{
  "index": "1941-A-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "3. A circle of radius a rolls in its plane along the \\( x \\)-axis. Show that the envelope of a diameter is a cycloid, coinciding with the cycloid traced out by a point on the circumference of a circle of diameter \\( a \\), likewise rolling in its plane along the \\( x \\)-axis.",
  "solution": "First Solution. Let the two circles roll along the upper side of the \\( x \\)-axis so that at time \\( t \\) they both touch the axis at (at, 0). Let \\( P \\) and \\( Q \\) be the points that start in contact with the origin \\( O \\), and let \\( D \\) and \\( E \\) be the centers of the larger and smaller circles, respectively. Let \\( P^{\\prime} \\) be the point opposite \\( P \\) on the larger circle. We may suppose that \\( P P^{\\prime} \\) is the moving diameter of the problem, since we can choose the origin along the \\( x \\)-axis at our convenience.\n\nThe point \\( Q \\) traces a cycloid \\( C \\), and we shall prove that \\( P P^{\\prime} \\) is always tangent to \\( C \\) at \\( Q \\). This will prove that the envelope of \\( P P^{\\prime} \\) is \\( C \\).\n\nSince arc \\( P T=\\operatorname{arc} Q T=O T=a t \\), while the circles are of radius \\( a \\) and \\( a / 2 \\), we see that \\( \\angle P D T=t, \\angle Q E T=2 t \\). Hence at time \\( t, P \\) is at \\( (a t-a \\sin t, a-a \\cos t), P^{\\prime} \\) is at \\( (a t+a \\sin t, a+a \\cos t) \\), and the coordinates of \\( Q \\) are\n\\[\n\\begin{array}{l}\nx=a t-\\frac{a}{2} \\sin 2 t \\\\\ny=\\frac{a}{2}-\\frac{a}{2} \\cos 2 t .\n\\end{array}\n\\]\n\nThese are parametric equations for the motion of \\( Q \\), that is, for the cycloid \\( C \\). This parametrization is non-singular when \\( t \\) is not a multiple of \\( \\pi \\) and singular when it is (for then both \\( d x / d t \\) and \\( d y / d t \\) are zero). The tangent to \\( \\boldsymbol{C} \\) at a point corresponding to a non-singular \\( t \\) has direction vector\n\\[\n(a-a \\cos 2 t, a \\sin 2 t)\n\\]\nso the tangent to \\( C \\) at \\( Q \\) has the representation\n\\[\nx=a t-\\frac{a}{2} \\sin 2 t+u(a-a \\cos 2 t)\n\\]\n\\[\ny=\\frac{a}{2}-\\frac{a}{2} \\cos 2 t+u(a \\sin 2 t)\n\\]\nin terms of the parameter \\( u \\). Putting\n\\[\nu=\\frac{\\cos t-1}{2 \\sin t} \\quad \\text { and } \\quad u=\\frac{\\cos t+1}{2 \\sin t}\n\\]\nin (1), we see that \\( P \\) and \\( P^{\\prime} \\) are both on the tangent line. This proves that \\( P P^{\\prime} \\) is tangent to \\( C \\) at \\( Q \\) for \\( t \\) not a multiple of \\( \\pi \\).\n\nWhen \\( t \\) is a multiple of \\( \\pi, Q \\) is on the \\( x \\)-axis, the curve \\( C \\) has a cusp at \\( Q \\) with a vertical tangent, and \\( P P^{\\prime} \\) stands vertical with either \\( P \\) or \\( P^{\\prime} \\) on the \\( x \\)-axis. Hence \\( P P^{\\prime} \\) is tangent to \\( C \\) in this case too.\n\nSecond Solution. A purely synthetic argument can be given as follows: As before, we see that \\( \\angle Q E T=2 \\angle P D T \\). Since \\( \\angle Q D T \\) is measured by half of arc \\( Q T \\), we have \\( \\frac{1}{2} \\angle Q E T=\\angle Q D T \\). It follows that \\( \\angle Q D T= \\) \\( \\angle P D T \\), so \\( Q \\) is on \\( P P^{\\prime} \\).\nAs the small circle rolls, the point \\( T \\) is instantaneously at rest; hence all other points of the moving circle are instantaneously in rotation about \\( T \\). This implies that the tangent at \\( Q \\) to the orbit of \\( Q \\) is perpendicular to \\( Q T \\). But \\( Q D=P P^{\\prime} \\) is this perpendicular since \\( \\angle T Q D \\) is inscribed in a semicircle.\n\nAs is often the case with such synthetic arguments, a slightly different analysis is required after \\( Q \\) attains its highest point.",
  "vars": [
    "x",
    "t",
    "P",
    "Q",
    "D",
    "E",
    "O",
    "C",
    "T",
    "u"
  ],
  "params": [
    "a"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizcoor",
        "t": "timeparm",
        "P": "pointpval",
        "Q": "pointqval",
        "D": "centerbig",
        "E": "centersml",
        "O": "originpt",
        "C": "cycloidcrv",
        "T": "contactpt",
        "u": "linparam",
        "a": "radvalue"
      },
      "question": "3. A circle of radius radvalue rolls in its plane along the \\( horizcoor \\)-axis. Show that the envelope of a diameter is a cycloid, coinciding with the cycloid traced out by a point on the circumference of a circle of diameter \\( radvalue \\), likewise rolling in its plane along the \\( horizcoor \\)-axis.",
      "solution": "First Solution. Let the two circles roll along the upper side of the \\( horizcoor \\)-axis so that at time \\( timeparm \\) they both touch the axis at (radvalue timeparm, 0). Let \\( pointpval \\) and \\( pointqval \\) be the points that start in contact with the origin \\( originpt \\), and let \\( centerbig \\) and \\( centersml \\) be the centers of the larger and smaller circles, respectively. Let \\( pointpval^{\\prime} \\) be the point opposite \\( pointpval \\) on the larger circle. We may suppose that \\( pointpval pointpval^{\\prime} \\) is the moving diameter of the problem, since we can choose the origin along the \\( horizcoor \\)-axis at our convenience.\n\nThe point \\( pointqval \\) traces a cycloid \\( cycloidcrv \\), and we shall prove that \\( pointpval pointpval^{\\prime} \\) is always tangent to \\( cycloidcrv \\) at \\( pointqval \\). This will prove that the envelope of \\( pointpval pointpval^{\\prime} \\) is \\( cycloidcrv \\).\n\nSince arc \\( pointpval contactpt=\\operatorname{arc} pointqval contactpt=originpt contactpt=radvalue timeparm \\), while the circles are of radius \\( radvalue \\) and \\( radvalue / 2 \\), we see that \\( \\angle pointpval centerbig contactpt=timeparm, \\angle pointqval centersml contactpt=2 timeparm \\). Hence at time \\( timeparm, pointpval \\) is at \\( (radvalue timeparm-radvalue \\sin timeparm, radvalue-radvalue \\cos timeparm), pointpval^{\\prime} \\) is at \\( (radvalue timeparm+radvalue \\sin timeparm, radvalue+radvalue \\cos timeparm) \\), and the coordinates of \\( pointqval \\) are\n\\[\n\\begin{array}{l}\nhorizcoor=radvalue timeparm-\\frac{radvalue}{2} \\sin 2 timeparm \\\\\ny=\\frac{radvalue}{2}-\\frac{radvalue}{2} \\cos 2 timeparm .\n\\end{array}\n\\]\n\nThese are parametric equations for the motion of \\( pointqval \\), that is, for the cycloid \\( cycloidcrv \\). This parametrization is non-singular when \\( timeparm \\) is not a multiple of \\( \\pi \\) and singular when it is (for then both \\( d horizcoor / d timeparm \\) and \\( d y / d timeparm \\) are zero). The tangent to \\( \\boldsymbol{cycloidcrv} \\) at a point corresponding to a non-singular \\( timeparm \\) has direction vector\n\\[\n(radvalue-radvalue \\cos 2 timeparm, radvalue \\sin 2 timeparm)\n\\]\nso the tangent to \\( cycloidcrv \\) at \\( pointqval \\) has the representation\n\\[\nhorizcoor=radvalue timeparm-\\frac{radvalue}{2} \\sin 2 timeparm+linparam(radvalue-radvalue \\cos 2 timeparm)\n\\]\n\\[\ny=\\frac{radvalue}{2}-\\frac{radvalue}{2} \\cos 2 timeparm+linparam(radvalue \\sin 2 timeparm)\n\\]\nin terms of the parameter \\( linparam \\). Putting\n\\[\nlinparam=\\frac{\\cos timeparm-1}{2 \\sin timeparm} \\quad \\text { and } \\quad linparam=\\frac{\\cos timeparm+1}{2 \\sin timeparm}\n\\]\nin (1), we see that \\( pointpval \\) and \\( pointpval^{\\prime} \\) are both on the tangent line. This proves that \\( pointpval pointpval^{\\prime} \\) is tangent to \\( cycloidcrv \\) at \\( pointqval \\) for \\( timeparm \\) not a multiple of \\( \\pi \\).\n\nWhen \\( timeparm \\) is a multiple of \\( \\pi, pointqval \\) is on the \\( horizcoor \\)-axis, the curve \\( cycloidcrv \\) has a cusp at \\( pointqval \\) with a vertical tangent, and \\( pointpval pointpval^{\\prime} \\) stands vertical with either \\( pointpval \\) or \\( pointpval^{\\prime} \\) on the \\( horizcoor \\)-axis. Hence \\( pointpval pointpval^{\\prime} \\) is tangent to \\( cycloidcrv \\) in this case too.\n\nSecond Solution. A purely synthetic argument can be given as follows: As before, we see that \\( \\angle pointqval centersml contactpt=2 \\angle pointpval centerbig contactpt \\). Since \\( \\angle pointqval centerbig contactpt \\) is measured by half of arc \\( pointqval contactpt \\), we have \\( \\frac{1}{2} \\angle pointqval centersml contactpt=\\angle pointqval centerbig contactpt \\). It follows that \\( \\angle pointqval centerbig contactpt= \\) \\( \\angle pointpval centerbig contactpt \\), so \\( pointqval \\) is on \\( pointpval pointpval^{\\prime} \\).\nAs the small circle rolls, the point \\( contactpt \\) is instantaneously at rest; hence all other points of the moving circle are instantaneously in rotation about \\( contactpt \\). This implies that the tangent at \\( pointqval \\) to the orbit of \\( pointqval \\) is perpendicular to \\( pointqval contactpt \\). But \\( pointqval centerbig=pointpval pointpval^{\\prime} \\) is this perpendicular since \\( \\angle contactpt pointqval centerbig \\) is inscribed in a semicircle.\n\nAs is often the case with such synthetic arguments, a slightly different analysis is required after \\( pointqval \\) attains its highest point."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "t": "pendulum",
        "P": "marigold",
        "Q": "dragonfly",
        "D": "parchment",
        "E": "sandstorm",
        "O": "raincloud",
        "C": "moonlight",
        "T": "riverbank",
        "u": "huckleberry",
        "a": "seashells"
      },
      "question": "3. A circle of radius seashells rolls in its plane along the \\( lighthouse \\)-axis. Show that the envelope of a diameter is a cycloid, coinciding with the cycloid traced out by a point on the circumference of a circle of diameter \\( seashells \\), likewise rolling in its plane along the \\( lighthouse \\)-axis.",
      "solution": "First Solution. Let the two circles roll along the upper side of the \\( lighthouse \\)-axis so that at time \\( pendulum \\) they both touch the axis at \\((seashells pendulum, 0)\\). Let \\( marigold \\) and \\( dragonfly \\) be the points that start in contact with the origin \\( raincloud \\), and let \\( parchment \\) and \\( sandstorm \\) be the centers of the larger and smaller circles, respectively. Let \\( marigold^{\\prime} \\) be the point opposite \\( marigold \\) on the larger circle. We may suppose that \\( marigold marigold^{\\prime} \\) is the moving diameter of the problem, since we can choose the origin along the \\( lighthouse \\)-axis at our convenience.\n\nThe point \\( dragonfly \\) traces a cycloid \\( moonlight \\), and we shall prove that \\( marigold marigold^{\\prime} \\) is always tangent to \\( moonlight \\) at \\( dragonfly \\). This will prove that the envelope of \\( marigold marigold^{\\prime} \\) is \\( moonlight \\).\n\nSince arc \\( marigold riverbank=\\operatorname{arc} dragonfly riverbank=raincloud riverbank=seashells pendulum \\), while the circles are of radius \\( seashells \\) and \\( seashells / 2 \\), we see that \\( \\angle marigold parchment riverbank=pendulum, \\angle dragonfly sandstorm riverbank=2 pendulum \\). Hence at time \\( pendulum, marigold \\) is at \\( (seashells pendulum-seashells \\sin pendulum, seashells-seashells \\cos pendulum), marigold^{\\prime} \\) is at \\( (seashells pendulum+seashells \\sin pendulum, seashells+seashells \\cos pendulum) \\), and the coordinates of \\( dragonfly \\) are\n\\[\n\\begin{array}{l}\nlighthouse=seashells pendulum-\\frac{seashells}{2} \\sin 2 pendulum \\\\\ny=\\frac{seashells}{2}-\\frac{seashells}{2} \\cos 2 pendulum .\n\\end{array}\n\\]\n\nThese are parametric equations for the motion of \\( dragonfly \\), that is, for the cycloid \\( moonlight \\). This parametrization is non-singular when \\( pendulum \\) is not a multiple of \\( \\pi \\) and singular when it is (for then both \\( d lighthouse / d pendulum \\) and \\( d y / d pendulum \\) are zero). The tangent to \\( \\boldsymbol{moonlight} \\) at a point corresponding to a non-singular \\( pendulum \\) has direction vector\n\\[\n(seashells-seashells \\cos 2 pendulum, seashells \\sin 2 pendulum)\n\\]\nso the tangent to \\( moonlight \\) at \\( dragonfly \\) has the representation\n\\[\nlighthouse=seashells pendulum-\\frac{seashells}{2} \\sin 2 pendulum+huckleberry(seashells-seashells \\cos 2 pendulum)\n\\]\n\\[\ny=\\frac{seashells}{2}-\\frac{seashells}{2} \\cos 2 pendulum+huckleberry(seashells \\sin 2 pendulum)\n\\]\nin terms of the parameter \\( huckleberry \\). Putting\n\\[\nhuckleberry=\\frac{\\cos pendulum-1}{2 \\sin pendulum} \\quad \\text { and } \\quad huckleberry=\\frac{\\cos pendulum+1}{2 \\sin pendulum}\n\\]\nin (1), we see that \\( marigold \\) and \\( marigold^{\\prime} \\) are both on the tangent line. This proves that \\( marigold marigold^{\\prime} \\) is tangent to \\( moonlight \\) at \\( dragonfly \\) for \\( pendulum \\) not a multiple of \\( \\pi \\).\n\nWhen \\( pendulum \\) is a multiple of \\( \\pi, dragonfly \\) is on the \\( lighthouse \\)-axis, the curve \\( moonlight \\) has a cusp at \\( dragonfly \\) with a vertical tangent, and \\( marigold marigold^{\\prime} \\) stands vertical with either \\( marigold \\) or \\( marigold^{\\prime} \\) on the \\( lighthouse \\)-axis. Hence \\( marigold marigold^{\\prime} \\) is tangent to \\( moonlight \\) in this case too.\n\nSecond Solution. A purely synthetic argument can be given as follows: As before, we see that \\( \\angle dragonfly sandstorm riverbank=2 \\angle marigold parchment riverbank \\). Since \\( \\angle dragonfly parchment riverbank \\) is measured by half of arc \\( dragonfly riverbank \\), we have \\( \\frac{1}{2} \\angle dragonfly sandstorm riverbank=\\angle dragonfly parchment riverbank \\). It follows that \\( \\angle dragonfly parchment riverbank= \\) \\( \\angle marigold parchment riverbank \\), so \\( dragonfly \\) is on \\( marigold marigold^{\\prime} \\).\nAs the small circle rolls, the point \\( riverbank \\) is instantaneously at rest; hence all other points of the moving circle are instantaneously in rotation about \\( riverbank \\). This implies that the tangent at \\( dragonfly \\) to the orbit of \\( dragonfly \\) is perpendicular to \\( dragonfly riverbank \\). But \\( dragonfly parchment=marigold marigold^{\\prime} \\) is this perpendicular since \\( \\angle riverbank dragonfly parchment \\) is inscribed in a semicircle.\n\nAs is often the case with such synthetic arguments, a slightly different analysis is required after \\( dragonfly \\) attains its highest point."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "t": "timeless",
        "P": "emptiness",
        "Q": "fullness",
        "D": "periphery",
        "E": "boundary",
        "O": "destination",
        "C": "straightline",
        "T": "breakpoint",
        "u": "constant",
        "a": "diameter"
      },
      "question": "3. A circle of radius diameter rolls in its plane along the \\( verticalaxis \\)-axis. Show that the envelope of a diameter is a cycloid, coinciding with the cycloid traced out by a point on the circumference of a circle of diameter \\( diameter \\), likewise rolling in its plane along the \\( verticalaxis \\)-axis.",
      "solution": "First Solution. Let the two circles roll along the upper side of the \\( verticalaxis \\)-axis so that at time \\( timeless \\) they both touch the axis at (diameter timeless, 0). Let \\( emptiness \\) and \\( fullness \\) be the points that start in contact with the origin \\( destination \\), and let \\( periphery \\) and \\( boundary \\) be the centers of the larger and smaller circles, respectively. Let \\( emptiness^{\\prime} \\) be the point opposite \\( emptiness \\) on the larger circle. We may suppose that \\( emptiness emptiness^{\\prime} \\) is the moving diameter of the problem, since we can choose the origin along the \\( verticalaxis \\)-axis at our convenience.\n\nThe point \\( fullness \\) traces a cycloid \\( straightline \\), and we shall prove that \\( emptiness emptiness^{\\prime} \\) is always tangent to \\( straightline \\) at \\( fullness \\). This will prove that the envelope of \\( emptiness emptiness^{\\prime} \\) is \\( straightline \\).\n\nSince arc \\( emptiness breakpoint=\\operatorname{arc} fullness breakpoint=destination breakpoint=diameter timeless \\), while the circles are of radius \\( diameter \\) and \\( diameter / 2 \\), we see that \\( \\angle emptiness periphery breakpoint=timeless, \\angle fullness boundary breakpoint=2 timeless \\). Hence at time \\( timeless, emptiness \\) is at \\( (diameter timeless-diameter \\sin timeless, diameter-diameter \\cos timeless), emptiness^{\\prime} \\) is at \\( (diameter timeless+diameter \\sin timeless, diameter+diameter \\cos timeless) \\), and the coordinates of \\( fullness \\) are\n\\[\n\\begin{array}{l}\nverticalaxis=diameter timeless-\\frac{diameter}{2} \\sin 2 timeless \\\\\ny=\\frac{diameter}{2}-\\frac{diameter}{2} \\cos 2 timeless .\n\\end{array}\n\\]\n\nThese are parametric equations for the motion of \\( fullness \\), that is, for the cycloid \\( straightline \\). This parametrization is non-singular when \\( timeless \\) is not a multiple of \\( \\pi \\) and singular when it is (for then both \\( d verticalaxis / d timeless \\) and \\( d y / d timeless \\) are zero). The tangent to \\( \\boldsymbol{straightline} \\) at a point corresponding to a non-singular \\( timeless \\) has direction vector\n\\[\n(diameter-diameter \\cos 2 timeless, diameter \\sin 2 timeless)\n\\]\nso the tangent to \\( straightline \\) at \\( fullness \\) has the representation\n\\[\nverticalaxis=diameter timeless-\\frac{diameter}{2} \\sin 2 timeless+constant(diameter-diameter \\cos 2 timeless)\n\\]\n\\[\ny=\\frac{diameter}{2}-\\frac{diameter}{2} \\cos 2 timeless+constant(diameter \\sin 2 timeless)\n\\]\nin terms of the parameter \\( constant \\). Putting\n\\[\nconstant=\\frac{\\cos timeless-1}{2 \\sin timeless} \\quad \\text { and } \\quad constant=\\frac{\\cos timeless+1}{2 \\sin timeless}\n\\]\nin (1), we see that \\( emptiness \\) and \\( emptiness^{\\prime} \\) are both on the tangent line. This proves that \\( emptiness emptiness^{\\prime} \\) is tangent to \\( straightline \\) at \\( fullness \\) for \\( timeless \\) not a multiple of \\( \\pi \\).\n\nWhen \\( timeless \\) is a multiple of \\( \\pi, fullness \\) is on the \\( verticalaxis \\)-axis, the curve \\( straightline \\) has a cusp at \\( fullness \\) with a vertical tangent, and \\( emptiness emptiness^{\\prime} \\) stands vertical with either \\( emptiness \\) or \\( emptiness^{\\prime} \\) on the \\( verticalaxis \\)-axis. Hence \\( emptiness emptiness^{\\prime} \\) is tangent to \\( straightline \\) in this case too.\n\nSecond Solution. A purely synthetic argument can be given as follows: As before, we see that \\( \\angle fullness boundary breakpoint=2 \\angle emptiness periphery breakpoint \\). Since \\( \\angle fullness periphery breakpoint \\) is measured by half of arc \\( fullness breakpoint \\), we have \\( \\frac{1}{2} \\angle fullness boundary breakpoint=\\angle fullness periphery breakpoint \\). It follows that \\( \\angle fullness periphery breakpoint= \\) \\( \\angle emptiness periphery breakpoint \\), so \\( fullness \\) is on \\( emptiness emptiness^{\\prime} \\).\nAs the small circle rolls, the point \\( breakpoint \\) is instantaneously at rest; hence all other points of the moving circle are instantaneously in rotation about \\( breakpoint \\). This implies that the tangent at \\( fullness \\) to the orbit of \\( fullness \\) is perpendicular to \\( fullness breakpoint \\). But \\( fullness periphery=emptiness emptiness^{\\prime} \\) is this perpendicular since \\( \\angle breakpoint fullness periphery \\) is inscribed in a semicircle.\n\nAs is often the case with such synthetic arguments, a slightly different analysis is required after \\( fullness \\) attains its highest point."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "P": "mgbxraez",
        "Q": "kpyvwols",
        "D": "uehrjxmc",
        "E": "ralydfze",
        "O": "stequfop",
        "C": "wqzsnmya",
        "T": "lcrevbdh",
        "u": "znmpqtav",
        "a": "vohwulzi"
      },
      "question": "A circle of radius vohwulzi rolls in its plane along the \\( qzxwvtnp \\)-axis. Show that the envelope of a diameter is a cycloid, coinciding with the cycloid traced out by a point on the circumference of a circle of diameter vohwulzi, likewise rolling in its plane along the \\( qzxwvtnp \\)-axis.",
      "solution": "First Solution. Let the two circles roll along the upper side of the \\( qzxwvtnp \\)-axis so that at time \\( hjgrksla \\) they both touch the axis at (vohwulzi hjgrksla, 0). Let \\( mgbxraez \\) and \\( kpyvwols \\) be the points that start in contact with the origin \\( stequfop \\), and let \\( uehrjxmc \\) and \\( ralydfze \\) be the centers of the larger and smaller circles, respectively. Let \\( mgbxraez^{\\prime} \\) be the point opposite \\( mgbxraez \\) on the larger circle. We may suppose that \\( mgbxraez mgbxraez^{\\prime} \\) is the moving diameter of the problem, since we can choose the origin along the \\( qzxwvtnp \\)-axis at our convenience.\n\nThe point \\( kpyvwols \\) traces a cycloid \\( wqzsnmya \\), and we shall prove that \\( mgbxraez mgbxraez^{\\prime} \\) is always tangent to \\( wqzsnmya \\) at \\( kpyvwols \\). This will prove that the envelope of \\( mgbxraez mgbxraez^{\\prime} \\) is \\( wqzsnmya \\).\n\nSince arc \\( mgbxraez lcrevbdh=\\operatorname{arc} kpyvwols lcrevbdh= stequfop lcrevbdh= vohwulzi hjgrksla \\), while the circles are of radius \\( vohwulzi \\) and \\( vohwulzi / 2 \\), we see that \\( \\angle mgbxraez uehrjxmc lcrevbdh=hjgrksla, \\angle kpyvwols ralydfze lcrevbdh=2 hjgrksla \\). Hence at time \\( hjgrksla, mgbxraez \\) is at \\( (vohwulzi hjgrksla- vohwulzi \\sin hjgrksla, vohwulzi- vohwulzi \\cos hjgrksla) \\), \\( mgbxraez^{\\prime} \\) is at \\( (vohwulzi hjgrksla+ vohwulzi \\sin hjgrksla, vohwulzi+ vohwulzi \\cos hjgrksla) \\), and the coordinates of \\( kpyvwols \\) are\n\\[\n\\begin{array}{l}\nqzxwvtnp= vohwulzi hjgrksla-\\frac{vohwulzi}{2} \\sin 2 hjgrksla \\\\\ny=\\frac{vohwulzi}{2}-\\frac{vohwulzi}{2} \\cos 2 hjgrksla .\n\\end{array}\n\\]\n\nThese are parametric equations for the motion of \\( kpyvwols \\), that is, for the cycloid \\( wqzsnmya \\). This parametrization is non-singular when \\( hjgrksla \\) is not a multiple of \\( \\pi \\) and singular when it is (for then both \\( d qzxwvtnp / d hjgrksla \\) and \\( d y / d hjgrksla \\) are zero). The tangent to \\( \\boldsymbol{wqzsnmya} \\) at a point corresponding to a non-singular \\( hjgrksla \\) has direction vector\n\\[\n(vohwulzi- vohwulzi \\cos 2 hjgrksla, vohwulzi \\sin 2 hjgrksla)\n\\]\nso the tangent to \\( wqzsnmya \\) at \\( kpyvwols \\) has the representation\n\\[\nqzxwvtnp= vohwulzi hjgrksla-\\frac{vohwulzi}{2} \\sin 2 hjgrksla+znmpqtav(vohwulzi- vohwulzi \\cos 2 hjgrksla)\n\\]\n\\[\ny=\\frac{vohwulzi}{2}-\\frac{vohwulzi}{2} \\cos 2 hjgrksla+znmpqtav(vohwulzi \\sin 2 hjgrksla)\n\\]\nin terms of the parameter \\( znmpqtav \\). Putting\n\\[\nznmpqtav=\\frac{\\cos hjgrksla-1}{2 \\sin hjgrksla} \\quad \\text { and } \\quad znmpqtav=\\frac{\\cos hjgrksla+1}{2 \\sin hjgrksla}\n\\]\nin (1), we see that \\( mgbxraez \\) and \\( mgbxraez^{\\prime} \\) are both on the tangent line. This proves that \\( mgbxraez mgbxraez^{\\prime} \\) is tangent to \\( wqzsnmya \\) at \\( kpyvwols \\) for \\( hjgrksla \\) not a multiple of \\( \\pi \\).\n\nWhen \\( hjgrksla \\) is a multiple of \\( \\pi, kpyvwols \\) is on the \\( qzxwvtnp \\)-axis, the curve \\( wqzsnmya \\) has a cusp at \\( kpyvwols \\) with a vertical tangent, and \\( mgbxraez mgbxraez^{\\prime} \\) stands vertical with either \\( mgbxraez \\) or \\( mgbxraez^{\\prime} \\) on the \\( qzxwvtnp \\)-axis. Hence \\( mgbxraez mgbxraez^{\\prime} \\) is tangent to \\( wqzsnmya \\) in this case too.\n\nSecond Solution. A purely synthetic argument can be given as follows: As before, we see that \\( \\angle kpyvwols ralydfze lcrevbdh=2 \\angle mgbxraez uehrjxmc lcrevbdh \\). Since \\( \\angle kpyvwols uehrjxmc lcrevbdh \\) is measured by half of arc \\( kpyvwols lcrevbdh \\), we have \\( \\frac{1}{2} \\angle kpyvwols ralydfze lcrevbdh=\\angle kpyvwols uehrjxmc lcrevbdh \\). It follows that \\( \\angle kpyvwols uehrjxmc lcrevbdh= \\) \\( \\angle mgbxraez uehrjxmc lcrevbdh \\), so \\( kpyvwols \\) is on \\( mgbxraez mgbxraez^{\\prime} \\).\nAs the small circle rolls, the point \\( lcrevbdh \\) is instantaneously at rest; hence all other points of the moving circle are instantaneously in rotation about \\( lcrevbdh \\). This implies that the tangent at \\( kpyvwols \\) to the orbit of \\( kpyvwols \\) is perpendicular to \\( kpyvwols lcrevbdh \\). But \\( kpyvwols uehrjxmc=mgbxraez mgbxraez^{\\prime} \\) is this perpendicular since \\( \\angle lcrevbdh kpyvwols uehrjxmc \\) is inscribed in a semicircle.\n\nAs is often the case with such synthetic arguments, a slightly different analysis is required after \\( kpyvwols \\) attains its highest point."
    },
    "kernel_variant": {
      "question": "(The statement below is identical to the one already supplied by the author, because only Section 5 of the solution required correction.  The problem itself is mathematically sound and already possesses the desired level of difficulty.)\n\nA homogeneous solid sphere $\\Sigma$ of radius $R$ rolls without slipping on the fixed horizontal plane  \n\n\\[\n\\Pi:\\; z=0 .\n\\]\n\nTake a right–handed Cartesian frame whose $(x,y)$–plane is $\\Pi$ and denote by  \n\n\\[\nC(t)=\\bigl(Rt,\\,0,\\,R\\bigr), \\qquad t\\in\\mathbb R\n\\]\n\nthe centre of the sphere at time $t$ (so the point of contact is $(Rt,0,0)$).  \nBecause there is no slip, the body has rotated through the angle $t$ about the {\\em negative} $y$–axis, hence the instantaneous angular–velocity vector is  \n\n\\[\n\\omega(t)=\\bigl(0,-1,0\\bigr).\n\\]\n\nFix an angle $\\beta$ with $0\\le \\beta\\le \\pi/2$ and consider the diameter $d_{0}=A_{0}B_{0}$ determined by the unit vector  \n\n\\[\nv_{0}=\\bigl(0,\\sin\\beta,\\cos\\beta\\bigr).\n\\]\n\nThus, at $t=0$  \n\n\\begin{itemize}\n\\item $d_{0}$ lies in the vertical plane $x=0$,\n\\item $d_{0}$ makes the angle $\\beta$ with the upward vertical,\n\\item the orthogonal projection of $d_{0}$ on $\\Pi$ is parallel to the $y$–axis.\n\\end{itemize}\n\nLet $A(t),B(t)$ be the images of $A_{0},B_{0}$ under the rigid motion of $\\Sigma$ at time $t$ and write  \n\n\\[\n\\ell(t)=A(t)B(t)\n\\]\n\nfor the (directed) moving line that joins them.  Because $v(t)=R_{y}(-t)\\,v_{0}$,  \n\n\\[\nv(t)=\\bigl(-\\cos\\beta\\sin t,\\;\\sin\\beta,\\;\\cos\\beta\\cos t\\bigr), \\qquad \\|v(t)\\|=1 ,\n\\]\n\nand every ruling is  \n\n\\[\n\\ell(t):\\;X(t,s)=C(t)+s\\,v(t), \\quad s\\in\\mathbb R . \\tag{1}\n\\]\n\n\\medskip\n\\noindent{\\bf 1.\\ Projection on $\\Pi$.}  \n(a)  Show that the projection $\\mathrm{proj}_{\\Pi}\\,\\ell(t)$ is the straight line  \n\n\\[\n\\bigl(x-Rt\\bigr)\\sin\\beta+y\\,\\cos\\beta\\sin t=0. \\tag{2}\n\\]\n\n(b)  For $0\\le\\beta<\\pi/2$ find the envelope of the one–parameter family {\\rm(2)} and prove that it is the plane curve  \n\n\\[\nE_{\\beta}:\\quad x(t)=R\\,[\\,t-\\tan t\\,],\\qquad y(t)=\\dfrac{R\\;\\tan\\beta}{\\cos t}, \\tag{3}\n\\]\n\ndefined for $t\\neq \\pi/2+k\\pi$.  \n\\begin{enumerate}\n\\item[(i)]  Show that when $\\beta=\\pi/2$ the family {\\rm(2)} possesses no envelope.  \n\\item[(ii)]  Establish that $E_{\\beta}$ has no finite vertical asymptote; instead, as $t\\to\\pi/2+k\\pi$ the branch of $E_{\\beta}$ runs off to infinity along the direction of slope $\\pm\\tan\\beta$.  \n\\item[(iii)]  Prove that $E_{\\beta}$ is symmetric with respect to the $y$–axis and that the translation formula  \n\n\\[\nx(t+\\pi)=x(t)+\\pi R,\\qquad y(t+\\pi)=-y(t)\n\\]\n\nholds; discuss the geometric consequences.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 2.\\ Spatial envelope of the rulings.}  \nLet  \n\n\\[\nD(t,s)=v(t)\\times\\frac{\\partial X}{\\partial t}(t,s)\n          =v(t)\\times\\bigl(C'(t)+s\\,v'(t)\\bigr)  \\tag{4}\n\\]\n\nbe the Darboux vector field of the ruling family.  \n\\begin{enumerate}\n\\item[(a)]  Compute  \n\n\\[\nC'(t)=(R,0,0),\\qquad \nv'(t)=\\bigl(-\\cos\\beta\\cos t,\\,0,\\,-\\cos\\beta\\sin t\\bigr),\n\\]\n\nand prove that  \n\n\\[\n\\boxed{\\,D(t,s)=\n\\bigl(-s\\sin\\beta\\cos\\beta\\sin t,\\;\n      R\\cos\\beta\\cos t-s\\cos^{2}\\beta,\\;\n      -R\\sin\\beta+s\\sin\\beta\\cos\\beta\\cos t\\bigr)\\,}. \\tag{5}\n\\]\n\nDeduce that, for $0<\\beta<\\pi/2$, $\\,D(t,s)=0$ iff $\\sin t=0$ and  \n\n\\[\ns=(-1)^{k}\\,\\dfrac{R}{\\cos\\beta}\\qquad (t=k\\pi). \\tag{6}\n\\]\n\nHence every point of space lies on at most one ruling, {\\em except} for the two discrete point–orbits  \n\n\\[\nP_{k}=C(k\\pi)+(-1)^{k}\\,\\dfrac{R}{\\cos\\beta}\\;v(k\\pi)\\qquad(k\\in\\mathbb Z), \\tag{7}\n\\]\n\nwhich are simultaneously carried by the perpendicular rulings $\\ell(k\\pi)$ and $\\ell(k\\pi\\pm0)$.  Conclude that, for $0<\\beta<\\pi/2$, the family $\\{\\ell(t)\\}$ admits no smooth envelope surface.  \n\n\\item[(b)]  Show that  \n\n\\[\n\\det\\bigl(v,v',C'\\bigr)=\\bigl[v,v',C'\\bigr]=-R\\sin\\beta\\cos\\beta\\sin t, \\tag{8}\n\\]\n\nso that the ruled surface  \n\n\\[\n\\Sigma_{\\beta}=\\{\\,X(t,s);\\,t,s\\in\\mathbb R\\}\n\\]\n\nis developable iff $\\beta=0$ or $\\beta=\\pi/2$.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 3.\\ Two degenerate developable limits.}\n\\begin{enumerate}\n\\item[(i)]  $\\beta=0$ (vertical diameter).  \nShow that every ruling $\\ell(t)$ lies in the plane $y=0$ and that the edge of regression is  \n\n\\[\n\\Gamma_{0}(t)=C(t)+R\\cos t\\,v(t)\n            =\\bigl(Rt-\\tfrac{1}{2}R\\sin2t,\\;0,\\;R(1+\\cos^{2}t)\\bigr). \\tag{9}\n\\]\n\nCompute its curvature $\\kappa(t)=1\\big/ \\bigl(2R\\lvert\\sin t\\rvert\\bigr)$ and torsion $\\tau(t)=0$, explaining the planar nature of the developable.  \n\n\\item[(ii)]  $\\beta=\\pi/2$ (horizontal diameter).  \nProve that each $\\ell(t)$ is the vertical line parallel to the $y$–axis through $(Rt,0,R)$; their union is the plane $z=R$, again a developable surface with no genuine edge of regression.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 4.\\ Gaussian curvature of the generic surface.}  \nFor $0<\\beta<\\pi/2$ verify that $v$ and $v'$ are linearly independent for all $t$ and that  \n\n\\[\nK(t,s)= -\\,\\frac{\\bigl[v,v',C'\\bigr]^{2}}\n              {\\lVert\\,v\\times\\bigl(C'(t)+s\\,v'(t)\\bigr)\\rVert^{4}}\n      = -\\,\\frac{R^{2}\\sin^{2}\\beta\\,\\cos^{2}\\beta\\,\\sin^{2}t}\n              {\\lVert\\,D(t,s)\\rVert^{4}}\\;\\le\\;0. \\tag{10}\n\\]\n\nShow that $K(t,s)<0$ whenever $\\sin t\\neq0$, and that $\\Sigma_{\\beta}$ possesses no conical point ($v\\times v'$ never vanishes).  \n\n\\medskip\n\\noindent{\\bf 5.\\ Extremal distance problem (bonus).}  \nFor $0<\\beta<\\pi/2$ let  \n\n\\[\nd(t)=\\mathrm{dist}\\bigl(O,\\ell(t)\\bigr)=\\lVert\\,C(t)\\times v(t)\\rVert .\n\\]\n\n\\begin{enumerate}\n\\item[(a)]  Show that  \n\n\\[\nd^{2}(t)=R^{2}t^{2}+R^{2}-R^{2}\\cos^{2}\\beta\\,\n         \\bigl(\\cos t-t\\sin t\\bigr)^{2}. \\tag{11}\n\\]\n\n\\item[(b)]  Prove that $t=0$ is the {\\em unique global minimiser} and that  \n\n\\[\nd_{\\min}=R\\sin\\beta ,\n\\]\n\nattained on the initial ruling $\\ell(0)$.  \n\\end{enumerate}",
      "solution": "The original solution is completely correct except for the sign analysis in Section 5.  \nOnly that part is replaced below; every other paragraph of the previous\nsolution remains valid.\n\n\\bigskip\n%----------------------------------------------------\n\\textbf{5.\\ Minimal distance to the origin (corrected)}\n\n\\smallskip\n\\textbf{(a)} was already correct and gives\n\n\\[\nd^{2}(t)=R^{2}\\Bigl[t^{2}+1-\\cos^{2}\\beta\\,\n                    \\bigl(\\cos t-t\\sin t\\bigr)^{2}\\Bigr].\n\\]\n\nDefine  \n\n\\[\nf(t)=\\frac{d^{2}(t)}{R^{2}},\\qquad \nc=\\cos\\beta\\in(0,1), \\qquad s(t)=\\cos t-t\\sin t .\n\\]\n\nThen  \n\n\\[\nf(t)=t^{2}+1-c^{2}s(t)^{2}.  \\tag{5.1}\n\\]\n\nBecause $f$ is even, it is enough to study $t\\ge0$.\n\n\\bigskip\n\\textbf{A sharp uniform bound on $s(t)$}\n\nWrite  \n\n\\[\n\\bigl(\\cos t,-\\sin t\\bigr)\\cdot\\bigl(1,t\\bigr)=\\cos t-t\\sin t=s(t).\n\\]\n\nThe vector $(\\cos t,-\\sin t)$ has norm $1$, whereas $(1,t)$ has norm\n$\\sqrt{1+t^{2}}$.  By Cauchy–Schwarz,\n\n\\[\n\\lvert s(t)\\rvert\\le\\sqrt{1+t^{2}}\\qquad\\forall\\,t\\in\\mathbb R. \\tag{5.2}\n\\]\n\n\\bigskip\n\\textbf{(b)  Global comparison with $f(0)$}\n\nAt $t=0$ one has $s(0)=1$ and therefore  \n\n\\[\nf(0)=1-c^{2}= \\sin^{2}\\beta .\n\\]\n\nFor general $t$ combine (5.1) and (5.2):\n\n\\[\n\\begin{aligned}\nf(t)-f(0)\n&=t^{2}-c^{2}\\bigl[s(t)^{2}-1\\bigr] \\\\[2pt]\n&\\ge t^{2}-c^{2}\\bigl[(1+t^{2})-1\\bigr]  \\\\[2pt]\n&=(1-c^{2})\\,t^{2}= \\sin^{2}\\beta\\;t^{2}\\;\\ge\\;0 .\n\\end{aligned} \\tag{5.3}\n\\]\n\nIf $t\\neq0$ the right–hand side of (5.3) is strictly positive\n(because $0<\\beta<\\pi/2$ implies $\\sin\\beta\\neq0$), hence  \n\n\\[\nf(t)>f(0)\\qquad(t\\neq0).\n\\]\n\nThus $f$ attains its global minimum uniquely at $t=0$, and\n\n\\[\nd_{\\min}=R\\sqrt{f(0)}=R\\sin\\beta .\n\\]\n\nSince $t=0$ corresponds to the initial ruling $\\ell(0)$, the bonus\nquestion is completely settled.\n\n\\hfill$\\blacksquare$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.380916",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension – the original problem is planar; the variant takes place in\n   ℝ³ and mixes planar and spatial geometry.\n\n•  Additional structures – the solution needs rigid‐body kinematics (rotation\n   matrices), classical roulettes (cycloid), and differential geometry (curvature,\n   torsion, developables, edge of regression, Gaussian curvature).\n\n•  Multiple interacting concepts – rolling without slipping, envelopes of lines,\n   tangent developables, helical space curves, conical degeneration.\n\n•  Deeper theory – parts 3–5 demand full use of Frenet–Serret apparatus and\n   developable‐surface theory that are far beyond the calculus and elementary\n   geometry sufficient for the original cycloid task.\n\n•  Length and subtlety – each item requires nontrivial computations (solving\n   simultaneous envelope equations, lengthy derivatives, determinant tests for\n   developability, parameter elimination); every step is a significant escalation\n   from the original two‐dimensional, single‐concept exercise."
      }
    },
    "original_kernel_variant": {
      "question": "(The statement below is identical to the one already supplied by the author, because only Section 5 of the solution required correction.  The problem itself is mathematically sound and already possesses the desired level of difficulty.)\n\nA homogeneous solid sphere $\\Sigma$ of radius $R$ rolls without slipping on the fixed horizontal plane  \n\n\\[\n\\Pi:\\; z=0 .\n\\]\n\nTake a right–handed Cartesian frame whose $(x,y)$–plane is $\\Pi$ and denote by  \n\n\\[\nC(t)=\\bigl(Rt,\\,0,\\,R\\bigr), \\qquad t\\in\\mathbb R\n\\]\n\nthe centre of the sphere at time $t$ (so the point of contact is $(Rt,0,0)$).  \nBecause there is no slip, the body has rotated through the angle $t$ about the {\\em negative} $y$–axis, hence the instantaneous angular–velocity vector is  \n\n\\[\n\\omega(t)=\\bigl(0,-1,0\\bigr).\n\\]\n\nFix an angle $\\beta$ with $0\\le \\beta\\le \\pi/2$ and consider the diameter $d_{0}=A_{0}B_{0}$ determined by the unit vector  \n\n\\[\nv_{0}=\\bigl(0,\\sin\\beta,\\cos\\beta\\bigr).\n\\]\n\nThus, at $t=0$  \n\n\\begin{itemize}\n\\item $d_{0}$ lies in the vertical plane $x=0$,\n\\item $d_{0}$ makes the angle $\\beta$ with the upward vertical,\n\\item the orthogonal projection of $d_{0}$ on $\\Pi$ is parallel to the $y$–axis.\n\\end{itemize}\n\nLet $A(t),B(t)$ be the images of $A_{0},B_{0}$ under the rigid motion of $\\Sigma$ at time $t$ and write  \n\n\\[\n\\ell(t)=A(t)B(t)\n\\]\n\nfor the (directed) moving line that joins them.  Because $v(t)=R_{y}(-t)\\,v_{0}$,  \n\n\\[\nv(t)=\\bigl(-\\cos\\beta\\sin t,\\;\\sin\\beta,\\;\\cos\\beta\\cos t\\bigr), \\qquad \\|v(t)\\|=1 ,\n\\]\n\nand every ruling is  \n\n\\[\n\\ell(t):\\;X(t,s)=C(t)+s\\,v(t), \\quad s\\in\\mathbb R . \\tag{1}\n\\]\n\n\\medskip\n\\noindent{\\bf 1.\\ Projection on $\\Pi$.}  \n(a)  Show that the projection $\\mathrm{proj}_{\\Pi}\\,\\ell(t)$ is the straight line  \n\n\\[\n\\bigl(x-Rt\\bigr)\\sin\\beta+y\\,\\cos\\beta\\sin t=0. \\tag{2}\n\\]\n\n(b)  For $0\\le\\beta<\\pi/2$ find the envelope of the one–parameter family {\\rm(2)} and prove that it is the plane curve  \n\n\\[\nE_{\\beta}:\\quad x(t)=R\\,[\\,t-\\tan t\\,],\\qquad y(t)=\\dfrac{R\\;\\tan\\beta}{\\cos t}, \\tag{3}\n\\]\n\ndefined for $t\\neq \\pi/2+k\\pi$.  \n\\begin{enumerate}\n\\item[(i)]  Show that when $\\beta=\\pi/2$ the family {\\rm(2)} possesses no envelope.  \n\\item[(ii)]  Establish that $E_{\\beta}$ has no finite vertical asymptote; instead, as $t\\to\\pi/2+k\\pi$ the branch of $E_{\\beta}$ runs off to infinity along the direction of slope $\\pm\\tan\\beta$.  \n\\item[(iii)]  Prove that $E_{\\beta}$ is symmetric with respect to the $y$–axis and that the translation formula  \n\n\\[\nx(t+\\pi)=x(t)+\\pi R,\\qquad y(t+\\pi)=-y(t)\n\\]\n\nholds; discuss the geometric consequences.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 2.\\ Spatial envelope of the rulings.}  \nLet  \n\n\\[\nD(t,s)=v(t)\\times\\frac{\\partial X}{\\partial t}(t,s)\n          =v(t)\\times\\bigl(C'(t)+s\\,v'(t)\\bigr)  \\tag{4}\n\\]\n\nbe the Darboux vector field of the ruling family.  \n\\begin{enumerate}\n\\item[(a)]  Compute  \n\n\\[\nC'(t)=(R,0,0),\\qquad \nv'(t)=\\bigl(-\\cos\\beta\\cos t,\\,0,\\,-\\cos\\beta\\sin t\\bigr),\n\\]\n\nand prove that  \n\n\\[\n\\boxed{\\,D(t,s)=\n\\bigl(-s\\sin\\beta\\cos\\beta\\sin t,\\;\n      R\\cos\\beta\\cos t-s\\cos^{2}\\beta,\\;\n      -R\\sin\\beta+s\\sin\\beta\\cos\\beta\\cos t\\bigr)\\,}. \\tag{5}\n\\]\n\nDeduce that, for $0<\\beta<\\pi/2$, $\\,D(t,s)=0$ iff $\\sin t=0$ and  \n\n\\[\ns=(-1)^{k}\\,\\dfrac{R}{\\cos\\beta}\\qquad (t=k\\pi). \\tag{6}\n\\]\n\nHence every point of space lies on at most one ruling, {\\em except} for the two discrete point–orbits  \n\n\\[\nP_{k}=C(k\\pi)+(-1)^{k}\\,\\dfrac{R}{\\cos\\beta}\\;v(k\\pi)\\qquad(k\\in\\mathbb Z), \\tag{7}\n\\]\n\nwhich are simultaneously carried by the perpendicular rulings $\\ell(k\\pi)$ and $\\ell(k\\pi\\pm0)$.  Conclude that, for $0<\\beta<\\pi/2$, the family $\\{\\ell(t)\\}$ admits no smooth envelope surface.  \n\n\\item[(b)]  Show that  \n\n\\[\n\\det\\bigl(v,v',C'\\bigr)=\\bigl[v,v',C'\\bigr]=-R\\sin\\beta\\cos\\beta\\sin t, \\tag{8}\n\\]\n\nso that the ruled surface  \n\n\\[\n\\Sigma_{\\beta}=\\{\\,X(t,s);\\,t,s\\in\\mathbb R\\}\n\\]\n\nis developable iff $\\beta=0$ or $\\beta=\\pi/2$.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 3.\\ Two degenerate developable limits.}\n\\begin{enumerate}\n\\item[(i)]  $\\beta=0$ (vertical diameter).  \nShow that every ruling $\\ell(t)$ lies in the plane $y=0$ and that the edge of regression is  \n\n\\[\n\\Gamma_{0}(t)=C(t)+R\\cos t\\,v(t)\n            =\\bigl(Rt-\\tfrac{1}{2}R\\sin2t,\\;0,\\;R(1+\\cos^{2}t)\\bigr). \\tag{9}\n\\]\n\nCompute its curvature $\\kappa(t)=1\\big/ \\bigl(2R\\lvert\\sin t\\rvert\\bigr)$ and torsion $\\tau(t)=0$, explaining the planar nature of the developable.  \n\n\\item[(ii)]  $\\beta=\\pi/2$ (horizontal diameter).  \nProve that each $\\ell(t)$ is the vertical line parallel to the $y$–axis through $(Rt,0,R)$; their union is the plane $z=R$, again a developable surface with no genuine edge of regression.  \n\\end{enumerate}\n\n\\medskip\n\\noindent{\\bf 4.\\ Gaussian curvature of the generic surface.}  \nFor $0<\\beta<\\pi/2$ verify that $v$ and $v'$ are linearly independent for all $t$ and that  \n\n\\[\nK(t,s)= -\\,\\frac{\\bigl[v,v',C'\\bigr]^{2}}\n              {\\lVert\\,v\\times\\bigl(C'(t)+s\\,v'(t)\\bigr)\\rVert^{4}}\n      = -\\,\\frac{R^{2}\\sin^{2}\\beta\\,\\cos^{2}\\beta\\,\\sin^{2}t}\n              {\\lVert\\,D(t,s)\\rVert^{4}}\\;\\le\\;0. \\tag{10}\n\\]\n\nShow that $K(t,s)<0$ whenever $\\sin t\\neq0$, and that $\\Sigma_{\\beta}$ possesses no conical point ($v\\times v'$ never vanishes).  \n\n\\medskip\n\\noindent{\\bf 5.\\ Extremal distance problem (bonus).}  \nFor $0<\\beta<\\pi/2$ let  \n\n\\[\nd(t)=\\mathrm{dist}\\bigl(O,\\ell(t)\\bigr)=\\lVert\\,C(t)\\times v(t)\\rVert .\n\\]\n\n\\begin{enumerate}\n\\item[(a)]  Show that  \n\n\\[\nd^{2}(t)=R^{2}t^{2}+R^{2}-R^{2}\\cos^{2}\\beta\\,\n         \\bigl(\\cos t-t\\sin t\\bigr)^{2}. \\tag{11}\n\\]\n\n\\item[(b)]  Prove that $t=0$ is the {\\em unique global minimiser} and that  \n\n\\[\nd_{\\min}=R\\sin\\beta ,\n\\]\n\nattained on the initial ruling $\\ell(0)$.  \n\\end{enumerate}",
      "solution": "The original solution is completely correct except for the sign analysis in Section 5.  \nOnly that part is replaced below; every other paragraph of the previous\nsolution remains valid.\n\n\\bigskip\n%----------------------------------------------------\n\\textbf{5.\\ Minimal distance to the origin (corrected)}\n\n\\smallskip\n\\textbf{(a)} was already correct and gives\n\n\\[\nd^{2}(t)=R^{2}\\Bigl[t^{2}+1-\\cos^{2}\\beta\\,\n                    \\bigl(\\cos t-t\\sin t\\bigr)^{2}\\Bigr].\n\\]\n\nDefine  \n\n\\[\nf(t)=\\frac{d^{2}(t)}{R^{2}},\\qquad \nc=\\cos\\beta\\in(0,1), \\qquad s(t)=\\cos t-t\\sin t .\n\\]\n\nThen  \n\n\\[\nf(t)=t^{2}+1-c^{2}s(t)^{2}.  \\tag{5.1}\n\\]\n\nBecause $f$ is even, it is enough to study $t\\ge0$.\n\n\\bigskip\n\\textbf{A sharp uniform bound on $s(t)$}\n\nWrite  \n\n\\[\n\\bigl(\\cos t,-\\sin t\\bigr)\\cdot\\bigl(1,t\\bigr)=\\cos t-t\\sin t=s(t).\n\\]\n\nThe vector $(\\cos t,-\\sin t)$ has norm $1$, whereas $(1,t)$ has norm\n$\\sqrt{1+t^{2}}$.  By Cauchy–Schwarz,\n\n\\[\n\\lvert s(t)\\rvert\\le\\sqrt{1+t^{2}}\\qquad\\forall\\,t\\in\\mathbb R. \\tag{5.2}\n\\]\n\n\\bigskip\n\\textbf{(b)  Global comparison with $f(0)$}\n\nAt $t=0$ one has $s(0)=1$ and therefore  \n\n\\[\nf(0)=1-c^{2}= \\sin^{2}\\beta .\n\\]\n\nFor general $t$ combine (5.1) and (5.2):\n\n\\[\n\\begin{aligned}\nf(t)-f(0)\n&=t^{2}-c^{2}\\bigl[s(t)^{2}-1\\bigr] \\\\[2pt]\n&\\ge t^{2}-c^{2}\\bigl[(1+t^{2})-1\\bigr]  \\\\[2pt]\n&=(1-c^{2})\\,t^{2}= \\sin^{2}\\beta\\;t^{2}\\;\\ge\\;0 .\n\\end{aligned} \\tag{5.3}\n\\]\n\nIf $t\\neq0$ the right–hand side of (5.3) is strictly positive\n(because $0<\\beta<\\pi/2$ implies $\\sin\\beta\\neq0$), hence  \n\n\\[\nf(t)>f(0)\\qquad(t\\neq0).\n\\]\n\nThus $f$ attains its global minimum uniquely at $t=0$, and\n\n\\[\nd_{\\min}=R\\sqrt{f(0)}=R\\sin\\beta .\n\\]\n\nSince $t=0$ corresponds to the initial ruling $\\ell(0)$, the bonus\nquestion is completely settled.\n\n\\hfill$\\blacksquare$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.328913",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension – the original problem is planar; the variant takes place in\n   ℝ³ and mixes planar and spatial geometry.\n\n•  Additional structures – the solution needs rigid‐body kinematics (rotation\n   matrices), classical roulettes (cycloid), and differential geometry (curvature,\n   torsion, developables, edge of regression, Gaussian curvature).\n\n•  Multiple interacting concepts – rolling without slipping, envelopes of lines,\n   tangent developables, helical space curves, conical degeneration.\n\n•  Deeper theory – parts 3–5 demand full use of Frenet–Serret apparatus and\n   developable‐surface theory that are far beyond the calculus and elementary\n   geometry sufficient for the original cycloid task.\n\n•  Length and subtlety – each item requires nontrivial computations (solving\n   simultaneous envelope equations, lengthy derivatives, determinant tests for\n   developability, parameter elimination); every step is a significant escalation\n   from the original two‐dimensional, single‐concept exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}