite point, there exists a non-zero scalar αᵢ such that\n\n  Aᵢ = αᵢ vᵢ  (αᵢ ≠ 0, i = 1,…,4). (1)\n\nIf n is any non-zero normal of Πₚ we have the affine equation\n\n  Πₚ : ⟨n,x⟩ = ⟨n,P⟩. (2)\n\nConsequently αᵢ = ⟨n,P⟩ / ⟨n,vᵢ⟩.  In particular αᵢ ≠ 0 implies ⟨n,vᵢ⟩ ≠ 0.\n\nStep 2.  Vector form of the orthocentre condition.  \nLet H be the orthocentre of the tetrahedron A₁A₂A₃A₄ that lies in Πₚ.  \nIn an affine 3-space, H is the common intersection of the four altitudes; equivalently\n\n  ⟨H − Aᵢ , Aⱼ − A_k⟩ = 0 for every choice of pairwise distinct indices i,j,k. (3)\n\nHypothesis (b) states H = P, so we substitute H = P and Aᵢ = αᵢ vᵢ in (3).\n\nStep 3.  Using the orthogonality of the vᵢ.  \nFix i and two distinct indices j,k different from i.  Because vᵢ ⟂ vⱼ and vᵢ ⟂ v_k we have\n\n  ⟨vᵢ , Aⱼ − A_k⟩ = 0. (4)\n\nInsert (4) and (1) into (3) to obtain\n\n  ⟨P , Aⱼ − A_k⟩ = 0 for every unordered pair {j,k}. (5)\n\nStep 4.  The tangent hyperplane is radial.  \nFor a non-degenerate tetrahedron the three vectors A₂−A₁, A₃−A₁, A₄−A₁ span the direction space of Πₚ; hence (5) means precisely\n\n  P ⟂ Πₚ. (6)\n\nThus P itself can be taken as a normal of Πₚ.  Choosing n = P in (2) gives\n\n  Πₚ : ⟨P,x⟩ = ⟨P,P⟩ = ‖P‖². (7)\n\nConsequently, the normal line of Σ at every point is its radial line through the origin.\n\nStep 5.  Σ lies on a sphere centred at the origin.  \nA smooth hypersurface Φ whose position vector is normal to every tangent space is necessarily contained in a level set of the function f(x)=‖x‖², because the 1-form\n\n  ω := x₁ dx₁ + x₂ dx₂ + x₃ dx₃ + x₄ dx₄ = ½ d‖x‖²\n\nvanishes on every tangent space of Φ.  On any connected component, ‖x‖² is therefore constant.\n\nSince P₀ ∈ Σ, we conclude that every x ∈ Σ satisfies\n\n  ‖x‖² = ‖P₀‖² = 4² + 1² + 1² + 1² = 19. (8)\n\nThus  \n\n  Σ ⊂ 𝕊³(√19) := { x ∈ ℝ⁴ : x₁² + x₂² + x₃² + x₄² = 19 }. (9)\n\nStep 6.  The hyperplane is never parallel to any ℓᵢ.  \nCondition (b) also requires that Πₚ meet every ℓᵢ in a finite point, i.e. Πₚ is not parallel to vᵢ.  By (7) this is equivalent to\n\n  ⟨P,vᵢ⟩ ≠ 0 (i = 1,2,3,4). (10)\n\nAt P₀ we compute\n\n  ⟨v₁,P₀⟩ = 7, ⟨v₂,P₀⟩ = 3, ⟨v₃,P₀⟩ = 3, ⟨v₄,P₀⟩ = 3,\n\nall of which are positive.  Because Σ is connected and each function x ↦ ⟨vᵢ,x⟩ has no zero on Σ by (10), every ⟨vᵢ,x⟩ keeps its sign throughout Σ; hence\n\n  ⟨vᵢ,x⟩ > 0 for i = 1,2,3,4 and all x ∈ Σ. (11)\n\nStep 7.  Re-establishing the orthocentre property for every point.  \nLet P be an arbitrary point of the set\n\n  𝕊³₊ := { x ∈ ℝ⁴ : ‖x‖² = 19 and ⟨vᵢ,x⟩ > 0 for i=1,2,3,4 }. (12)\n\nDefine\n\n  αᵢ := ‖P‖² / ⟨P,vᵢ⟩ = 19 / ⟨P,vᵢ⟩  (positive by (11)), and Aᵢ := αᵢ vᵢ. (13)\n\nBecause Πₚ is given by (7), each Aᵢ lies on Πₚ (indeed ⟨P,Aᵢ⟩ = ‖P‖²), so Aᵢ is the unique intersection point Πₚ ∩ ℓᵢ.\n\nNow fix distinct indices i,j,k.  Using (13) and the orthogonality of the vᵢ we compute\n\n  ⟨P , Aⱼ − A_k⟩  \n   = αⱼ⟨P , vⱼ⟩ − α_k⟨P , v_k⟩  \n   = ‖P‖² − ‖P‖² = 0. (14)\n\nBecause ⟨vᵢ , Aⱼ − A_k⟩ = 0, we finally get\n\n  ⟨P − Aᵢ , Aⱼ − A_k⟩ = ⟨P , Aⱼ − A_k⟩ − αᵢ⟨vᵢ , Aⱼ − A_k⟩ = 0 − 0 = 0. (15)\n\nHence P is indeed the orthocentre of the tetrahedron A₁A₂A₃A₄.  This verifies property (b) for every point of 𝕊³₊.\n\nStep 8.  Maximality.  \nThe subset 𝕊³₊ of the sphere (9) is connected (it is the intersection of the sphere with four open hemispheres).  We have shown that every point of 𝕊³₊ satisfies (a) and (b), and Σ itself is contained in 𝕊³₊ by Steps 5–6.  Property (c) (maximality) therefore forces\n\n  Σ = 𝕊³₊. (16)\n\nFinal description.  \nA point x = (x₁,x₂,x₃,x₄) lies on Σ iff\n\n  x₁² + x₂² + x₃² + x₄² = 19           (equal­ity)  \n  x₁ + x₂ + x₃ + x₄      > 0  \n  x₁ − x₂ + x₃ − x₄     > 0  \n  x₁ + x₂ − x₃ − x₄     > 0  \n  x₁ − x₂ − x₃ + x₄     > 0.       (in­e­qual­ities)\n\nConsequently Σ is the “positive vᵢ-orthant’’ of the 3-sphere of radius √19.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.391492",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The problem moves from ℝ³ (surface of codimension 1) to ℝ⁴ (hypersurface), obliging the solver to juggle four coordinates, a three-dimensional tangent space, and a tetrahedral rather than triangular configuration.  \n\n• More variables and objects.  Instead of three axis-parallel lines we have four lines in arbitrary (yet orthogonal) directions, producing a tetrahedron inside the tangent hyperplane.  \n\n• Sophisticated geometry.  Understanding concurrency of altitudes in a tetrahedron and translating it into inner-product identities (step 3) is markedly subtler than the planar orthocentre of a triangle.  \n\n• Linear-algebraic subtlety.  The solver must recognise that radial normal vectors imply constancy of the squared norm via the differential 1-form ω = ½ d(‖x‖²), a step absent from the original exercise.  \n\n• Sign analysis.  Additional inequalities (13) are required to guarantee that the intersections with the lines are finite and distinct, something the original problem could avoid by working in the positive octant only.  \n\n• Maximality argument.  After finding the sphere the solver must still determine which connected component is selected by the sign constraints and justify that no further enlargement is possible, integrating topology (connected components) with geometry.  \n\nThese layers demand substantially deeper geometric insight, broader linear-algebraic competence, and more elaborate reasoning than the original and the first kernel variant, fulfilling the brief of a significantly harder problem."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}