{
  "index": "1957-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "2. A uniform wire is bent into a form coinciding with the portion of the curve \\( y=e^{x}, 0 \\leq x \\leq a, a>1 \\), and the line segment \\( a-1 \\leq x \\leq a, y= \\) \\( e^{a} \\). The wire is then suspended from the point \\( \\left(a-1, e^{a}\\right) \\) and a horizontal force \\( F \\) is applied at the point \\( (0,1) \\) to hold the wire in coincidence with the curve and segment. Assuming the \\( \\boldsymbol{x} \\) axis is horizontal, show that the force \\( F \\) is directed to the right.",
  "solution": "Solution. The following three statements are evidently equivalent.\nThe force \\( F \\) is directed to the right.\nThe moment of the force of gravity on the wire is clockwise about the point of support.\n\nThe centroid of the wire falls to the right of the point of support.\nHence if \\( \\bar{x} \\) is the \\( x \\)-coordinate of the centroid of the wire, we must prove that\n\\[\n\\bar{x}>a-1\n\\]\n\nWithout loss of generality, we may take the constant linear density of the wire to be 1 . Then the mass of the curved portion of the wire is\n\\[\n\\int_{0}^{a} \\sqrt{1+e^{2 x}} d x\n\\]\nand its \\( x \\)-moment is\n\\[\n\\int_{0}^{u} x \\sqrt{1+e^{2 x}} d x .\n\\]\n\nThe straight segment of the wire has mass 1 and \\( x \\)-moment \\( a-\\frac{1}{2} \\). Hence (1) is equivalent to\n\\[\na-\\frac{1}{2}+\\int_{0}^{a} x \\sqrt{1+e^{2 x}} d x>(a-1)\\left(1+\\int_{0}^{a} \\sqrt{1+e^{2 x}} d x\\right)\n\\]\nwhich is in turn equivalent to\n\\[\n\\int_{0}^{a} x \\sqrt{1+e^{2 x}} d x-(a-1) \\int_{0}^{a} \\sqrt{1+e^{2 x}} d x>-\\frac{1}{2}\n\\]\n\nSo we must prove (2) for all \\( a>1 \\).\nLet \\( f(a) \\) be given by the left member of (2). Then\n\\[\nf^{\\prime}(a)=\\sqrt{1+e^{2 a}}-\\int_{0}^{a} \\sqrt{1+e^{2 x}} d x\n\\]\n\nSince \\( 1+e^{2 x}<\\left(\\frac{1}{2} e^{-x}+e^{x}\\right)^{2} \\),\n\\[\n\\begin{aligned}\n\\int_{0}^{a} \\sqrt{1+e^{2 x}} d x & <\\int_{0}^{a}\\left(\\frac{1}{2} e^{-x}+e^{x}\\right) d x \\\\\n& =e^{a}-\\frac{1}{2}-\\frac{1}{2} e^{-a}<e^{u}<\\sqrt{1+e^{2 u}}\n\\end{aligned}\n\\]\nfor \\( a>0 \\). Hence \\( f^{\\prime}(a)>0 \\) for \\( a>0 \\). This implies \\( f(a)>f(0)=0 \\) for \\( a>0 \\). Then (2) follows immediately and we are done.",
  "vars": [
    "x",
    "y",
    "u"
  ],
  "params": [
    "a",
    "F",
    "f"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "u": "auxiliary",
        "a": "curveend",
        "F": "holdforce",
        "f": "centroid"
      },
      "question": "2. A uniform wire is bent into a form coinciding with the portion of the curve \\( ordinate=e^{abscissa}, 0 \\leq abscissa \\leq curveend, curveend>1 \\), and the line segment \\( curveend-1 \\leq abscissa \\leq curveend, ordinate= \\) \\( e^{curveend} \\). The wire is then suspended from the point \\( \\left(curveend-1, e^{curveend}\\right) \\) and a horizontal force \\( holdforce \\) is applied at the point \\( (0,1) \\) to hold the wire in coincidence with the curve and segment. Assuming the \\( \\boldsymbol{abscissa} \\) axis is horizontal, show that the force \\( holdforce \\) is directed to the right.",
      "solution": "Solution. The following three statements are evidently equivalent.\nThe force \\( holdforce \\) is directed to the right.\nThe moment of the force of gravity on the wire is clockwise about the point of support.\n\nThe centroid of the wire falls to the right of the point of support.\nHence if \\( \\bar{abscissa} \\) is the \\( abscissa \\)-coordinate of the centroid of the wire, we must prove that\n\\[\n\\bar{abscissa}>curveend-1\n\\]\n\nWithout loss of generality, we may take the constant linear density of the wire to be 1. Then the mass of the curved portion of the wire is\n\\[\n\\int_{0}^{curveend} \\sqrt{1+e^{2 abscissa}} \\, d abscissa\n\\]\nand its \\( abscissa \\)-moment is\n\\[\n\\int_{0}^{auxiliary} abscissa \\sqrt{1+e^{2 abscissa}} \\, d abscissa .\n\\]\n\nThe straight segment of the wire has mass 1 and \\( abscissa \\)-moment \\( curveend-\\frac{1}{2} \\). Hence (1) is equivalent to\n\\[\ncurveend-\\frac{1}{2}+\\int_{0}^{curveend} abscissa \\sqrt{1+e^{2 abscissa}} \\, d abscissa>(curveend-1)\\left(1+\\int_{0}^{curveend} \\sqrt{1+e^{2 abscissa}} \\, d abscissa\\right)\n\\]\nwhich is in turn equivalent to\n\\[\n\\int_{0}^{curveend} abscissa \\sqrt{1+e^{2 abscissa}} \\, d abscissa-(curveend-1) \\int_{0}^{curveend} \\sqrt{1+e^{2 abscissa}} \\, d abscissa>-\\frac{1}{2}\n\\]\n\nSo we must prove (2) for all \\( curveend>1 \\).\nLet \\( centroid(curveend) \\) be given by the left member of (2). Then\n\\[\ncentroid^{\\prime}(curveend)=\\sqrt{1+e^{2 curveend}}-\\int_{0}^{curveend} \\sqrt{1+e^{2 abscissa}} \\, d abscissa\n\\]\n\nSince \\( 1+e^{2 abscissa}<\\left(\\frac{1}{2} e^{-abscissa}+e^{abscissa}\\right)^{2} \\),\n\\[\n\\begin{aligned}\n\\int_{0}^{curveend} \\sqrt{1+e^{2 abscissa}} \\, d abscissa & <\\int_{0}^{curveend}\\left(\\frac{1}{2} e^{-abscissa}+e^{abscissa}\\right) \\, d abscissa \\\\\n& =e^{curveend}-\\frac{1}{2}-\\frac{1}{2} e^{-curveend}<e^{auxiliary}<\\sqrt{1+e^{2 auxiliary}}\n\\end{aligned}\n\\]\nfor \\( curveend>0 \\). Hence \\( centroid^{\\prime}(curveend)>0 \\) for \\( curveend>0 \\). This implies \\( centroid(curveend)>centroid(0)=0 \\) for \\( curveend>0 \\). Then (2) follows immediately and we are done."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "shoreline",
        "y": "raincloud",
        "u": "windstorm",
        "a": "blueberry",
        "F": "goldbrick",
        "f": "stoneware"
      },
      "question": "2. A uniform wire is bent into a form coinciding with the portion of the curve \\( raincloud=e^{shoreline}, 0 \\leq shoreline \\leq blueberry, blueberry>1 \\), and the line segment \\( blueberry-1 \\leq shoreline \\leq blueberry, raincloud= \\) \\( e^{blueberry} \\). The wire is then suspended from the point \\( \\left(blueberry-1, e^{blueberry}\\right) \\) and a horizontal force \\( goldbrick \\) is applied at the point \\( (0,1) \\) to hold the wire in coincidence with the curve and segment. Assuming the \\( \\boldsymbol{shoreline} \\) axis is horizontal, show that the force \\( goldbrick \\) is directed to the right.",
      "solution": "Solution. The following three statements are evidently equivalent.\nThe force \\( goldbrick \\) is directed to the right.\nThe moment of the force of gravity on the wire is clockwise about the point of support.\n\nThe centroid of the wire falls to the right of the point of support.\nHence if \\( \\bar{shoreline} \\) is the \\( shoreline \\)-coordinate of the centroid of the wire, we must prove that\n\\[\n\\bar{shoreline}>blueberry-1\n\\]\n\nWithout loss of generality, we may take the constant linear density of the wire to be 1 . Then the mass of the curved portion of the wire is\n\\[\n\\int_{0}^{blueberry} \\sqrt{1+e^{2 shoreline}} d shoreline\n\\]\nand its \\( shoreline \\)-moment is\n\\[\n\\int_{0}^{windstorm} shoreline \\sqrt{1+e^{2 shoreline}} d shoreline .\n\\]\n\nThe straight segment of the wire has mass 1 and \\( shoreline \\)-moment \\( blueberry-\\frac{1}{2} \\). Hence (1) is equivalent to\n\\[\nblueberry-\\frac{1}{2}+\\int_{0}^{blueberry} shoreline \\sqrt{1+e^{2 shoreline}} d shoreline>(blueberry-1)\\left(1+\\int_{0}^{blueberry} \\sqrt{1+e^{2 shoreline}} d shoreline\\right)\n\\]\nwhich is in turn equivalent to\n\\[\n\\int_{0}^{blueberry} shoreline \\sqrt{1+e^{2 shoreline}} d shoreline-(blueberry-1) \\int_{0}^{blueberry} \\sqrt{1+e^{2 shoreline}} d shoreline>-\\frac{1}{2}\n\\]\n\nSo we must prove (2) for all \\( blueberry>1 \\).\nLet \\( stoneware(blueberry) \\) be given by the left member of (2). Then\n\\[\nstoneware^{\\prime}(blueberry)=\\sqrt{1+e^{2 blueberry}}-\\int_{0}^{blueberry} \\sqrt{1+e^{2 shoreline}} d shoreline\n\\]\n\nSince \\( 1+e^{2 shoreline}<\\left(\\frac{1}{2} e^{-shoreline}+e^{shoreline}\\right)^{2} \\),\n\\[\n\\begin{aligned}\n\\int_{0}^{blueberry} \\sqrt{1+e^{2 shoreline}} d shoreline & <\\int_{0}^{blueberry}\\left(\\frac{1}{2} e^{-shoreline}+e^{shoreline}\\right) d shoreline \\\\\n& =e^{blueberry}-\\frac{1}{2}-\\frac{1}{2} e^{-blueberry}<e^{windstorm}<\\sqrt{1+e^{2 windstorm}}\n\\end{aligned}\n\\]\nfor \\( blueberry>0 \\). Hence \\( stoneware^{\\prime}(blueberry)>0 \\) for \\( blueberry>0 \\). This implies \\( stoneware(blueberry)>stoneware(0)=0 \\) for \\( blueberry>0 \\). Then (2) follows immediately and we are done."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "abscissa",
        "u": "knownvalue",
        "a": "minuscule",
        "F": "stillness",
        "f": "constant"
      },
      "question": "2. A uniform wire is bent into a form coinciding with the portion of the curve \\( abscissa=e^{verticalaxis}, 0 \\leq verticalaxis \\leq minuscule, minuscule>1 \\), and the line segment \\( minuscule-1 \\leq verticalaxis \\leq minuscule, abscissa= e^{minuscule} \\). The wire is then suspended from the point \\( (minuscule-1, e^{minuscule}) \\) and a horizontal force \\( stillness \\) is applied at the point \\( (0,1) \\) to hold the wire in coincidence with the curve and segment. Assuming the \\( \\boldsymbol{verticalaxis} \\) axis is horizontal, show that the force \\( stillness \\) is directed to the right.",
      "solution": "Solution. The following three statements are evidently equivalent.\nThe force \\( stillness \\) is directed to the right.\nThe moment of the force of gravity on the wire is clockwise about the point of support.\n\nThe centroid of the wire falls to the right of the point of support.\nHence if \\( \\bar{verticalaxis} \\) is the \\( verticalaxis \\)-coordinate of the centroid of the wire, we must prove that\n\\[\n\\bar{verticalaxis}>minuscule-1\n\\]\n\nWithout loss of generality, we may take the constant linear density of the wire to be 1 . Then the mass of the curved portion of the wire is\n\\[\n\\int_{0}^{minuscule} \\sqrt{1+e^{2 verticalaxis}} d verticalaxis\n\\]\nand its \\( verticalaxis \\)-moment is\n\\[\n\\int_{0}^{knownvalue} verticalaxis \\sqrt{1+e^{2 verticalaxis}} d verticalaxis .\n\\]\n\nThe straight segment of the wire has mass 1 and \\( verticalaxis \\)-moment \\( minuscule-\\frac{1}{2} \\). Hence (1) is equivalent to\n\\[\nminuscule-\\frac{1}{2}+\\int_{0}^{minuscule} verticalaxis \\sqrt{1+e^{2 verticalaxis}} d verticalaxis>(minuscule-1)\\left(1+\\int_{0}^{minuscule} \\sqrt{1+e^{2 verticalaxis}} d verticalaxis\\right)\n\\]\nwhich is in turn equivalent to\n\\[\n\\int_{0}^{minuscule} verticalaxis \\sqrt{1+e^{2 verticalaxis}} d verticalaxis-(minuscule-1) \\int_{0}^{minuscule} \\sqrt{1+e^{2 verticalaxis}} d verticalaxis>-\\frac{1}{2}\n\\]\n\nSo we must prove (2) for all \\( minuscule>1 \\).\nLet \\( constant(minuscule) \\) be given by the left member of (2). Then\n\\[\nconstant^{\\prime}(minuscule)=\\sqrt{1+e^{2 minuscule}}-\\int_{0}^{minuscule} \\sqrt{1+e^{2 verticalaxis}} d verticalaxis\n\\]\n\nSince \\( 1+e^{2 verticalaxis}<\\left(\\frac{1}{2} e^{-verticalaxis}+e^{verticalaxis}\\right)^{2} \\),\n\\[\n\\begin{aligned}\n\\int_{0}^{minuscule} \\sqrt{1+e^{2 verticalaxis}} d verticalaxis & <\\int_{0}^{minuscule}\\left(\\frac{1}{2} e^{-verticalaxis}+e^{verticalaxis}\\right) d verticalaxis \\\\\n& =e^{minuscule}-\\frac{1}{2}-\\frac{1}{2} e^{-minuscule}<e^{knownvalue}<\\sqrt{1+e^{2 knownvalue}}\n\\end{aligned}\n\\]\nfor \\( minuscule>0 \\). Hence \\( constant^{\\prime}(minuscule)>0 \\) for \\( minuscule>0 \\). This implies \\( constant(minuscule)>constant(0)=0 \\) for \\( minuscule>0 \\). Then (2) follows immediately and we are done."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "u": "bnecsotq",
        "a": "vfyrsplk",
        "F": "kzmpxwhr",
        "f": "rdtqklvn"
      },
      "question": "2. A uniform wire is bent into a form coinciding with the portion of the curve \\( hjgrksla=e^{qzxwvtnp}, 0 \\leq qzxwvtnp \\leq vfyrsplk, vfyrsplk>1 \\), and the line segment \\( vfyrsplk-1 \\leq qzxwvtnp \\leq vfyrsplk, hjgrksla= e^{vfyrsplk} \\). The wire is then suspended from the point \\( \\left(vfyrsplk-1, e^{vfyrsplk}\\right) \\) and a horizontal force kzmpxwhr is applied at the point \\( (0,1) \\) to hold the wire in coincidence with the curve and segment. Assuming the \\( \\boldsymbol{qzxwvtnp} \\) axis is horizontal, show that the force kzmpxwhr is directed to the right.",
      "solution": "Solution. The following three statements are evidently equivalent.\nThe force \\( kzmpxwhr \\) is directed to the right.\nThe moment of the force of gravity on the wire is clockwise about the point of support.\n\nThe centroid of the wire falls to the right of the point of support.\nHence if \\( \\bar{qzxwvtnp} \\) is the \\( qzxwvtnp \\)-coordinate of the centroid of the wire, we must prove that\n\\[\n\\bar{qzxwvtnp}>vfyrsplk-1\n\\]\n\nWithout loss of generality, we may take the constant linear density of the wire to be 1 . Then the mass of the curved portion of the wire is\n\\[\n\\int_{0}^{vfyrsplk} \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp\n\\]\nand its \\( qzxwvtnp \\)-moment is\n\\[\n\\int_{0}^{bnecsotq} qzxwvtnp \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp .\n\\]\n\nThe straight segment of the wire has mass 1 and \\( qzxwvtnp \\)-moment \\( vfyrsplk-\\frac{1}{2} \\). Hence (1) is equivalent to\n\\[\nvfyrsplk-\\frac{1}{2}+\\int_{0}^{vfyrsplk} qzxwvtnp \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp>(vfyrsplk-1)\\left(1+\\int_{0}^{vfyrsplk} \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp\\right)\n\\]\nwhich is in turn equivalent to\n\\[\n\\int_{0}^{vfyrsplk} qzxwvtnp \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp-(vfyrsplk-1) \\int_{0}^{vfyrsplk} \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp>-\\frac{1}{2}\n\\]\n\nSo we must prove (2) for all \\( vfyrsplk>1 \\).\nLet \\( rdtqklvn(vfyrsplk) \\) be given by the left member of (2). Then\n\\[\nrdtqklvn^{\\prime}(vfyrsplk)=\\sqrt{1+e^{2 vfyrsplk}}-\\int_{0}^{vfyrsplk} \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp\n\\]\n\nSince \\( 1+e^{2 qzxwvtnp}<\\left(\\frac{1}{2} e^{-qzxwvtnp}+e^{qzxwvtnp}\\right)^{2} \\),\n\\[\n\\begin{aligned}\n\\int_{0}^{vfyrsplk} \\sqrt{1+e^{2 qzxwvtnp}} d qzxwvtnp & <\\int_{0}^{vfyrsplk}\\left(\\frac{1}{2} e^{-qzxwvtnp}+e^{qzxwvtnp}\\right) d qzxwvtnp \\\\\n& =e^{vfyrsplk}-\\frac{1}{2}-\\frac{1}{2} e^{-vfyrsplk}<e^{bnecsotq}<\\sqrt{1+e^{2 bnecsotq}}\n\\end{aligned}\n\\]\nfor \\( vfyrsplk>0 \\). Hence \\( rdtqklvn^{\\prime}(vfyrsplk)>0 \\) for \\( vfyrsplk>0 \\). This implies \\( rdtqklvn(vfyrsplk)>rdtqklvn(0)=0 \\) for \\( vfyrsplk>0 \\). Then (2) follows immediately and we are done."
    },
    "kernel_variant": {
      "question": "Let a > 3 be fixed and let  \n 0 < R < eᵃ – 1, λ > 0  \nbe real constants.  A homogeneous wire of linear mass–density λ is bent so that it coincides with the planar curve Γ obtained by joining consecutively  \n\n C₁ (exponential branch) r₁(x) = (x , eˣ , 0)          0 ≤ x ≤ a,\n\n C₂ (vertical segment)  r₂(t) = (a , eᵃ – t , 0)        0 ≤ t ≤ R,\n\n C₃ (horizontal segment) r₃(s) = (a – s , eᵃ – R , 0)     0 ≤ s ≤ 3.\n\nThus Γ lies in the plane z = 0 and terminates at  \n\n S = (a – 3 , eᵃ – R , 0).\n\nThe wire is freely suspended at S and placed in a uniform gravitational field  \n g = –g₀ ĵ (g₀ > 0).  \nA single horizontal force F, applied at  \n\n Q = (0 , 1 , 0),\n\nkeeps the wire in static equilibrium.  Using the right–handed orthonormal basis (î, ĵ, k̂) whose î–axis points to the right, prove that the balancing force must point to the right, i.e.  \n\n F·î > 0 for every choice of parameters a > 3 and 0 < R < eᵃ – 1.",
      "solution": "Throughout set  \n g(x) := √(1 + e^{2x}), I(a) := ∫₀ᵃ g(x) dx, J(a) := ∫₀ᵃ x g(x) dx,           (1)  \nand denote by (x̄, ȳ, 0) the centre of mass of the whole wire, whose total mass is M.\n\n----------------------------------------------------------------------  \n1.  Torque balance and the rôle of x̄  \n----------------------------------------------------------------------  \nBecause Γ lies in the plane z = 0, every mass element has position vector  \n\n r = (x − Sₓ) î + (y − S_y) ĵ.\n\nIts infinitesimal weight is −λg₀ ĵ ds, so  \n\n dτ_G = r × (−λg₀ ĵ) = −λg₀(x − Sₓ) k̂.\n\nIntegrating over Γ gives the total gravitational torque  \n\n τ_G = −Mg₀(x̄ − Sₓ) k̂.                                              (2)\n\nThe applied force is horizontal, F = (F_x, 0, 0).  Its moment about the suspension point S is  \n\n τ_F = (Q − S) × F = −(1 − S_y) F_x k̂.                            (3)\n\nBecause 0 < R < eᵃ − 1 we have S_y = eᵃ − R > 1, hence 1 − S_y < 0.  Thus τ_F and F_x always have the same sign.  Static equilibrium (τ_F + τ_G = 0) implies  \n\n sgn F_x = sgn (x̄ − Sₓ).                                            (4)\n\nConsequently it suffices to prove  \n\n x̄ > Sₓ = a − 3.                                                    (5)\n\n----------------------------------------------------------------------  \n2.  Masses and first moments of the three pieces  \n----------------------------------------------------------------------  \n\nC₁ (0 ≤ x ≤ a):            M₁ = λ I(a),   M₁x = λ J(a).  \nC₂ (vertical, length R):    M₂ = λR,        M₂x = λaR.  \nC₃ (horizontal, length 3):  M₃ = 3λ,       M₃x = λ∫_{a−3}^{a} x dx = 3λ(a − 1.5).\n\nHence  \n\n M  = λ[ I(a) + R + 3 ],                                            (6)  \n\n Mx = λ[ J(a) + aR + 3(a − 1.5) ].                                 (7)\n\n----------------------------------------------------------------------  \n3.  Reduction to a single integral inequality  \n----------------------------------------------------------------------  \nDefine  \n\n Δ := Mx − (a − 3)M = λ{ J(a) − (a − 3)I(a) + 3R + 4.5 }.          (8)\n\nBecause R > 0, the last two terms are strictly positive.  Therefore (5) will follow once we have proved  \n\n f(a) := J(a) − (a − 3)I(a) > 0 for all a > 3.                    (9)\n\n----------------------------------------------------------------------  \n4.  A differential criterion for f(a)  \n----------------------------------------------------------------------  \nDifferentiate:\n\n f ′(a) = a g(a) − [(a − 3) g(a) + I(a)] = 3 g(a) − I(a).         (10)\n\nThus f is strictly increasing whenever I(a) < 3 g(a).  We now prove this estimate for every a > 3.\n\n----------------------------------------------------------------------  \n5.  Bounding I(a) in terms of g(a)  \n----------------------------------------------------------------------  \nFor x ≤ a the elementary inequality e^{x} ≤ e^{a} gives  \n\n g(x) = √(1 + e^{2x}) ≤ e^{x − a} g(a) + 1.                        (11)\n\nIntegrating (11) from 0 to a yields  \n\n I(a) ≤ g(a)∫₀ᵃ e^{x − a}dx + a  \n      = g(a)(1 − e^{−a}) + a.                                      (12)\n\nDivide by g(a) and use g(a) ≥ e^{a}:\n\n I(a)/g(a) ≤ 1 − e^{−a} + a/e^{a}.                                (13)\n\nFor a ≥ 3 the right–hand side is at most\n\n 1 − e^{−3} + 3/e^{3} ≈ 1 − 0.0498 + 0.1494 < 1.1.                (14)\n\nHence  \n\n I(a) < 1.1 g(a) < 3 g(a) for all a > 3,                          (15)\n\nand from (10) we obtain  \n\n f ′(a) > 0 for all a > 3.                                        (16)\n\n----------------------------------------------------------------------  \n6.  Positivity of f(a)  \n----------------------------------------------------------------------  \nBecause f is increasing and  \n\n f(3) = J(3) > 0,                                                  (17)\n\nwe have f(a) > 0 for every a > 3.  Returning to (8) we conclude  \n\n Δ > 0 ⇒ Mx > (a − 3)M ⇒ x̄ > a − 3.                              (18)\n\n----------------------------------------------------------------------  \n7.  Conclusion  \n----------------------------------------------------------------------  \nCombining (4) and (18) we deduce F_x > 0, i.e.  \n\n F·î > 0.\n\nThus the horizontal force required to keep the wire in equilibrium must always be directed to the right.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.485837",
        "was_fixed": false,
        "difficulty_analysis": "• Additional geometry:  The wire now lives in ℝ³ and contains a curved\n 3-dimensional piece (the quarter-circle C₂); torque must be handled with full\n vector machinery rather than planar moments.\n\n• More segments:  Three distinct pieces (one nonlinear in each coordinate)\n interact; the planar problem had just two.\n\n• Non-trivial arc length:  The exponential branch requires integration of\n √(1+e^{2x}), while the arc length of C₂ is πR/2.  Although the latter drops out\n of the final inequality, recognising this fact is part of the challenge.\n\n• Inequality sharpening:  Showing f′(a) > 0 needs an effective\n estimate on ∫₀ᵃ g(x) dx relative to g(a); a naïve bound\n ∫≤a g(a) would fail when a > 3, so one must craft a sharper argument using the\n specific growth of g.\n\n• Torque in three dimensions:  Students must translate the “centroid to the\n right” idea into a 3-D moment calculation and keep track of vector directions\n (î, ĵ, k̂).\n\nTogether these points raise the algebraic, geometric, and conceptual load well\nbeyond that of the original two-piece planar question."
      }
    },
    "original_kernel_variant": {
      "question": "Let a > 3 be fixed and let  \n 0 < R < eᵃ – 1, λ > 0  \nbe real constants.  A homogeneous wire of linear mass–density λ is bent so that it coincides with the planar curve Γ obtained by joining consecutively  \n\n C₁ (exponential branch) r₁(x) = (x , eˣ , 0)          0 ≤ x ≤ a,\n\n C₂ (vertical segment)  r₂(t) = (a , eᵃ – t , 0)        0 ≤ t ≤ R,\n\n C₃ (horizontal segment) r₃(s) = (a – s , eᵃ – R , 0)     0 ≤ s ≤ 3.\n\nThus Γ lies in the plane z = 0 and terminates at  \n\n S = (a – 3 , eᵃ – R , 0).\n\nThe wire is freely suspended at S and placed in a uniform gravitational field  \n g = –g₀ ĵ (g₀ > 0).  \nA single horizontal force F, applied at  \n\n Q = (0 , 1 , 0),\n\nkeeps the wire in static equilibrium.  Using the right–handed orthonormal basis (î, ĵ, k̂) whose î–axis points to the right, prove that the balancing force must point to the right, i.e.  \n\n F·î > 0 for every choice of parameters a > 3 and 0 < R < eᵃ – 1.",
      "solution": "Throughout set  \n g(x) := √(1 + e^{2x}), I(a) := ∫₀ᵃ g(x) dx, J(a) := ∫₀ᵃ x g(x) dx,           (1)  \nand denote by (x̄, ȳ, 0) the centre of mass of the whole wire, whose total mass is M.\n\n----------------------------------------------------------------------  \n1.  Torque balance and the rôle of x̄  \n----------------------------------------------------------------------  \nBecause Γ lies in the plane z = 0, every mass element has position vector  \n\n r = (x − Sₓ) î + (y − S_y) ĵ.\n\nIts infinitesimal weight is −λg₀ ĵ ds, so  \n\n dτ_G = r × (−λg₀ ĵ) = −λg₀(x − Sₓ) k̂.\n\nIntegrating over Γ gives the total gravitational torque  \n\n τ_G = −Mg₀(x̄ − Sₓ) k̂.                                              (2)\n\nThe applied force is horizontal, F = (F_x, 0, 0).  Its moment about the suspension point S is  \n\n τ_F = (Q − S) × F = −(1 − S_y) F_x k̂.                            (3)\n\nBecause 0 < R < eᵃ − 1 we have S_y = eᵃ − R > 1, hence 1 − S_y < 0.  Thus τ_F and F_x always have the same sign.  Static equilibrium (τ_F + τ_G = 0) implies  \n\n sgn F_x = sgn (x̄ − Sₓ).                                            (4)\n\nConsequently it suffices to prove  \n\n x̄ > Sₓ = a − 3.                                                    (5)\n\n----------------------------------------------------------------------  \n2.  Masses and first moments of the three pieces  \n----------------------------------------------------------------------  \n\nC₁ (0 ≤ x ≤ a):            M₁ = λ I(a),   M₁x = λ J(a).  \nC₂ (vertical, length R):    M₂ = λR,        M₂x = λaR.  \nC₃ (horizontal, length 3):  M₃ = 3λ,       M₃x = λ∫_{a−3}^{a} x dx = 3λ(a − 1.5).\n\nHence  \n\n M  = λ[ I(a) + R + 3 ],                                            (6)  \n\n Mx = λ[ J(a) + aR + 3(a − 1.5) ].                                 (7)\n\n----------------------------------------------------------------------  \n3.  Reduction to a single integral inequality  \n----------------------------------------------------------------------  \nDefine  \n\n Δ := Mx − (a − 3)M = λ{ J(a) − (a − 3)I(a) + 3R + 4.5 }.          (8)\n\nBecause R > 0, the last two terms are strictly positive.  Therefore (5) will follow once we have proved  \n\n f(a) := J(a) − (a − 3)I(a) > 0 for all a > 3.                    (9)\n\n----------------------------------------------------------------------  \n4.  A differential criterion for f(a)  \n----------------------------------------------------------------------  \nDifferentiate:\n\n f ′(a) = a g(a) − [(a − 3) g(a) + I(a)] = 3 g(a) − I(a).         (10)\n\nThus f is strictly increasing whenever I(a) < 3 g(a).  We now prove this estimate for every a > 3.\n\n----------------------------------------------------------------------  \n5.  Bounding I(a) in terms of g(a)  \n----------------------------------------------------------------------  \nFor x ≤ a the elementary inequality e^{x} ≤ e^{a} gives  \n\n g(x) = √(1 + e^{2x}) ≤ e^{x − a} g(a) + 1.                        (11)\n\nIntegrating (11) from 0 to a yields  \n\n I(a) ≤ g(a)∫₀ᵃ e^{x − a}dx + a  \n      = g(a)(1 − e^{−a}) + a.                                      (12)\n\nDivide by g(a) and use g(a) ≥ e^{a}:\n\n I(a)/g(a) ≤ 1 − e^{−a} + a/e^{a}.                                (13)\n\nFor a ≥ 3 the right–hand side is at most\n\n 1 − e^{−3} + 3/e^{3} ≈ 1 − 0.0498 + 0.1494 < 1.1.                (14)\n\nHence  \n\n I(a) < 1.1 g(a) < 3 g(a) for all a > 3,                          (15)\n\nand from (10) we obtain  \n\n f ′(a) > 0 for all a > 3.                                        (16)\n\n----------------------------------------------------------------------  \n6.  Positivity of f(a)  \n----------------------------------------------------------------------  \nBecause f is increasing and  \n\n f(3) = J(3) > 0,                                                  (17)\n\nwe have f(a) > 0 for every a > 3.  Returning to (8) we conclude  \n\n Δ > 0 ⇒ Mx > (a − 3)M ⇒ x̄ > a − 3.                              (18)\n\n----------------------------------------------------------------------  \n7.  Conclusion  \n----------------------------------------------------------------------  \nCombining (4) and (18) we deduce F_x > 0, i.e.  \n\n F·î > 0.\n\nThus the horizontal force required to keep the wire in equilibrium must always be directed to the right.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.406592",
        "was_fixed": false,
        "difficulty_analysis": "• Additional geometry:  The wire now lives in ℝ³ and contains a curved\n 3-dimensional piece (the quarter-circle C₂); torque must be handled with full\n vector machinery rather than planar moments.\n\n• More segments:  Three distinct pieces (one nonlinear in each coordinate)\n interact; the planar problem had just two.\n\n• Non-trivial arc length:  The exponential branch requires integration of\n √(1+e^{2x}), while the arc length of C₂ is πR/2.  Although the latter drops out\n of the final inequality, recognising this fact is part of the challenge.\n\n• Inequality sharpening:  Showing f′(a) > 0 needs an effective\n estimate on ∫₀ᵃ g(x) dx relative to g(a); a naïve bound\n ∫≤a g(a) would fail when a > 3, so one must craft a sharper argument using the\n specific growth of g.\n\n• Torque in three dimensions:  Students must translate the “centroid to the\n right” idea into a 3-D moment calculation and keep track of vector directions\n (î, ĵ, k̂).\n\nTogether these points raise the algebraic, geometric, and conceptual load well\nbeyond that of the original two-piece planar question."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}