{
  "index": "1938-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "11. Given the parabola \\( y^{2}=2 m x \\), what is the length of the shortest chord that is normal to the curve at one end?",
  "solution": "Solution. Any point on the parabola has coordinates of the form \\( \\left(2 m t^{2}, 2 m t\\right) \\). Let \\( A B \\) be a chord normal to the parabola at \\( A \\). Say \\( A= \\) \\( \\left(2 m t^{2}, 2 m t\\right) \\) and \\( B=\\left(2 m s^{2}, 2 m s\\right) \\). The slope of \\( A B \\) is \\( 1 /(s+t) \\), and the slope of the tangent at \\( A \\) is \\( 1 /(2 t) \\). Hence \\( s+t=-1 /(2 t) \\).\n\nThe length \\( L \\) of \\( A B \\) is given by\n\\[\nL^{2}=4 m^{2}\\left[\\left(s^{2}-t^{2}\\right)^{2}+(s-t)^{2}\\right]=4 m^{2}(s-t)^{2}\\left[(s+t)^{2}+1\\right]\n\\]\n\nSubstituting \\( s=-t-1 /(2 t) \\) we have\n\\[\nL^{2}=4 m^{2}\\left(\\frac{4 t^{2}+1}{2 t}\\right)^{2} \\frac{1+4 t^{2}}{4 t^{2}}=\\frac{m^{2}}{4} \\frac{\\left(4 t^{2}+1\\right)^{3}}{t^{4}}\n\\]\n\nWe seek the value of \\( t \\) which minimizes \\( L \\), so we may just as well choose \\( t \\) to minimize\n\\[\n\\frac{4 t^{2}+1}{t^{4 / 3}}=4 t^{2 / 3}+t^{-4 / 3}\n\\]\n\nSetting the derivative equal to zero, we find two critical points, \\( t= \\) \\( \\pm \\sqrt{2} / 2 \\). Since \\( L \\rightarrow \\infty \\) as \\( t \\rightarrow 0, \\pm \\infty \\), these two critical values both give minima. Either of the two shortest chords is of length \\( 3 \\sqrt{3}|m| \\), from (1).",
  "vars": [
    "y",
    "x",
    "A",
    "B",
    "t",
    "s",
    "L"
  ],
  "params": [
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "y": "ordinate",
        "x": "abscissa",
        "A": "pointa",
        "B": "pointb",
        "t": "parameter",
        "s": "helper",
        "L": "chordlen",
        "m": "parabcoef"
      },
      "question": "11. Given the parabola \\( ordinate^{2}=2 parabcoef abscissa \\), what is the length of the shortest chord that is normal to the curve at one end?",
      "solution": "Solution. Any point on the parabola has coordinates of the form \\( \\left(2 parabcoef parameter^{2}, 2 parabcoef parameter\\right) \\). Let \\( pointa pointb \\) be a chord normal to the parabola at \\( pointa \\). Say \\( pointa= \\left(2 parabcoef parameter^{2}, 2 parabcoef parameter\\right) \\) and \\( pointb=\\left(2 parabcoef helper^{2}, 2 parabcoef helper\\right) \\). The slope of \\( pointa pointb \\) is \\( 1 /(helper+parameter) \\), and the slope of the tangent at \\( pointa \\) is \\( 1 /(2 parameter) \\). Hence \\( helper+parameter=-1 /(2 parameter) \\).\n\nThe length \\( chordlen \\) of \\( pointa pointb \\) is given by\n\\[\nchordlen^{2}=4 parabcoef^{2}\\left[\\left(helper^{2}-parameter^{2}\\right)^{2}+(helper-parameter)^{2}\\right]=4 parabcoef^{2}(helper-parameter)^{2}\\left[(helper+parameter)^{2}+1\\right]\n\\]\n\nSubstituting \\( helper=-parameter-1 /(2 parameter) \\) we have\n\\[\nchordlen^{2}=4 parabcoef^{2}\\left(\\frac{4 parameter^{2}+1}{2 parameter}\\right)^{2} \\frac{1+4 parameter^{2}}{4 parameter^{2}}=\\frac{parabcoef^{2}}{4} \\frac{\\left(4 parameter^{2}+1\\right)^{3}}{parameter^{4}}\n\\]\n\nWe seek the value of \\( parameter \\) which minimizes \\( chordlen \\), so we may just as well choose \\( parameter \\) to minimize\n\\[\n\\frac{4 parameter^{2}+1}{parameter^{4 / 3}}=4 parameter^{2 / 3}+parameter^{-4 / 3}\n\\]\n\nSetting the derivative equal to zero, we find two critical points, \\( parameter= \\pm \\sqrt{2} / 2 \\). Since \\( chordlen \\rightarrow \\infty \\) as \\( parameter \\rightarrow 0, \\pm \\infty \\), these two critical values both give minima. Either of the two shortest chords is of length \\( 3 \\sqrt{3}|parabcoef| \\), from (1)."
    },
    "descriptive_long_confusing": {
      "map": {
        "y": "lanterns",
        "x": "cushions",
        "A": "magnetism",
        "B": "sunflower",
        "t": "whirlwind",
        "s": "pinecones",
        "L": "candlestick",
        "m": "butterfly"
      },
      "question": "11. Given the parabola \\( lanterns^{2}=2 butterfly cushions \\), what is the length of the shortest chord that is normal to the curve at one end?",
      "solution": "Solution. Any point on the parabola has coordinates of the form \\( \\left(2 butterfly whirlwind^{2}, 2 butterfly whirlwind\\right) \\). Let \\( magnetism sunflower \\) be a chord normal to the parabola at \\( magnetism \\). Say \\( magnetism= \\) \\( \\left(2 butterfly whirlwind^{2}, 2 butterfly whirlwind\\right) \\) and \\( sunflower=\\left(2 butterfly pinecones^{2}, 2 butterfly pinecones\\right) \\). The slope of \\( magnetism sunflower \\) is \\( 1 /(pinecones+whirlwind) \\), and the slope of the tangent at \\( magnetism \\) is \\( 1 /(2 whirlwind) \\). Hence \\( pinecones+whirlwind=-1 /(2 whirlwind) \\).\n\nThe length \\( candlestick \\) of \\( magnetism sunflower \\) is given by\n\\[\ncandlestick^{2}=4 butterfly^{2}\\left[\\left(pinecones^{2}-whirlwind^{2}\\right)^{2}+(pinecones-whirlwind)^{2}\\right]=4 butterfly^{2}(pinecones-whirlwind)^{2}\\left[(pinecones+whirlwind)^{2}+1\\right]\n\\]\n\nSubstituting \\( pinecones=-whirlwind-1 /(2 whirlwind) \\) we have\n\\[\ncandlestick^{2}=4 butterfly^{2}\\left(\\frac{4 whirlwind^{2}+1}{2 whirlwind}\\right)^{2} \\frac{1+4 whirlwind^{2}}{4 whirlwind^{2}}=\\frac{butterfly^{2}}{4} \\frac{\\left(4 whirlwind^{2}+1\\right)^{3}}{whirlwind^{4}}\n\\]\n\nWe seek the value of \\( whirlwind \\) which minimizes \\( candlestick \\), so we may just as well choose \\( whirlwind \\) to minimize\n\\[\n\\frac{4 whirlwind^{2}+1}{whirlwind^{4 / 3}}=4 whirlwind^{2 / 3}+whirlwind^{-4 / 3}\n\\]\n\nSetting the derivative equal to zero, we find two critical points, \\( whirlwind= \\pm \\sqrt{2} / 2 \\). Since \\( candlestick \\rightarrow \\infty \\) as \\( whirlwind \\rightarrow 0, \\pm \\infty \\), these two critical values both give minima. Either of the two shortest chords is of length \\( 3 \\sqrt{3}|butterfly| \\), from (1)."
    },
    "descriptive_long_misleading": {
      "map": {
        "y": "horizontal",
        "x": "vertical",
        "A": "voidpoint",
        "B": "nullpoint",
        "t": "constanty",
        "s": "steadfast",
        "L": "brevitude",
        "m": "fluctual"
      },
      "question": "11. Given the parabola \\( horizontal^{2}=2 fluctual vertical \\), what is the length of the shortest chord that is normal to the curve at one end?",
      "solution": "Solution. Any point on the parabola has coordinates of the form \\( \\left(2 fluctual constanty^{2}, 2 fluctual constanty\\right) \\). Let \\( voidpoint nullpoint \\) be a chord normal to the parabola at \\( voidpoint \\). Say \\( voidpoint= \\) \\( \\left(2 fluctual constanty^{2}, 2 fluctual constanty\\right) \\) and \\( nullpoint=\\left(2 fluctual steadfast^{2}, 2 fluctual steadfast\\right) \\). The slope of \\( voidpoint nullpoint \\) is \\( 1 /(steadfast+constanty) \\), and the slope of the tangent at \\( voidpoint \\) is \\( 1 /(2 constanty) \\). Hence \\( steadfast+constanty=-1 /(2 constanty) \\).\n\nThe length \\( brevitude \\) of \\( voidpoint nullpoint \\) is given by\n\\[\nbrevitude^{2}=4 fluctual^{2}\\left[\\left(steadfast^{2}-constanty^{2}\\right)^{2}+(steadfast-constanty)^{2}\\right]=4 fluctual^{2}(steadfast-constanty)^{2}\\left[(steadfast+constanty)^{2}+1\\right]\n\\]\n\nSubstituting \\( steadfast=-constanty-1 /(2 constanty) \\) we have\n\\[\nbrevitude^{2}=4 fluctual^{2}\\left(\\frac{4 constanty^{2}+1}{2 constanty}\\right)^{2} \\frac{1+4 constanty^{2}}{4 constanty^{2}}=\\frac{fluctual^{2}}{4} \\frac{\\left(4 constanty^{2}+1\\right)^{3}}{constanty^{4}}\n\\]\n\nWe seek the value of \\( constanty \\) which minimizes \\( brevitude \\), so we may just as well choose \\( constanty \\) to minimize\n\\[\n\\frac{4 constanty^{2}+1}{constanty^{4 / 3}}=4 constanty^{2 / 3}+constanty^{-4 / 3}\n\\]\n\nSetting the derivative equal to zero, we find two critical points, \\( constanty= \\) \\( \\pm \\sqrt{2} / 2 \\). Since \\( brevitude \\rightarrow \\infty \\) as \\( constanty \\rightarrow 0, \\pm \\infty \\), these two critical values both give minima. Either of the two shortest chords is of length \\( 3 \\sqrt{3}|fluctual| \\), from (1)."
    },
    "garbled_string": {
      "map": {
        "y": "qzxwvtnp",
        "x": "hjgrksla",
        "A": "dfghjklq",
        "B": "zmxncbva",
        "t": "rtyuiope",
        "s": "lkjhgfdz",
        "L": "poiuytre",
        "m": "asdfghjk"
      },
      "question": "<<<\n11. Given the parabola \\( qzxwvtnp^{2}=2 asdfghjk hjgrksla \\), what is the length of the shortest chord that is normal to the curve at one end?\n>>>",
      "solution": "<<<\nSolution. Any point on the parabola has coordinates of the form \\( \\left(2 asdfghjk rtyuiope^{2}, 2 asdfghjk rtyuiope\\right) \\). Let \\( dfghjklq zmxncbva \\) be a chord normal to the parabola at \\( dfghjklq \\). Say \\( dfghjklq= \\) \\( \\left(2 asdfghjk rtyuiope^{2}, 2 asdfghjk rtyuiope\\right) \\) and \\( zmxncbva=\\left(2 asdfghjk lkjhgfdz^{2}, 2 asdfghjk lkjhgfdz\\right) \\). The slope of \\( dfghjklq zmxncbva \\) is \\( 1 /(lkjhgfdz+rtyuiope) \\), and the slope of the tangent at \\( dfghjklq \\) is \\( 1 /(2 rtyuiope) \\). Hence \\( lkjhgfdz+rtyuiope=-1 /(2 rtyuiope) \\).\n\nThe length \\( poiuytre \\) of \\( dfghjklq zmxncbva \\) is given by\n\\[\npoiuytre^{2}=4 asdfghjk^{2}\\left[\\left(lkjhgfdz^{2}-rtyuiope^{2}\\right)^{2}+(lkjhgfdz-rtyuiope)^{2}\\right]=4 asdfghjk^{2}(lkjhgfdz-rtyuiope)^{2}\\left[(lkjhgfdz+rtyuiope)^{2}+1\\right]\n\\]\n\nSubstituting \\( lkjhgfdz=-rtyuiope-1 /(2 rtyuiope) \\) we have\n\\[\npoiuytre^{2}=4 asdfghjk^{2}\\left(\\frac{4 rtyuiope^{2}+1}{2 rtyuiope}\\right)^{2} \\frac{1+4 rtyuiope^{2}}{4 rtyuiope^{2}}=\\frac{asdfghjk^{2}}{4} \\frac{\\left(4 rtyuiope^{2}+1\\right)^{3}}{rtyuiope^{4}}\n\\]\n\nWe seek the value of \\( rtyuiope \\) which minimizes \\( poiuytre \\), so we may just as well choose \\( rtyuiope \\) to minimize\n\\[\n\\frac{4 rtyuiope^{2}+1}{rtyuiope^{4 / 3}}=4 rtyuiope^{2 / 3}+rtyuiope^{-4 / 3}\n\\]\n\nSetting the derivative equal to zero, we find two critical points, \\( rtyuiope= \\) \\( \\pm \\sqrt{2} / 2 \\). Since \\( poiuytre \\rightarrow \\infty \\) as \\( rtyuiope \\rightarrow 0, \\pm \\infty \\), these two critical values both give minima. Either of the two shortest chords is of length \\( 3 \\sqrt{3}|asdfghjk| \\), from (1).\n>>>"
    },
    "kernel_variant": {
      "question": "Let p ≠ 0 be a real constant and consider the circular paraboloid  \n\n  Σ : y² + z² = 4 p x.  \n\nA line segment AB with end-points on Σ is called a one-sided normal chord if the straight line AB is perpendicular to the tangent plane to Σ at A (no condition is imposed at B).\n\n(a)  Show that every one-sided normal chord can be written in the form  \n  B = A + λ ∇F(A) with λ > 0,  \nwhere F(x, y, z) := 4 p x − y² − z².\n\n(b)  If r = √(y_A² + z_A²) is the radial distance of A from the x-axis, express the squared length |AB|² solely in terms of r and p.\n\n(c)  Prove that there is a unique value r = r₀ > 0 that minimises |AB| and find r₀ explicitly.\n\n(d)  Hence obtain the minimum possible length L_min of a one-sided normal chord in terms of p.\n\n(e)  For every shortest one-sided normal chord determine  \n  (i) the coordinates of its midpoint, and  \n  (ii) the acute angle that the chord makes with the axis Ox.  \n  Show that the set of all such midpoints is the circle  \n\n   x = 5 p, y² + z² = 2 p².",
      "solution": "Prerequisites: multivariable calculus (gradients, tangent planes), elementary optimisation and three-dimensional analytic geometry.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (a) – Vector description of a one-sided normal chord  \nRegard Σ as the level surface F(x, y, z)=0 with  \n\n F(x, y, z)=4 p x−y²−z².\n\nAt A=(x_A, y_A, z_A)∈Σ the normal vector is  \n\n n_A = ∇F(A) = (4 p, −2 y_A, −2 z_A).\n\nBecause AB is normal to Σ at A, AB is parallel to n_A; hence there exists λ>0 such that  \n\n B = A + λ n_A.                                        (1)\n\nConversely, if B is given by (1) with λ>0 and satisfies F(B)=0, then AB is a chord of Σ perpendicular to the tangent plane at A; so every one-sided normal chord has the required representation.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (b) – Length in terms of r  \nIntroduce cylindrical coordinates about the x-axis:\n\n y_A = r cos θ, z_A = r sin θ, r>0, θ∈[0,2π).\n\nBecause A ∈ Σ,\n\n x_A = r²/(4 p).                                      (2)\n\nWith n_A as above, (1) gives componentwise\n\n x_B = r²/(4 p) + 4 p λ,  \n y_B = r cos θ − 2 r λ cos θ = r cos θ(1−2λ),  \n z_B = r sin θ − 2 r λ sin θ = r sin θ(1−2λ).\n\nImpose F(B)=0:\n\n 4 p x_B − y_B² − z_B² = 0  \n ⇔ 4 p(r²/4 p + 4 p λ) − r²(1−2λ)² = 0  \n ⇔ r² + 16 p² λ − r²(1−2λ)² = 0  \n ⇔ 4 λ(4 p² + r² − r² λ) = 0.\n\nSince λ>0,  \n\n λ = 1 + 4 p²/r².                                    (3)\n\nThe norm of the normal vector is  \n\n |n_A| = √[(4 p)² + (−2 y_A)² + (−2 z_A)²]  \n        = √(16 p² + 4 r²) = 2√(4 p² + r²).\n\nTherefore\n\n |AB| = λ|n_A| = 2(1 + 4 p²/r²)√(4 p² + r²),  \n |AB|² = 4(1 + 4 p²/r²)²(4 p² + r²).              (4)\n\nExpression (4) depends only on r (and the parameter p).\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (c) – Optimising in r  \nLet  \n\n G(r) := |AB|² = 4(1 + 4 p²/r²)²(4 p² + r²) (r>0).\n\nTo minimise G it suffices to minimise  \n\n g(r) := ln G(r)/2 = ln(1 + 4 p²/r²) + ½ ln(4 p² + r²).\n\nDifferentiate:\n\n d/dr [ln(1 + 4 p²/r²)] = (−8 p²/r³)/(1 + 4 p²/r²)  \n                        = −8 p² / [r(r² + 4 p²)],\n\n d/dr[½ ln(4 p² + r²)]  = r/(4 p² + r²).\n\nHence  \n\n g′(r) = −8 p² / [r(r² + 4 p²)] + r/(4 p² + r²)  \n        = (−8 p² + r²) / [r(r² + 4 p²)].\n\nSetting g′(r)=0 gives the unique positive critical point  \n\n r² = 8 p² ⇒ r₀ = 2√2 |p|.                        (5)\n\nBecause G(r) → ∞ as r→0⁺ or r→∞, this critical point indeed yields the global minimum.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (d) – The minimum length  \nInsert r₀ into (3) and (4):\n\n λ₀ = 1 + 4 p²/r₀² = 1 + 4 p²/(8 p²) = 3/2,\n\n |AB|_min = 2·(3/2)·√(4 p² + 8 p²) = 3·2√3 |p| = 6√3 |p|.\n\nThus  \n\n L_min = 6√3 |p|.                                   (6)\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (e) – Midpoints and fixed angle  \n\n(i) Midpoint. The midpoint M of AB is  \n\n M = ½(A + B) = A + ½ λ₀ n_A.\n\nFrom (2), (5) and λ₀ = 3/2 we obtain\n\n x_M = r₀²/(4 p) + (½ λ₀)(4 p) = (8 p²)/(4 p) + (3/4)(4 p)  \n    = 2 p + 3 p = 5 p,\n\n y_M = r₀ cos θ + (½ λ₀)(−2 r₀ cos θ) = r₀ cos θ(1 − 3/2) = −½r₀ cos θ,\n\n z_M = r₀ sin θ + (½ λ₀)(−2 r₀ sin θ) = −½r₀ sin θ.\n\nSince r₀ = 2√2 |p|,\n\n y_M² + z_M² = (r₀²/4) = 2 p².\n\nThus every midpoint satisfies  \n\n x = 5 p, y² + z² = 2 p²,\n\ni.e. all midpoints lie on the circle described in the statement.\n\n(ii) Angle with the x-axis. The direction vector of AB is  \n\n AB = λ₀ n_A = (3/2)(4 p, −2 y_A, −2 z_A).\n\nHence its x-component has absolute value 6 |p|, while |AB| = 6√3 |p|.  \nTherefore  \n\n cos ϕ = |6 p| / (6√3 |p|) = 1/√3, ϕ = arccos(1/√3)≈54.7°.\n\nEvery shortest one-sided normal chord thus makes the fixed acute angle ϕ = arccos(1/√3) with the axis Ox.\n\n–––––––––––––––––––––––––––––––––––––––––",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.344292",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimensional escalation – the problem moves from a plane curve to the three-dimensional paraboloid y²+z²=4 p x.  The geometry of normals, tangent planes and chords in 3-space is intrinsically more involved than in two dimensions.\n\n2. Advanced tools – the solution requires multivariable calculus (gradients, tangent planes), vector algebra, cylindrical coordinates and logarithmic differentiation; none of these is needed in the original problem.\n\n3. Higher algebraic complexity – eliminating variables yields a non-linear optimisation problem in a single radial parameter r that must be handled with care (logarithmic differentiation, analysis of end–behaviour).\n\n4. Additional geometric conclusions – beyond finding the minimum length, the solver must locate the entire set of midpoints and compute a constant angle, showing deeper understanding of the spatial configuration.\n\n5. Multiple interacting concepts – the task intertwines surface theory, optimisation, and solid-geometry loci, demanding several lines of reasoning rather than the single-parameter calculus of the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let p ≠ 0 be a real constant and consider the circular paraboloid  \n\n  Σ : y² + z² = 4 p x.  \n\nA line segment AB with end-points on Σ is called a one-sided normal chord if the straight line AB is perpendicular to the tangent plane to Σ at A (no condition is imposed at B).\n\n(a)  Show that every one-sided normal chord can be written in the form  \n  B = A + λ ∇F(A) with λ > 0,  \nwhere F(x, y, z) := 4 p x − y² − z².\n\n(b)  If r = √(y_A² + z_A²) is the radial distance of A from the x-axis, express the squared length |AB|² solely in terms of r and p.\n\n(c)  Prove that there is a unique value r = r₀ > 0 that minimises |AB| and find r₀ explicitly.\n\n(d)  Hence obtain the minimum possible length L_min of a one-sided normal chord in terms of p.\n\n(e)  For every shortest one-sided normal chord determine  \n  (i) the coordinates of its midpoint, and  \n  (ii) the acute angle that the chord makes with the axis Ox.  \n  Show that the set of all such midpoints is the circle  \n\n   x = 5 p, y² + z² = 2 p².",
      "solution": "Prerequisites: multivariable calculus (gradients, tangent planes), elementary optimisation and three-dimensional analytic geometry.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (a) – Vector description of a one-sided normal chord  \nRegard Σ as the level surface F(x, y, z)=0 with  \n\n F(x, y, z)=4 p x−y²−z².\n\nAt A=(x_A, y_A, z_A)∈Σ the normal vector is  \n\n n_A = ∇F(A) = (4 p, −2 y_A, −2 z_A).\n\nBecause AB is normal to Σ at A, AB is parallel to n_A; hence there exists λ>0 such that  \n\n B = A + λ n_A.                                        (1)\n\nConversely, if B is given by (1) with λ>0 and satisfies F(B)=0, then AB is a chord of Σ perpendicular to the tangent plane at A; so every one-sided normal chord has the required representation.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (b) – Length in terms of r  \nIntroduce cylindrical coordinates about the x-axis:\n\n y_A = r cos θ, z_A = r sin θ, r>0, θ∈[0,2π).\n\nBecause A ∈ Σ,\n\n x_A = r²/(4 p).                                      (2)\n\nWith n_A as above, (1) gives componentwise\n\n x_B = r²/(4 p) + 4 p λ,  \n y_B = r cos θ − 2 r λ cos θ = r cos θ(1−2λ),  \n z_B = r sin θ − 2 r λ sin θ = r sin θ(1−2λ).\n\nImpose F(B)=0:\n\n 4 p x_B − y_B² − z_B² = 0  \n ⇔ 4 p(r²/4 p + 4 p λ) − r²(1−2λ)² = 0  \n ⇔ r² + 16 p² λ − r²(1−2λ)² = 0  \n ⇔ 4 λ(4 p² + r² − r² λ) = 0.\n\nSince λ>0,  \n\n λ = 1 + 4 p²/r².                                    (3)\n\nThe norm of the normal vector is  \n\n |n_A| = √[(4 p)² + (−2 y_A)² + (−2 z_A)²]  \n        = √(16 p² + 4 r²) = 2√(4 p² + r²).\n\nTherefore\n\n |AB| = λ|n_A| = 2(1 + 4 p²/r²)√(4 p² + r²),  \n |AB|² = 4(1 + 4 p²/r²)²(4 p² + r²).              (4)\n\nExpression (4) depends only on r (and the parameter p).\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (c) – Optimising in r  \nLet  \n\n G(r) := |AB|² = 4(1 + 4 p²/r²)²(4 p² + r²) (r>0).\n\nTo minimise G it suffices to minimise  \n\n g(r) := ln G(r)/2 = ln(1 + 4 p²/r²) + ½ ln(4 p² + r²).\n\nDifferentiate:\n\n d/dr [ln(1 + 4 p²/r²)] = (−8 p²/r³)/(1 + 4 p²/r²)  \n                        = −8 p² / [r(r² + 4 p²)],\n\n d/dr[½ ln(4 p² + r²)]  = r/(4 p² + r²).\n\nHence  \n\n g′(r) = −8 p² / [r(r² + 4 p²)] + r/(4 p² + r²)  \n        = (−8 p² + r²) / [r(r² + 4 p²)].\n\nSetting g′(r)=0 gives the unique positive critical point  \n\n r² = 8 p² ⇒ r₀ = 2√2 |p|.                        (5)\n\nBecause G(r) → ∞ as r→0⁺ or r→∞, this critical point indeed yields the global minimum.\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (d) – The minimum length  \nInsert r₀ into (3) and (4):\n\n λ₀ = 1 + 4 p²/r₀² = 1 + 4 p²/(8 p²) = 3/2,\n\n |AB|_min = 2·(3/2)·√(4 p² + 8 p²) = 3·2√3 |p| = 6√3 |p|.\n\nThus  \n\n L_min = 6√3 |p|.                                   (6)\n\n–––––––––––––––––––––––––––––––––––––––––\nStep (e) – Midpoints and fixed angle  \n\n(i) Midpoint. The midpoint M of AB is  \n\n M = ½(A + B) = A + ½ λ₀ n_A.\n\nFrom (2), (5) and λ₀ = 3/2 we obtain\n\n x_M = r₀²/(4 p) + (½ λ₀)(4 p) = (8 p²)/(4 p) + (3/4)(4 p)  \n    = 2 p + 3 p = 5 p,\n\n y_M = r₀ cos θ + (½ λ₀)(−2 r₀ cos θ) = r₀ cos θ(1 − 3/2) = −½r₀ cos θ,\n\n z_M = r₀ sin θ + (½ λ₀)(−2 r₀ sin θ) = −½r₀ sin θ.\n\nSince r₀ = 2√2 |p|,\n\n y_M² + z_M² = (r₀²/4) = 2 p².\n\nThus every midpoint satisfies  \n\n x = 5 p, y² + z² = 2 p²,\n\ni.e. all midpoints lie on the circle described in the statement.\n\n(ii) Angle with the x-axis. The direction vector of AB is  \n\n AB = λ₀ n_A = (3/2)(4 p, −2 y_A, −2 z_A).\n\nHence its x-component has absolute value 6 |p|, while |AB| = 6√3 |p|.  \nTherefore  \n\n cos ϕ = |6 p| / (6√3 |p|) = 1/√3, ϕ = arccos(1/√3)≈54.7°.\n\nEvery shortest one-sided normal chord thus makes the fixed acute angle ϕ = arccos(1/√3) with the axis Ox.\n\n–––––––––––––––––––––––––––––––––––––––––",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.301584",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimensional escalation – the problem moves from a plane curve to the three-dimensional paraboloid y²+z²=4 p x.  The geometry of normals, tangent planes and chords in 3-space is intrinsically more involved than in two dimensions.\n\n2. Advanced tools – the solution requires multivariable calculus (gradients, tangent planes), vector algebra, cylindrical coordinates and logarithmic differentiation; none of these is needed in the original problem.\n\n3. Higher algebraic complexity – eliminating variables yields a non-linear optimisation problem in a single radial parameter r that must be handled with care (logarithmic differentiation, analysis of end–behaviour).\n\n4. Additional geometric conclusions – beyond finding the minimum length, the solver must locate the entire set of midpoints and compute a constant angle, showing deeper understanding of the spatial configuration.\n\n5. Multiple interacting concepts – the task intertwines surface theory, optimisation, and solid-geometry loci, demanding several lines of reasoning rather than the single-parameter calculus of the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}