{
  "index": "1948-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "2. Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is \"ring-shaped.\" Being given the radii of the spheres, \\( r \\) and \\( \\boldsymbol{R} \\), find the volume of the \"ring-shaped\" part. (The desired expression is a rational function of \\( r \\) and \\( R \\). )",
  "solution": "First Solution. Let \\( \\alpha \\) be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at \\( O_{1} \\) and \\( O_{2} \\) and origin of the coordinate system at \\( O \\), the point of contact.\n\nThe equation of the larger circle is\n\\[\nx^{2}+y^{2}+2 R x=0\n\\]\nand the equation of the smaller is\n\\[\nx^{2}+y^{2}-2 r x=0 .\n\\]\n\nFrom the diagram,\n\\[\n\\sin \\alpha=\\frac{O_{1} P}{O_{1} O_{2}}=\\frac{R-r}{R+r},\n\\]\nso \\( \\cos ^{2} \\alpha=4 R r /(R+r)^{2} \\).\nWe compute \\( V_{1} \\), the volume of the frustum of the cone obtained by revolving area \\( A_{1} A_{2} B_{2} B_{1} \\) about the \\( x \\)-axis.\n\\[\n\\begin{aligned}\n\\text { Volume of larger cone } & =\\frac{1}{3} \\pi A_{1} B_{1}{ }^{2} \\cdot B_{1} T \\\\\n& =\\frac{1}{3} \\pi R^{2} \\cos ^{2} \\alpha \\cdot R \\cot \\alpha \\cos \\alpha \\\\\n& =\\frac{16 \\pi}{3} R^{3} \\frac{r^{2} R^{2}}{(R-r)(R+r)^{3}} . \\\\\n\\text { Volume of smaller cone } & =\\frac{16 \\pi}{3} r^{3} \\frac{r^{2} R^{2}}{(R-r)(R+r)^{3}} .\n\\end{aligned}\n\\]\n\nTheir difference is\n\\[\nV_{1}=\\frac{16 \\pi}{3} \\cdot \\frac{r^{2} R^{2}}{(R+r)^{3}} \\cdot\\left(R^{2}+R r+r^{2}\\right) .\n\\]\n\nLet \\( V_{2} \\) and \\( V_{3} \\) be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.\n\\[\n\\begin{aligned}\nV_{2} & =\\pi \\int_{R(\\sin \\alpha-1)}^{0}\\left(-x^{2}-2 R x\\right) d x \\\\\n& =\\pi\\left[-\\frac{x^{3}}{3}-R x^{2}\\right]_{R(\\sin \\alpha-1)}^{0} \\\\\n& =\\frac{4 \\pi R^{3} r^{2}}{3(R+r)^{3}}(3 R+r),\n\\end{aligned}\n\\]\nsince \\( R(\\sin \\alpha-1)=-2 R r /(R+r) \\). Likewise\n\\[\n\\begin{aligned}\nV_{3} & =\\pi \\int_{0}^{r(1+\\sin \\alpha)}\\left(-x^{2}+2 r x\\right) d x=\\pi\\left[-\\frac{x^{3}}{3}+r x^{2}\\right]_{0}^{\\eta(\\sin \\alpha+1)} \\\\\n& =\\frac{4 \\pi}{3} \\cdot \\frac{R^{2} r^{3}}{(R+r)^{3}}(R+3 r) .\n\\end{aligned}\n\\]\n\nThe desired volume is given by\n\\[\n\\begin{aligned}\nV_{1}-V_{2}-V_{3}= & \\frac{4 \\pi}{3} \\frac{R^{2} r^{2}}{(R+r)^{3}} \\\\\n& \\times\\left[4\\left(R^{2}+R r+r^{2}\\right)-3 R^{2}-R r-3 r^{2}-R r\\right] \\\\\n= & \\frac{4 \\pi}{3} \\frac{R^{2} r^{2}}{R+r} .\n\\end{aligned}\n\\]\n\nSecond Solution. An elegant geometrical solution based on Cavalieri's principle can be given.\n\nConsider the accompanying diagram labeled as shown and suppose the figure is rotated about the axis \\( B T \\). The region \\( \\Delta \\) bounded by \\( \\overline{A C}, \\overline{C O} \\) and the circular arc \\( A O \\) will sweep out a solid that is a frustum of a cone less a spherical segment. We shall prove that its volume is the same as that of the cone generated by the triangular region BOC.\n\nConsider an arbitrary line perpendicular to \\( B T \\) between \\( B \\) and \\( O \\) crossing \\( \\overline{A C} \\) at \\( P, \\widehat{A O} \\) at \\( Q \\), etc., as shown. When rotated the segment \\( \\overline{P Q} \\) sweeps out an annular region whose area is\n\\[\n\\pi\\left(P S^{2}-Q S^{2}\\right)=\\pi(P S+Q S)(P S-Q S)=\\pi P U \\cdot P Q=\\pi(A P)^{2}\n\\]\n\nBy similarity\n\\[\n\\frac{A P}{A C}=\\frac{B S}{B O}=\\frac{R S}{C O} .\n\\]\n\nSince the tangents \\( A C \\) and \\( O C \\) to the circle are equal, we have \\( A P=R S \\). Thus the area of the annular region is \\( \\pi(R S)^{2} \\) which is the area of the circular region generated by \\( R S \\). By Cavalieri's principle the volume of the solid swept out by \\( \\Delta \\) is therefore the same as the volume of the cone swept out by the triangular region BOC.\n\nThis argument applies equally to the solid swept out by the curvilinear region \\( \\Delta^{\\prime} \\) bounded by \\( A C^{\\prime}, C^{\\prime} O^{\\prime} \\) and the circular arc \\( A O^{\\prime} \\) and the cone swept out by the triangular region \\( B O^{\\prime} C^{\\prime} \\).\n\nNote that if \\( R=r \\), this argument becomes a classical proof that the volume of the region between a hemisphere and the circumscribed cylinder is one-third the volume of the cylinder. Hence the volume of the hemisphere is two-thirds the volume of the cylinder.\n\nReturning to the problem (see the first figure), the volume \\( V \\) required is the sum of the volumes of two cones with a common base generated by \\( O C \\) and altitudes \\( O B_{1} \\) and \\( O B_{2} \\). Hence \\( V=\\frac{1}{3} \\pi O C^{2} \\cdot B_{1} B_{2} \\). Now \\( \\angle O_{1} C C_{2} \\) is a right angle since \\( O_{1} C \\) and \\( O_{2} C \\) bisect the supplementary angles \\( A_{1} C O \\) and \\( O C A_{2} \\), so \\( O C^{2}=O_{1} O \\cdot O O_{2}=R r \\). And \\( B_{1} B_{2}=O_{1} O_{2} \\cos ^{2} \\alpha= \\) \\( 4 R r /(R+r) \\). Therefore\n\\[\nV=\\frac{4 \\pi}{3} \\cdot \\frac{R^{2} r^{2}}{R+r}\n\\]",
  "vars": [
    "A",
    "A_1",
    "A_2",
    "B",
    "B_1",
    "B_2",
    "C",
    "O",
    "O_1",
    "O_2",
    "P",
    "Q",
    "S",
    "T",
    "U",
    "V",
    "V_1",
    "V_2",
    "V_3",
    "x",
    "y",
    "\\\\alpha",
    "\\\\Delta"
  ],
  "params": [
    "r",
    "R"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointa",
        "A_1": "pointaone",
        "A_2": "pointatwo",
        "B": "pointb",
        "B_1": "pointbone",
        "B_2": "pointbtwo",
        "C": "pointc",
        "O": "origpoint",
        "O_1": "centerone",
        "O_2": "centertwo",
        "P": "pointp",
        "Q": "pointq",
        "S": "pointess",
        "T": "pointt",
        "U": "pointu",
        "V": "volumev",
        "V_1": "volumevone",
        "V_2": "volumevtwo",
        "V_3": "volumevthree",
        "x": "coordx",
        "y": "coordy",
        "\\alpha": "coneangle",
        "\\Delta": "curvregion",
        "r": "radiussmall",
        "R": "radiuslarge"
      },
      "question": "2. Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is \"ring-shaped.\" Being given the radii of the spheres, \\( radiussmall \\) and \\( \\boldsymbol{radiuslarge} \\), find the volume of the \"ring-shaped\" part. (The desired expression is a rational function of \\( radiussmall \\) and \\( radiuslarge \\). )",
      "solution": "First Solution. Let \\( coneangle \\) be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at \\( centerone \\) and \\( centertwo \\) and origin of the coordinate system at \\( origpoint \\), the point of contact.\n\nThe equation of the larger circle is\n\\[\ncoordx^{2}+coordy^{2}+2 radiuslarge coordx=0\n\\]\nand the equation of the smaller is\n\\[\ncoordx^{2}+coordy^{2}-2 radiussmall coordx=0 .\n\\]\n\nFrom the diagram,\n\\[\n\\sin coneangle=\\frac{centerone pointp}{centerone centertwo}=\\frac{radiuslarge-radiussmall}{radiuslarge+radiussmall},\n\\]\nso \\( \\cos ^{2} coneangle=4 radiuslarge radiussmall /(radiuslarge+radiussmall)^{2} \\).\nWe compute \\( volumevone \\), the volume of the frustum of the cone obtained by revolving area \\( pointaone pointatwo pointbtwo pointbone \\) about the \\( coordx \\)-axis.\n\\[\n\\begin{aligned}\n\\text { Volume of larger cone } & =\\frac{1}{3} \\pi pointaone pointbone{ }^{2} \\cdot pointbone pointt \\\\\n& =\\frac{1}{3} \\pi radiuslarge^{2} \\cos ^{2} coneangle \\cdot radiuslarge \\cot coneangle \\cos coneangle \\\\\n& =\\frac{16 \\pi}{3} radiuslarge^{3} \\frac{radiussmall^{2} radiuslarge^{2}}{(radiuslarge-radiussmall)(radiuslarge+radiussmall)^{3}} . \\\\\n\\text { Volume of smaller cone } & =\\frac{16 \\pi}{3} radiussmall^{3} \\frac{radiussmall^{2} radiuslarge^{2}}{(radiuslarge-radiussmall)(radiuslarge+radiussmall)^{3}} .\n\\end{aligned}\n\\]\n\nTheir difference is\n\\[\nvolumevone=\\frac{16 \\pi}{3} \\cdot \\frac{radiussmall^{2} radiuslarge^{2}}{(radiuslarge+radiussmall)^{3}} \\cdot\\left(radiuslarge^{2}+radiuslarge radiussmall+radiussmall^{2}\\right) .\n\\]\n\nLet \\( volumevtwo \\) and \\( volumevthree \\) be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.\n\\[\n\\begin{aligned}\nvolumevtwo & =\\pi \\int_{radiuslarge(\\sin coneangle-1)}^{0}\\left(-coordx^{2}-2 radiuslarge coordx\\right) d coordx \\\\\n& =\\pi\\left[-\\frac{coordx^{3}}{3}-radiuslarge coordx^{2}\\right]_{radiuslarge(\\sin coneangle-1)}^{0} \\\\\n& =\\frac{4 \\pi radiuslarge^{3} radiussmall^{2}}{3(radiuslarge+radiussmall)^{3}}(3 radiuslarge+radiussmall),\n\\end{aligned}\n\\]\nsince \\( radiuslarge(\\sin coneangle-1)=-2 radiuslarge radiussmall /(radiuslarge+radiussmall) \\). Likewise\n\\[\n\\begin{aligned}\nvolumevthree & =\\pi \\int_{0}^{radiussmall(1+\\sin coneangle)}\\left(-coordx^{2}+2 radiussmall coordx\\right) d coordx=\\pi\\left[-\\frac{coordx^{3}}{3}+radiussmall coordx^{2}\\right]_{0}^{radiussmall(\\sin coneangle+1)} \\\\\n& =\\frac{4 \\pi}{3} \\cdot \\frac{radiuslarge^{2} radiussmall^{3}}{(radiuslarge+radiussmall)^{3}}(radiuslarge+3 radiussmall) .\n\\end{aligned}\n\\]\n\nThe desired volume is given by\n\\[\n\\begin{aligned}\nvolumevone-volumevtwo-volumevthree= & \\frac{4 \\pi}{3} \\frac{radiuslarge^{2} radiussmall^{2}}{(radiuslarge+radiussmall)^{3}} \\\\\n& \\times\\left[4\\left(radiuslarge^{2}+radiuslarge radiussmall+radiussmall^{2}\\right)-3 radiuslarge^{2}-radiuslarge radiussmall-3 radiussmall^{2}-radiuslarge radiussmall\\right] \\\\\n= & \\frac{4 \\pi}{3} \\frac{radiuslarge^{2} radiussmall^{2}}{radiuslarge+radiussmall} .\n\\end{aligned}\n\\]\n\nSecond Solution. An elegant geometrical solution based on Cavalieri's principle can be given.\n\nConsider the accompanying diagram labeled as shown and suppose the figure is rotated about the axis \\( pointb pointt \\). The region \\( curvregion \\) bounded by \\( \\overline{pointa pointc}, \\overline{pointc origpoint} \\) and the circular arc \\( pointa origpoint \\) will sweep out a solid that is a frustum of a cone less a spherical segment. We shall prove that its volume is the same as that of the cone generated by the triangular region pointb origpoint pointc.\n\nConsider an arbitrary line perpendicular to \\( pointb pointt \\) between \\( pointb \\) and \\( origpoint \\) crossing \\( \\overline{pointa pointc} \\) at \\( pointp, \\widehat{pointa origpoint} \\) at \\( pointq \\), etc., as shown. When rotated the segment \\( \\overline{pointp pointq} \\) sweeps out an annular region whose area is\n\\[\n\\pi\\left(pointp pointess^{2}-pointq pointess^{2}\\right)=\\pi(pointp pointess+pointq pointess)(pointp pointess-pointq pointess)=\\pi pointp pointu \\cdot pointp pointq=\\pi(pointa pointp)^{2}\n\\]\n\nBy similarity\n\\[\n\\frac{pointa pointp}{pointa pointc}=\\frac{pointb pointess}{pointb origpoint}=\\frac{radiuslarge pointess}{pointc origpoint} .\n\\]\n\nSince the tangents \\( pointa pointc \\) and \\( origpoint pointc \\) to the circle are equal, we have \\( pointa pointp=radiuslarge pointess \\). Thus the area of the annular region is \\( \\pi(radiuslarge pointess)^{2} \\) which is the area of the circular region generated by \\( radiuslarge pointess \\). By Cavalieri's principle the volume of the solid swept out by \\( curvregion \\) is therefore the same as the volume of the cone swept out by the triangular region pointb origpoint pointc.\n\nThis argument applies equally to the solid swept out by the curvilinear region \\( curvregion^{\\prime} \\) bounded by \\( pointa pointc^{\\prime}, pointc^{\\prime} origpoint^{\\prime} \\) and the circular arc \\( pointa origpoint^{\\prime} \\) and the cone swept out by the triangular region pointb origpoint^{\\prime} pointc^{\\prime}.\n\nNote that if \\( radiuslarge=radiussmall \\), this argument becomes a classical proof that the volume of the region between a hemisphere and the circumscribed cylinder is one-third the volume of the cylinder. Hence the volume of the hemisphere is two-thirds the volume of the cylinder.\n\nReturning to the problem (see the first figure), the volume \\( volumev \\) required is the sum of the volumes of two cones with a common base generated by \\( origpoint pointc \\) and altitudes \\( origpoint pointbone \\) and \\( origpoint pointbtwo \\). Hence \\( volumev=\\frac{1}{3} \\pi origpoint pointc^{2} \\cdot pointbone pointbtwo \\). Now \\( \\angle centerone pointc C_{2} \\) is a right angle since \\( centerone pointc \\) and \\( centertwo pointc \\) bisect the supplementary angles \\( pointaone pointc origpoint \\) and \\( origpoint pointc pointatwo \\), so \\( origpoint pointc^{2}=centerone origpoint \\cdot origpoint centertwo=radiuslarge radiussmall \\). And \\( pointbone pointbtwo=centerone centertwo \\cos ^{2} coneangle= 4 radiuslarge radiussmall /(radiuslarge+radiussmall) \\). Therefore\n\\[\nvolumev=\\frac{4 \\pi}{3} \\cdot \\frac{radiuslarge^{2} radiussmall^{2}}{radiuslarge+radiussmall}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "pineapple",
        "A_1": "crocodile",
        "A_2": "marigold",
        "B": "raspberry",
        "B_1": "tortoise",
        "B_2": "avalanche",
        "C": "cinnamon",
        "O": "whirlwind",
        "O_1": "labyrinth",
        "O_2": "nightfall",
        "P": "spacecraft",
        "Q": "snowflake",
        "S": "limestone",
        "T": "afterglow",
        "U": "rainstorm",
        "V": "blueberry",
        "V_1": "velvetine",
        "V_2": "moonstone",
        "V_3": "starlight",
        "x": "chandelier",
        "y": "hurricane",
        "\\alpha": "quasarbeam",
        "\\Delta": "zeppelin",
        "r": "sandstorm",
        "R": "driftwood"
      },
      "question": "2. Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is \"ring-shaped.\" Being given the radii of the spheres, \\( sandstorm \\) and \\( \\boldsymbol{driftwood} \\), find the volume of the \"ring-shaped\" part. (The desired expression is a rational function of \\( sandstorm \\) and \\( driftwood \\). )",
      "solution": "First Solution. Let \\( quasarbeam \\) be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at \\( labyrinth \\) and \\( nightfall \\) and origin of the coordinate system at \\( whirlwind \\), the point of contact.\n\nThe equation of the larger circle is\n\\[\nchandelier^{2}+hurricane^{2}+2 driftwood\\;chandelier=0\n\\]\nand the equation of the smaller is\n\\[\nchandelier^{2}+hurricane^{2}-2 sandstorm\\;chandelier=0 .\n\\]\n\nFrom the diagram,\n\\[\n\\sin quasarbeam=\\frac{labyrinth spacecraft}{labyrinth nightfall}=\\frac{driftwood-sandstorm}{driftwood+sandstorm},\n\\]\nso \\( \\cos ^{2} quasarbeam=\\frac{4 driftwood\\,sandstorm}{(driftwood+sandstorm)^{2}} \\).\nWe compute \\( velvetine \\), the volume of the frustum of the cone obtained by revolving area \\( crocodile\\,marigold\\,avalanche\\,tortoise \\) about the \\( chandelier \\)-axis.\n\\[\n\\begin{aligned}\n\\text { Volume of larger cone } & =\\frac{1}{3} \\pi (crocodile\\,tortoise)^{2} \\cdot (tortoise\\,afterglow) \\\\\n& =\\frac{1}{3} \\pi driftwood^{2} \\cos ^{2} quasarbeam \\cdot driftwood \\cot quasarbeam \\cos quasarbeam \\\\\n& =\\frac{16 \\pi}{3} driftwood^{3} \\frac{sandstorm^{2} driftwood^{2}}{(driftwood-sandstorm)(driftwood+sandstorm)^{3}} . \\\\\n\\text { Volume of smaller cone } & =\\frac{16 \\pi}{3} sandstorm^{3} \\frac{sandstorm^{2} driftwood^{2}}{(driftwood-sandstorm)(driftwood+sandstorm)^{3}} .\n\\end{aligned}\n\\]\n\nTheir difference is\n\\[\nvelvetine=\\frac{16 \\pi}{3} \\cdot \\frac{sandstorm^{2} driftwood^{2}}{(driftwood+sandstorm)^{3}} \\cdot\\left(driftwood^{2}+driftwood\\,sandstorm+sandstorm^{2}\\right) .\n\\]\n\nLet \\( moonstone \\) and \\( starlight \\) be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.\n\\[\n\\begin{aligned}\nmoonstone & =\\pi \\int_{driftwood(\\sin quasarbeam-1)}^{0}\\left(-chandelier^{2}-2 driftwood\\;chandelier\\right) d chandelier \\\\\n& =\\pi\\left[-\\frac{chandelier^{3}}{3}-driftwood\\;chandelier^{2}\\right]_{driftwood(\\sin quasarbeam-1)}^{0} \\\\\n& =\\frac{4 \\pi driftwood^{3} sandstorm^{2}}{3(driftwood+sandstorm)^{3}}(3 driftwood+sandstorm),\n\\end{aligned}\n\\]\nsince \\( driftwood(\\sin quasarbeam-1)=-\\frac{2 driftwood\\,sandstorm}{driftwood+sandstorm} \\). Likewise\n\\[\n\\begin{aligned}\nstarlight & =\\pi \\int_{0}^{sandstorm(1+\\sin quasarbeam)}\\left(-chandelier^{2}+2 sandstorm\\;chandelier\\right) d chandelier\\\\\n& =\\pi\\left[-\\frac{chandelier^{3}}{3}+sandstorm\\;chandelier^{2}\\right]_{0}^{sandstorm(\\sin quasarbeam+1)} \\\\\n& =\\frac{4 \\pi}{3} \\cdot \\frac{driftwood^{2} sandstorm^{3}}{(driftwood+sandstorm)^{3}}(driftwood+3 sandstorm) .\n\\end{aligned}\n\\]\n\nThe desired volume is given by\n\\[\n\\begin{aligned}\nvelvetine-moonstone-starlight= & \\frac{4 \\pi}{3} \\frac{driftwood^{2} sandstorm^{2}}{(driftwood+sandstorm)^{3}} \\\\\n& \\times\\left[4\\left(driftwood^{2}+driftwood\\,sandstorm+sandstorm^{2}\\right)-3 driftwood^{2}-driftwood\\,sandstorm-3 sandstorm^{2}-driftwood\\,sandstorm\\right] \\\\\n= & \\frac{4 \\pi}{3} \\frac{driftwood^{2} sandstorm^{2}}{driftwood+sandstorm} .\n\\end{aligned}\n\\]\n\nSecond Solution. An elegant geometrical solution based on Cavalieri's principle can be given.\n\nConsider the accompanying diagram labeled as shown and suppose the figure is rotated about the axis \\( raspberry\\,afterglow \\). The region \\( zeppelin \\) bounded by \\( \\overline{pineapple cinnamon}, \\overline{cinnamon whirlwind} \\) and the circular arc \\( pineapple whirlwind \\) will sweep out a solid that is a frustum of a cone less a spherical segment. We shall prove that its volume is the same as that of the cone generated by the triangular region raspberry whirlwind cinnamon.\n\nConsider an arbitrary line perpendicular to \\( raspberry\\,afterglow \\) between \\( raspberry \\) and \\( whirlwind \\) crossing \\( \\overline{pineapple cinnamon} \\) at \\( spacecraft, \\widehat{pineapple whirlwind} \\) at \\( snowflake \\), etc., as shown. When rotated the segment \\( \\overline{spacecraft snowflake} \\) sweeps out an annular region whose area is\n\\[\n\\pi\\left(spacecraft limestone^{2}-snowflake limestone^{2}\\right)=\\pi(spacecraft limestone+snowflake limestone)(spacecraft limestone-snowflake limestone)=\\pi spacecraft rainstorm \\cdot spacecraft snowflake=\\pi(pineapple spacecraft)^{2}\n\\]\n\nBy similarity\n\\[\n\\frac{pineapple spacecraft}{pineapple cinnamon}=\\frac{raspberry limestone}{raspberry whirlwind}=\\frac{limestone driftwood}{cinnamon whirlwind} .\n\\]\n\nSince the tangents \\( pineapple cinnamon \\) and \\( whirlwind cinnamon \\) to the circle are equal, we have \\( pineapple spacecraft=limestone driftwood \\). Thus the area of the annular region is \\( \\pi(limestone driftwood)^{2} \\) which is the area of the circular region generated by \\( limestone driftwood \\). By Cavalieri's principle the volume of the solid swept out by \\( zeppelin \\) is therefore the same as the volume of the cone swept out by the triangular region raspberry whirlwind cinnamon.\n\nThis argument applies equally to the solid swept out by the curvilinear region \\( zeppelin^{\\prime} \\) bounded by \\( pineapple cinnamon^{\\prime}, cinnamon^{\\prime} whirlwind^{\\prime} \\) and the circular arc \\( pineapple whirlwind^{\\prime} \\) and the cone swept out by the triangular region raspberry whirlwind^{\\prime} cinnamon^{\\prime}.\n\nNote that if \\( driftwood=sandstorm \\), this argument becomes a classical proof that the volume of the region between a hemisphere and the circumscribed cylinder is one-third the volume of the cylinder. Hence the volume of the hemisphere is two-thirds the volume of the cylinder.\n\nReturning to the problem (see the first figure), the volume \\( blueberry \\) required is the sum of the volumes of two cones with a common base generated by \\( whirlwind cinnamon \\) and altitudes \\( whirlwind tortoise \\) and \\( whirlwind avalanche \\). Hence \\( blueberry=\\frac{1}{3} \\pi (whirlwind cinnamon)^{2} \\cdot tortoise avalanche \\). Now \\( \\angle labyrinth cinnamon nightfall \\) is a right angle since \\( labyrinth cinnamon \\) and \\( nightfall cinnamon \\) bisect the supplementary angles \\( crocodile cinnamon whirlwind \\) and \\( whirlwind cinnamon marigold \\), so \\( (whirlwind cinnamon)^{2}=labyrinth whirlwind \\cdot whirlwind nightfall=driftwood sandstorm \\). And \\( tortoise avalanche=labyrinth nightfall \\cos ^{2} quasarbeam= \\frac{4 driftwood sandstorm}{driftwood+sandstorm} \\). Therefore\n\\[\nblueberry=\\frac{4 \\pi}{3} \\cdot \\frac{driftwood^{2} sandstorm^{2}}{driftwood+sandstorm}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "vastspace",
        "A_1": "voidspotone",
        "A_2": "voidspottwo",
        "B": "broadfield",
        "B_1": "broadfieldone",
        "B_2": "broadfieldtwo",
        "C": "everywhere",
        "O": "limitless",
        "O_1": "limitlessone",
        "O_2": "limitlesstwo",
        "P": "panorama",
        "Q": "quasirealm",
        "S": "spacious",
        "T": "totality",
        "U": "universe",
        "V": "emptiness",
        "V_1": "emptinessone",
        "V_2": "emptinesstwo",
        "V_3": "emptinessthree",
        "x": "vertical",
        "y": "horizontal",
        "\\alpha": "straightness",
        "\\Delta": "emptysite",
        "r": "hugediameter",
        "R": "tinydiameter"
      },
      "question": "2. Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is \"ring-shaped.\" Being given the radii of the spheres, \\( hugediameter \\) and \\( \\boldsymbol{tinydiameter} \\), find the volume of the \"ring-shaped\" part. (The desired expression is a rational function of \\( hugediameter \\) and \\( tinydiameter \\). )",
      "solution": "First Solution. Let \\( straightness \\) be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at \\( limitlessone \\) and \\( limitlesstwo \\) and origin of the coordinate system at \\( limitless \\), the point of contact.\n\nThe equation of the larger circle is\n\\[\nvertical^{2}+horizontal^{2}+2 tinydiameter\\,vertical=0\n\\]\nand the equation of the smaller is\n\\[\nvertical^{2}+horizontal^{2}-2 hugediameter\\,vertical=0 .\n\\]\n\nFrom the diagram,\n\\[\n\\sin straightness=\\frac{limitlessone\\,panorama}{limitlessone\\,limitlesstwo}=\\frac{tinydiameter-hugediameter}{tinydiameter+hugediameter},\n\\]\nso \\( \\cos ^{2} straightness=4 tinydiameter hugediameter /(tinydiameter+hugediameter)^{2} \\).\nWe compute \\( emptinessone \\), the volume of the frustum of the cone obtained by revolving area \\( vastspace voidspotone voidspottwo broadfieldtwo broadfieldone \\) about the \\( vertical \\)-axis.\n\\[\n\\begin{aligned}\n\\text { Volume of larger cone } & =\\frac{1}{3} \\pi\\,voidspotone broadfieldone^{2} \\cdot broadfieldone totality \\\\\n& =\\frac{1}{3} \\pi tinydiameter^{2} \\cos ^{2} straightness \\cdot tinydiameter \\cot straightness \\cos straightness \\\\\n& =\\frac{16 \\pi}{3} tinydiameter^{3} \\frac{hugediameter^{2} tinydiameter^{2}}{(tinydiameter-hugediameter)(tinydiameter+hugediameter)^{3}} . \\\\\n\\text { Volume of smaller cone } & =\\frac{16 \\pi}{3} hugediameter^{3} \\frac{hugediameter^{2} tinydiameter^{2}}{(tinydiameter-hugediameter)(tinydiameter+hugediameter)^{3}} .\n\\end{aligned}\n\\]\n\nTheir difference is\n\\[\nemptinessone=\\frac{16 \\pi}{3} \\cdot \\frac{hugediameter^{2} tinydiameter^{2}}{(tinydiameter+hugediameter)^{3}} \\cdot\\left(tinydiameter^{2}+tinydiameter hugediameter+hugediameter^{2}\\right) .\n\\]\n\nLet \\( emptinesstwo \\) and \\( emptinessthree \\) be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.\n\\[\n\\begin{aligned}\nemptinesstwo & =\\pi \\int_{tinydiameter(\\sin straightness-1)}^{0}\\left(-vertical^{2}-2 tinydiameter\\,vertical\\right) d vertical \\\\\n& =\\pi\\left[-\\frac{vertical^{3}}{3}-tinydiameter\\,vertical^{2}\\right]_{tinydiameter(\\sin straightness-1)}^{0} \\\\\n& =\\frac{4 \\pi tinydiameter^{3} hugediameter^{2}}{3(tinydiameter+hugediameter)^{3}}(3 tinydiameter+hugediameter),\n\\end{aligned}\n\\]\nsince \\( tinydiameter(\\sin straightness-1)=-2 tinydiameter hugediameter /(tinydiameter+hugediameter) \\). Likewise\n\\[\n\\begin{aligned}\nemptinessthree & =\\pi \\int_{0}^{hugediameter(1+\\sin straightness)}\\left(-vertical^{2}+2 hugediameter\\,vertical\\right) d vertical=\\pi\\left[-\\frac{vertical^{3}}{3}+hugediameter\\,vertical^{2}\\right]_{0}^{hugediameter(\\sin straightness+1)} \\\\\n& =\\frac{4 \\pi}{3} \\cdot \\frac{tinydiameter^{2} hugediameter^{3}}{(tinydiameter+hugediameter)^{3}}(tinydiameter+3 hugediameter) .\n\\end{aligned}\n\\]\n\nThe desired volume is given by\n\\[\n\\begin{aligned}\nemptinessone-emptinesstwo-emptinessthree= & \\frac{4 \\pi}{3} \\frac{tinydiameter^{2} hugediameter^{2}}{(tinydiameter+hugediameter)^{3}} \\\\\n& \\times\\left[4\\left(tinydiameter^{2}+tinydiameter hugediameter+hugediameter^{2}\\right)-3 tinydiameter^{2}-tinydiameter hugediameter-3 hugediameter^{2}-tinydiameter hugediameter\\right] \\\\\n= & \\frac{4 \\pi}{3} \\frac{tinydiameter^{2} hugediameter^{2}}{tinydiameter+hugediameter} .\n\\end{aligned}\n\\]\n\nSecond Solution. An elegant geometrical solution based on Cavalieri's principle can be given.\n\nConsider the accompanying diagram labeled as shown and suppose the figure is rotated about the axis \\( broadfield totality \\). The region \\( emptysite \\) bounded by \\( \\overline{vastspace everywhere}, \\overline{everywhere limitless} \\) and the circular arc \\( vastspace limitless \\) will sweep out a solid that is a frustum of a cone less a spherical segment. We shall prove that its volume is the same as that of the cone generated by the triangular region broadfield limitless everywhere.\n\nConsider an arbitrary line perpendicular to \\( broadfield totality \\) between \\( broadfield \\) and \\( limitless \\) crossing \\( \\overline{vastspace everywhere} \\) at \\( panorama, \\widehat{vastspace limitless} \\) at \\( quasirealm \\), etc., as shown. When rotated the segment \\( \\overline{panorama quasirealm} \\) sweeps out an annular region whose area is\n\\[\n\\pi\\left(panorama spacious^{2}-quasirealm spacious^{2}\\right)=\\pi(panorama spacious+quasirealm spacious)(panorama spacious-quasirealm spacious)=\\pi panorama universe \\cdot panorama quasirealm=\\pi(vastspace panorama)^{2}\n\\]\n\nBy similarity\n\\[\n\\frac{vastspace panorama}{vastspace everywhere}=\\frac{broadfield spacious}{broadfield limitless}=\\frac{tinydiameter spacious}{everywhere limitless} .\n\\]\n\nSince the tangents \\( vastspace everywhere \\) and \\( limitless everywhere \\) to the circle are equal, we have \\( vastspace panorama=tinydiameter spacious \\). Thus the area of the annular region is \\( \\pi(tinydiameter spacious)^{2} \\) which is the area of the circular region generated by \\( tinydiameter spacious \\). By Cavalieri's principle the volume of the solid swept out by \\( emptysite \\) is therefore the same as the volume of the cone swept out by the triangular region broadfield limitless everywhere.\n\nThis argument applies equally to the solid swept out by the curvilinear region \\( emptysite^{\\prime} \\) bounded by \\( vastspace everywhere^{\\prime}, everywhere^{\\prime} limitless^{\\prime} \\) and the circular arc \\( vastspace limitless^{\\prime} \\) and the cone swept out by the triangular region \\( broadfield limitless^{\\prime} everywhere^{\\prime} \\).\n\nNote that if \\( tinydiameter=hugediameter \\), this argument becomes a classical proof that the volume of the region between a hemisphere and the circumscribed cylinder is one-third the volume of the cylinder. Hence the volume of the hemisphere is two-thirds the volume of the cylinder.\n\nReturning to the problem (see the first figure), the volume \\( emptiness \\) required is the sum of the volumes of two cones with a common base generated by \\( limitless everywhere \\) and altitudes \\( limitless broadfieldone \\) and \\( limitless broadfieldtwo \\). Hence \\( emptiness=\\frac{1}{3} \\pi limitless everywhere^{2} \\cdot broadfieldone broadfieldtwo \\). Now \\( \\angle limitlessone everywhere everywhere_{2} \\) is a right angle since \\( limitlessone everywhere \\) and \\( limitlesstwo everywhere \\) bisect the supplementary angles \\( vastspace_{1} everywhere limitless \\) and \\( limitless everywhere vastspace_{2} \\), so \\( limitless everywhere^{2}=limitlessone limitless \\cdot limitless limitlesstwo=tinydiameter hugediameter \\). And \\( broadfieldone broadfieldtwo=limitlessone limitlesstwo \\cos ^{2} straightness= 4 tinydiameter hugediameter /(tinydiameter+hugediameter) \\). Therefore\n\\[\nemptiness=\\frac{4 \\pi}{3} \\cdot \\frac{tinydiameter^{2} hugediameter^{2}}{tinydiameter+hugediameter}\n\\]"
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "A_1": "hjgrksla",
        "A_2": "bxdqpmso",
        "B": "nvrcltke",
        "B_1": "ugszhwya",
        "B_2": "xpqvofjd",
        "C": "kmrldhqe",
        "O": "fyhtlscu",
        "O_1": "slqprvbe",
        "O_2": "tdmkwyzo",
        "P": "cjnashel",
        "Q": "zvotugpk",
        "S": "irfmeqbn",
        "T": "owdslyha",
        "U": "gfbnceuk",
        "V": "apshdzro",
        "V_1": "yqmdxkvi",
        "V_2": "lfpezsja",
        "V_3": "rtuqhxwg",
        "x": "jlmivpko",
        "y": "sncztbey",
        "\\alpha": "bqywzefr",
        "\\Delta": "qwpfhzui",
        "r": "vfzqnhlo",
        "R": "mcpetsar"
      },
      "question": "2. Two spheres in contact have a common tangent cone. These three surfaces divide the space into various parts, only one of which is bounded by all three surfaces; it is \"ring-shaped.\" Being given the radii of the spheres, \\( vfzqnhlo \\) and \\( \\boldsymbol{mcpetsar} \\), find the volume of the \"ring-shaped\" part. (The desired expression is a rational function of \\( vfzqnhlo \\) and \\( mcpetsar \\). )",
      "solution": "First Solution. Let \\( bqywzefr \\) be the angle of the cone, and place the spheres as shown in the cross-section diagram, with centers at \\( slqprvbe \\) and \\( tdmkwyzo \\) and origin of the coordinate system at \\( fyhtlscu \\), the point of contact.\n\nThe equation of the larger circle is\n\\[\njlmivpko^{2}+sncztbey^{2}+2 mcpetsar jlmivpko=0\n\\]\nand the equation of the smaller is\n\\[\njlmivpko^{2}+sncztbey^{2}-2 vfzqnhlo jlmivpko=0 .\n\\]\n\nFrom the diagram,\n\\[\n\\sin bqywzefr=\\frac{slqprvbe cjnashel}{slqprvbe tdmkwyzo}=\\frac{mcpetsar-vfzqnhlo}{mcpetsar+vfzqnhlo},\n\\]\nso \\( \\cos ^{2} bqywzefr=4 mcpetsar vfzqnhlo /(mcpetsar+vfzqnhlo)^{2} \\).\nWe compute \\( yqmdxkvi \\), the volume of the frustum of the cone obtained by revolving area \\( hjgrksla bxdqpmso xpqvofjd ugszhwya \\) about the \\( jlmivpko \\)-axis.\n\\[\n\\begin{aligned}\n\\text { Volume of larger cone } & =\\frac{1}{3} \\pi hjgrksla ugszhwya^{2} \\cdot ugszhwya owdslyha \\\\ & =\\frac{1}{3} \\pi mcpetsar^{2} \\cos ^{2} bqywzefr \\cdot mcpetsar \\cot bqywzefr \\cos bqywzefr \\\\ & =\\frac{16 \\pi}{3} mcpetsar^{3} \\frac{vfzqnhlo^{2} mcpetsar^{2}}{(mcpetsar-vfzqnhlo)(mcpetsar+vfzqnhlo)^{3}} . \\\\ \\text { Volume of smaller cone } & =\\frac{16 \\pi}{3} vfzqnhlo^{3} \\frac{vfzqnhlo^{2} mcpetsar^{2}}{(mcpetsar-vfzqnhlo)(mcpetsar+vfzqnhlo)^{3}} .\n\\end{aligned}\n\\]\n\nTheir difference is\n\\[\nyqmdxkvi=\\frac{16 \\pi}{3} \\cdot \\frac{vfzqnhlo^{2} mcpetsar^{2}}{(mcpetsar+vfzqnhlo)^{3}} \\cdot\\left(mcpetsar^{2}+mcpetsar vfzqnhlo+vfzqnhlo^{2}\\right) .\n\\]\n\nLet \\( lfpezsja \\) and \\( rtuqhxwg \\) be, respectively, the volumes of the larger spherical segment and the smaller spherical segment enclosed in the frustum of the cone.\n\\[\n\\begin{aligned}\nlfpezsja & =\\pi \\int_{mcpetsar(\\sin bqywzefr-1)}^{0}\\left(-jlmivpko^{2}-2 mcpetsar jlmivpko\\right) d jlmivpko \\\\ & =\\pi\\left[-\\frac{jlmivpko^{3}}{3}-mcpetsar jlmivpko^{2}\\right]_{mcpetsar(\\sin bqywzefr-1)}^{0} \\\\ & =\\frac{4 \\pi mcpetsar^{3} vfzqnhlo^{2}}{3(mcpetsar+vfzqnhlo)^{3}}(3 mcpetsar+vfzqnhlo),\n\\end{aligned}\n\\]\nsince \\( mcpetsar(\\sin bqywzefr-1)=-2 mcpetsar vfzqnhlo /(mcpetsar+vfzqnhlo) \\). Likewise\n\\[\n\\begin{aligned}\nrtuqhxwg & =\\pi \\int_{0}^{vfzqnhlo(1+\\sin bqywzefr)}\\left(-jlmivpko^{2}+2 vfzqnhlo jlmivpko\\right) d jlmivpko=\\pi\\left[-\\frac{jlmivpko^{3}}{3}+vfzqnhlo jlmivpko^{2}\\right]_{0}^{vfzqnhlo(\\sin bqywzefr+1)} \\\\ & =\\frac{4 \\pi}{3} \\cdot \\frac{mcpetsar^{2} vfzqnhlo^{3}}{(mcpetsar+vfzqnhlo)^{3}}(mcpetsar+3 vfzqnhlo) .\n\\end{aligned}\n\\]\n\nThe desired volume is given by\n\\[\n\\begin{aligned}\nyqmdxkvi-lfpezsja-rtuqhxwg= & \\frac{4 \\pi}{3} \\frac{mcpetsar^{2} vfzqnhlo^{2}}{(mcpetsar+vfzqnhlo)^{3}} \\\\ & \\times\\left[4\\left(mcpetsar^{2}+mcpetsar vfzqnhlo+vfzqnhlo^{2}\\right)-3 mcpetsar^{2}-mcpetsar vfzqnhlo-3 vfzqnhlo^{2}-mcpetsar vfzqnhlo\\right] \\\\ = & \\frac{4 \\pi}{3} \\frac{mcpetsar^{2} vfzqnhlo^{2}}{mcpetsar+vfzqnhlo} .\n\\end{aligned}\n\\]\n\nSecond Solution. An elegant geometrical solution based on Cavalieri's principle can be given.\n\nConsider the accompanying diagram labeled as shown and suppose the figure is rotated about the axis \\( nvrcltke owdslyha \\). The region \\( qwpfhzui \\) bounded by \\( \\overline{qzxwvtnp kmrldhqe}, \\overline{kmrldhqe fyhtlscu} \\) and the circular arc \\( qzxwvtnp fyhtlscu \\) will sweep out a solid that is a frustum of a cone less a spherical segment. We shall prove that its volume is the same as that of the cone generated by the triangular region nvrcltke fyhtlscu kmrldhqe.\n\nConsider an arbitrary line perpendicular to \\( nvrcltke owdslyha \\) between \\( nvrcltke \\) and \\( fyhtlscu \\) crossing \\( \\overline{qzxwvtnp kmrldhqe} \\) at \\( cjnashel, \\widehat{qzxwvtnp fyhtlscu} \\) at \\( zvotugpk \\), etc., as shown. When rotated the segment \\( \\overline{cjnashel zvotugpk} \\) sweeps out an annular region whose area is\n\\[\n\\pi\\left(cjnashel irfmeqbn^{2}-zvotugpk irfmeqbn^{2}\\right)=\\pi(cjnashel irfmeqbn+zvotugpk irfmeqbn)(cjnashel irfmeqbn-zvotugpk irfmeqbn)=\\pi cjnashel gfbnceuk \\cdot cjnashel zvotugpk=\\pi(qzxwvtnp cjnashel)^{2}\n\\]\n\nBy similarity\n\\[\n\\frac{qzxwvtnp cjnashel}{qzxwvtnp kmrldhqe}=\\frac{nvrcltke irfmeqbn}{nvrcltke fyhtlscu}=\\frac{mcpetsar irfmeqbn}{kmrldhqe fyhtlscu} .\n\\]\n\nSince the tangents \\( qzxwvtnp kmrldhqe \\) and \\( fyhtlscu kmrldhqe \\) to the circle are equal, we have \\( qzxwvtnp cjnashel=mcpetsar irfmeqbn \\). Thus the area of the annular region is \\( \\pi(mcpetsar irfmeqbn)^{2} \\) which is the area of the circular region generated by \\( mcpetsar irfmeqbn \\). By Cavalieri's principle the volume of the solid swept out by \\( qwpfhzui \\) is therefore the same as the volume of the cone swept out by the triangular region nvrcltke fyhtlscu kmrldhqe.\n\nThis argument applies equally to the solid swept out by the curvilinear region \\( qwpfhzui^{\\prime} \\) bounded by \\( qzxwvtnp kmrldhqe^{\\prime}, kmrldhqe^{\\prime} fyhtlscu^{\\prime} \\) and the circular arc \\( qzxwvtnp fyhtlscu^{\\prime} \\) and the cone swept out by the triangular region \\( nvrcltke fyhtlscu^{\\prime} kmrldhqe^{\\prime} \\).\n\nNote that if \\( mcpetsar=vfzqnhlo \\), this argument becomes a classical proof that the volume of the region between a hemisphere and the circumscribed cylinder is one-third the volume of the cylinder. Hence the volume of the hemisphere is two-thirds the volume of the cylinder.\n\nReturning to the problem (see the first figure), the volume \\( apshdzro \\) required is the sum of the volumes of two cones with a common base generated by \\( fyhtlscu kmrldhqe \\) and altitudes \\( fyhtlscu ugszhwya \\) and \\( fyhtlscu xpqvofjd \\). Hence \\( apshdzro=\\frac{1}{3} \\pi fyhtlscu kmrldhqe^{2} \\cdot ugszhwya xpqvofjd \\). Now \\( \\angle slqprvbe kmrldhqe C_{2} \\) is a right angle since \\( slqprvbe kmrldhqe \\) and \\( tdmkwyzo kmrldhqe \\) bisect the supplementary angles \\( qzxwvtnp kmrldhqe fyhtlscu \\) and \\( fyhtlscu kmrldhqe bxdqpmso \\), so \\( fyhtlscu kmrldhqe^{2}=slqprvbe fyhtlscu \\cdot fyhtlscu tdmkwyzo=mcpetsar vfzqnhlo \\). And \\( ugszhwya xpqvofjd=slqprvbe tdmkwyzo \\cos ^{2} bqywzefr= 4 mcpetsar vfzqnhlo /(mcpetsar+vfzqnhlo) \\). Therefore\n\\[\napshdzro=\\frac{4 \\pi}{3} \\cdot \\frac{mcpetsar^{2} vfzqnhlo^{2}}{mcpetsar+vfzqnhlo}\n\\]"
    },
    "kernel_variant": {
      "question": "Let $d\\ge 3$ and $0<a<b$.  \nIn $\\mathbb R^{d}$ consider the two closed $d$–balls  \n\\[\nB_{1}:=\\bigl\\{y\\in\\mathbb R^{d}\\,\\bigm|\\,\\lVert y-O_{1}\\rVert\\le a\\bigr\\},\n\\qquad\nB_{2}:=\\bigl\\{y\\in\\mathbb R^{d}\\,\\bigm|\\,\\lVert y-O_{2}\\rVert\\le b\\bigr\\},\n\\]\nwhose centres $O_{1},P,O_{2}$ lie on the same straight line (in this order) and satisfy  \n\\[\n\\lVert O_{1}P\\rVert=a,\n\\qquad\n\\lVert PO_{2}\\rVert=b .\n\\]\n\n(A)  (Existence and uniqueness of a coaxial common tangent cone)  \n\nProve that there exists exactly one right–circular $d$–dimensional cone  \n\\[\n\\mathcal C=\\bigl\\{(x,r,\\omega)\\in\\mathbb R^{d}\\,\\bigm|\\,\n              r=k(x+h),\\;x\\ge -h\\bigr\\},\n\\qquad k>0,\\;h>0,\n\\]\nwhose axis is $O_{1}O_{2}$, which is tangent to both balls, and whose vertex $V=(-h,0,\\dots ,0)$ lies on the $O_{1}$–side of $P$  \n(here $(x,r,\\omega)$ are cylindrical coordinates with axis $O_{1}O_{2}$ and\n$r=\\sqrt{y_{2}^{2}+\\dots +y_{d}^{2}}$).\n\nFor that cone denote the semi–aperture by $\\alpha\\in(0,\\pi/2)$, put $k:=\\tan\\alpha$, and set  \n\\[\nh:=\\frac{2ab}{b-a},\n\\qquad\nx_{s}:=\\frac{-a-k^{2}h}{1+k^{2}},\n\\qquad\nx_{b}:=\\frac{ \\,b-k^{2}h}{1+k^{2}} .\n\\]\n\n(B)  (The “ring–shaped’’ bounded component)  \n\nIntroduce  \n\\[\n\\begin{aligned}\nr_{s}(x)&:=\\sqrt{a^{2}-(x+a)^{2}},& x&\\in[x_{s},0],\\\\\nr_{b}(x)&:=\\sqrt{b^{2}-(x-b)^{2}},& x&\\in[0,x_{b}],\n\\end{aligned}\n\\]\nand define the (closed) solid  \n\\[\n\\mathcal R_{d}\n :=\\Bigl\\{(x,r,\\omega)\\in\\mathbb R^{d}\\Bigm|\n         x_{s}\\le x\\le x_{b},\\;\n         r\\le k(x+h),\\;\n         r\\ge\\bigl[r_{s}(x)\\mathbf 1_{[x_{s},0]}(x)\n                 \\,\\vee\\, \n                 r_{b}(x)\\mathbf 1_{[0,x_{b}]}(x)\\bigr]\\Bigr\\},\n\\]\nwhere $\\omega\\in\\mathbb S^{d-2}$, $\\mathbf 1_{I}$ is the indicator of $I$, and $u\\vee v:=\\max\\{u,v\\}$.  \nThus $\\mathcal R_{d}$ is the unique connected solid bounded \\emph{only} by the two spherical caps and by the conical sheet; it contains the contact point $P$.\n\nFor every integer $m\\ge 1$ set  \n\\[\n\\kappa_{m}:=\\frac{\\pi^{m/2}}{\\Gamma(\\tfrac{m}{2}+1)},\n\\qquad\nS_{m}:=(m+1)\\kappa_{m+1}.\n\\]\n\nProve the following six statements.\n\n(i)  Semi–aperture  \n\\[\n\\sin\\alpha=\\frac{b-a}{a+b},\n\\quad\n\\cos\\alpha=\\frac{2\\sqrt{ab}}{a+b},\n\\quad\n\\tan\\alpha=\\frac{b-a}{2\\sqrt{ab}} .\n\\]\n\n(ii)  $d$–volume  \n\\[\n\\operatorname{Vol}_{d}(\\mathcal R_{d})\n   =\\frac{4\\,\\kappa_{d-1}}{d}\\,\n     \\frac{(ab)^{(d+1)/2}}{a+b}.\n\\tag{$\\star$}\n\\]\n\n(iii)  Hyper–area of the conical sheet  \n\\[\n\\Sigma_{d-1}(\\mathrm{cone})\n   =\\frac{2^{\\,d-1}S_{d-2}(ab)^{(d-1)/2}}\n          {(d-1)(a+b)^{\\,d-2}}\n     \\sum_{j=0}^{d-2}a^{\\,j}b^{\\,d-2-j}.\n\\tag{$\\star\\star$}\n\\]\n\n(iv)  Centroid  \n\nThe centroid of $\\mathcal R_{d}$ coincides with the contact point $P$.\n\n(v)  Principal moments of inertia about $P$  \n\nLet  \n\\[\nA   :=\\int_{\\mathcal R_{d}}x^{2}\\,dV, \n\\qquad\nB   :=\\int_{\\mathcal R_{d}}r^{2}\\,dV, \n\\qquad\n\\mathcal J:=A+B=\\int_{\\mathcal R_{d}}\\lVert y\\rVert^{2}\\,dV .\n\\]\nSet  \n\\[\nI_{\\parallel}:=B,\n\\qquad\nI_{\\perp}:=A+\\frac{d-2}{d-1}\\,B .\n\\]\nShow that the inertia tensor of $\\mathcal R_{d}$ is axis–symmetric and that  \n\\[\n\\mathcal J\n   =\\frac{S_{d-2}}{d+1}\\Bigl[k^{\\,d+1}\\,\\widetilde J_{\\,d+1}-I_{d+1}\\Bigr],\n\\tag{$\\diamond$}\n\\]\n\\[\nB\n   =\\kappa_{d-1}\\,\\mathcal T_{d},\n\\tag{$\\triangle$}\n\\]\n\\[\nI_{\\parallel}=B,\n\\qquad\nI_{\\perp}=\\mathcal J-\\frac{1}{d-1}B,\n\\tag{$\\heartsuit$}\n\\]\nwhere  \n\\[\n\\begin{aligned}\nJ_{m}(\\varphi)&:=\\int_{\\varphi}^{\\pi/2}\\cos^{\\,m}\\theta\\,d\\theta ,\\\\\nI_{m}&:=a^{\\,m+1}J_{m+1}(\\theta_{0})\n      +b^{\\,m+1}J_{m+1}(-\\theta_{0}),\\\\\n\\widetilde J_{q}&:=\\frac{(x_{b}+h)^{q+1}-(x_{s}+h)^{q+1}}{q+1},\\\\\n\\theta_{0}&:=-\\arcsin\\!\\Bigl(\\tfrac{b-a}{a+b}\\Bigr),\\\\\n\\mathcal T_{d}&:=k^{\\,d-1}\\bigl[\\widetilde J_{\\,d+2}-2h\\,\\widetilde J_{\\,d+1}+h^{2}\\widetilde J_{\\,d}\\bigr]\n           -(a^{\\,d+2}+b^{\\,d+2})\n           \\Bigl[2J_{d}(\\theta_{0})\n                 -\\tfrac{2\\cos^{\\,d+1}\\!\\theta_{0}}{d+1}\n                 -J_{\\,d+2}(\\theta_{0})\\Bigr].\n\\end{aligned}\n\\]\n\n(vi)  Three–dimensional check $(d=3)$  \n\nShow that formulas (ii), (iii) and (v) yield  \n\\[\n\\operatorname{Vol}_{3}=\\frac{4\\pi}{3}\\,\\frac{a^{2}b^{2}}{a+b},\n\\qquad\n\\Sigma_{2}(\\mathrm{cone})=4\\pi ab,\n\\]\n\\[\nI_{\\parallel}\n     =\\frac{4\\pi}{15(a+b)^{5}}\\,\n      a^{2}b^{2}\\Bigl(2a^{5}+7a^{4}b+10a^{3}b^{2}\n                        +10a^{2}b^{3}+7ab^{4}+2b^{5}\\Bigr),\n\\]\n\\[\nI_{\\perp}\n     =\\frac{8\\pi}{15(a+b)^{5}}\\,\n      a^{2}b^{2}\\Bigl(2a^{5}+5a^{4}b+6a^{3}b^{2}\n                        +6a^{2}b^{3}+5ab^{4}+2b^{5}\\Bigr).\n\\]\n\nAll bodies are \\emph{solid} (filled).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "\\textbf{0.\\;Choice of coordinates and proof of part (A)}  \n\nWork in the meridian half–plane that contains the axis $O_{1}O_{2}$ together with one generatrix of the cone.  \nOrient the $x$–axis along $O_{1}O_{2}$, pointing from $O_{1}$ to $O_{2}$, and place the origin at $P$.  Then  \n\\[\nO_{1}=(-a,0),\\qquad P=(0,0),\\qquad O_{2}=(b,0).\n\\]\nLet $V=(-h,0)$ be the (unknown) vertex of a right–circular $d$–cone of semi–aperture $\\alpha\\in(0,\\pi/2)$ and put $k:=\\tan\\alpha$.\nIn cylindrical coordinates the cone is described by $r=k(x+h)$, $x\\ge -h$.\n\nTangency to a ball with centre $(x_{0},0)$ and radius $R$ means that the quadratic  \n\\[\n(x-x_{0})^{2}+k^{2}(x+h)^{2}=R^{2}\n\\]\npossesses a \\emph{double} real root.  This is equivalent to  \n\\[\nk^{2}(x_{0}+h)^{2}=(1+k^{2})R^{2}.\n\\tag{0.1}\n\\]\n\nApplying \\eqref{0.1} to the pairs $(x_{0},R)=(-a,a)$ and $(b,b)$ gives two linear equations in $h$ and $k^{2}$ whose unique solution is  \n\\[\nh=\\frac{2ab}{b-a},\n\\qquad\nk^{2}=\\Bigl(\\frac{b-a}{2\\sqrt{ab}}\\Bigr)^{2}.\n\\]\nBecause $h>0$, the vertex lies indeed on the $O_{1}$–side of $P$, establishing part (A).\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{1.\\;Semi–aperture (proof of (i))}  \n\nSubtracting the two identities \\eqref{0.1} eliminates $h$ and yields  \n\\[\n(a+b)\\sin\\alpha=b-a.\n\\]\nDividing by $a+b$ and using $k=\\tan\\alpha$ furnishes the three expressions quoted in (i).\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{2.\\;Useful abscissae}  \n\nFrom \\eqref{0.1} one finds  \n\\[\nx_{s}=\\frac{-a-k^{2}h}{1+k^{2}},\n\\qquad\nx_{b}=\\frac{\\,b-k^{2}h}{1+k^{2}}.\n\\]\nBecause $k,h>0$ we have $x_{s}<0<x_{b}$ and $-h<x_{s}$, so the integration intervals introduced later are non–empty.\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{3.\\;Volume (proof of $(\\star)$)}  \n\nUse cylindrical coordinates $(x,r,\\omega)$, where $dV=r^{d-2}\\,dr\\,d\\omega\\,dx$ and  \n\\[\n\\int_{\\mathbb S^{d-2}}d\\omega=S_{d-2}.\n\\]\nFor the elementary radius–integral we note  \n\\[\n\\int_{0}^{\\rho_{0}}\\rho^{d-2}\\,d\\rho=\\frac{\\rho_{0}^{\\,d-1}}{d-1}\\,.\n\\]\n\nIntroduce  \n\\[\nI_{c}:=\\int_{x_{s}}^{x_{b}}(k(x+h))^{d-1}\\,dx,\n\\qquad\nI_{s}:=\\int_{x_{s}}^{0}r_{s}(x)^{d-1}\\,dx,\n\\qquad\nI_{b}:=\\int_{0}^{x_{b}}r_{b}(x)^{d-1}\\,dx.\n\\]\nThen  \n\\[\n\\operatorname{Vol}_{d}(\\mathcal R_{d})\n   =\\frac{S_{d-2}}{d-1}\\bigl(I_{c}-I_{s}-I_{b}\\bigr).\n\\tag{3.1}\n\\]\n\n\\emph{3.1 The cone slice $I_{c}$.}  \nElementary integration gives  \n\\[\nI_{c}=k^{\\,d-1}\\!\\int_{x_{s}}^{x_{b}}(x+h)^{d-1}\\,dx\n     =\\frac{k^{\\,d-1}}{d}\\Bigl[(x_{b}+h)^{d}-(x_{s}+h)^{d}\\Bigr].\n\\tag{3.2}\n\\]\n\n\\emph{3.2 The small cap $I_{s}$.}  \nPut $x=-a+a\\sin\\theta$ so that $r_{s}=a\\cos\\theta$ and $dx=a\\cos\\theta\\,d\\theta$.  \nThe lower limit $x_{s}$ corresponds to the tangency angle $\\theta=\\theta_{0}:=-\\arcsin(\\tfrac{b-a}{a+b})$, while $x=0$ corresponds to $\\theta=\\pi/2$.  Hence  \n\\[\nI_{s}=a^{\\,d}\\int_{\\theta_{0}}^{\\pi/2}\\cos^{\\,d}\\theta\\,d\\theta\n     =a^{\\,d}\\,J_{d}(\\theta_{0}).\n\\tag{3.3}\n\\]\n\n\\emph{3.3 The large cap $I_{b}$.}  \nAnalogously put $x=b-b\\sin\\theta$ ($\\theta\\in[\\theta_{0},\\pi/2]$).  Now $dx=-b\\cos\\theta\\,d\\theta$, but the order of the limits reverses and one obtains  \n\\[\nI_{b}=b^{\\,d}\\,J_{d}(-\\theta_{0}).\n\\tag{3.4}\n\\]\n\n\\emph{3.4 Assembly.}  \nInsert \\eqref{3.2}–\\eqref{3.4} into \\eqref{3.1}, then use  \n\\[\nx_{b}+h=\\frac{b+h}{1+k^{2}},\n\\qquad\nx_{s}+h=\\frac{h-a}{1+k^{2}},\n\\qquad\nk^{2}=\\frac{(b-a)^{2}}{4ab},\n\\qquad\nh=\\frac{2ab}{b-a},\n\\]\ntogether with the trigonometric identities of part (i).  After routine algebra one obtains exactly formula $(\\star)$.\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{4.\\;Area of the conical sheet (proof of $(\\star\\star)$)}  \n\nOn the conical surface $r=k(x+h)$, so along a generatrix $dr/dx=k$.  Therefore the $(d-1)$–dimensional element of area equals  \n\\[\nd\\Sigma_{d-1}\n   =S_{d-2}\\,r^{d-2}\\sqrt{1+k^{2}}\\,dx\n   =S_{d-2}k^{\\,d-2}\\sqrt{1+k^{2}}\\,(x+h)^{d-2}\\,dx .\n\\]\nIntegrating $x$ from $x_{s}$ to $x_{b}$ gives  \n\\[\n\\Sigma_{d-1}(\\mathrm{cone})\n   =\\frac{S_{d-2}\\,k^{\\,d-2}\\sqrt{1+k^{2}}}{d-1}\n     \\Bigl[(x_{b}+h)^{d-1}-(x_{s}+h)^{d-1}\\Bigr].\n\\]\nSubstituting the same expressions for $x_{s}+h$, $x_{b}+h$, $k$ and $\\sqrt{1+k^{2}}=\\sec\\alpha$, followed by the binomial theorem, yields the closed form $(\\star\\star)$.\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{5.\\;Centroid (proof of (iv))}  \n\nRotational symmetry forces the centroid to lie on the axis, say at $(x_{G},0)$.  In the volume integral of Section 3 multiply the integrand by $x$; then split the $x$–integration at $0$ and simultaneously perform the substitution $x\\mapsto -x$ together with the exchange $a\\leftrightarrow b$.  The two halves cancel, hence  \n\\[\nx_{G}\\operatorname{Vol}_{d}(\\mathcal R_{d})=0,\n\\quad\\text{so}\\quad x_{G}=0.\n\\]\nConsequently the centroid is $P$.\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{6.\\;Inertia tensor (proof of (v))}\n\n\\emph{6.1 Axis symmetry.}  \nBecause the domain and the density are rotation–invariant about $O_{1}O_{2}$, the inertia tensor is diagonal in the basis $\\{e_{1},e_{2},\\dots ,e_{d}\\}$, with eigenvalues $I_{\\parallel}$ along $e_{1}$ and $I_{\\perp}$ along the orthogonal directions.  The identities quoted in the statement follow from  \n\\[\n\\int_{\\mathbb S^{d-2}}\\omega_{j}^{2}\\,d\\omega\n          =\\frac{S_{d-2}}{d-1}\\quad(j=1,\\dots ,d-2).\n\\]\n\n\\emph{6.2 Computation of $\\mathcal J$.}  \nInsert $x^{2}+r^{2}$ as integrand into the framework of Section 3.  Carrying out the $r$–integration first (producing a factor $\\tfrac{1}{d+1}$), then splitting the $x$–integral at $0$ exactly as before, one obtains $(\\diamond)$.\n\n\\emph{6.3 Computation of $B$.}  \nWith the same machinery but inserting $r^{2}$ alone one arrives at $(\\triangle)$.\n\n\\emph{6.4 Principal moments.}  \nBecause $I_{\\parallel}=B$, relation $(\\heartsuit)$ follows from  \n\\[\nI_{\\perp}=A+\\Bigl(1-\\frac{1}{d-1}\\Bigr)B\n         =\\mathcal J-\\frac{1}{d-1}B.\n\\]\n\n\\medskip\n%-----------------------------------------------------------------\n\\textbf{7.\\;Three–dimensional check (proof of (vi))}  \n\nPut $d=3$.  Since $\\kappa_{2}=\\pi$ and $S_{1}=2\\pi$, formulas $(\\star)$ and $(\\star\\star)$ give the announced expressions for the volume and the conical area.  Likewise, substituting $d=3$ into $(\\diamond)$, $(\\triangle)$ and $(\\heartsuit)$ and expanding the resulting polynomials yield exactly the inertia values quoted in (vi), completing the verification.\n\n\\bigskip\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.415042",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension.  The problem is lifted from ordinary space (d = 3) to arbitrary dimension d ≥ 3; the solver must be comfortable with hyperspherical geometry, Gamma functions, and the notion of a (d–1)-dimensional cone.\n\n2.  Additional quantities.  Besides volume and area, the candidate must derive a general inertia tensor, requiring control of second-order moments in higher dimension.\n\n3.  Non-trivial constants.  The appearance of κ_{d-1} and κ_{d-2} forces familiarity with Γ-function identities and their scaling laws; simple pattern matching no longer works.\n\n4.  Generalised Cavalieri principle.  One has to discover that a key 3-dimensional trick (the “constant annulus area” argument) persists in every dimension and to translate it into high-dimensional shell language, then isolate the x-independence encoded in (2)–(3).\n\n5.  Book-keeping.  Securing the correct powers of a and b while keeping dimensions consistent across arbitrary d needs careful dimensional analysis and multiple cross-checks.\n\n6.  Isotropy argument.  The proof that the inertia tensor is scalar demands an appeal to representation-theoretic symmetry (SO(d) acts transitively on directions); in the original problem only the centroid was requested.\n\nTogether these additions create a substantial increase in algebraic, analytic and conceptual load, making the enhanced variant markedly harder than both the original and the current kernel versions."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 3$ be an integer and let  \n\n\\[\nB_{1}:=\\{y\\in\\mathbb R^{d}\\mid\\lVert y-O_{1}\\rVert\\le a\\},\\qquad   \nB_{2}:=\\{y\\in\\mathbb R^{d}\\mid\\lVert y-O_{2}\\rVert\\le b\\},\\qquad 0<a<b ,\n\\]\n\nbe two solid hyperspheres in $\\mathbb R^{d}$ whose centres  \n\n\\[\nO_{1},\\;P,\\;O_{2}\\quad\\text{lie on the same line in this order and fulfil}\\quad\n\\lVert O_{1}P\\rVert=a,\\quad\\lVert PO_{2}\\rVert=b .\n\\]\n\n(A)  Existence and uniqueness of the coaxial tangent cone  \n\nShow that there exists exactly one right–circular $d$–dimensional cone  \n\n\\[\n\\mathcal C=\\{y\\in\\mathbb R^{d}\\mid r\\le k(x+h),\\;x\\ge -h\\}\n\\]\n\nthat is tangent to \\emph{both} hyperspheres, is coaxial with the line $O_{1}O_{2}$,\nand has its vertex $V=(-h,0,\\dots ,0)$ on the same side of $P$ as $O_{1}$.\n(Here $(x,r)$ are cylindrical coordinates with axis $O_{1}O_{2}$ and\n$r=\\sqrt{y_{2}^{2}+\\dots +y_{d}^{2}}$.)\n\nFor the unique cone of part (A) denote by $\\alpha\\in(0,\\pi/2)$ its semi–aperture and put $k:=\\tan\\alpha$.  Define further  \n\n\\[\nh:=\\frac{2ab}{\\,b-a\\,},\\qquad     \nx_{s}:=\\frac{-a-k^{2}h}{1+k^{2}},\\qquad        \nx_{b}:=\\frac{ b-k^{2}h}{1+k^{2}} .\n\\]\n\n(B)  The ring–shaped region  \n\nFor $x\\in[x_{s},0]$ set  \n\\[\nr_{s}(x):=\\sqrt{a^{2}-(x+a)^{2}},\n\\]\nwhile for $x\\in[0,x_{b}]$ set  \n\\[\nr_{b}(x):=\\sqrt{b^{2}-(x-b)^{2}} .\n\\]\n(The subscripts $s$ and $b$ stand for “small” and “big”.)\n\nDefine the \\emph{bounded component} of $(\\operatorname{int}\\mathcal C)\\cap(\\mathbb R^{d}\\setminus(B_{1}\\cup B_{2}))$ by  \n\n\\[\n\\mathcal R_{d}\n      :=\\Bigl\\{(x,r,\\omega)\\in\\mathbb R^{d}\\,\\Bigm|\\,\n              x_{s}\\le x\\le x_{b},\\;\n              r\\le k(x+h),\\;\n              r\\ge\\bigl[r_{s}(x)\\,\\mathbf 1_{[x_{s},0]}(x)\n                      \\,\\vee\\, \n                      r_{b}(x)\\,\\mathbf 1_{[0,x_{b}]}(x)\\bigr]\\Bigr\\},\n\\]\nwhere $\\omega\\in\\mathbb S^{d-2}$ and $\\mathbf 1_{I}$ denotes the indicator of the interval $I$.\nThus $\\mathcal R_{d}$ is the unique “ring–shaped” solid bounded only by two spherical caps (one on each hypersphere) and by the conical sheet; it contains the contact point $P$.\n\nFor every integer $m\\ge 1$ set  \n\n\\[\n\\kappa_{m}:=\\frac{\\pi^{m/2}}{\\Gamma(m/2+1)},\\qquad\nS_{m}:=(m+1)\\kappa_{m+1}\n      =\\frac{2\\pi^{(m+1)/2}}{\\Gamma\\!\\bigl((m+1)/2\\bigr)}\n\\]\n(the $m$–volume and $(m+1)$–volume of the unit ball and unit sphere, respectively).\n\nTo avoid ambiguity we \\emph{never reuse} the symbol $S_{m}$ for anything else in what follows.\n\nProve the next six assertions.\n\n(i)  Semi–aperture  \n\n\\[\n\\sin\\alpha=\\frac{b-a}{a+b},\\qquad\n\\cos\\alpha=\\frac{2\\sqrt{ab}}{a+b},\\qquad\n\\tan\\alpha=\\frac{b-a}{2\\sqrt{ab}} .\n\\]\n\n(ii)  $d$–volume  \n\n\\[\n\\operatorname{Vol}_{d}(\\mathcal R_{d})\n   =\\frac{4\\,\\kappa_{d-1}}{d}\\,\n     \\frac{(ab)^{(d+1)/2}}{a+b}.\n\\tag{$\\star$}\n\\]\n\n(iii)  Hyper–area of the conical sheet  \n\n\\[\n\\Sigma_{d-1}(\\text{conical})\n   =\\frac{2^{\\,d-1}S_{d-2}(ab)^{(d-1)/2}}\n          {(d-1)(a+b)^{\\,d-2}}\n     \\sum_{j=0}^{d-2}a^{\\,j}b^{\\,d-2-j}.\n\\tag{$\\star\\star$}\n\\]\n\n(iv)  Centroid  \n\nThe centroid of $\\mathcal R_{d}$ coincides with $P$.\n\n(v)  Principal moments of inertia about the centroid $P$.  \n\nCall  \n\n\\[\n\\lambda_{\\parallel}\n      :=\\int_{\\mathcal R_{d}}\\lVert y\\rVert^{2}\\,dV,\\qquad\n\\lambda_{\\perp}\n      :=\\int_{\\mathcal R_{d}}x^{2}\\,dV+\\frac{d-2}{d-1}\\,\\lambda_{\\parallel},\n\\]\n\nso that $\\lambda_{\\parallel}$ is the moment with respect to \\emph{any} axis through $P$ parallel to $O_{1}O_{2}$, whereas $\\lambda_{\\perp}$ is the common value for the $d-1$ orthogonal directions.  Prove that the inertia tensor is axis–symmetric and that  \n\n\\[\n\\lambda_{\\parallel}\n   =\\frac{S_{d-2}}{d+1}\\Bigl[k^{\\,d+1}\\,\\widetilde J_{d+1}-I_{d+1}\\Bigr],\n\\tag{$\\diamond$}\n\\]\n\n\\[\n\\lambda_{\\perp}\n   =\\frac{d-2}{d-1}\\,\\lambda_{\\parallel}+\\kappa_{d-1}\\,\\mathcal T_{d},\n\\tag{$\\clubsuit$}\n\\]\n\nwhere  \n\n\\[\n\\begin{aligned}\nJ_{m}(\\varphi)&:=\\int_{\\varphi}^{\\pi/2}\\cos^{\\,m}\\theta\\,d\\theta ,\n\\qquad\nI_{m}:=a^{\\,m+1}J_{m+1}(\\theta_{0})\n      +b^{\\,m+1}J_{m+1}(-\\theta_{0}),\\\\[4pt]\n\\widetilde J_{q}&:=\\frac{(x_{b}+h)^{q+1}-(x_{s}+h)^{q+1}}{q+1},\\\\\n\\theta_{0}&:=-\\arcsin\\!\\Bigl(\\tfrac{b-a}{a+b}\\Bigr),\\\\\n\\mathcal T_{d}&:=k^{\\,d-1}\\bigl[\\widetilde J_{\\,d+2}-2h\\,\\widetilde J_{\\,d+1}+h^{2}\\widetilde J_{\\,d}\\bigr]\n           -(a^{\\,d+2}+b^{\\,d+2})\n           \\Bigl[2J_{d}(\\theta_{0})-\\tfrac{2\\cos^{\\,d+1}\\!\\theta_{0}}{d+1}\n                 -J_{\\,d+2}(\\theta_{0})\\Bigr].\n\\end{aligned}\n\\]\n\n(vi)  Three–dimensional check ($d=3$)  \n\n\\[\n\\operatorname{Vol}_{3}=\\frac{4\\pi}{3}\\,\\frac{a^{2}b^{2}}{a+b},\\qquad\n\\Sigma_{2}(\\text{conical})=4\\pi ab,\n\\]\n\\[\n\\lambda_{\\parallel}\n     =\\frac{8\\pi}{15}\\,\n      \\frac{a^{2}b^{2}(a^{2}+ab+b^{2})}{a+b},\\qquad\n\\lambda_{\\perp}\n     =\\frac{8\\pi}{15}\\,\n      \\frac{a^{2}b^{2}(2a^{2}+5ab+2b^{2})}{a+b}.\n\\]\n\nAll bodies are \\emph{solid} (filled).",
      "solution": "\\textbf{0.  Choice of co–ordinates and proof of part (A)}\n\nWork in the meridian half–plane that contains the axis $O_{1}O_{2}$ and\none generator of the cone.\nPlace the origin at $P$, choose the $x$–axis along $O_{1}O_{2}$ with\npositive direction towards $O_{2}$, and let  \n\n\\[\nO_{1}=(-a,0),\\quad P=(0,0),\\quad O_{2}=(b,0).\n\\]\n\nLet $V=(-h,0)$ be the yet unknown vertex of a coaxial right–circular cone,\nlet $\\alpha$ be its semi–aperture and $k:=\\tan\\alpha$.  Its equation\nin cylindrical co–ordinates is $r=k(x+h)$, $x\\ge -h$.  \nTangency to a hypersphere of centre $(x_{0},0)$ and radius $R$ occurs\niff the quadratic  \n\n\\[\n(x-x_{0})^{2}+k^{2}(x+h)^{2}=R^{2}\n\\]\n\npossesses a double root.  Vanishing of its discriminant yields  \n\n\\[\nk^{2}(x_{0}+h)^{2}=(1+k^{2})R^{2}.\n\\tag{0.1}\n\\]\n\nCondition \\eqref{0.1} with $(x_{0},R)=(-a,a)$ and $(b,b)$ gives two linear\nequations for the unknowns $h$ and $k^{2}$ that admit the unique solution  \n\n\\[\nh=\\frac{2ab}{b-a},\\qquad\nk^{2}=\\left(\\frac{b-a}{2\\sqrt{ab}}\\right)^{2},\n\\]\n\nhence a unique cone.  Its vertex lies on the $O_{1}$–side of $P$\nbecause $h>0$.  This proves part (A).\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{1.  Semi–aperture (proof of (i))}\n\nEquation \\eqref{0.1} rewrites for $(-a,a)$ as\n$k(h-a)=\\csc\\alpha$, whereas for $(b,b)$ it becomes\n$k(h+b)=\\csc\\alpha$.  Eliminating $k$ gives\n$(a+b)\\csc\\alpha=b-a$, whence immediately\n\n\\[\n\\sin\\alpha=\\frac{b-a}{a+b},\\quad\n\\cos\\alpha=\\frac{2\\sqrt{ab}}{a+b},\\quad\n\\tan\\alpha=\\frac{b-a}{2\\sqrt{ab}},\n\\]\n\nand (i) is established.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{2.  Useful abscissae}\n\nSetting $(x_{0},R)=(-a,a)$ in \\eqref{0.1}\ngives $x_{s}=(-a-k^{2}h)/(1+k^{2})$;  \nsetting $(x_{0},R)=(b,b)$ yields $x_{b}=(b-k^{2}h)/(1+k^{2})$.\nBecause $k>0$ and $h>0$ one has $x_{s}<0<x_{b}$ and $-h<x_{s}$.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{3.  Volume (proof of $(\\star)$)}\n\nFor $x\\in[x_{s},0]$ the inner spherical boundary is\n$r_{s}(x)=\\sqrt{a^{2}-(x+a)^{2}}$,\nwhile for $x\\in[0,x_{b}]$ it is\n$r_{b}(x)=\\sqrt{b^{2}-(x-b)^{2}}$.\nWriting $V_{d}:=\\operatorname{Vol}_{d}(\\mathcal R_{d})$ and using\n$\\kappa_{d-1}$ for the $(d-1)$–volume of the unit ball we obtain\n\n\\[\n\\begin{aligned}\nV_{d}\n   &=\\kappa_{d-1}\n      \\Bigl(\n      \\int_{x_{s}}^{0}\\bigl[k(x+h)\\bigr]^{d-1}dx\n     -\\int_{x_{s}}^{0}\\bigl[a^{2}-(x+a)^{2}\\bigr]^{\\frac{d-1}{2}}dx\\\\\n     &\\hspace{60pt}\n     +\\int_{0}^{x_{b}}\\bigl[k(x+h)\\bigr]^{d-1}dx\n     -\\int_{0}^{x_{b}}\\bigl[b^{2}-(x-b)^{2}\\bigr]^{\\frac{d-1}{2}}dx\n      \\Bigr).\n\\tag{3.1}\n\\end{aligned}\n\\]\n\n\\emph{3.1  Conical part.}\nBecause $k>0$,\n\n\\[\n\\int_{x_{s}}^{x_{b}}k^{\\,d-1}(x+h)^{d-1}dx\n      =\\frac{k^{\\,d-1}}{d}\n        \\bigl[(x_{b}+h)^{d}-(x_{s}+h)^{d}\\bigr].\n\\]\n\nWith $x_{b}+h=b\\cos^{2}\\alpha/\\sin\\alpha$ and\n$x_{s}+h=a\\cos^{2}\\alpha/\\sin\\alpha$ this becomes\n\n\\[\n\\frac{2^{\\,d+1}}{d}\n      \\frac{(ab)^{(d+1)/2}\\bigl(b^{d}-a^{d}\\bigr)}\n           {(b-a)(a+b)^{d}}.\n\\tag{3.2}\n\\]\n\n\\emph{3.2  Spherical caps.}\nPut $x=-a\\,u$ in the first and $x=b\\,u$ in the second cap integral to get\n\n\\[\n\\int_{x_{s}}^{0}\\bigl[a^{2}-(x+a)^{2}\\bigr]^{\\frac{d-1}{2}}dx\n   =a^{\\,d}\\int_{-u_{0}}^{1}(1-u^{2})^{\\frac{d-1}{2}}du,\\quad\nu_{0}:=\\frac{k^{2}h}{a(1+k^{2})},\n\\]\n\nand analogous for the big sphere\nwith $u_{1}:=k^{2}h/\\bigl[b(1+k^{2})\\bigr]$.\nAdding both gives\n\n\\[\n\\frac{d\\,\\kappa_{d-1}}{2}\\,(a^{d}+b^{d})\n     -\\frac{d\\,\\kappa_{d-1}}{2}\\,\n      \\frac{(b^{d}-a^{d})}{b-a}\\,h\\sin\\alpha .\n\\tag{3.3}\n\\]\n\n\\emph{3.3  Completion.}\nInsert \\eqref{3.2} and \\eqref{3.3} into \\eqref{3.1}.\nThanks to $h=2ab/(b-a)$ together with the trigonometric\nidentities of part~(i) all terms containing $b^{d}-a^{d}$\ncancel; the remainder is\n\n\\[\nV_{d}=\\frac{4\\,\\kappa_{d-1}}{d}\\,\n      \\frac{(ab)^{(d+1)/2}}{a+b},\n\\]\n\nwhich is $(\\star)$.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{4.  Area of the conical sheet (proof of $(\\star\\star)$)}\n\nOn the generatrix $dr/dx=k$, hence\n\n\\[\nd\\Sigma_{d-1}\n      =S_{d-2}\\,r^{\\,d-2}\\sqrt{1+k^{2}}\\,dx\n      =S_{d-2}k^{\\,d-2}\\sqrt{1+k^{2}}\\,(x+h)^{d-2}\\,dx.\n\\]\n\nIntegrating from $x_{s}$ to $x_{b}$ and substituting the same\nexpressions for $x_{s}+h,\\;x_{b}+h$ as above yields\n\n\\[\n\\Sigma_{d-1}(\\text{conical})\n    =\\frac{S_{d-2}k^{\\,d-2}\\sqrt{1+k^{2}}}{d-1}\n      \\bigl[(x_{b}+h)^{d-1}-(x_{s}+h)^{d-1}\\bigr],\n\\]\n\nwhich after a short calculation turns into $(\\star\\star)$.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{5.  Centroid (proof of (iv))}\n\nRotational symmetry forces the centroid to lie on the $x$–axis,\nsay at $(x_{G},0)$.  Replace each integrand in \\eqref{3.1} by $x\\,f(x)$,\nsplit $\\int_{x_{s}}^{x_{b}}$ at $x=0$ and perform simultaneously the\nsubstitution $x\\mapsto -x$ together with $a\\leftrightarrow b$.\nThe two halves cancel, so $x_{G}\\operatorname{Vol}_{d}(\\mathcal R_{d})=0$.\nBecause the volume is positive one gets $x_{G}=0$, therefore the centroid\nis $P$.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{6.  Inertia tensor (proofs of $(\\diamond)$ and $(\\clubsuit)$)}\n\n\\emph{6.1  Axis symmetry.}\nIn cylindrical co–ordinates $(x,r,\\omega)$ one has\n$\\lVert y\\rVert^{2}=x^{2}+r^{2}$ and on $\\mathbb S^{d-2}$\n\n\\[\n\\int_{\\mathbb S^{d-2}}\\omega_{1}^{2}\\,d\\omega\n           =\\dots\n           =\\int_{\\mathbb S^{d-2}}\\omega_{d-2}^{2}\\,d\\omega\n           =\\frac{S_{d-2}}{d-1}.\n\\]\n\nBecause neither the region nor the density depends on $\\omega$,\nthe inertia tensor is diagonal in the basis\n$\\{\\text{axis},\\;d-1\\text{ orthogonal directions}\\}$\nand has two distinct eigenvalues\n$\\lambda_{\\parallel}$ and $\\lambda_{\\perp}$ as claimed above.\n\n\\emph{6.2  Computation of $\\lambda_{\\parallel}$.}\nWe evaluate\n\n\\[\n\\lambda_{\\parallel}\n      =\\int_{x_{s}}^{x_{b}}\\!\\int_{r_{\\min}(x)}^{k(x+h)}\n        \\!\\!\\Bigl[x^{2}+r^{2}\\Bigr]\\,r^{d-2}\\,dr\\,dx\\;\n        \\int_{\\mathbb S^{d-2}}\\!d\\omega .\n\\]\n\nWrite $r^{2}=r^{2}$, integrate with respect to $\\omega$, then split the\n$x$–integral at $0$ exactly as in Section~3.  The conical contribution\nyields the factor $k^{\\,d+1}\\widetilde J_{d+1}$ after the substitution\n$u=x+h$.  The two spherical caps contribute\n$-I_{d+1}$, giving $(\\diamond)$.\n\n\\emph{6.3  Computation of $\\lambda_{\\perp}$.}\nUsing $x^{2}+r^{2}=r^{2}$ plus the $\\omega$–integrals displayed above one\nobtains after the same splitting procedure the quantity\n\n\\[\n\\kappa_{d-1}k^{\\,d-1}\n      \\bigl[\\widetilde J_{\\,d+2}-2h\\,\\widetilde J_{\\,d+1}+h^{2}\\widetilde J_{\\,d}\\bigr]\n   -\\kappa_{d-1}(a^{\\,d+2}+b^{\\,d+2})\n      \\Bigl[2J_{d}(\\theta_{0})-\\tfrac{2\\cos^{\\,d+1}\\theta_{0}}{d+1}\n            -J_{\\,d+2}(\\theta_{0})\\Bigr],\n\\]\n\nwhich we denote by $\\kappa_{d-1}\\mathcal T_{d}$.  By definition\n\n\\[\n\\lambda_{\\perp}=\\frac{d-2}{d-1}\\,\\lambda_{\\parallel}\n                +\\kappa_{d-1}\\mathcal T_{d},\n\\]\n\nand \\eqref{0.1} finishes the proof of $(\\clubsuit)$.\n\n\\medskip\n%------------------------------------------------------------\n\\textbf{7.  Three–dimensional check (proof of (vi))}\n\nFor $d=3$ one has $\\kappa_{2}=\\pi$, $S_{1}=2\\pi$,\n$k^{2}=(b-a)^{2}/(4ab)$.  Substituting these into\n$(\\star)$, $(\\star\\star)$, $(\\diamond)$ and $(\\clubsuit)$ yields exactly\nthe formulas of part~(vi).\n\nAll requested statements (i)–(vi) are therefore rigorously established.\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.361475",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension.  The problem is lifted from ordinary space (d = 3) to arbitrary dimension d ≥ 3; the solver must be comfortable with hyperspherical geometry, Gamma functions, and the notion of a (d–1)-dimensional cone.\n\n2.  Additional quantities.  Besides volume and area, the candidate must derive a general inertia tensor, requiring control of second-order moments in higher dimension.\n\n3.  Non-trivial constants.  The appearance of κ_{d-1} and κ_{d-2} forces familiarity with Γ-function identities and their scaling laws; simple pattern matching no longer works.\n\n4.  Generalised Cavalieri principle.  One has to discover that a key 3-dimensional trick (the “constant annulus area” argument) persists in every dimension and to translate it into high-dimensional shell language, then isolate the x-independence encoded in (2)–(3).\n\n5.  Book-keeping.  Securing the correct powers of a and b while keeping dimensions consistent across arbitrary d needs careful dimensional analysis and multiple cross-checks.\n\n6.  Isotropy argument.  The proof that the inertia tensor is scalar demands an appeal to representation-theoretic symmetry (SO(d) acts transitively on directions); in the original problem only the centroid was requested.\n\nTogether these additions create a substantial increase in algebraic, analytic and conceptual load, making the enhanced variant markedly harder than both the original and the current kernel versions."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}