{
  "index": "1994-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $A$ be the area of the region in the first quadrant bounded by the\nline $y = \\frac{1}{2} x$, the $x$-axis, and the ellipse $\\frac{1}{9} x^2\n+ y^2 = 1$. Find the positive number $m$ such that $A$ is equal to the\narea of the region in the first quadrant bounded by the line $y = mx$,\nthe $y$-axis, and the ellipse $\\frac{1}{9} x^2 + y^2 = 1$.",
  "solution": "Solution 1. The linear transformation given by \\( x_{1}=x / 3, y_{1}=y \\) transforms the region \\( R \\) bounded by \\( y=x / 2 \\), the \\( x \\)-axis, and the ellipse \\( x^{2} / 9+y^{2}=1 \\) into the region \\( R^{\\prime} \\) bounded by \\( y_{1}=3 x_{1} / 2 \\), the \\( x_{1} \\)-axis, and the circle \\( x_{1}^{2}+y_{1}^{2}=1 \\); it also transforms the region \\( S \\) bounded by \\( y=m x \\), the \\( y \\)-axis, and \\( x^{2} / 9+y^{2}=1 \\) into the region \\( S^{\\prime \\prime} \\) bounded by \\( y_{1}=3 m x_{1} \\), the \\( y_{1} \\)-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the linear transformation, \\( R \\) and \\( S \\) have the same area if and only if \\( R^{\\prime} \\) and \\( S^{\\prime} \\) have the same area. However, we can see by symmetry about the line \\( y_{1}=x_{1} \\) that this happens if and only if \\( 3 m=2 / 3 \\), that is, \\( m=2 / 9 \\).\n\nSolution 2 (Noam Elkies). Apply the linear transformation \\( (x, y) \\rightarrow(3 y, x / 3) \\). This preserves area, and the ellipse \\( x^{2} / 9+y^{2}=1 \\). It switches the \\( x \\) and \\( y \\) axes, and takes \\( y=x / 2 \\) to the desired line, \\( x / 3=(3 y / 2) \\), i.e. \\( y=(2 / 9) x \\). Thus \\( m=2 / 9 \\).",
  "vars": [
    "x",
    "x_1",
    "y",
    "y_1"
  ],
  "params": [
    "A",
    "R",
    "S",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizvar",
        "x_1": "horizprime",
        "y": "vertvar",
        "y_1": "vertprime",
        "A": "regionarea",
        "R": "regionorig",
        "S": "regionswap",
        "m": "slopevalue"
      },
      "question": "Let $regionarea$ be the area of the region in the first quadrant bounded by the\nline $vertvar = \\frac{1}{2} horizvar$, the $horizvar$-axis, and the ellipse $\\frac{1}{9} horizvar^2\n+ vertvar^2 = 1$. Find the positive number $slopevalue$ such that $regionarea$ is equal to the\narea of the region in the first quadrant bounded by the line $vertvar = slopevalue horizvar$,\nthe $vertvar$-axis, and the ellipse $\\frac{1}{9} horizvar^2 + vertvar^2 = 1$.",
      "solution": "Solution 1. The linear transformation given by \\( horizprime=horizvar / 3, vertprime=vertvar \\) transforms the region \\( regionorig \\) bounded by \\( vertvar=horizvar / 2 \\), the \\( horizvar \\)-axis, and the ellipse \\( horizvar^{2} / 9+vertvar^{2}=1 \\) into the region \\( regionorig^{\\prime} \\) bounded by \\( vertprime=3 horizprime / 2 \\), the \\( horizprime \\)-axis, and the circle \\( horizprime^{2}+vertprime^{2}=1 \\); it also transforms the region \\( regionswap \\) bounded by \\( vertvar=slopevalue horizvar \\), the \\( vertvar \\)-axis, and \\( horizvar^{2} / 9+vertvar^{2}=1 \\) into the region \\( regionswap^{\\prime \\prime} \\) bounded by \\( vertprime=3 slopevalue horizprime \\), the \\( vertprime \\)-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the linear transformation, \\( regionorig \\) and \\( regionswap \\) have the same area if and only if \\( regionorig^{\\prime} \\) and \\( regionswap^{\\prime} \\) have the same area. However, we can see by symmetry about the line \\( vertprime=horizprime \\) that this happens if and only if \\( 3 slopevalue=2 / 3 \\), that is, \\( slopevalue=2 / 9 \\).\n\nSolution 2 (Noam Elkies). Apply the linear transformation \\( (horizvar, vertvar) \\rightarrow(3 vertvar, horizvar / 3) \\). This preserves area, and the ellipse \\( horizvar^{2} / 9+vertvar^{2}=1 \\). It switches the \\( horizvar \\) and \\( vertvar \\) axes, and takes \\( vertvar=horizvar / 2 \\) to the desired line, \\( horizvar / 3=(3 vertvar / 2) \\), i.e. \\( vertvar=(2 / 9) horizvar \\). Thus \\( slopevalue=2 / 9 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "buttercup",
        "x_1": "weathervane",
        "y": "tangerine",
        "y_1": "hummingbird",
        "A": "pendulum",
        "R": "windshield",
        "S": "dockyard",
        "m": "raincloud"
      },
      "question": "Let $pendulum$ be the area of the region in the first quadrant bounded by the\nline $tangerine = \\frac{1}{2} buttercup$, the $buttercup$-axis, and the ellipse $\\frac{1}{9} buttercup^2\n+ tangerine^2 = 1$. Find the positive number $raincloud$ such that $pendulum$ is equal to the\narea of the region in the first quadrant bounded by the line $tangerine = raincloud buttercup$,\nthe $tangerine$-axis, and the ellipse $\\frac{1}{9} buttercup^2 + tangerine^2 = 1$.",
      "solution": "Solution 1. The linear transformation given by \\( weathervane = buttercup / 3, hummingbird = tangerine \\) transforms the region \\( windshield \\) bounded by \\( tangerine = buttercup / 2 \\), the \\( buttercup \\)-axis, and the ellipse \\( buttercup^{2} / 9+tangerine^{2}=1 \\) into the region \\( windshield^{\\prime} \\) bounded by \\( hummingbird = 3 weathervane / 2 \\), the \\( weathervane \\)-axis, and the circle \\( weathervane^{2}+hummingbird^{2}=1 \\); it also transforms the region \\( dockyard \\) bounded by \\( tangerine = raincloud buttercup \\), the \\( tangerine \\)-axis, and \\( buttercup^{2} / 9+tangerine^{2}=1 \\) into the region \\( dockyard^{\\prime \\prime} \\) bounded by \\( hummingbird = 3 raincloud weathervane \\), the \\( hummingbird \\)-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the linear transformation, \\( windshield \\) and \\( dockyard \\) have the same area if and only if \\( windshield^{\\prime} \\) and \\( dockyard^{\\prime} \\) have the same area. However, we can see by symmetry about the line \\( hummingbird = weathervane \\) that this happens if and only if \\( 3 raincloud = 2 / 3 \\), that is, \\( raincloud = 2 / 9 \\).\n\nSolution 2 (Noam Elkies). Apply the linear transformation \\( (buttercup, tangerine) \\rightarrow (3 tangerine, buttercup / 3) \\). This preserves area, and the ellipse \\( buttercup^{2} / 9+tangerine^{2}=1 \\). It switches the \\( buttercup \\) and \\( tangerine \\) axes, and takes \\( tangerine = buttercup / 2 \\) to the desired line, \\( buttercup / 3 = (3 tangerine / 2) \\), i.e. \\( tangerine = (2 / 9) buttercup \\). Thus \\( raincloud = 2 / 9 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "x_1": "verticalprime",
        "y": "horizontalaxis",
        "y_1": "horizontalprime",
        "A": "perimeter",
        "R": "complement",
        "S": "exterior",
        "m": "intercept"
      },
      "question": "Let $perimeter$ be the area of the region in the first quadrant bounded by the\nline $horizontalaxis = \\frac{1}{2} verticalaxis$, the $verticalaxis$-axis, and the ellipse $\\frac{1}{9} verticalaxis^2\n+ horizontalaxis^2 = 1$. Find the positive number $intercept$ such that $perimeter$ is equal to the\narea of the region in the first quadrant bounded by the line $horizontalaxis = intercept\\,verticalaxis$,\nthe $horizontalaxis$-axis, and the ellipse $\\frac{1}{9} verticalaxis^2 + horizontalaxis^2 = 1$.",
      "solution": "Solution 1. The linear transformation given by \\( verticalprime = verticalaxis / 3, horizontalprime = horizontalaxis \\) transforms the region \\( complement \\) bounded by \\( horizontalaxis = verticalaxis / 2 \\), the \\( verticalaxis \\)-axis, and the ellipse \\( verticalaxis^{2} / 9 + horizontalaxis^{2} = 1 \\) into the region \\( complement^{\\prime} \\) bounded by \\( horizontalprime = 3 verticalprime / 2 \\), the \\( verticalprime \\)-axis, and the circle \\( verticalprime^{2} + horizontalprime^{2} = 1 \\); it also transforms the region \\( exterior \\) bounded by \\( horizontalaxis = intercept \\, verticalaxis \\), the \\( horizontalaxis \\)-axis, and \\( verticalaxis^{2} / 9 + horizontalaxis^{2} = 1 \\) into the region \\( exterior^{\\prime \\prime} \\) bounded by \\( horizontalprime = 3 intercept\\, verticalprime \\), the \\( horizontalprime \\)-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the linear transformation, \\( complement \\) and \\( exterior \\) have the same area if and only if \\( complement^{\\prime} \\) and \\( exterior^{\\prime} \\) have the same area. However, we can see by symmetry about the line \\( horizontalprime = verticalprime \\) that this happens if and only if \\( 3 intercept = 2 / 3 \\), that is, \\( intercept = 2 / 9 \\).\n\nSolution 2 (Noam Elkies). Apply the linear transformation \\( (verticalaxis, horizontalaxis) \\rightarrow(3 horizontalaxis, verticalaxis / 3) \\). This preserves area, and the ellipse \\( verticalaxis^{2} / 9 + horizontalaxis^{2} = 1 \\). It switches the \\( verticalaxis \\) and \\( horizontalaxis \\) axes, and takes \\( horizontalaxis = verticalaxis / 2 \\) to the desired line, \\( verticalaxis / 3 = (3 horizontalaxis / 2) \\), i.e. \\( horizontalaxis = (2 / 9) verticalaxis \\). Thus \\( intercept = 2 / 9 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "x_1": "hjgrksla",
        "y": "mnbvcxzs",
        "y_1": "plokijuh",
        "A": "asdfghjk",
        "R": "zxcvbnml",
        "S": "qwertyui",
        "m": "lkjhgfds"
      },
      "question": "Let $asdfghjk$ be the area of the region in the first quadrant bounded by the line $mnbvcxzs = \\frac{1}{2} qzxwvtnp$, the $qzxwvtnp$-axis, and the ellipse $\\frac{1}{9} qzxwvtnp^2 + mnbvcxzs^2 = 1$. Find the positive number $lkjhgfds$ such that $asdfghjk$ is equal to the area of the region in the first quadrant bounded by the line $mnbvcxzs = lkjhgfds qzxwvtnp$, the $mnbvcxzs$-axis, and the ellipse $\\frac{1}{9} qzxwvtnp^2 + mnbvcxzs^2 = 1$.",
      "solution": "Solution 1. The linear transformation given by \\( hjgrksla=qzxwvtnp / 3, plokijuh=mnbvcxzs \\) transforms the region \\( zxcvbnml \\) bounded by \\( mnbvcxzs=qzxwvtnp / 2 \\), the \\( qzxwvtnp \\)-axis, and the ellipse \\( qzxwvtnp^{2} / 9+mnbvcxzs^{2}=1 \\) into the region \\( zxcvbnml^{\\prime} \\) bounded by \\( plokijuh=3 hjgrksla / 2 \\), the \\( hjgrksla \\)-axis, and the circle \\( hjgrksla^{2}+plokijuh^{2}=1 \\); it also transforms the region \\( qwertyui \\) bounded by \\( mnbvcxzs=lkjhgfds qzxwvtnp \\), the \\( mnbvcxzs \\)-axis, and \\( qzxwvtnp^{2} / 9+mnbvcxzs^{2}=1 \\) into the region \\( qwertyui^{\\prime \\prime} \\) bounded by \\( plokijuh=3 lkjhgfds hjgrksla \\), the \\( plokijuh \\)-axis, and the circle. Since all areas are multiplied by the same (nonzero) factor under the linear transformation, \\( zxcvbnml \\) and \\( qwertyui \\) have the same area if and only if \\( zxcvbnml^{\\prime} \\) and \\( qwertyui^{\\prime} \\) have the same area. However, we can see by symmetry about the line \\( plokijuh=hjgrksla \\) that this happens if and only if \\( 3 lkjhgfds=2 / 3 \\), that is, \\( lkjhgfds=2 / 9 \\).\n\nSolution 2 (Noam Elkies). Apply the linear transformation \\( (qzxwvtnp, mnbvcxzs) \\rightarrow(3 mnbvcxzs, qzxwvtnp / 3) \\). This preserves area, and the ellipse \\( qzxwvtnp^{2} / 9+mnbvcxzs^{2}=1 \\). It switches the \\( qzxwvtnp \\) and \\( mnbvcxzs \\) axes, and takes \\( mnbvcxzs=qzxwvtnp / 2 \\) to the desired line, \\( qzxwvtnp / 3=(3 mnbvcxzs / 2) \\), i.e. \\( mnbvcxzs=(2 / 9) qzxwvtnp \\). Thus \\( lkjhgfds=2 / 9 \\)."
    },
    "kernel_variant": {
      "question": "Let 𝔈 be the ellipse in the plane defined by  \n  5x² + 6xy + 5y² = 20.    \nAll regions are to be taken in the first quadrant (x ≥ 0, y ≥ 0).\n\n• Let 𝓡 be the region bounded by  \n  (i) the x-axis,  \n  (ii) the ray y = 4x,  \n  (iii) the ellipse 𝔈.\n\n• For m > 0 let 𝓢ₘ be the region bounded by  \n  (i) the y-axis,  \n  (ii) the ray y = m x,  \n  (iii) the ellipse 𝔈.\n\nDetermine the unique positive real number m for which Area(𝓡) = Area(𝓢ₘ).\n\n────────────────────────────────────────────────────────",
      "solution": "Step 1.  Remove the cross–term by a 45° rotation.  \nWrite the conic in the form Ax² + 2Bxy + Cy² = 20 with  \nA = C = 5, B = 3.  Because A = C, tan 2θ = ∞ and the principal axes are obtained by the rotation\n\n  (u,v) = ( x – y, x + y ) / √2,    i.e.   \n  x = (u – v)/√2,  y = (u + v)/√2.\n\nSubstituting gives  \n 5x² + 6xy + 5y² = 8u² + 2v² = 20,  \nor                             4u² + v² = 10.    (1)\n\nThe Jacobian of the rotation is |det R| = 1, so areas are unchanged.\n\nStep 2.  Express the bounding rays in (u,v).\n\n• x-axis (y = 0):  v = –u.  \n• y = 4x        :  (u + v) = 4(u – v)  ⇒ 5v = 3u  ⇒ v = (3/5)u.  \n• y-axis (x = 0):  v =  u.  \n• y = m x        :  (u + v) = m(u – v)  ⇒  (1–m)u + (1+m)v = 0  \n                      ⇒  v = κu with κ = (m–1)/(m+1).    (2)\n\nStep 3.  Rescale to a circle.  \nFrom (1), define U = 2u/√10, V = v/√10; then U² + V² = 1, a unit circle.  \nThe inverse map has Jacobian |J| = 5, so\n\n Area in (U,V) × 5  =  Area in (u,v)  =  original area.\n\nStep 4.  Convert each ray to an angle on the unit circle.  \nFor a line v = ku the corresponding slope in (U,V) is V/U = k/2, hence its polar\nangle is θ(k) = arctan(k/2).\n\nThe relevant rays and their angles are\n\n x-axis      k = –1    ⇒ θ₁ = θ(–1) = arctan(–½);  \n y = 4x      k = 3/5  ⇒ θ₂ = θ(3/5) = arctan(3/10);  \n y-axis      k =  1    ⇒ θ₃ = θ( 1) = arctan(½);  \n y = m x     k = κ    ⇒ θ₄ = θ(κ) = arctan(κ/2).\n\nStep 5.  Write the two areas in circular form.\n\nInside the unit circle a sector of angular width Δθ has area ½Δθ.  Multiplying by 5\nfor the Jacobian gives area  (5/2) Δθ.\n\n• 𝓡: between θ₁ and θ₂  ⇒  Area(𝓡) = (5/2)(θ₂ – θ₁).  \n• 𝓢ₘ: between θ₄ and θ₃ (note θ₄ < θ₃)  ⇒  Area(𝓢ₘ) = (5/2)(θ₃ – θ₄).\n\nSet them equal:\n\n θ₂ – θ₁ = θ₃ – θ₄.                                                (3)\n\nStep 6.  Insert the explicit angles.  \nUsing θ(k) = arctan(k/2),\n\n arctan(3/10) – arctan(–½) = arctan(½) – arctan(κ/2).\n\nCompute the left–hand side:\n\n arctan(3/10) – arctan(–½) = arctan(3/10) + arctan(½)\n                  = 0.29145679… + 0.46364761…  \n                  = 0.75510440….\n\nHence                                                   \n\n arctan(κ/2) = arctan(½) – 0.75510440…  \n         = –0.29145679… = –arctan(3/10).\n\nTherefore κ/2 = –3/10, so κ = –3/5.\n\nStep 7.  Recover m from κ using (2):\n\n (m – 1)/(m + 1) = –3/5  \n 5(m – 1) = –3(m + 1)  \n 5m – 5 = –3m – 3  \n 8m = 2  \n m = 1/4.\n\nConsequently\n\n m = 1/4.\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.734588",
        "was_fixed": false,
        "difficulty_analysis": "• Rotated Conic:  Unlike the original axis-aligned ellipse, the cross-term 6xy forces a 45° coordinate rotation and identification of principal axes.  \n• Sequential Linear Maps:  A second anisotropic scaling is required to change the rotated ellipse into a unit circle, introducing an additional Jacobian determinant.  \n• Geometry on the Unit Circle:  Areas are converted into angular widths of sectors, demanding fluency with polar geometry and trigonometric identities.  \n• Interacting Constraints:  Four different bounding rays transform to four distinct angles, and the equality condition becomes a transcendental trigonometric equation before reverting to an elementary rational solution.  \n• Multi-step Reasoning:  The solver must combine linear algebra (diagonalising a quadratic form), multivariable calculus (Jacobian determinants), analytic geometry (ray transformations), and trigonometry (sector areas and inverse-tangent equations).  \nThese layers of technical detail and the requisite synthesis of several advanced concepts make the enhanced variant substantially more challenging than either the original problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let 𝔈 be the ellipse in the plane defined by  \n  5x² + 6xy + 5y² = 20.    \nAll regions are to be taken in the first quadrant (x ≥ 0, y ≥ 0).\n\n• Let 𝓡 be the region bounded by  \n  (i) the x-axis,  \n  (ii) the ray y = 4x,  \n  (iii) the ellipse 𝔈.\n\n• For m > 0 let 𝓢ₘ be the region bounded by  \n  (i) the y-axis,  \n  (ii) the ray y = m x,  \n  (iii) the ellipse 𝔈.\n\nDetermine the unique positive real number m for which Area(𝓡) = Area(𝓢ₘ).\n\n────────────────────────────────────────────────────────",
      "solution": "Step 1.  Remove the cross–term by a 45° rotation.  \nWrite the conic in the form Ax² + 2Bxy + Cy² = 20 with  \nA = C = 5, B = 3.  Because A = C, tan 2θ = ∞ and the principal axes are obtained by the rotation\n\n  (u,v) = ( x – y, x + y ) / √2,    i.e.   \n  x = (u – v)/√2,  y = (u + v)/√2.\n\nSubstituting gives  \n 5x² + 6xy + 5y² = 8u² + 2v² = 20,  \nor                             4u² + v² = 10.    (1)\n\nThe Jacobian of the rotation is |det R| = 1, so areas are unchanged.\n\nStep 2.  Express the bounding rays in (u,v).\n\n• x-axis (y = 0):  v = –u.  \n• y = 4x        :  (u + v) = 4(u – v)  ⇒ 5v = 3u  ⇒ v = (3/5)u.  \n• y-axis (x = 0):  v =  u.  \n• y = m x        :  (u + v) = m(u – v)  ⇒  (1–m)u + (1+m)v = 0  \n                      ⇒  v = κu with κ = (m–1)/(m+1).    (2)\n\nStep 3.  Rescale to a circle.  \nFrom (1), define U = 2u/√10, V = v/√10; then U² + V² = 1, a unit circle.  \nThe inverse map has Jacobian |J| = 5, so\n\n Area in (U,V) × 5  =  Area in (u,v)  =  original area.\n\nStep 4.  Convert each ray to an angle on the unit circle.  \nFor a line v = ku the corresponding slope in (U,V) is V/U = k/2, hence its polar\nangle is θ(k) = arctan(k/2).\n\nThe relevant rays and their angles are\n\n x-axis      k = –1    ⇒ θ₁ = θ(–1) = arctan(–½);  \n y = 4x      k = 3/5  ⇒ θ₂ = θ(3/5) = arctan(3/10);  \n y-axis      k =  1    ⇒ θ₃ = θ( 1) = arctan(½);  \n y = m x     k = κ    ⇒ θ₄ = θ(κ) = arctan(κ/2).\n\nStep 5.  Write the two areas in circular form.\n\nInside the unit circle a sector of angular width Δθ has area ½Δθ.  Multiplying by 5\nfor the Jacobian gives area  (5/2) Δθ.\n\n• 𝓡: between θ₁ and θ₂  ⇒  Area(𝓡) = (5/2)(θ₂ – θ₁).  \n• 𝓢ₘ: between θ₄ and θ₃ (note θ₄ < θ₃)  ⇒  Area(𝓢ₘ) = (5/2)(θ₃ – θ₄).\n\nSet them equal:\n\n θ₂ – θ₁ = θ₃ – θ₄.                                                (3)\n\nStep 6.  Insert the explicit angles.  \nUsing θ(k) = arctan(k/2),\n\n arctan(3/10) – arctan(–½) = arctan(½) – arctan(κ/2).\n\nCompute the left–hand side:\n\n arctan(3/10) – arctan(–½) = arctan(3/10) + arctan(½)\n                  = 0.29145679… + 0.46364761…  \n                  = 0.75510440….\n\nHence                                                   \n\n arctan(κ/2) = arctan(½) – 0.75510440…  \n         = –0.29145679… = –arctan(3/10).\n\nTherefore κ/2 = –3/10, so κ = –3/5.\n\nStep 7.  Recover m from κ using (2):\n\n (m – 1)/(m + 1) = –3/5  \n 5(m – 1) = –3(m + 1)  \n 5m – 5 = –3m – 3  \n 8m = 2  \n m = 1/4.\n\nConsequently\n\n m = 1/4.\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.569064",
        "was_fixed": false,
        "difficulty_analysis": "• Rotated Conic:  Unlike the original axis-aligned ellipse, the cross-term 6xy forces a 45° coordinate rotation and identification of principal axes.  \n• Sequential Linear Maps:  A second anisotropic scaling is required to change the rotated ellipse into a unit circle, introducing an additional Jacobian determinant.  \n• Geometry on the Unit Circle:  Areas are converted into angular widths of sectors, demanding fluency with polar geometry and trigonometric identities.  \n• Interacting Constraints:  Four different bounding rays transform to four distinct angles, and the equality condition becomes a transcendental trigonometric equation before reverting to an elementary rational solution.  \n• Multi-step Reasoning:  The solver must combine linear algebra (diagonalising a quadratic form), multivariable calculus (Jacobian determinants), analytic geometry (ray transformations), and trigonometry (sector areas and inverse-tangent equations).  \nThese layers of technical detail and the requisite synthesis of several advanced concepts make the enhanced variant substantially more challenging than either the original problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}