{
  "index": "1955-B-1",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "1. A sphere rolls along two intersecting straight lines. Find the locus of its center.",
  "solution": "First Solution. Let \\( l_{1} \\) and \\( l_{2} \\) be the given intersecting lines and let \\( \\pi \\) be their plane. Let lines \\( m_{1} \\) and \\( m_{2} \\) be the bisectors of the angles formed by \\( l_{1} \\) and \\( l_{2} \\) and let \\( \\sigma_{1} \\) and \\( \\sigma_{2} \\) be the planes perpendicular to \\( \\pi \\) containing \\( m_{1} \\) and \\( m_{2} \\), respectively. Then \\( \\sigma_{1} \\cup \\sigma_{2} \\) is the locus of all points equidistant from \\( l_{1} \\) and \\( l_{2} \\).\n\nLet \\( \\mathfrak{C} \\) be the right circular cylinder of radius \\( r \\) and axis \\( l_{1} \\). Then \\( \\mathfrak{C} \\) is the locus of all points having distance \\( r \\) from \\( l_{1} \\).\n\nA sphere of radius \\( r \\) is tangent to \\( l_{1} \\) and \\( l_{2} \\) if and only if its center is at distance \\( r \\) from both \\( l_{1} \\) and \\( l_{2} \\). Hence the desired locus is \\( \\mathbb{C} \\cap\\left(\\sigma_{1} \\cup \\sigma_{2}\\right) \\) \\( =\\left(\\mathbb{C} \\cap \\sigma_{1}\\right) \\cup\\left(\\mathbb{C} \\cap \\sigma_{2}\\right) \\). Now the intersection of a right circular cylinder with a plane neither parallel nor perpendicular to its axis is an ellipse, so the locus is the union of two ellipses. These two ellipses have a common minor axis of length \\( 2 r \\) lying on the line \\( \\sigma_{1} \\cap \\sigma_{2} \\) and major axes lying on the lines \\( \\boldsymbol{m}_{1} \\) and \\( \\boldsymbol{m}_{2} \\).\n\nSecond Solution. We choose axes so that the given lines lie in the \\( x, y \\) plane and the \\( x \\)-axis bisects the angle between them. Then the normal forms of the equations of the lines are\n\\[\nx \\sin \\theta-y \\cos \\theta=0\n\\]\nand\n\\[\nx \\sin \\theta+y \\cos \\theta=0\n\\]\nwhere \\( 0<\\theta<\\pi^{\\prime} 2 \\).\n\nThe squared distance from the point \\( \\langle x, y, z\\rangle \\) to these lines is \\( z^{2}+ \\) \\( (x \\sin \\theta-y \\cos \\theta)^{2} \\) and \\( z^{2}+(x \\sin \\theta+y \\cos \\theta)^{2} \\). The center of the sphere must be at distance \\( r \\) from both lines; hence the desired locus is given by the two equations\n\\[\n\\begin{array}{l}\nz^{2}+(x \\sin \\theta-y \\cos \\theta)^{2}=r^{2} \\\\\nz^{2}+(x \\sin \\theta+y \\cos \\theta)^{2}=r^{2}\n\\end{array}\n\\]\n\nSubtracting these equations, we obtain \\( 4 x y \\sin \\theta \\cos \\theta=0 \\). Therefore the locus lies on the union of the planes \\( x=0 \\) and \\( y=0 \\), i.e., the \\( y, z \\)-plane and the \\( x, z \\)-plane. The part on the \\( y, z \\)-plane has the equation \\( z^{2}+y^{2} \\cos ^{2} \\theta \\) \\( =r^{2} \\), so it is an ellipse having minor axis on the \\( z \\)-axis of length \\( 2 r \\) and having major axis on the \\( y \\)-axis of length \\( 2 r \\sec \\theta \\). The part on the \\( x, z \\)-plane has the equation \\( z^{2}+x^{2} \\sin ^{2} \\theta=r^{2} \\). It is therefore an ellipse having minor axis of length \\( 2 r \\) lying on the \\( z \\)-axis and major axis of length \\( 2 r \\operatorname{cosec} \\theta \\) lying on the \\( x \\)-axis. The locus is the union of these two ellipses.",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "r",
    "\\\\theta",
    "l_1",
    "l_2",
    "m_1",
    "m_2",
    "\\\\sigma_1",
    "\\\\sigma_2"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "z": "elevation",
        "r": "spherrad",
        "\\theta": "angleval",
        "l_1": "lineone",
        "l_2": "linetwo",
        "m_1": "bisectorone",
        "m_2": "bisectortwo",
        "\\sigma_1": "perpplaneone",
        "\\sigma_2": "perpplanetwo"
      },
      "question": "1. A sphere rolls along two intersecting straight lines. Find the locus of its center.",
      "solution": "First Solution. Let \\( lineone \\) and \\( linetwo \\) be the given intersecting lines and let \\( \\pi \\) be their plane. Let lines \\( bisectorone \\) and \\( bisectortwo \\) be the bisectors of the angles formed by \\( lineone \\) and \\( linetwo \\) and let \\( perpplaneone \\) and \\( perpplanetwo \\) be the planes perpendicular to \\( \\pi \\) containing \\( bisectorone \\) and \\( bisectortwo \\), respectively. Then \\( perpplaneone \\cup perpplanetwo \\) is the locus of all points equidistant from \\( lineone \\) and \\( linetwo \\).\n\nLet \\( \\mathfrak{C} \\) be the right circular cylinder of radius \\( spherrad \\) and axis \\( lineone \\). Then \\( \\mathfrak{C} \\) is the locus of all points having distance \\( spherrad \\) from \\( lineone \\).\n\nA sphere of radius \\( spherrad \\) is tangent to \\( lineone \\) and \\( linetwo \\) if and only if its center is at distance \\( spherrad \\) from both \\( lineone \\) and \\( linetwo \\). Hence the desired locus is \\( \\mathbb{C} \\cap\\left(perpplaneone \\cup perpplanetwo\\right) =\\left(\\mathbb{C} \\cap perpplaneone\\right) \\cup\\left(\\mathbb{C} \\cap perpplanetwo\\right) \\). Now the intersection of a right circular cylinder with a plane neither parallel nor perpendicular to its axis is an ellipse, so the locus is the union of two ellipses. These two ellipses have a common minor axis of length \\( 2 spherrad \\) lying on the line \\( perpplaneone \\cap perpplanetwo \\) and major axes lying on the lines \\( \\boldsymbol{bisectorone} \\) and \\( \\boldsymbol{bisectortwo} \\).\n\nSecond Solution. We choose axes so that the given lines lie in the \\( abscissa, ordinate \\) plane and the \\( abscissa \\)-axis bisects the angle between them. Then the normal forms of the equations of the lines are\n\\[\nabscissa \\sin angleval-ordinate \\cos angleval=0\n\\]\nand\n\\[\nabscissa \\sin angleval+ordinate \\cos angleval=0\n\\]\nwhere \\( 0<angleval<\\pi^{\\prime} 2 \\).\n\nThe squared distance from the point \\( \\langle abscissa, ordinate, elevation\\rangle \\) to these lines is \\( elevation^{2}+ (abscissa \\sin angleval-ordinate \\cos angleval)^{2} \\) and \\( elevation^{2}+(abscissa \\sin angleval+ordinate \\cos angleval)^{2} \\). The center of the sphere must be at distance \\( spherrad \\) from both lines; hence the desired locus is given by the two equations\n\\[\n\\begin{array}{l}\nelevation^{2}+(abscissa \\sin angleval-ordinate \\cos angleval)^{2}=spherrad^{2} \\\\\nelevation^{2}+(abscissa \\sin angleval+ordinate \\cos angleval)^{2}=spherrad^{2}\n\\end{array}\n\\]\n\nSubtracting these equations, we obtain \\( 4 abscissa\\, ordinate \\sin angleval \\cos angleval=0 \\). Therefore the locus lies on the union of the planes \\( abscissa=0 \\) and \\( ordinate=0 \\), i.e., the \\( ordinate, elevation \\)-plane and the \\( abscissa, elevation \\)-plane. The part on the \\( ordinate, elevation \\)-plane has the equation \\( elevation^{2}+ordinate^{2} \\cos ^{2} angleval =spherrad^{2} \\), so it is an ellipse having minor axis on the \\( elevation \\)-axis of length \\( 2 spherrad \\) and having major axis on the \\( ordinate \\)-axis of length \\( 2 spherrad \\sec angleval \\). The part on the \\( abscissa, elevation \\)-plane has the equation \\( elevation^{2}+abscissa^{2} \\sin ^{2} angleval=spherrad^{2} \\). It is therefore an ellipse having minor axis of length \\( 2 spherrad \\) lying on the \\( elevation \\)-axis and major axis of length \\( 2 spherrad \\operatorname{cosec} angleval \\) lying on the \\( abscissa \\)-axis. The locus is the union of these two ellipses."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "labyrinth",
        "y": "chandelier",
        "z": "harmonica",
        "r": "snowflake",
        "\\theta": "windchime",
        "l_1": "sunflower",
        "l_2": "toothbrush",
        "m_1": "skateboard",
        "m_2": "microphone",
        "\\sigma_1": "watermelon",
        "\\sigma_2": "dragonfly"
      },
      "question": "1. A sphere rolls along two intersecting straight lines. Find the locus of its center.",
      "solution": "First Solution. Let \\( sunflower \\) and \\( toothbrush \\) be the given intersecting lines and let \\( \\pi \\) be their plane. Let lines \\( skateboard \\) and \\( microphone \\) be the bisectors of the angles formed by \\( sunflower \\) and \\( toothbrush \\) and let \\( watermelon \\) and \\( dragonfly \\) be the planes perpendicular to \\( \\pi \\) containing \\( skateboard \\) and \\( microphone \\), respectively. Then \\( watermelon \\cup dragonfly \\) is the locus of all points equidistant from \\( sunflower \\) and \\( toothbrush \\).\n\nLet \\( \\mathfrak{C} \\) be the right circular cylinder of radius \\( snowflake \\) and axis \\( sunflower \\). Then \\( \\mathfrak{C} \\) is the locus of all points having distance \\( snowflake \\) from \\( sunflower \\).\n\nA sphere of radius \\( snowflake \\) is tangent to \\( sunflower \\) and \\( toothbrush \\) if and only if its center is at distance \\( snowflake \\) from both \\( sunflower \\) and \\( toothbrush \\). Hence the desired locus is \\( \\mathbb{C} \\cap\\left(watermelon \\cup dragonfly\\right) \\) \\( =\\left(\\mathbb{C} \\cap watermelon\\right) \\cup\\left(\\mathbb{C} \\cap dragonfly\\right) \\). Now the intersection of a right circular cylinder with a plane neither parallel nor perpendicular to its axis is an ellipse, so the locus is the union of two ellipses. These two ellipses have a common minor axis of length \\( 2 snowflake \\) lying on the line \\( watermelon \\cap dragonfly \\) and major axes lying on the lines \\( \\boldsymbol{skateboard} \\) and \\( \\boldsymbol{microphone} \\).\n\nSecond Solution. We choose axes so that the given lines lie in the \\( labyrinth, chandelier \\) plane and the \\( labyrinth \\)-axis bisects the angle between them. Then the normal forms of the equations of the lines are\n\\[\nlabyrinth \\sin windchime-chandelier \\cos windchime=0\n\\]\nand\n\\[\nlabyrinth \\sin windchime+chandelier \\cos windchime=0\n\\]\nwhere \\( 0<windchime<\\pi^{\\prime} 2 \\).\n\nThe squared distance from the point \\( \\langle labyrinth, chandelier, harmonica\\rangle \\) to these lines is \\( harmonica^{2}+ \\) \\( (labyrinth \\sin windchime-chandelier \\cos windchime)^{2} \\) and \\( harmonica^{2}+(labyrinth \\sin windchime+chandelier \\cos windchime)^{2} \\). The center of the sphere must be at distance \\( snowflake \\) from both lines; hence the desired locus is given by the two equations\n\\[\n\\begin{array}{l}\nharmonica^{2}+(labyrinth \\sin windchime-chandelier \\cos windchime)^{2}=snowflake^{2} \\\\\nharmonica^{2}+(labyrinth \\sin windchime+chandelier \\cos windchime)^{2}=snowflake^{2}\n\\end{array}\n\\]\n\nSubtracting these equations, we obtain \\( 4 labyrinth chandelier \\sin windchime \\cos windchime=0 \\). Therefore the locus lies on the union of the planes \\( labyrinth=0 \\) and \\( chandelier=0 \\), i.e., the \\( chandelier, harmonica \\)-plane and the \\( labyrinth, harmonica \\)-plane. The part on the \\( chandelier, harmonica \\)-plane has the equation \\( harmonica^{2}+chandelier^{2} \\cos ^{2} windchime \\) \\( =snowflake^{2} \\), so it is an ellipse having minor axis on the \\( harmonica \\)-axis of length \\( 2 snowflake \\) and having major axis on the \\( chandelier \\)-axis of length \\( 2 snowflake \\sec windchime \\). The part on the \\( labyrinth, harmonica \\)-plane has the equation \\( harmonica^{2}+labyrinth^{2} \\sin ^{2} windchime=snowflake^{2} \\). It is therefore an ellipse having minor axis of length \\( 2 snowflake \\) lying on the \\( harmonica \\)-axis and major axis of length \\( 2 snowflake \\operatorname{cosec} windchime \\) lying on the \\( labyrinth \\)-axis. The locus is the union of these two ellipses."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "z": "planaraxis",
        "r": "diametervalue",
        "\\theta": "straightmeasure",
        "l_1": "curvedpathone",
        "l_{1}": "curvedpathone",
        "l_2": "curvedpathtwo",
        "l_{2}": "curvedpathtwo",
        "m_1": "nonbisectorone",
        "m_{1}": "nonbisectorone",
        "m_2": "nonbisectortwo",
        "m_{2}": "nonbisectortwo",
        "\\sigma_1": "nonplanarone",
        "\\sigma_{1}": "nonplanarone",
        "\\sigma_2": "nonplanartwo",
        "\\sigma_{2}": "nonplanartwo"
      },
      "question": "1. A sphere rolls along two intersecting straight lines. Find the locus of its center.",
      "solution": "First Solution. Let \\( curvedpathone \\) and \\( curvedpathtwo \\) be the given intersecting lines and let \\( \\pi \\) be their plane. Let lines \\( nonbisectorone \\) and \\( nonbisectortwo \\) be the bisectors of the angles formed by \\( curvedpathone \\) and \\( curvedpathtwo \\) and let \\( nonplanarone \\) and \\( nonplanartwo \\) be the planes perpendicular to \\( \\pi \\) containing \\( nonbisectorone \\) and \\( nonbisectortwo \\), respectively. Then \\( nonplanarone \\cup nonplanartwo \\) is the locus of all points equidistant from \\( curvedpathone \\) and \\( curvedpathtwo \\).\n\nLet \\( \\mathfrak{C} \\) be the right circular cylinder of radius \\( diametervalue \\) and axis \\( curvedpathone \\). Then \\( \\mathfrak{C} \\) is the locus of all points having distance \\( diametervalue \\) from \\( curvedpathone \\).\n\nA sphere of radius \\( diametervalue \\) is tangent to \\( curvedpathone \\) and \\( curvedpathtwo \\) if and only if its center is at distance \\( diametervalue \\) from both \\( curvedpathone \\) and \\( curvedpathtwo \\). Hence the desired locus is \\( \\mathbb{C} \\cap\\left(nonplanarone \\cup nonplanartwo\\right) \\) \\( =\\left(\\mathbb{C} \\cap nonplanarone\\right) \\cup\\left(\\mathbb{C} \\cap nonplanartwo\\right) \\). Now the intersection of a right circular cylinder with a plane neither parallel nor perpendicular to its axis is an ellipse, so the locus is the union of two ellipses. These two ellipses have a common minor axis of length \\( 2 diametervalue \\) lying on the line \\( nonplanarone \\cap nonplanartwo \\) and major axes lying on the lines \\( \\boldsymbol{nonbisectorone} \\) and \\( \\boldsymbol{nonbisectortwo} \\).\n\nSecond Solution. We choose axes so that the given lines lie in the \\( verticalaxis, horizontalaxis \\) plane and the \\( verticalaxis \\)-axis bisects the angle between them. Then the normal forms of the equations of the lines are\n\\[\nverticalaxis \\sin straightmeasure-horizontalaxis \\cos straightmeasure=0\n\\]\nand\n\\[\nverticalaxis \\sin straightmeasure+horizontalaxis \\cos straightmeasure=0\n\\]\nwhere \\( 0<straightmeasure<\\pi^{\\prime} 2 \\).\n\nThe squared distance from the point \\( \\langle verticalaxis, horizontalaxis, planaraxis\\rangle \\) to these lines is \\( planaraxis^{2}+ (verticalaxis \\sin straightmeasure-horizontalaxis \\cos straightmeasure)^{2} \\) and \\( planaraxis^{2}+(verticalaxis \\sin straightmeasure+horizontalaxis \\cos straightmeasure)^{2} \\). The center of the sphere must be at distance \\( diametervalue \\) from both lines; hence the desired locus is given by the two equations\n\\[\n\\begin{array}{l}\nplanaraxis^{2}+(verticalaxis \\sin straightmeasure-horizontalaxis \\cos straightmeasure)^{2}=diametervalue^{2} \\\\\nplanaraxis^{2}+(verticalaxis \\sin straightmeasure+horizontalaxis \\cos straightmeasure)^{2}=diametervalue^{2}\n\\end{array}\n\\]\n\nSubtracting these equations, we obtain \\( 4 verticalaxis horizontalaxis \\sin straightmeasure \\cos straightmeasure=0 \\). Therefore the locus lies on the union of the planes \\( verticalaxis=0 \\) and \\( horizontalaxis=0 \\), i.e., the \\( horizontalaxis, planaraxis \\)-plane and the \\( verticalaxis, planaraxis \\)-plane. The part on the \\( horizontalaxis, planaraxis \\)-plane has the equation \\( planaraxis^{2}+horizontalaxis^{2} \\cos ^{2} straightmeasure =diametervalue^{2} \\), so it is an ellipse having minor axis on the \\( planaraxis \\)-axis of length \\( 2 diametervalue \\) and having major axis on the \\( horizontalaxis \\)-axis of length \\( 2 diametervalue \\sec straightmeasure \\). The part on the \\( verticalaxis, planaraxis \\)-plane has the equation \\( planaraxis^{2}+verticalaxis^{2} \\sin ^{2} straightmeasure=diametervalue^{2} \\). It is therefore an ellipse having minor axis of length \\( 2 diametervalue \\) lying on the \\( planaraxis \\)-axis and major axis of length \\( 2 diametervalue \\operatorname{cosec} straightmeasure \\) lying on the \\( verticalaxis \\)-axis. The locus is the union of these two ellipses."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "bqpmndce",
        "r": "vgtcflsz",
        "\\theta": "mzlarikd",
        "l_1": "pfjndtek",
        "l_2": "xmcqzoln",
        "m_1": "rhskpwye",
        "m_2": "vswbjgfa",
        "\\sigma_1": "nghptzra",
        "\\sigma_2": "zbnsvear"
      },
      "question": "1. A sphere rolls along two intersecting straight lines. Find the locus of its center.",
      "solution": "First Solution. Let \\( pfjndtek \\) and \\( xmcqzoln \\) be the given intersecting lines and let \\( \\pi \\) be their plane. Let lines \\( rhskpwye \\) and \\( vswbjgfa \\) be the bisectors of the angles formed by \\( pfjndtek \\) and \\( xmcqzoln \\) and let \\( nghptzra \\) and \\( zbnsvear \\) be the planes perpendicular to \\( \\pi \\) containing \\( rhskpwye \\) and \\( vswbjgfa \\), respectively. Then \\( nghptzra \\cup zbnsvear \\) is the locus of all points equidistant from \\( pfjndtek \\) and \\( xmcqzoln \\).\n\nLet \\( \\mathfrak{C} \\) be the right circular cylinder of radius \\( vgtcflsz \\) and axis \\( pfjndtek \\). Then \\( \\mathfrak{C} \\) is the locus of all points having distance \\( vgtcflsz \\) from \\( pfjndtek \\).\n\nA sphere of radius \\( vgtcflsz \\) is tangent to \\( pfjndtek \\) and \\( xmcqzoln \\) if and only if its center is at distance \\( vgtcflsz \\) from both \\( pfjndtek \\) and \\( xmcqzoln \\). Hence the desired locus is \\( \\mathbb{C} \\cap\\left(nghptzra \\cup zbnsvear\\right) \\) \\( =\\left(\\mathbb{C} \\cap nghptzra\\right) \\cup\\left(\\mathbb{C} \\cap zbnsvear\\right) \\). Now the intersection of a right circular cylinder with a plane neither parallel nor perpendicular to its axis is an ellipse, so the locus is the union of two ellipses. These two ellipses have a common minor axis of length \\( 2 vgtcflsz \\) lying on the line \\( nghptzra \\cap zbnsvear \\) and major axes lying on the lines \\( \\boldsymbol{rhskpwye} \\) and \\( \\boldsymbol{vswbjgfa} \\).\n\nSecond Solution. We choose axes so that the given lines lie in the \\( qzxwvtnp, hjgrksla \\) plane and the \\( qzxwvtnp \\)-axis bisects the angle between them. Then the normal forms of the equations of the lines are\n\\[\nqzxwvtnp \\sin mzlarikd-hjgrksla \\cos mzlarikd=0\n\\]\nand\n\\[\nqzxwvtnp \\sin mzlarikd+hjgrksla \\cos mzlarikd=0\n\\]\nwhere \\( 0<mzlarikd<\\pi^{\\prime} 2 \\).\n\nThe squared distance from the point \\( \\langle qzxwvtnp, hjgrksla, bqpmndce\\rangle \\) to these lines is \\( bqpmndce^{2}+ \\) \\( (qzxwvtnp \\sin mzlarikd-hjgrksla \\cos mzlarikd)^{2} \\) and \\( bqpmndce^{2}+(qzxwvtnp \\sin mzlarikd+hjgrksla \\cos mzlarikd)^{2} \\). The center of the sphere must be at distance \\( vgtcflsz \\) from both lines; hence the desired locus is given by the two equations\n\\[\n\\begin{array}{l}\nbqpmndce^{2}+(qzxwvtnp \\sin mzlarikd-hjgrksla \\cos mzlarikd)^{2}=vgtcflsz^{2} \\\\\nbqpmndce^{2}+(qzxwvtnp \\sin mzlarikd+hjgrksla \\cos mzlarikd)^{2}=vgtcflsz^{2}\n\\end{array}\n\\]\n\nSubtracting these equations, we obtain \\( 4 qzxwvtnp hjgrksla \\sin mzlarikd \\cos mzlarikd=0 \\). Therefore the locus lies on the union of the planes \\( qzxwvtnp=0 \\) and \\( hjgrksla=0 \\), i.e., the \\( hjgrksla, bqpmndce \\)-plane and the \\( qzxwvtnp, bqpmndce \\)-plane. The part on the \\( hjgrksla, bqpmndce \\)-plane has the equation \\( bqpmndce^{2}+hjgrksla^{2} \\cos ^{2} mzlarikd \\) \\( =vgtcflsz^{2} \\), so it is an ellipse having minor axis on the \\( bqpmndce \\)-axis of length \\( 2 vgtcflsz \\) and having major axis on the \\( hjgrksla \\)-axis of length \\( 2 vgtcflsz \\sec mzlarikd \\). The part on the \\( qzxwvtnp, bqpmndce \\)-plane has the equation \\( bqpmndce^{2}+qzxwvtnp^{2} \\sin ^{2} mzlarikd=vgtcflsz^{2} \\). It is therefore an ellipse having minor axis of length \\( 2 vgtcflsz \\) lying on the \\( bqpmndce \\)-axis and major axis of length \\( 2 vgtcflsz \\operatorname{cosec} mzlarikd \\) lying on the \\( qzxwvtnp \\)-axis. The locus is the union of these two ellipses."
    },
    "kernel_variant": {
      "question": "Let g and h be two straight lines that intersect at the origin and lie in the y z-plane.  Assume that the (acute) angle between them is 2φ (0 < φ < π∕2) and that the y-axis is the interior angle-bisector; hence unit direction vectors of the lines may be chosen as  \n  u_g = (0, cos φ,  sin φ), u_h = (0, cos φ, −sin φ).  \nA solid sphere of fixed radius a > 0 rolls so that it is always tangent to both lines.  Describe completely the locus of the centre of the sphere.",
      "solution": "First (synthetic) solution.\n1.  Every point that is equidistant from the two intersecting lines g and h lies in one of the two planes that (i) are perpendicular to the plane of the lines (here the y z-plane) and (ii) contain an angle-bisector of g and h.  Because the interior bisector is the y-axis and the exterior bisector is the z-axis, these planes are\n       Π₁ : y = 0   (the x z-plane),\n       Π₂ : z = 0   (the x y-plane).\n2.  Fix one of the lines, say h.  The locus of points whose distance from h is exactly a is the right circular cylinder 𝒞 of radius a and axis h.\n3.  A point P can serve as the centre of the rolling sphere iff its distances to g and to h are both a; equivalently,\n       P ∈ 𝒞  and  dist(P, g) = a.\n   Point 1 shows that the condition dist(P, g) = dist(P, h) forces P to lie in Π₁ ∪ Π₂, so the required locus is\n       𝒞 ∩ (Π₁ ∪ Π₂) = (𝒞 ∩ Π₁) ∪ (𝒞 ∩ Π₂).\n4.  Each plane Πᵢ (i = 1, 2) meets the cylinder obliquely (the plane is neither parallel nor perpendicular to the axis h), hence each intersection is an ellipse.\n5.  Therefore the desired locus is the union of the two ellipses\n       𝓔₁ = 𝒞 ∩ Π₁, 𝓔₂ = 𝒞 ∩ Π₂.\n   Both ellipses have their minor axis on the x-axis with the same semi-minor length a, so they share the two vertices (±a, 0, 0).  Apart from these two points the curves are disjoint.\n\nSecond (analytic) solution.\nIntroduce coordinates as in the statement.  For P = (x, y, z) write p = (x, y, z) and keep\n       u_g = (0, cos φ,  sin φ), u_h = (0, cos φ, −sin φ).\nFor a point p and a line with unit direction u, the squared distance is\n       dist²(p, ℓ) = |p|² − (p·u)².\nHence\n       d_g² = x² + y² + z² − (y cos φ + z sin φ)²,\n       d_h² = x² + y² + z² − (y cos φ − z sin φ)².\nBecause the sphere is tangent to both lines, we require d_g² = d_h² = a².  Subtracting the two expressions gives\n       0 = d_g² − d_h² = −4 y z cos φ sin φ.\nSince sin φ and cos φ are non–zero, we get y z = 0, so P lies in Π₁ ∪ Π₂.\n\n(i)  On Π₁ (y = 0) we have\n       d_h² = x² + z² − (−z sin φ)² = x² + z² cos² φ = a².\n   Dividing by a² gives\n       x²/a² + z²/(a² ∕ cos² φ) = 1.\n   Thus 𝓔₁ is an ellipse in the x z-plane whose semi-minor axis is a (along Ox) and whose semi-major axis is a sec φ (along Oz).\n\n(ii)  On Π₂ (z = 0) we obtain\n       d_h² = x² + y² − (y cos φ)² = x² + y² sin² φ = a²,\n   hence\n       x²/a² + y²/(a² ∕ sin² φ) = 1.\n   Consequently 𝓔₂ is an ellipse in the x y-plane with semi-minor axis a (again along Ox) and semi-major axis a csc φ (along Oy).\n\nThe locus of the centre of the rolling sphere is therefore\n       𝓔₁ ∪ 𝓔₂,\nwhich consists of two perpendicular ellipses lying in the coordinate planes y = 0 and z = 0, intersecting only at the two common vertices (±a, 0, 0).",
      "_meta": {
        "core_steps": [
          "Points equidistant from two intersecting lines lie on the two planes through the angle-bisector lines that are perpendicular to the lines’ plane.",
          "Points at fixed distance r from one chosen line form a right circular cylinder of radius r with that line as axis.",
          "The sphere’s center must satisfy both conditions, so the locus is (bisector-planes) ∩ (cylinder).",
          "Intersecting an oblique plane with a right circular cylinder produces an ellipse.",
          "Hence the overall locus is the union of the two resulting ellipses."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Numerical size of the sphere / cylinder radius",
            "original": "r"
          },
          "slot2": {
            "description": "Which of the two given lines is chosen as the cylinder’s axis",
            "original": "l1"
          },
          "slot3": {
            "description": "Size of the angle between the two lines (appears as θ in the analytic approach)",
            "original": "θ  (0 < θ < π/2)"
          },
          "slot4": {
            "description": "Orientation of the coordinate system (e.g., taking the x-axis along an angle bisector to make resulting planes x=0 and y=0)",
            "original": "Specific x-, y-, z-axes used in the second solution"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}