{
  "index": "1938-B-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "12. From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve.",
  "solution": "Solution. Let the axes be the asymptotes, so that \\( x y=a^{2} \\) is the equation of the given hyperbola. Let the point \\( (h, k) \\) be on the hyperbola. Then \\( h k=a^{2} \\) and the equation of the tangent line at \\( (h, k) \\) is \\( h y+k x-2 h k=0 \\).\n\nThe \\( x \\) and \\( y \\) intercepts of this tangent line are \\( 2 h \\) and \\( 2 k \\) respectively. Let \\( (r, \\theta) \\) be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then \\( 2 h \\cos \\theta=r \\) and \\( 2 k \\sin \\theta=r \\), and hence \\( r^{2}=4 h k \\sin \\theta \\cos \\theta \\) or\n\\[\nr^{2}=2 a^{2} \\sin 2 \\theta\n\\]\n\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, \\( P=(r, \\theta) \\), which must be in either the first or third quadrant, the equations \\( 2 h \\cos \\theta=r \\) and \\( 2 k \\sin \\theta=r \\) determine a point \\( (h, k) \\) on the hyperbola and \\( P \\) is the foot of the perpendicular on the tangent at \\( (h, k) \\).\n\nThe locus is the well-known lemniscate of Bernoulli.",
  "vars": [
    "x",
    "y",
    "h",
    "k",
    "r",
    "\\\\theta",
    "P"
  ],
  "params": [
    "a"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizont",
        "y": "vertical",
        "h": "tangentx",
        "k": "tangenty",
        "r": "radialpt",
        "\\theta": "angletheta",
        "P": "footpoint",
        "a": "constant"
      },
      "question": "From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve.",
      "solution": "Solution. Let the axes be the asymptotes, so that \\( horizont vertical=constant^{2} \\) is the equation of the given hyperbola. Let the point \\( (tangentx, tangenty) \\) be on the hyperbola. Then \\( tangentx tangenty=constant^{2} \\) and the equation of the tangent line at \\( (tangentx, tangenty) \\) is \\( tangentx vertical+tangenty horizont-2 tangentx tangenty=0 \\).\n\nThe \\( horizont \\) and \\( vertical \\) intercepts of this tangent line are \\( 2 tangentx \\) and \\( 2 tangenty \\) respectively. Let \\( (radialpt, angletheta) \\) be the polar coordinates of the footpoint of the perpendicular from the origin to the tangent line. Then \\( 2 tangentx \\cos angletheta=radialpt \\) and \\( 2 tangenty \\sin angletheta=radialpt \\), and hence \\( radialpt^{2}=4 tangentx tangenty \\sin angletheta \\cos angletheta \\) or\n\\[\nradialpt^{2}=2 constant^{2} \\sin 2 angletheta\n\\]\n\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, \\( footpoint=(radialpt, angletheta) \\), which must be in either the first or third quadrant, the equations \\( 2 tangentx \\cos angletheta=radialpt \\) and \\( 2 tangenty \\sin angletheta=radialpt \\) determine a point \\( (tangentx, tangenty) \\) on the hyperbola and \\( footpoint \\) is the foot of the perpendicular on the tangent at \\( (tangentx, tangenty) \\).\n\nThe locus is the well-known lemniscate of Bernoulli."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orchardapple",
        "y": "lanternshade",
        "h": "cascadefrost",
        "k": "juniperbloom",
        "r": "quiverstone",
        "\\theta": "whisperwind",
        "P": "zephyrpoint",
        "a": "rivershadow"
      },
      "question": "From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve.",
      "solution": "Let the axes be the asymptotes, so that \\( orchardapple lanternshade=rivershadow^{2} \\) is the equation of the given hyperbola. Let the point \\( (cascadefrost, juniperbloom) \\) be on the hyperbola. Then \\( cascadefrost juniperbloom=rivershadow^{2} \\) and the equation of the tangent line at \\( (cascadefrost, juniperbloom) \\) is \\( cascadefrost lanternshade+juniperbloom orchardapple-2 cascadefrost juniperbloom=0 \\).\n\nThe \\( orchardapple \\) and \\( lanternshade \\) intercepts of this tangent line are \\( 2 cascadefrost \\) and \\( 2 juniperbloom \\) respectively. Let \\( (quiverstone, whisperwind) \\) be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then \\( 2 cascadefrost \\cos whisperwind=quiverstone \\) and \\( 2 juniperbloom \\sin whisperwind=quiverstone \\), and hence \\( quiverstone^{2}=4 cascadefrost juniperbloom \\sin whisperwind \\cos whisperwind \\) or\n\\[\nquiverstone^{2}=2 rivershadow^{2} \\sin 2 whisperwind\n\\]\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, \\( zephyrpoint=(quiverstone, whisperwind) \\), which must be in either the first or third quadrant, the equations \\( 2 cascadefrost \\cos whisperwind=quiverstone \\) and \\( 2 juniperbloom \\sin whisperwind=quiverstone \\) determine a point \\( (cascadefrost, juniperbloom) \\) on the hyperbola and \\( zephyrpoint \\) is the foot of the perpendicular on the tangent at \\( (cascadefrost, juniperbloom) \\).\n\nThe locus is the well-known lemniscate of Bernoulli."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "h": "depthvalue",
        "k": "valleyvalue",
        "r": "proximity",
        "\\theta": "alignment",
        "P": "planararea",
        "a": "variableness"
      },
      "question": "From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve.",
      "solution": "Solution. Let the axes be the asymptotes, so that \\( verticalaxis horizontalaxis=variableness^{2} \\) is the equation of the given hyperbola. Let the point \\( (depthvalue, valleyvalue) \\) be on the hyperbola. Then \\( depthvalue valleyvalue=variableness^{2} \\) and the equation of the tangent line at \\( (depthvalue, valleyvalue) \\) is \\( depthvalue horizontalaxis+valleyvalue verticalaxis-2 depthvalue valleyvalue=0 \\).\n\nThe \\( verticalaxis \\) and \\( horizontalaxis \\) intercepts of this tangent line are \\( 2 depthvalue \\) and \\( 2 valleyvalue \\) respectively. Let \\( (proximity, alignment) \\) be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then \\( 2 depthvalue \\cos alignment=proximity \\) and \\( 2 valleyvalue \\sin alignment=proximity \\), and hence \\( proximity^{2}=4 depthvalue valleyvalue \\sin alignment \\cos alignment \\) or\n\\[\nproximity^{2}=2 variableness^{2} \\sin 2 alignment\n\\]\n\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, \\( planararea=(proximity, alignment) \\), which must be in either the first or third quadrant, the equations \\( 2 depthvalue \\cos alignment=proximity \\) and \\( 2 valleyvalue \\sin alignment=proximity \\) determine a point \\( (depthvalue, valleyvalue) \\) on the hyperbola and \\( planararea \\) is the foot of the perpendicular on the tangent at \\( (depthvalue, valleyvalue) \\).\n\nThe locus is the well-known lemniscate of Bernoulli."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "h": "mnrpqlos",
        "k": "dfghjkwe",
        "r": "bncvxzlu",
        "\\theta": "asdfghjkq",
        "P": "qwertyui",
        "a": "plmoknij"
      },
      "question": "12. From the center of a rectangular hyperbola a perpendicular is dropped upon a variable tangent. Find the locus of the foot of the perpendicular. Obtain the equation of the locus in polar coordinates, and sketch the curve.",
      "solution": "Solution. Let the axes be the asymptotes, so that \\( qzxwvtnp hjgrksla=plmoknij^{2} \\) is the equation of the given hyperbola. Let the point \\( (mnrpqlos, dfghjkwe) \\) be on the hyperbola. Then \\( mnrpqlos dfghjkwe=plmoknij^{2} \\) and the equation of the tangent line at \\( (mnrpqlos, dfghjkwe) \\) is \\( mnrpqlos hjgrksla+dfghjkwe qzxwvtnp-2 mnrpqlos dfghjkwe=0 \\).\n\nThe \\( qzxwvtnp \\) and \\( hjgrksla \\) intercepts of this tangent line are \\( 2 mnrpqlos \\) and \\( 2 dfghjkwe \\) respectively. Let \\( (bncvxzlu, asdfghjkq) \\) be the polar coordinates of the foot of the perpendicular from the origin to the tangent line. Then \\( 2 mnrpqlos \\cos asdfghjkq=bncvxzlu \\) and \\( 2 dfghjkwe \\sin asdfghjkq=bncvxzlu \\), and hence \\( bncvxzlu^{2}=4 mnrpqlos dfghjkwe \\sin asdfghjkq \\cos asdfghjkq \\) or\n\\[\nbncvxzlu^{2}=2 plmoknij^{2} \\sin 2 asdfghjkq\n\\]\n\nThis is the polar equation of the desired locus. We have shown that the foot of every perpendicular lies on (1).\n\nConversely, every point satisfying (1), except the pole, is the foot of some perpendicular: Given such a point, \\( qwertyui=(bncvxzlu, asdfghjkq) \\), which must be in either the first or third quadrant, the equations \\( 2 mnrpqlos \\cos asdfghjkq=bncvxzlu \\) and \\( 2 dfghjkwe \\sin asdfghjkq=bncvxzlu \\) determine a point \\( (mnrpqlos, dfghjkwe) \\) on the hyperbola and \\( qwertyui \\) is the foot of the perpendicular on the tangent at \\( (mnrpqlos, dfghjkwe) \\).\n\nThe locus is the well-known lemniscate of Bernoulli."
    },
    "kernel_variant": {
      "question": "Let  \n\n    𝓗 : x² + y² − z² = a²    (a > 0)               (1)\n\nbe the one-sheet hyperboloid with centre O = (0,0,0) in ℝ³.  \nFor a variable point  \n\n    P = (x₀ , y₀ , z₀) ∈ 𝓗\n\ndenote by Π the tangent plane to 𝓗 at P and by H the (Euclidean) foot of the perpendicular dropped from O onto Π.\n\n(a) Show that the locus of H is the quartic algebraic surface  \n\n     (X² + Y² + Z²)² = a²( X² + Y² − Z² ).              (2)\n\n(b) Introduce spherical coordinates  \n\n    X = ρ sinφ cosθ, Y = ρ sinφ sinθ, Z = ρ cosφ  \n    (ρ ≥ 0, 0 ≤ θ < 2π, 0 ≤ φ ≤ π)\n\nand deduce the polar equation of the locus\n\n    ρ = a √(sin²φ − cos²φ) = a √(−cos 2φ).             (3)\n\nHence prove that real points occur precisely for\n\n    π/4 ≤ φ ≤ 3π/4,                         (4)\n\nand describe geometrically the resulting single, axis-symmetric “belt’’ in ℝ³.\n\n(c) Conversely, prove that every point of the quartic surface (2) different from the origin arises as H for one and only one tangent plane of 𝓗.\n\n(d) Determine the intersection of the locus with the equatorial plane z = 0, show that it is the planar quartic  \n\n    (x² + y²)² = a²(x² + y²),                    (5)\n\nand analyse its geometry.  In particular locate all singular points of (2), decide whether the surface is connected, and describe its qualitative shape.",
      "solution": "Step 0. Preliminaries  \nPut F(x,y,z) := x² + y² − z² − a².  Then 𝓗 = {F = 0}.  The normal vector at  \nP = (x₀, y₀, z₀) is  \n\n    ∇F(P) = (2x₀, 2y₀, −2z₀) =: n.                                         (6)\n\nHence Π has equation  \n\n    n·(x,y,z) = n·P ⇔ x₀x + y₀y − z₀z = a².                              (7)\n\nStep 1. Foot of the perpendicular from O onto Π  \nFor a plane A x + B y + C z = d through distance d/√(A²+B²+C²) from O, the perpendicular foot is  \n\n    H = (d/(A²+B²+C²))(A,B,C).                                            (8)\n\nIn (7) we have A = x₀, B = y₀, C = −z₀, d = a²; therefore  \n\n    H = (a²/(x₀² + y₀² + z₀²))(x₀, y₀, −z₀).                             (9)\n\nWrite H = (X,Y,Z) and r² := x₀² + y₀² + z₀².  Then  \n\n    X = a²x₀/r², Y = a²y₀/r², Z = −a²z₀/r².                              (10)\n\nStep 2. Elimination of the parameters  \nSolve (10) for x₀, y₀, z₀:\n\n    x₀ = r²X/a², y₀ = r²Y/a², z₀ = −r²Z/a².                              (11)\n\nInsert these in r² = x₀² + y₀² + z₀²:\n\n    r² = (r⁴/a⁴)(X² + Y² + Z²) ⇒ r² = a⁴/(X² + Y² + Z²).                (12)\n\nFinally substitute (11) in the surface condition x₀² + y₀² − z₀² = a²:\n\n    (a⁸/(X² + Y² + Z²)²)(X² + Y² − Z²)/a⁴ = a²  \n    ⇔ (X² + Y² + Z²)² = a²(X² + Y² − Z²).                                (2)\n\nThus part (a) is proved.\n\nStep 3. Passage to spherical coordinates  \nPut ρ² := X² + Y² + Z².  Then (2) reads  \n\n    ρ⁴ = a²(ρ² − 2Z²).                                                    (13)\n\nWith X,Y,Z expressed by (ρ,θ,φ),\n\n    Z = ρ cosφ, X²+Y² = ρ² sin²φ,                                        (14)\n\nso that X² + Y² − Z² = ρ²(sin²φ − cos²φ) = −ρ² cos2φ.  \nEquation (13) becomes\n\n    ρ⁴ = a²ρ²(−cos 2φ) ⇒ ρ² = a²(−cos 2φ) ⇒ ρ = a√(−cos 2φ).    (3)\n\nReal solutions require −cos 2φ ≥ 0, i.e. cos 2φ ≤ 0, whence  \n\n    π/4 ≤ φ ≤ 3π/4.                                                       (4)\n\nFor the endpoints φ = π/4, 3π/4 the radical vanishes and the surface contracts to the single point O.  \nFor φ = π/2 one has ρ = a, giving the greatest distance from O; rotating the meridian curve  \nφ ∈ (π/4,3π/4) ↦ (ρ(φ),φ) about the Z–axis yields a smooth, axis-symmetric “belt’’ which closes at O.  Outside O the surface is regular; O is the unique singular (pinch) point.\n\nStep 4. Converse statement (part c)  \nLet H ≠ O satisfy (2).  Define r by (12) and P by (11).  Then r² > 0, and substituting (11) into x₀² + y₀² − z₀² gives a², so P ∈ 𝓗.  Equation (10) shows that H is the perpendicular projection of O onto the tangent plane at P.  Conversely, (11) determines P uniquely because r² is fixed by (12); hence every non-origin point of the quartic is obtained exactly once.  The origin itself is not produced, because no tangent plane of 𝓗 passes through O (see (7)).\n\nStep 5. Intersection with the plane z = 0 and global topology  \nSet Z = 0 in (2):\n\n    (X² + Y²)² = a²(X² + Y²).                                             (15)\n\nWriting x := X, y := Y, this is precisely (x² + y²)² = a²(x² + y²), i.e. (5).  \nThus the intersection consists of\n\n  • the circle x² + y² = a² lying in z = 0, and  \n  • the isolated point O (double point).\n\nTo locate singularities of the quartic surface (2) compute its gradient  \nG := ∇[(X²+Y²+Z²)² − a²(X²+Y²−Z²)] = 4(X²+Y²+Z²)(X,Y,Z) − 2a²(X,Y,−Z). (16)\n\nG vanishes iff X = Y = Z = 0, so O is the only singular point.  \nBecause (θ,φ) ↦ (ρ(φ) sinφ cosθ, ρ(φ) sinφ sinθ, ρ(φ) cosφ) with φ ∈ (π/4,3π/4), θ ∈ [0,2π) is a continuous, surjective parametrisation of the quartic minus O, that set is connected.  Adding the limit point O keeps the surface connected.  Topologically the quartic is a smooth sphere pinched at O; it possesses a circular “waist’’ at z = 0 of radius a and no self-intersection.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.345892",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a plane conic to a 3-dimensional quadratic surface; tangents become tangent planes, and the locus is a surface, not a curve.\n2. Additional algebraic complexity: eliminating the parameters of a point on 𝓗 produces a quartic equation in three variables, requiring careful manipulation of symmetric expressions.\n3. Multiple interacting concepts: differential geometry (normal vectors), linear algebra (orthogonal projections), algebraic geometry (classification of a degree-4 surface), and multivariable calculus (spherical coordinates) are all invoked.\n4. Deeper theoretical requirements: step (c) demands a converse existence-and-uniqueness argument involving normals to quadrics and the geometry of cones of directions.\n5. Richer qualitative analysis: the candidate must study reality conditions in spherical coordinates, identify exceptional sets, and describe the topology of intersection curves.\n\nThese layers of difficulty—dimensional elevation, quartic elimination, and geometric–topological analysis—make the enhanced variant significantly more challenging than both the original textbook problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n    𝓗 : x² + y² − z² = a²    (a > 0)               (1)\n\nbe the one-sheet hyperboloid with centre O = (0,0,0) in ℝ³.  \nFor a variable point  \n\n    P = (x₀ , y₀ , z₀) ∈ 𝓗\n\ndenote by Π the tangent plane to 𝓗 at P and by H the (Euclidean) foot of the perpendicular dropped from O onto Π.\n\n(a) Show that the locus of H is the quartic algebraic surface  \n\n     (X² + Y² + Z²)² = a²( X² + Y² − Z² ).              (2)\n\n(b) Introduce spherical coordinates  \n\n    X = ρ sinφ cosθ, Y = ρ sinφ sinθ, Z = ρ cosφ  \n    (ρ ≥ 0, 0 ≤ θ < 2π, 0 ≤ φ ≤ π)\n\nand deduce the polar equation of the locus\n\n    ρ = a √(sin²φ − cos²φ) = a √(−cos 2φ).             (3)\n\nHence prove that real points occur precisely for\n\n    π/4 ≤ φ ≤ 3π/4,                         (4)\n\nand describe geometrically the resulting single, axis-symmetric “belt’’ in ℝ³.\n\n(c) Conversely, prove that every point of the quartic surface (2) different from the origin arises as H for one and only one tangent plane of 𝓗.\n\n(d) Determine the intersection of the locus with the equatorial plane z = 0, show that it is the planar quartic  \n\n    (x² + y²)² = a²(x² + y²),                    (5)\n\nand analyse its geometry.  In particular locate all singular points of (2), decide whether the surface is connected, and describe its qualitative shape.",
      "solution": "Step 0. Preliminaries  \nPut F(x,y,z) := x² + y² − z² − a².  Then 𝓗 = {F = 0}.  The normal vector at  \nP = (x₀, y₀, z₀) is  \n\n    ∇F(P) = (2x₀, 2y₀, −2z₀) =: n.                                         (6)\n\nHence Π has equation  \n\n    n·(x,y,z) = n·P ⇔ x₀x + y₀y − z₀z = a².                              (7)\n\nStep 1. Foot of the perpendicular from O onto Π  \nFor a plane A x + B y + C z = d through distance d/√(A²+B²+C²) from O, the perpendicular foot is  \n\n    H = (d/(A²+B²+C²))(A,B,C).                                            (8)\n\nIn (7) we have A = x₀, B = y₀, C = −z₀, d = a²; therefore  \n\n    H = (a²/(x₀² + y₀² + z₀²))(x₀, y₀, −z₀).                             (9)\n\nWrite H = (X,Y,Z) and r² := x₀² + y₀² + z₀².  Then  \n\n    X = a²x₀/r², Y = a²y₀/r², Z = −a²z₀/r².                              (10)\n\nStep 2. Elimination of the parameters  \nSolve (10) for x₀, y₀, z₀:\n\n    x₀ = r²X/a², y₀ = r²Y/a², z₀ = −r²Z/a².                              (11)\n\nInsert these in r² = x₀² + y₀² + z₀²:\n\n    r² = (r⁴/a⁴)(X² + Y² + Z²) ⇒ r² = a⁴/(X² + Y² + Z²).                (12)\n\nFinally substitute (11) in the surface condition x₀² + y₀² − z₀² = a²:\n\n    (a⁸/(X² + Y² + Z²)²)(X² + Y² − Z²)/a⁴ = a²  \n    ⇔ (X² + Y² + Z²)² = a²(X² + Y² − Z²).                                (2)\n\nThus part (a) is proved.\n\nStep 3. Passage to spherical coordinates  \nPut ρ² := X² + Y² + Z².  Then (2) reads  \n\n    ρ⁴ = a²(ρ² − 2Z²).                                                    (13)\n\nWith X,Y,Z expressed by (ρ,θ,φ),\n\n    Z = ρ cosφ, X²+Y² = ρ² sin²φ,                                        (14)\n\nso that X² + Y² − Z² = ρ²(sin²φ − cos²φ) = −ρ² cos2φ.  \nEquation (13) becomes\n\n    ρ⁴ = a²ρ²(−cos 2φ) ⇒ ρ² = a²(−cos 2φ) ⇒ ρ = a√(−cos 2φ).    (3)\n\nReal solutions require −cos 2φ ≥ 0, i.e. cos 2φ ≤ 0, whence  \n\n    π/4 ≤ φ ≤ 3π/4.                                                       (4)\n\nFor the endpoints φ = π/4, 3π/4 the radical vanishes and the surface contracts to the single point O.  \nFor φ = π/2 one has ρ = a, giving the greatest distance from O; rotating the meridian curve  \nφ ∈ (π/4,3π/4) ↦ (ρ(φ),φ) about the Z–axis yields a smooth, axis-symmetric “belt’’ which closes at O.  Outside O the surface is regular; O is the unique singular (pinch) point.\n\nStep 4. Converse statement (part c)  \nLet H ≠ O satisfy (2).  Define r by (12) and P by (11).  Then r² > 0, and substituting (11) into x₀² + y₀² − z₀² gives a², so P ∈ 𝓗.  Equation (10) shows that H is the perpendicular projection of O onto the tangent plane at P.  Conversely, (11) determines P uniquely because r² is fixed by (12); hence every non-origin point of the quartic is obtained exactly once.  The origin itself is not produced, because no tangent plane of 𝓗 passes through O (see (7)).\n\nStep 5. Intersection with the plane z = 0 and global topology  \nSet Z = 0 in (2):\n\n    (X² + Y²)² = a²(X² + Y²).                                             (15)\n\nWriting x := X, y := Y, this is precisely (x² + y²)² = a²(x² + y²), i.e. (5).  \nThus the intersection consists of\n\n  • the circle x² + y² = a² lying in z = 0, and  \n  • the isolated point O (double point).\n\nTo locate singularities of the quartic surface (2) compute its gradient  \nG := ∇[(X²+Y²+Z²)² − a²(X²+Y²−Z²)] = 4(X²+Y²+Z²)(X,Y,Z) − 2a²(X,Y,−Z). (16)\n\nG vanishes iff X = Y = Z = 0, so O is the only singular point.  \nBecause (θ,φ) ↦ (ρ(φ) sinφ cosθ, ρ(φ) sinφ sinθ, ρ(φ) cosφ) with φ ∈ (π/4,3π/4), θ ∈ [0,2π) is a continuous, surjective parametrisation of the quartic minus O, that set is connected.  Adding the limit point O keeps the surface connected.  Topologically the quartic is a smooth sphere pinched at O; it possesses a circular “waist’’ at z = 0 of radius a and no self-intersection.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.302274",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a plane conic to a 3-dimensional quadratic surface; tangents become tangent planes, and the locus is a surface, not a curve.\n2. Additional algebraic complexity: eliminating the parameters of a point on 𝓗 produces a quartic equation in three variables, requiring careful manipulation of symmetric expressions.\n3. Multiple interacting concepts: differential geometry (normal vectors), linear algebra (orthogonal projections), algebraic geometry (classification of a degree-4 surface), and multivariable calculus (spherical coordinates) are all invoked.\n4. Deeper theoretical requirements: step (c) demands a converse existence-and-uniqueness argument involving normals to quadrics and the geometry of cones of directions.\n5. Richer qualitative analysis: the candidate must study reality conditions in spherical coordinates, identify exceptional sets, and describe the topology of intersection curves.\n\nThese layers of difficulty—dimensional elevation, quartic elimination, and geometric–topological analysis—make the enhanced variant significantly more challenging than both the original textbook problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}