{
  "index": "1939-B-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "8. From the vertex \\( (0, c) \\) of the catenary\n\\[\ny=c \\cosh \\frac{x}{c}\n\\]\na line \\( L \\) is drawn perpendicular to the tangent to the catenary at a point \\( P \\). Prove that the length of \\( L \\) intercepted by the axes is equal to the ordinate \\( y \\) of the point \\( P \\).",
  "solution": "Solution. At the point \\( \\left(x_{1}, c \\cosh \\left(x_{1} / c\\right)\\right) \\) the slope of the given catenary is \\( \\sinh \\left(x_{1} / c\\right) \\). Hence the equation of the line \\( L \\) is\n\\[\ny-c=\\frac{-x}{\\sinh \\left(x_{1} / c\\right)}\n\\]\nand this line intersects the \\( x \\)-axis at \\( \\left(c \\sinh \\left(x_{1} / c\\right), 0\\right) \\). Therefore the length of the segment of \\( L \\) between the axes is \\( \\sqrt{c^{2} \\sinh ^{2}\\left(x_{1} / c\\right)+c^{2}}=c \\cosh \\) \\( \\left(x_{1} / c\\right) \\), which is indeed the ordinate of the point \\( P \\).",
  "vars": [
    "x",
    "y",
    "x_1"
  ],
  "params": [
    "c",
    "L",
    "P"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "x_1": "shiftedx",
        "c": "catenpar",
        "L": "perpline",
        "P": "contactp"
      },
      "question": "8. From the vertex \\( (0, catenpar) \\) of the catenary\n\\[\nordinate = catenpar \\cosh \\frac{abscissa}{catenpar}\n\\]\na line \\( perpline \\) is drawn perpendicular to the tangent to the catenary at a point \\( contactp \\). Prove that the length of \\( perpline \\) intercepted by the axes is equal to the ordinate \\( ordinate \\) of the point \\( contactp \\).",
      "solution": "Solution. At the point \\( \\left( shiftedx, catenpar \\cosh \\left( \\frac{shiftedx}{catenpar} \\right) \\right) \\) the slope of the given catenary is \\( \\sinh \\left( \\frac{shiftedx}{catenpar} \\right) \\). Hence the equation of the line \\( perpline \\) is\n\\[\nordinate - catenpar = \\frac{-\\,abscissa}{\\sinh \\left( \\frac{shiftedx}{catenpar} \\right)}\n\\]\nand this line intersects the \\( abscissa \\)-axis at \\( \\left( catenpar \\sinh \\left( \\frac{shiftedx}{catenpar} \\right), 0 \\right) \\). Therefore the length of the segment of \\( perpline \\) between the axes is\n\\[\n\\sqrt{ catenpar^{2} \\sinh^{2} \\left( \\frac{shiftedx}{catenpar} \\right) + catenpar^{2} } = catenpar \\cosh \\left( \\frac{shiftedx}{catenpar} \\right),\n\\]\nwhich is indeed the ordinate of the point \\( contactp \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "watermelon",
        "x_1": "strawberry",
        "c": "bluebutton",
        "L": "peppermint",
        "P": "honeysuckle"
      },
      "question": "From the vertex \\( (0, bluebutton) \\) of the catenary\n\\[\nwatermelon = bluebutton \\cosh \\frac{lighthouse}{bluebutton}\n\\]\na line \\( peppermint \\) is drawn perpendicular to the tangent to the catenary at a point \\( honeysuckle \\). Prove that the length of \\( peppermint \\) intercepted by the axes is equal to the ordinate \\( watermelon \\) of the point \\( honeysuckle \\).",
      "solution": "Solution. At the point \\( \\left(strawberry, bluebutton \\cosh \\left(strawberry / bluebutton\\right)\\right) \\) the slope of the given catenary is \\( \\sinh \\left(strawberry / bluebutton\\right) \\). Hence the equation of the line \\( peppermint \\) is\n\\[\nwatermelon-bluebutton=\\frac{-lighthouse}{\\sinh \\left(strawberry / bluebutton\\right)}\n\\]\nand this line intersects the \\( lighthouse \\)-axis at \\( \\left(bluebutton \\sinh \\left(strawberry / bluebutton\\right), 0\\right) \\). Therefore the length of the segment of \\( peppermint \\) between the axes is \\( \\sqrt{bluebutton^{2} \\sinh ^{2}\\left(strawberry / bluebutton\\right)+bluebutton^{2}}=bluebutton \\cosh \\) \\( \\left(strawberry / bluebutton\\right) \\), which is indeed the ordinate of the point \\( honeysuckle \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoord",
        "y": "horizontalcoord",
        "x_1": "finalcoordinate",
        "c": "variablefactor",
        "L": "curvepath",
        "P": "straightline"
      },
      "question": "8. From the vertex \\( (0, variablefactor) \\) of the catenary\n\\[\nhorizontalcoord=variablefactor \\cosh \\frac{verticalcoord}{variablefactor}\n\\]\na line \\( curvepath \\) is drawn perpendicular to the tangent to the catenary at a point \\( straightline \\). Prove that the length of \\( curvepath \\) intercepted by the axes is equal to the ordinate \\( horizontalcoord \\) of the point \\( straightline \\).",
      "solution": "Solution. At the point \\( \\left(finalcoordinate, variablefactor \\cosh \\left(finalcoordinate / variablefactor\\right)\\right) \\) the slope of the given catenary is \\( \\sinh \\left(finalcoordinate / variablefactor\\right) \\). Hence the equation of the line \\( curvepath \\) is\n\\[\nhorizontalcoord-variablefactor=\\frac{-verticalcoord}{\\sinh \\left(finalcoordinate / variablefactor\\right)}\n\\]\nand this line intersects the \\( verticalcoord \\)-axis at \\( \\left(variablefactor \\sinh \\left(finalcoordinate / variablefactor\\right), 0\\right) \\). Therefore the length of the segment of \\( curvepath \\) between the axes is \\( \\sqrt{variablefactor^{2} \\sinh ^{2}\\left(finalcoordinate / variablefactor\\right)+variablefactor^{2}}=variablefactor \\cosh \\left(finalcoordinate / variablefactor\\right) \\), which is indeed the ordinate of the point \\( straightline \\)."
    },
    "garbled_string": {
      "map": {
        "x": "zqtrpnfa",
        "y": "hvysgjml",
        "x_1": "jkqlnvra",
        "c": "brqtonwy",
        "L": "vxqmrnpz",
        "P": "kdlysmuf"
      },
      "question": "8. From the vertex \\( (0, brqtonwy) \\) of the catenary\n\\[\nhvysgjml=brqtonwy \\cosh \\frac{zqtrpnfa}{brqtonwy}\n\\]\na line \\( vxqmrnpz \\) is drawn perpendicular to the tangent to the catenary at a point \\( kdlysmuf \\). Prove that the length of \\( vxqmrnpz \\) intercepted by the axes is equal to the ordinate \\( hvysgjml \\) of the point \\( kdlysmuf \\).",
      "solution": "Solution. At the point \\( \\left(jkqlnvra, brqtonwy \\cosh \\left(jkqlnvra / brqtonwy\\right)\\right) \\) the slope of the given catenary is \\( \\sinh \\left(jkqlnvra / brqtonwy\\right) \\). Hence the equation of the line \\( vxqmrnpz \\) is\n\\[\nhvysgjml-brqtonwy=\\frac{-zqtrpnfa}{\\sinh \\left(jkqlnvra / brqtonwy\\right)}\n\\]\nand this line intersects the \\( zqtrpnfa \\)-axis at \\( \\left(brqtonwy \\sinh \\left(jkqlnvra / brqtonwy\\right), 0\\right) \\). Therefore the length of the segment of \\( vxqmrnpz \\) between the axes is \\( \\sqrt{brqtonwy^{2} \\sinh ^{2}\\left(jkqlnvra / brqtonwy\\right)+brqtonwy^{2}}=brqtonwy \\cosh \\left(jkqlnvra / brqtonwy\\right) \\), which is indeed the ordinate of the point \\( kdlysmuf \\)."
    },
    "kernel_variant": {
      "question": "Fix λ > 0 and any integer n ≥ 2.  Write r(x₁,…,x_{n-1})=√(x₁²+⋯+x_{n-1}²).  \nConsider the (n–1)-dimensional “radial catenary”  \n  x_n = λ cosh(r/λ)  (so the vertex is V=(0,…,0,λ)).  \n\nFor a chosen radius ρ ≥ 0 set Q = (ρu, λ cosh(ρ/λ)), where u is a unit vector in ℝ^{n-1}.  \nLet H be the hyper-plane through V that is orthogonal to the tangent hyper-plane of the surface at Q.  \nProve that the segment of H cut off by the hyper-planes x_n=0 and x₁=⋯=x_{n-1}=0 has length λ cosh(ρ/λ).\n\n────────────────────────",
      "solution": "(≈ 83 words)  \nAt Q the surface is the graph of f(r)=λ cosh(r/λ); its gradient is  \n ∇f(Q)=sinh(ρ/λ) u.  \nHence a normal to the tangent hyper-plane is N=(sinh(ρ/λ) u, –1).  \nThe required line through V in direction N is  \n V+τN = (τ sinh(ρ/λ) u, λ–τ).  \n\n1.  Intersection with x_n=0 occurs at τ=λ, giving P=(λ sinh(ρ/λ) u, 0).  \n2.  Intersection with x₁=⋯=x_{n-1}=0 is simply V itself.  \n\nThus |VP| = √(λ² sinh²(ρ/λ)+λ²)=λ cosh(ρ/λ).  \nWhen ρ=0 the vector N is (0,–1), P coincides with (0,…,0), and |VP|=λ=λ cosh 0, so the claim holds in all cases.\n\n────────────────────────",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.143933",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}