{
  "index": "1974-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "A-2. A circle stands in a plane perpendicular to the ground and a point \\( A \\) lies in this plane exterior to the circle and higher than its bottom. A particle starting from rest at \\( A \\) slides without friction down an inclined straight line until it reaches the circle. Which straight line allows descent in the shortest time? [Assume that the force of gravity is constant over the region involved, there are no relativistic effects, etc.]\n\nThe starting point A and the circle are fixed; the stopping point \\( B \\) is allowed to vary over the circle.\n\nNote. The answer may be given in any form which specifies the line of descent in an unambiguous manner; it is not required to find the coordinates of the point \\( B \\).",
  "solution": "A-2.\nLet \\( C \\) be the other point of intersection of line \\( A B \\) with the circle and let \\( \\theta \\) be the inclination of \\( A B \\). Let \\( \\overrightarrow{A B}=b \\) and \\( \\overrightarrow{A C}=c \\). The square of the time of descent is proportional to \\( b / \\sin \\theta \\) and hence to \\( 1 /(c \\sin \\theta) \\), since it is well known that \\( b c \\) is constant with respect to \\( \\theta \\). The time is minimized by maximizing \\( c \\sin \\theta \\); this is done by choosing \\( C \\) as the bottom of the circle.",
  "vars": [
    "B",
    "C",
    "\\\\theta",
    "b",
    "c"
  ],
  "params": [
    "A"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "fixedpoint",
        "B": "stoppoint",
        "C": "secondpoint",
        "\\theta": "inclination",
        "b": "vectorab",
        "c": "vectorac"
      },
      "question": "A-2. A circle stands in a plane perpendicular to the ground and a point \\( fixedpoint \\) lies in this plane exterior to the circle and higher than its bottom. A particle starting from rest at \\( fixedpoint \\) slides without friction down an inclined straight line until it reaches the circle. Which straight line allows descent in the shortest time? [Assume that the force of gravity is constant over the region involved, there are no relativistic effects, etc.]\n\nThe starting point fixedpoint and the circle are fixed; the stopping point \\( stoppoint \\) is allowed to vary over the circle.\n\nNote. The answer may be given in any form which specifies the line of descent in an unambiguous manner; it is not required to find the coordinates of the point \\( stoppoint \\).",
      "solution": "A-2.\nLet \\( secondpoint \\) be the other point of intersection of line \\( fixedpoint stoppoint \\) with the circle and let \\( inclination \\) be the inclination of \\( fixedpoint stoppoint \\). Let \\( \\overrightarrow{fixedpoint stoppoint}=vectorab \\) and \\( \\overrightarrow{fixedpoint secondpoint}=vectorac \\). The square of the time of descent is proportional to \\( vectorab / \\sin inclination \\) and hence to \\( 1 /(vectorac \\sin inclination) \\), since it is well known that \\( vectorab\\, vectorac \\) is constant with respect to \\( inclination \\). The time is minimized by maximizing \\( vectorac \\sin inclination \\); this is done by choosing \\( secondpoint \\) as the bottom of the circle."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "waterfall",
        "B": "rainstorm",
        "C": "butterfly",
        "\\theta": "moonlight",
        "b": "starlight",
        "c": "landscape"
      },
      "question": "A circle stands in a plane perpendicular to the ground and a point \\( waterfall \\) lies in this plane exterior to the circle and higher than its bottom. A particle starting from rest at \\( waterfall \\) slides without friction down an inclined straight line until it reaches the circle. Which straight line allows descent in the shortest time? [Assume that the force of gravity is constant over the region involved, there are no relativistic effects, etc.]\n\nThe starting point waterfall and the circle are fixed; the stopping point \\( rainstorm \\) is allowed to vary over the circle.",
      "solution": "Let \\( butterfly \\) be the other point of intersection of line \\( waterfall rainstorm \\) with the circle and let \\( moonlight \\) be the inclination of \\( waterfall rainstorm \\). Let \\( \\overrightarrow{waterfall rainstorm}=starlight \\) and \\( \\overrightarrow{waterfall butterfly}=landscape \\). The square of the time of descent is proportional to \\( starlight / \\sin moonlight \\) and hence to \\( 1 /(landscape \\sin moonlight) \\), since it is well known that \\( starlight\\, landscape \\) is constant with respect to \\( moonlight \\). The time is minimized by maximizing \\( landscape \\sin moonlight \\); this is done by choosing \\( butterfly \\) as the bottom of the circle."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "finishpoint",
        "B": "startpoint",
        "C": "apexpoint",
        "\\theta": "flatness",
        "b": "briefness",
        "c": "shortness"
      },
      "question": "A circle stands in a plane perpendicular to the ground and a point \\( finishpoint \\) lies in this plane exterior to the circle and higher than its bottom. A particle starting from rest at \\( finishpoint \\) slides without friction down an inclined straight line until it reaches the circle. Which straight line allows descent in the shortest time? [Assume that the force of gravity is constant over the region involved, there are no relativistic effects, etc.]\n\nThe starting point finishpoint and the circle are fixed; the stopping point \\( startpoint \\) is allowed to vary over the circle.",
      "solution": "A-2.\nLet \\( apexpoint \\) be the other point of intersection of line \\( finishpoint startpoint \\) with the circle and let \\( flatness \\) be the inclination of \\( finishpoint startpoint \\). Let \\( \\overrightarrow{finishpoint startpoint}=briefness \\) and \\( \\overrightarrow{finishpoint apexpoint}=shortness \\). The square of the time of descent is proportional to \\( briefness / \\sin flatness \\) and hence to \\( 1 /(shortness \\sin flatness) \\), since it is well known that \\( briefness shortness \\) is constant with respect to \\( flatness \\). The time is minimized by maximizing \\( shortness \\sin flatness \\); this is done by choosing \\( apexpoint \\) as the bottom of the circle."
    },
    "garbled_string": {
      "map": {
        "B": "cxmjavde",
        "C": "qztplwri",
        "\\theta": "vgksmpqe",
        "b": "fzhtkrlm",
        "c": "wyjdnxso",
        "A": "rplnhqsk"
      },
      "question": "A circle stands in a plane perpendicular to the ground and a point \\( rplnhqsk \\) lies in this plane exterior to the circle and higher than its bottom. A particle starting from rest at \\( rplnhqsk \\) slides without friction down an inclined straight line until it reaches the circle. Which straight line allows descent in the shortest time? [Assume that the force of gravity is constant over the region involved, there are no relativistic effects, etc.]\n\nThe starting point rplnhqsk and the circle are fixed; the stopping point \\( cxmjavde \\) is allowed to vary over the circle.\n\nNote. The answer may be given in any form which specifies the line of descent in an unambiguous manner; it is not required to find the coordinates of the point \\( cxmjavde \\).",
      "solution": "A-2.\nLet \\( qztplwri \\) be the other point of intersection of line \\( rplnhqsk\\,cxmjavde \\) with the circle and let \\( vgksmpqe \\) be the inclination of \\( rplnhqsk\\,cxmjavde \\). Let \\( \\overrightarrow{rplnhqsk\\,cxmjavde}=fzhtkrlm \\) and \\( \\overrightarrow{rplnhqsk\\,qztplwri}=wyjdnxso \\). The square of the time of descent is proportional to \\( fzhtkrlm / \\sin vgksmpqe \\) and hence to \\( 1 /(wyjdnxso \\sin vgksmpqe) \\), since it is well known that \\( fzhtkrlm wyjdnxso \\) is constant with respect to \\( vgksmpqe \\). The time is minimized by maximizing \\( wyjdnxso \\sin vgksmpqe \\); this is done by choosing \\( qztplwri \\) as the bottom of the circle."
    },
    "kernel_variant": {
      "question": "A uniform downward acceleration field of magnitude \\(\\gamma = 4\\text{ m/s}^2\\) prevails on a certain planet.  In a vertical plane of this field a rigid circular hoop of radius \\(r = 7\\text{ m}\\) is fixed.  A point \\(S\\) in the same plane lies outside the hoop and strictly higher than its lowest point.  From rest at \\(S\\) a small bead is released and constrained to slide without friction along a straight rod that meets the hoop at the first intersection point \\(T\\).  Over all possible rods through \\(S\\) that intersect the hoop, which rod minimises the travel time from \\(S\\) to \\(T\\)?  (An explicit coordinate description of \\(T\\) is not required; it is enough to specify the optimal rod unambiguously.)",
      "solution": "Let the straight rod through S strike the hoop first at T and again at U (S–T–U lie on the same line).  Write ST = b, SU = c, and let θ be the rod’s inclination above the horizontal.\n\n1. Time along an incline.  The component of the constant acceleration along the rod is γ sin θ.  Starting from rest, the time t required to traverse the distance b satisfies\n   b = ½(γ sin θ) t²  ⇒  t² = 2b/(γ sin θ).\n   Because 2/γ is a fixed numerical factor (γ = 4 m/s²), minimizing t is equivalent to minimizing\n   t² ∝ b/ sin θ.\n\n2. Power of a point.  For any line through the exterior point S intersecting the hoop,\n   ST·SU = b c = constant.\n   Eliminating b from t² ∝ b/ sin θ with b c = constant gives\n   t² ∝ 1/(c sin θ).\n\n3. What must be maximized.  Thus the descent time is minimized exactly when the product c sin θ is maximized.  Notice that c sin θ is the vertical drop from S down to U, because c is the length SU and sin θ projects that length onto the vertical.\n\n4. The best choice of U.  As the rod pivots about S, the second intersection U ranges over the hoop.  The vertical drop from S to U is largest when U is the lowest point of the hoop.  Therefore the optimal rod is the one that joins S to that lowest point L of the hoop.  The bead first encounters the hoop at T, the other intersection of SL with the circle.\n\nHence:  The quickest descent occurs along the unique straight rod through S that also passes through the hoop’s lowest point.",
      "_meta": {
        "core_steps": [
          "Use t² ∝ (length of incline)/(sin θ) for friction-free motion under gravity.",
          "Apply the power-of-a-point theorem: for any line through A, AB·AC is constant.",
          "Replace AB by (const)/AC to get t² ∝ 1/(AC·sin θ).",
          "Minimizing time ⇔ maximizing AC·sin θ, i.e. the vertical drop from A to C.",
          "That vertical drop is largest when C is the circle’s lowest point; take the line through A and that point (B is the first intersection)."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Numerical value of gravitational acceleration; it factors out of the optimization.",
            "original": "g (a positive constant)"
          },
          "slot2": {
            "description": "Size (radius) of the circle; only the power-of-a-point constant changes.",
            "original": "the fixed radius R of the given circle"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}