{
  "index": "1971-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "B-4. A \"spherical ellipse\" with foci \\( A, B \\) on a given sphere is defined as the set of all points \\( P \\) on the sphere such that \\( \\overparen{P A}+\\overparen{P B}= \\) constant. Here \\( \\overparen{P A} \\) denotes the shortest distance on the sphere between \\( P \\) and \\( A \\). Determine the entire class of real spherical ellipses which are circles.",
  "solution": "B-4 We take the radius of the sphere as unity and denote the constant sum \\( \\overparen{P A}+\\overparen{P B} \\) by \\( 2 a \\). To avoid trivial and degenerate cases we assume that \\( 0<\\overparen{A B}<\\pi \\) and that \\( \\overparen{A B}<2 a<2 \\pi-\\overparen{A B} \\).\n\nThe case \\( 2 a>\\pi \\) can be reduced to the case \\( 2 a<\\pi \\). For, if \\( A^{\\prime} \\) and \\( B^{\\prime} \\) are the points diametrically opposite to \\( A \\) and \\( B \\) then \\( \\overparen{P A}+\\overparen{P B}=2 a \\) if and only if \\( \\widehat{P A}^{\\prime}+\\widehat{P B}^{\\prime}=2 \\pi-2 a \\); that is, the spherical ellipses \\( \\overparen{P A}+\\overparen{P B}=2 a \\) and \\( \\widehat{P A}^{\\prime} \\) \\( +\\overparen{P B^{\\prime}}=2 \\pi-2 a \\) are identical. Since \\( \\min (2 a, 2 \\pi-2 a) \\leqq \\pi \\), we may assume without loss of generality that \\( 2 a \\leqq \\pi \\).\n\nLet \\( A \\) and \\( B \\) lie on the equator. There are two points \\( V_{1} \\) and \\( V_{2} \\) (the \"vertices\") on the equator which lie on the spherical ellipse. Obviously, \\( V_{1} V_{2}=2 a \\). The \"center\" of the spherical ellipse (common midpoint of the arcs \\( \\overparen{A B} \\) and \\( \\overparen{V_{1} V_{2}} \\) ) will be denoted by \\( C \\).\n\nFig. 1\n\nFig. 2\n\nWe first treat the case \\( 2 a<\\pi \\) and show that in this case the spherical ellipse\ncannot be a circle. Assume it were a circle; call it \\( \\Gamma \\) (see Figure 1). \\( \\Gamma \\) would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. \\( \\Gamma \\) would also have to pass through the vertices. Therefore its spherical diameter would be \\( \\widehat{V_{1} V_{2}}=2 a \\) and its spherical radius would be equal to \\( a \\). The spherical center of \\( \\Gamma \\) would be \\( C \\), the center of the ellipse. Let \\( M \\) be one of the two points on \\( \\Gamma \\) which lie half-way between the two vertices. Then, since \\( M \\) is supposed to be a point on the spherical ellipse, \\( 2 a=\\widehat{M A}+\\hat{M} \\widehat{B}>2 \\dot{M} \\grave{C}=2 a \\) (note that \\( M A C \\) is a right spherical triangle with the right angle at \\( C \\) and with side \\( \\left.\\mathscr{M} \\check{C}=a<\\frac{1}{2} \\pi\\right) \\). Contradiction shows that the only possible spherical ellipses which are circles must occur when \\( 2 a=\\pi \\).\n\nIn case \\( 2 a=\\pi, V_{1} \\) and \\( V_{2} \\) are diametrically opposite points on the equator. We shall show that the great circle \\( \\Gamma \\) through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse \\( \\overparen{P A}+\\overparen{P B}=\\pi \\). To see this, let \\( B^{*} \\) be the reflection of \\( B \\) about the plane of \\( \\Gamma . B^{*} \\) is on the equator diametrically opposite to \\( A \\) (see Fig. 2). Let \\( P \\) be an arbitrary point on the sphere, and draw the great circle through \\( A, P \\) and \\( B^{*} \\). Then \\( \\overparen{P A}+\\widehat{P B}^{*}=\\pi \\). Hence, \\( \\overparen{P A}+\\overparen{P B}=\\pi \\) if and only if \\( \\overparen{P B}=\\overparen{P B}^{*} \\), that is, if and only if \\( P \\) is on \\( \\Gamma \\). This shows that \\( \\Gamma \\) is the spherical ellipse \\( \\overparen{P A}+\\overparen{P B}=\\pi \\), as stated above.\n\nThus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle \\( \\Gamma \\) the foci can be any two points \\( A \\) and \\( B \\) which lie on the same great circle perpendicular to \\( \\Gamma \\), on the same side of \\( \\Gamma \\) and at equal distances from \\( \\Gamma \\). The equation of any such spherical ellipse is \\( \\overparen{P A}+\\overparen{P B}=\\pi \\).",
  "vars": [
    "P",
    "M"
  ],
  "params": [
    "A",
    "B",
    "C",
    "a",
    "V_1",
    "V_2",
    "\\\\Gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "pointpt",
        "M": "midpoint",
        "A": "focusone",
        "B": "focustwo",
        "C": "ellipscen",
        "a": "semisum",
        "V_1": "vertexone",
        "V_2": "vertextwo",
        "\\Gamma": "circlebig"
      },
      "question": "B-4. A \"spherical ellipse\" with foci \\( focusone, focustwo \\) on a given sphere is defined as the set of all points \\( pointpt \\) on the sphere such that \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}= \\) constant. Here \\( \\overparen{pointpt focusone} \\) denotes the shortest distance on the sphere between \\( pointpt \\) and \\( focusone \\). Determine the entire class of real spherical ellipses which are circles.",
      "solution": "B-4 We take the radius of the sphere as unity and denote the constant sum \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo} \\) by \\( 2 semisum \\). To avoid trivial and degenerate cases we assume that \\( 0<\\overparen{focusone focustwo}<\\pi \\) and that \\( \\overparen{focusone focustwo}<2 semisum<2 \\pi-\\overparen{focusone focustwo} \\).\n\nThe case \\( 2 semisum>\\pi \\) can be reduced to the case \\( 2 semisum<\\pi \\). For, if \\( focusone^{\\prime} \\) and \\( focustwo^{\\prime} \\) are the points diametrically opposite to \\( focusone \\) and \\( focustwo \\) then \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=2 semisum \\) if and only if \\( \\widehat{pointpt focusone}^{\\prime}+\\widehat{pointpt focustwo}^{\\prime}=2 \\pi-2 semisum \\); that is, the spherical ellipses \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=2 semisum \\) and \\( \\widehat{pointpt focusone}^{\\prime} \\) \\( +\\overparen{pointpt focustwo^{\\prime}}=2 \\pi-2 semisum \\) are identical. Since \\( \\min (2 semisum, 2 \\pi-2 semisum) \\leqq \\pi \\), we may assume without loss of generality that \\( 2 semisum \\leqq \\pi \\).\n\nLet \\( focusone \\) and \\( focustwo \\) lie on the equator. There are two points \\( vertexone \\) and \\( vertextwo \\) (the \"vertices\") on the equator which lie on the spherical ellipse. Obviously, \\( vertexone vertextwo=2 semisum \\). The \"center\" of the spherical ellipse (common midpoint of the arcs \\( \\overparen{focusone focustwo} \\) and \\( \\overparen{vertexone vertextwo} \\) ) will be denoted by \\( ellipscen \\).\n\nFig. 1\n\nFig. 2\n\nWe first treat the case \\( 2 semisum<\\pi \\) and show that in this case the spherical ellipse cannot be a circle. Assume it were a circle; call it \\( circlebig \\) (see Figure 1). \\( circlebig \\) would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. \\( circlebig \\) would also have to pass through the vertices. Therefore its spherical diameter would be \\( \\widehat{vertexone vertextwo}=2 semisum \\) and its spherical radius would be equal to \\( semisum \\). The spherical center of \\( circlebig \\) would be \\( ellipscen \\), the center of the ellipse. Let \\( midpoint \\) be one of the two points on \\( circlebig \\) which lie half-way between the two vertices. Then, since \\( midpoint \\) is supposed to be a point on the spherical ellipse, \\( 2 semisum=\\widehat{midpoint focusone}+\\hat{midpoint} \\widehat{focustwo}>2 \\dot{midpoint} \\grave{ellipscen}=2 semisum \\) (note that \\( midpoint focusone ellipscen \\) is a right spherical triangle with the right angle at \\( ellipscen \\) and with side \\( \\left.\\mathscr{midpoint} \\check{ellipscen}=semisum<\\frac{1}{2} \\pi\\right) \\). Contradiction shows that the only possible spherical ellipses which are circles must occur when \\( 2 semisum=\\pi \\).\n\nIn case \\( 2 semisum=\\pi, vertexone \\) and \\( vertextwo \\) are diametrically opposite points on the equator. We shall show that the great circle \\( circlebig \\) through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=\\pi \\). To see this, let \\( focustwo^{*} \\) be the reflection of \\( focustwo \\) about the plane of \\( circlebig . focustwo^{*} \\) is on the equator diametrically opposite to \\( focusone \\) (see Fig. 2). Let \\( pointpt \\) be an arbitrary point on the sphere, and draw the great circle through \\( focusone, pointpt \\) and \\( focustwo^{*} \\). Then \\( \\overparen{pointpt focusone}+\\widehat{pointpt focustwo}^{*}=\\pi \\). Hence, \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=\\pi \\) if and only if \\( \\overparen{pointpt focustwo}=\\overparen{pointpt focustwo}^{*} \\), that is, if and only if \\( pointpt \\) is on \\( circlebig \\). This shows that \\( circlebig \\) is the spherical ellipse \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=\\pi \\), as stated above.\n\nThus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle \\( circlebig \\) the foci can be any two points \\( focusone \\) and \\( focustwo \\) which lie on the same great circle perpendicular to \\( circlebig \\), on the same side of \\( circlebig \\) and at equal distances from \\( circlebig \\). The equation of any such spherical ellipse is \\( \\overparen{pointpt focusone}+\\overparen{pointpt focustwo}=\\pi \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "butterscotch",
        "M": "paperclip",
        "A": "dragonfly",
        "B": "tapestry",
        "C": "lighthouse",
        "a": "hummingbird",
        "V_1": "marshmallow",
        "V_2": "buttercup",
        "\\Gamma": "pineapple"
      },
      "question": "B-4. A \"spherical ellipse\" with foci \\( dragonfly, tapestry \\) on a given sphere is defined as the set of all points \\( butterscotch \\) on the sphere such that \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}= \\) constant. Here \\( \\overparen{butterscotch dragonfly} \\) denotes the shortest distance on the sphere between \\( butterscotch \\) and \\( dragonfly \\). Determine the entire class of real spherical ellipses which are circles.",
      "solution": "B-4 We take the radius of the sphere as unity and denote the constant sum \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry} \\) by \\( 2 hummingbird \\). To avoid trivial and degenerate cases we assume that \\( 0<\\overparen{dragonfly tapestry}<\\pi \\) and that \\( \\overparen{dragonfly tapestry}<2 hummingbird<2 \\pi-\\overparen{dragonfly tapestry} \\).\n\nThe case \\( 2 hummingbird>\\pi \\) can be reduced to the case \\( 2 hummingbird<\\pi \\). For, if \\( dragonfly^{\\prime} \\) and \\( tapestry^{\\prime} \\) are the points diametrically opposite to \\( dragonfly \\) and \\( tapestry \\) then \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=2 hummingbird \\) if and only if \\( \\widehat{butterscotch dragonfly}^{\\prime}+\\widehat{butterscotch tapestry}^{\\prime}=2 \\pi-2 hummingbird \\); that is, the spherical ellipses \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=2 hummingbird \\) and \\( \\widehat{butterscotch dragonfly}^{\\prime}+\\overparen{butterscotch tapestry^{\\prime}}=2 \\pi-2 hummingbird \\) are identical. Since \\( \\min (2 hummingbird, 2 \\pi-2 hummingbird) \\leqq \\pi \\), we may assume without loss of generality that \\( 2 hummingbird \\leqq \\pi \\).\n\nLet \\( dragonfly \\) and \\( tapestry \\) lie on the equator. There are two points \\( marshmallow \\) and \\( buttercup \\) (the \"vertices\") on the equator which lie on the spherical ellipse. Obviously, \\( marshmallow buttercup=2 hummingbird \\). The \"center\" of the spherical ellipse (common midpoint of the arcs \\( \\overparen{dragonfly tapestry} \\) and \\( \\overparen{marshmallow buttercup} \\) ) will be denoted by \\( lighthouse \\).\n\nFig. 1\n\nFig. 2\n\nWe first treat the case \\( 2 hummingbird<\\pi \\) and show that in this case the spherical ellipse\ncannot be a circle. Assume it were a circle; call it \\( pineapple \\) (see Figure 1). \\( pineapple \\) would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. \\( pineapple \\) would also have to pass through the vertices. Therefore its spherical diameter would be \\( \\widehat{marshmallow buttercup}=2 hummingbird \\) and its spherical radius would be equal to \\( hummingbird \\). The spherical center of \\( pineapple \\) would be \\( lighthouse \\), the center of the ellipse. Let \\( paperclip \\) be one of the two points on \\( pineapple \\) which lie half-way between the two vertices. Then, since \\( paperclip \\) is supposed to be a point on the spherical ellipse, \\( 2 hummingbird=\\widehat{paperclip dragonfly}+\\hat{paperclip} \\widehat{tapestry}>2 \\dot{paperclip} \\grave{lighthouse}=2 hummingbird \\) (note that \\( paperclip dragonfly lighthouse \\) is a right spherical triangle with the right angle at \\( lighthouse \\) and with side \\( \\left.\\mathscr{paperclip} \\check{lighthouse}=hummingbird<\\frac{1}{2} \\pi\\right) \\). Contradiction shows that the only possible spherical ellipses which are circles must occur when \\( 2 hummingbird=\\pi \\).\n\nIn case \\( 2 hummingbird=\\pi, marshmallow \\) and \\( buttercup \\) are diametrically opposite points on the equator. We shall show that the great circle \\( pineapple \\) through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=\\pi \\). To see this, let \\( tapestry^{*} \\) be the reflection of \\( tapestry \\) about the plane of \\( pineapple . tapestry^{*} \\) is on the equator diametrically opposite to \\( dragonfly \\) (see Fig. 2). Let \\( butterscotch \\) be an arbitrary point on the sphere, and draw the great circle through \\( dragonfly, butterscotch \\) and \\( tapestry^{*} \\). Then \\( \\overparen{butterscotch dragonfly}+\\widehat{butterscotch tapestry}^{*}=\\pi \\). Hence, \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=\\pi \\) if and only if \\( \\overparen{butterscotch tapestry}=\\overparen{butterscotch tapestry}^{*} \\), that is, if and only if \\( butterscotch \\) is on \\( pineapple \\). This shows that \\( pineapple \\) is the spherical ellipse \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=\\pi \\), as stated above.\n\nThus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle \\( pineapple \\) the foci can be any two points \\( dragonfly \\) and \\( tapestry \\) which lie on the same great circle perpendicular to \\( pineapple \\), on the same side of \\( pineapple \\) and at equal distances from \\( pineapple \\). The equation of any such spherical ellipse is \\( \\overparen{butterscotch dragonfly}+\\overparen{butterscotch tapestry}=\\pi \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "fixedpoint",
        "M": "edgepoint",
        "A": "blurpoint",
        "B": "uniformpoint",
        "C": "offcenter",
        "a": "fluctuant",
        "V_1": "basepointone",
        "V_2": "basepointtwo",
        "\\Gamma": "straightpath"
      },
      "question": "B-4. A \"spherical ellipse\" with foci \\( blurpoint, uniformpoint \\) on a given sphere is defined as the set of all points \\( fixedpoint \\) on the sphere such that \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}= \\) constant. Here \\( \\overparen{fixedpoint blurpoint} \\) denotes the shortest distance on the sphere between \\( fixedpoint \\) and \\( blurpoint \\). Determine the entire class of real spherical ellipses which are circles.",
      "solution": "B-4 We take the radius of the sphere as unity and denote the constant sum \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint} \\) by \\( 2 fluctuant \\). To avoid trivial and degenerate cases we assume that \\( 0<\\overparen{blurpoint uniformpoint}<\\pi \\) and that \\( \\overparen{blurpoint uniformpoint}<2 fluctuant<2 \\pi-\\overparen{blurpoint uniformpoint} \\).\n\nThe case \\( 2 fluctuant>\\pi \\) can be reduced to the case \\( 2 fluctuant<\\pi \\). For, if \\( blurpoint^{\\prime} \\) and \\( uniformpoint^{\\prime} \\) are the points diametrically opposite to \\( blurpoint \\) and \\( uniformpoint \\) then \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=2 fluctuant \\) if and only if \\( \\widehat{fixedpoint blurpoint}^{\\prime}+\\widehat{fixedpoint uniformpoint}^{\\prime}=2 \\pi-2 fluctuant \\); that is, the spherical ellipses \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=2 fluctuant \\) and \\( \\widehat{fixedpoint blurpoint}^{\\prime} \\) \\( +\\overparen{fixedpoint uniformpoint^{\\prime}}=2 \\pi-2 fluctuant \\) are identical. Since \\( \\min (2 fluctuant, 2 \\pi-2 fluctuant) \\leqq \\pi \\), we may assume without loss of generality that \\( 2 fluctuant \\leqq \\pi \\).\n\nLet \\( blurpoint \\) and \\( uniformpoint \\) lie on the equator. There are two points \\( basepointone \\) and \\( basepointtwo \\) (the \"vertices\") on the equator which lie on the spherical ellipse. Obviously, \\( basepointone basepointtwo=2 fluctuant \\). The \"center\" of the spherical ellipse (common midpoint of the arcs \\( \\overparen{blurpoint uniformpoint} \\) and \\( \\overparen{basepointone basepointtwo} \\) ) will be denoted by \\( offcenter \\).\n\nFig. 1\n\nFig. 2\n\nWe first treat the case \\( 2 fluctuant<\\pi \\) and show that in this case the spherical ellipse\ncannot be a circle. Assume it were a circle; call it \\( straightpath \\) (see Figure 1). \\( straightpath \\) would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. \\( straightpath \\) would also have to pass through the vertices. Therefore its spherical diameter would be \\( \\widehat{basepointone basepointtwo}=2 fluctuant \\) and its spherical radius would be equal to \\( fluctuant \\). The spherical center of \\( straightpath \\) would be \\( offcenter \\), the center of the ellipse. Let \\( edgepoint \\) be one of the two points on \\( straightpath \\) which lie half-way between the two vertices. Then, since \\( edgepoint \\) is supposed to be a point on the spherical ellipse, \\( 2 fluctuant=\\widehat{edgepoint blurpoint}+\\hat{edgepoint} \\widehat{uniformpoint}>2 \\dot{edgepoint} \\grave{offcenter}=2 fluctuant \\) (note that \\( edgepoint blurpoint offcenter \\) is a right spherical triangle with the right angle at \\( offcenter \\) and with side \\( \\left.\\mathscr{edgepoint} \\check{offcenter}=fluctuant<\\frac{1}{2} \\pi\\right) \\). Contradiction shows that the only possible spherical ellipses which are circles must occur when \\( 2 fluctuant=\\pi \\).\n\nIn case \\( 2 fluctuant=\\pi, basepointone \\) and \\( basepointtwo \\) are diametrically opposite points on the equator. We shall show that the great circle \\( straightpath \\) through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=\\pi \\). To see this, let \\( uniformpoint^{*} \\) be the reflection of \\( uniformpoint \\) about the plane of \\( straightpath . uniformpoint^{*} \\) is on the equator diametrically opposite to \\( blurpoint \\) (see Fig. 2). Let \\( fixedpoint \\) be an arbitrary point on the sphere, and draw the great circle through \\( blurpoint, fixedpoint \\) and \\( uniformpoint^{*} \\). Then \\( \\overparen{fixedpoint blurpoint}+\\widehat{fixedpoint uniformpoint}^{*}=\\pi \\). Hence, \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=\\pi \\) if and only if \\( \\overparen{fixedpoint uniformpoint}=\\overparen{fixedpoint uniformpoint}^{*} \\), that is, if and only if \\( fixedpoint \\) is on \\( straightpath \\). This shows that \\( straightpath \\) is the spherical ellipse \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=\\pi \\), as stated above.\n\nThus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle \\( straightpath \\) the foci can be any two points \\( blurpoint \\) and \\( uniformpoint \\) which lie on the same great circle perpendicular to \\( straightpath \\), on the same side of \\( straightpath \\) and at equal distances from \\( straightpath \\). The equation of any such spherical ellipse is \\( \\overparen{fixedpoint blurpoint}+\\overparen{fixedpoint uniformpoint}=\\pi \\)."
    },
    "garbled_string": {
      "map": {
        "P": "velqspzj",
        "M": "jfkwhbzt",
        "A": "rpsdqmah",
        "B": "ticvazoq",
        "C": "ghyenclu",
        "a": "xfsarlob",
        "V_1": "leumkhaz",
        "V_2": "wexpidny",
        "\\\\Gamma": "qxtyrmnb"
      },
      "question": "<<<\nB-4. A \"spherical ellipse\" with foci \\( rpsdqmah, ticvazoq \\) on a given sphere is defined as the set of all points \\( velqspzj \\) on the sphere such that \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}= \\) constant. Here \\( \\overparen{velqspzj rpsdqmah} \\) denotes the shortest distance on the sphere between \\( velqspzj \\) and \\( rpsdqmah \\). Determine the entire class of real spherical ellipses which are circles.\n>>>",
      "solution": "<<<\nB-4 We take the radius of the sphere as unity and denote the constant sum \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq} \\) by \\( 2 xfsarlob \\). To avoid trivial and degenerate cases we assume that \\( 0<\\overparen{rpsdqmah ticvazoq}<\\pi \\) and that \\( \\overparen{rpsdqmah ticvazoq}<2 xfsarlob<2 \\pi-\\overparen{rpsdqmah ticvazoq} \\).\n\nThe case \\( 2 xfsarlob>\\pi \\) can be reduced to the case \\( 2 xfsarlob<\\pi \\). For, if \\( rpsdqmah^{\\prime} \\) and \\( ticvazoq^{\\prime} \\) are the points diametrically opposite to \\( rpsdqmah \\) and \\( ticvazoq \\) then \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=2 xfsarlob \\) if and only if \\( \\widehat{velqspzj rpsdqmah}^{\\prime}+\\widehat{velqspzj ticvazoq}^{\\prime}=2 \\pi-2 xfsarlob \\); that is, the spherical ellipses \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=2 xfsarlob \\) and \\( \\widehat{velqspzj rpsdqmah}^{\\prime}+\\overparen{velqspzj ticvazoq^{\\prime}}=2 \\pi-2 xfsarlob \\) are identical. Since \\( \\min (2 xfsarlob, 2 \\pi-2 xfsarlob) \\leqq \\pi \\), we may assume without loss of generality that \\( 2 xfsarlob \\leqq \\pi \\).\n\nLet \\( rpsdqmah \\) and \\( ticvazoq \\) lie on the equator. There are two points \\( leumkhaz \\) and \\( wexpidny \\) (the \"vertices\") on the equator which lie on the spherical ellipse. Obviously, \\( leumkhaz wexpidny=2 xfsarlob \\). The \"center\" of the spherical ellipse (common midpoint of the arcs \\( \\overparen{rpsdqmah ticvazoq} \\) and \\( \\overparen{leumkhaz wexpidny} \\) ) will be denoted by \\( ghyenclu \\).\n\nFig. 1\n\nFig. 2\n\nWe first treat the case \\( 2 xfsarlob<\\pi \\) and show that in this case the spherical ellipse\ncannot be a circle. Assume it were a circle; call it \\( qxtyrmnb \\) (see Figure 1). \\( qxtyrmnb \\) would have to be symmetric with respect to the equatorial plane, thus lie in a plane perpendicular to the equatorial plane. \\( qxtyrmnb \\) would also have to pass through the vertices. Therefore its spherical diameter would be \\( \\widehat{leumkhaz wexpidny}=2 xfsarlob \\) and its spherical radius would be equal to \\( xfsarlob \\). The spherical center of \\( qxtyrmnb \\) would be \\( ghyenclu \\), the center of the ellipse. Let \\( jfkwhbzt \\) be one of the two points on \\( qxtyrmnb \\) which lie half-way between the two vertices. Then, since \\( jfkwhbzt \\) is supposed to be a point on the spherical ellipse, \\( 2 xfsarlob=\\widehat{jfkwhbzt rpsdqmah}+\\widehat{jfkwhbzt ticvazoq}>2 \\dot{jfkwhbzt} \\grave{ghyenclu}=2 xfsarlob \\) (note that \\( jfkwhbzt rpsdqmah ghyenclu \\) is a right spherical triangle with the right angle at \\( ghyenclu \\) and with side \\( \\mathscr{jfkwhbzt} \\check{ghyenclu}=xfsarlob<\\frac{1}{2} \\pi \\)). Contradiction shows that the only possible spherical ellipses which are circles must occur when \\( 2 xfsarlob=\\pi \\).\n\nIn case \\( 2 xfsarlob=\\pi, leumkhaz \\) and \\( wexpidny \\) are diametrically opposite points on the equator. We shall show that the great circle \\( qxtyrmnb \\) through the vertices and perpendicular to the equatorial plane is identical with the spherical ellipse \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=\\pi \\). To see this, let \\( ticvazoq^{*} \\) be the reflection of \\( ticvazoq \\) about the plane of \\( qxtyrmnb . ticvazoq^{*} \\) is on the equator diametrically opposite to \\( rpsdqmah \\) (see Fig. 2). Let \\( velqspzj \\) be an arbitrary point on the sphere, and draw the great circle through \\( rpsdqmah, velqspzj \\) and \\( ticvazoq^{*} \\). Then \\( \\overparen{velqspzj rpsdqmah}+\\widehat{velqspzj ticvazoq}^{*}=\\pi \\). Hence, \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=\\pi \\) if and only if \\( \\overparen{velqspzj ticvazoq}=\\overparen{velqspzj ticvazoq}^{*} \\), that is, if and only if \\( velqspzj \\) is on \\( qxtyrmnb \\). This shows that \\( qxtyrmnb \\) is the spherical ellipse \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=\\pi \\), as stated above.\n\nThus the only circles on the sphere that are spherical ellipses are the great circles. For any given great circle \\( qxtyrmnb \\) the foci can be any two points \\( rpsdqmah \\) and \\( ticvazoq \\) which lie on the same great circle perpendicular to \\( qxtyrmnb \\), on the same side of \\( qxtyrmnb \\) and at equal distances from \\( qxtyrmnb \\). The equation of any such spherical ellipse is \\( \\overparen{velqspzj rpsdqmah}+\\overparen{velqspzj ticvazoq}=\\pi \\).\n>>>"
    },
    "kernel_variant": {
      "question": "Let Σ ⊂ ℝ³ be the sphere of radius 2 and centre O.  \nFor two distinct points A , B ∈ Σ and for a real number 2s with 0 < 2s < 4π define\n  E_{A,B}(2s)= {P∈Σ | d(P,A)+d(P,B)=2s},\nwhere d(·,·) denotes the length of the shorter great-circle arc between the two points (the spherical distance on Σ).\nThe set E_{A,B}(2s) is called the spherical ellipse with foci A , B and (half-)parameter s.\n\nDetermine all triples (A , B , 2s) for which the locus E_{A,B}(2s) is a (Euclidean) circle, i.e. when it equals Σ ∩ Π for some plane Π.",
      "solution": "Throughout let R = 2 be the fixed radius of the sphere.  Thus every great circle has length 4π and the largest possible spherical distance between two points is 2π.\n\nStep 0 (Restricting the parameter).\nFor any point P ∈ Σ and for the antipodes A′, B′ of A, B we have\n d(P,A)+d(P,B)=2s  ⇔  d(P,A′)+d(P,B′)=4π−2s.\nHence E_{A,B}(2s)=E_{A′,B′}(4π−2s).  Replacing (A , B , 2s) by its antipodal triple if necessary we may assume\n  0 < 2s ≤ 2π.                                                        (1)\n\nStep 1 (Normal position and the vertices).\nBy a rotation of the sphere place A and B on the meridian\n G := Σ ∩ {y = 0},\nso that d(A,B)=2c with 0 < 2c ≤ 2π.  \nIf 2s>2c the ellipse possesses exactly two points V₁ , V₂ on G—its vertices—satisfying\n d(V₁,V₂)=2s,   d(V₁,A)=d(V₂,B)=s−c.                                (2)\n(The case 2s = 2c is degenerate: the ellipse is the shorter arc AB ⊂ G and is not a Euclidean circle.)  Hence from now on\n  2c < 2s ≤ 2π.                                                      (3)\nLet C be the midpoint (in the spherical sense) of the shorter arcs AB and V₁V₂; thus C ∈ G.\n\nStep 2 (A Euclidean circle with 2s < 2π is impossible).\nAssume, for a contradiction, that 2s < 2π and that Γ := E_{A,B}(2s) is a Euclidean circle, Γ = Σ ∩ Π.\n\n2.1 (The spherical centre of Γ).\nWrite the equation of Π as\n ⟨x , n⟩ = h, |n| = 1, |h| < R,                             (4)\nwhere n is chosen so that Π is perpendicular to the plane y = 0.  Hence n lies in the xz–plane.  Let\n D := R n (the intersection of the ray O + t n, t > 0, with Σ ).      (5)\nBecause ⟨P , n⟩ = h for every P ∈ Γ, we have for such P\n P·D = R h = const.                                             (6)\nOn a sphere the spherical distance between two points U,V satisfies\n cos[d(U,V)/R] = (U·V)/R².\nTherefore (6) implies that the quantity d(P,D) is constant for P ∈ Γ; i.e. D is the spherical centre of Γ and its spherical radius is\n ρ := d(D,Γ) = arccos(h/R) < π.                                 (7)\n\n2.2 (The centre D coincides with C).\nBoth vertices V₁, V₂ lie on Γ, so\n d(D,V₁) = d(D,V₂) = ρ.                                          (8)\nAll four points D, V₁, V₂, O lie in the plane y = 0, hence on the great circle G.  Along G there are exactly two points that are equidistant from V₁ and V₂, namely the spherical midpoint C of V₁V₂ and its antipode C*.  Since D and C lie on the same side of the plane Π ( Π separates D from −D while C and −C lie on different sides of Π ), we must have D = C.  Consequently\n ρ = d(C,V₁) = s    (because of (2)).                              (9)\n\n2.3 (Variation of the constant–sum function along Γ).\nIntroduce the notations\n ĉ := c/2,     ρ̂ := ρ/2 = s/2.\nUsing C as ‘north pole’ and G as zero longitude, every point of Γ can be written in spherical coordinates as\n P(λ) := (colatitude ρ, longitude λ),     −π ≤ λ ≤ π.\nWith the spherical law of cosines one obtains for\n f(λ) := d(P(λ),A) + d(P(λ),B)\nexactly the expression and derivative derived in the original solution:\n f′(λ)=−2 sinĉ sinρ̂ sinλ \n        ×\n        [1/√(1−g₂(λ)²) − 1/√(1−g₁(λ)²)],                            (10)\nwith non-vanishing g₁(λ)−g₂(λ) for all λ∈(0,π)\\{π/2}. Hence f′(λ)≠0 and f is not constant on Γ, contradicting the assumption Γ = E_{A,B}(2s).\n\nTherefore E_{A,B}(2s) cannot be a Euclidean circle when 2s < 2π.\n\nStep 3 (The remaining possibility is 2s = 2π).\nBy (1) and Step 2 the only remaining value is\n  2s = 2π.                                                            (11)\nFrom now on suppose Γ := E_{A,B}(2π) is a Euclidean circle.\n\nStep 4 (Locating the foci when 2s = 2π).\nAs before, Π is perpendicular to y = 0 and contains V₁, V₂.  Because of (2) and (11) we have d(V₁,V₂)=2π, so V₁ and V₂ are antipodes and Π passes through O; hence Γ is a great circle.\n\nLet ρ denote reflection in Π and put B*:=ρ(B).  Since Π fixes every P∈Γ,\n d(P,B)=d(P,B*)    for all P∈Γ.                                       (12)\nConsequently Γ is also the ellipse with foci A and B* and constant 2π.\nAt the vertex V₁,\n d(V₁,A)+d(V₁,B*)=2π.                                                (13)\nThe three points V₁, A, B* lie on G.  Writing α:=d(V₁,A), β:=d(V₁,B*) we have α+β=2π.  There are exactly two geodesic arcs joining A and B* along G, each of length 2π; therefore\n d(A,B*)=2π.                                                        (14)\nHence B* is the antipode A*.  Reflecting once more gives ρ(A)=B*, i.e. the reflection of either focus in Π is the antipode of the other.  Thus A and B lie on the same great circle perpendicular to Π, on the same side of Π and at equal spherical distance from Π.\n\nStep 5 (Sufficiency).\nConversely, assume\n (i) 2s = 2π, and\n (ii) the reflection of B in a plane Π through O equals the antipode A* of A (and hence vice versa).\nLet Γ:=Σ ∩ Π (a great circle).  For any P∈Γ we have d(P,B)=d(P,A*).  If θ denotes the smaller angle ∠AOP (measured in radians), then ∠A*OP=π−θ and\n d(P,A)=2θ,   d(P,A*)=2(π−θ),\nso d(P,A)+d(P,B)=2θ+2(π−θ)=2π.  Thus P∈E_{A,B}(2π), and therefore\n E_{A,B}(2π)=Γ,\nwhich is indeed a Euclidean circle.\n\nStep 6 (Conclusion).\nA spherical ellipse E_{A,B}(2s) on the sphere of radius 2 is a Euclidean circle if and only if\n (a) 2s = 2π, and\n (b) the reflection of either focus in the plane of the circle is the antipode of the other focus (equivalently, the two foci lie on the same great circle perpendicular to Π, on the same side of Π and at equal distance from Π).\nIn that case E_{A,B}(2π)=Σ ∩ Π, which is a great circle.\n\nHence the only spherical ellipses that are Euclidean circles are the great circles, the required constant sum of distances is 2π, and the foci are precisely the pairs described in (b).",
      "_meta": {
        "core_steps": [
          "Reduce sums >½ circumference by replacing each focus with its antipode so that 2a ≤ π",
          "Rotate sphere so A,B lie on a chosen great circle; mark vertices V₁,V₂ and their midpoint C",
          "Assume 2a < π gives a circular ellipse; use midpoint M and a right spherical triangle to show d(P,A)+d(P,B) < 2a, a contradiction",
          "When 2a = π, reflect one focus across the plane of the candidate circle and show d(P,A)+d(P,B)=π ⇔ P lies on that great circle",
          "Conclude only great circles work; their foci are any symmetric pair on a perpendicular great circle and the constant sum is π"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Numerical scale chosen for distances on the sphere",
            "original": "Radius normalised to 1"
          },
          "slot2": {
            "description": "Reference great circle used to place the foci (called the equator in the text)",
            "original": "Equator"
          },
          "slot3": {
            "description": "Symbol used for half the constant focal‐distance sum",
            "original": "a in 2a"
          },
          "slot4": {
            "description": "Numeric threshold equal to half a great‐circle circumference",
            "original": "π"
          },
          "slot5": {
            "description": "Orientation of the plane containing the candidate circle (taken perpendicular to the reference great circle)",
            "original": "Plane perpendicular to the equatorial plane"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}