{
  "index": "1942-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "5. A circle of radius \\( a \\) is revolved through \\( 180^{\\circ} \\) about a line in its plane, distant \\( b \\) from the center of the circle, where \\( b>a \\). For what value of the ratio \\( b / a \\) does the center of gravity of the solid thus generated lie on the surface of the solid?",
  "solution": "Solution. We choose axes so that the generating circle starts in the \\( x-z \\) plane and is revolved about the \\( z \\)-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).\n\nIt is clear from symmetry that the centroid lies at a point \\( (0, \\bar{y}, 0) \\) on the \\( y \\)-axis, and the requirement of the problem is that \\( \\bar{y}=b-a \\).\n\nTo find the centroid we introduce polar coordinates in the \\( x-y \\) plane. Corresponding to the element of area \\( r d r d \\theta \\) in the plane there is the element of volume\n\\[\n2 \\sqrt{a^{2}-(r-b)^{2}} r d r d \\theta\n\\]\nwhich contributes\n\\[\n2 r \\sin \\theta \\sqrt{a^{2}-(r-b)^{2}} r d r d \\theta\n\\]\nto the moment \\( M_{y} \\) of the solid in the \\( y \\) direction: We have \\( \\bar{y}=M_{y} / V \\) where \\( V \\) is the volume of the semi-toroid.\n\\[\n\\begin{aligned}\nV & =\\int_{0}^{\\pi} \\int_{b-a}^{b+a} 2 \\sqrt{a^{2}}-\\overline{(r-b)^{2}} r d r d \\theta \\\\\n& =2 \\pi \\int_{b-a}^{b+a} \\sqrt{a^{2}-(r-\\bar{b})^{2}} r d r \\\\\n& =2 \\pi a^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos \\phi(b+a \\sin \\phi) \\cos \\phi d \\phi \\\\\n& =\\pi^{2} a^{2} b . \\\\\nM_{y} & =\\int_{0}^{\\pi} \\int_{b-a}^{b+a} 2 \\sqrt{a^{2}-(r-b)^{2}} r^{2} d r \\sin \\theta d \\theta \\\\\n& =4 \\int_{b-a}^{b+a} \\sqrt{a^{2}-\\frac{1}{(r-b)^{2}} r^{2} d r} \\\\\n& =4 a^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos \\phi(b+a \\sin \\phi)^{2} \\cos \\phi d \\phi \\\\\n& =2 \\pi a^{2} b^{2}+\\frac{\\pi}{2} a^{4} .\n\\end{aligned}\n\\]\n\nIn both integrals we used the substitution \\( r=b+a \\sin \\phi \\).\nHence\n\\[\n\\bar{y}=\\frac{M_{y}}{V}=\\frac{a^{2}+4 b^{2}}{2 \\pi b} .\n\\]\n\nBut we require \\( \\bar{y}=b-a \\), so\n\\[\n2 \\pi b^{2}-2 \\pi a b=a^{2}+4 b^{2} .\n\\]\n\nIf \\( c=b / a \\), then\n\\[\n(2 \\pi-4) c^{2}-2 \\pi c-1=0\n\\]\nand\n\\[\nc=\\frac{\\pi+\\sqrt{\\pi^{2}+2 \\pi-4}}{2 \\pi-4} .\n\\]\n\nWe chose the positive sign since \\( \\boldsymbol{c} \\) must be positive.\nRemarks. The volume of the semi-toroid could have been obtained from Pappus's Theorem that the volume of a solid of revolution is the product of the area of the generating region times the length of the circle traversed by the centroid of the area.\n\nThe critical ratio is about 2.91 .",
  "vars": [
    "x",
    "z",
    "y",
    "r",
    "\\\\theta",
    "\\\\phi",
    "M_y",
    "V",
    "c"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoord",
        "z": "zcoord",
        "y": "ycoord",
        "r": "radial",
        "\\theta": "angletheta",
        "\\phi": "anglephi",
        "M_y": "momenty",
        "V": "volume",
        "c": "ratiofactor",
        "a": "radius",
        "b": "axisdist"
      },
      "question": "5. A circle of radius \\( radius \\) is revolved through \\( 180^{\\circ} \\) about a line in its plane, distant \\( axisdist \\) from the center of the circle, where \\( axisdist>radius \\). For what value of the ratio \\( axisdist / radius \\) does the center of gravity of the solid thus generated lie on the surface of the solid?",
      "solution": "Solution. We choose axes so that the generating circle starts in the \\( xcoord-zcoord \\) plane and is revolved about the \\( zcoord \\)-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).\n\nIt is clear from symmetry that the centroid lies at a point \\( (0, \\bar{ycoord}, 0) \\) on the \\( ycoord \\)-axis, and the requirement of the problem is that \\( \\bar{ycoord}=axisdist-radius \\).\n\nTo find the centroid we introduce polar coordinates in the \\( xcoord-ycoord \\) plane. Corresponding to the element of area \\( radial \\, d radial \\, d angletheta \\) in the plane there is the element of volume\n\\[\n2 \\sqrt{radius^{2}-(radial-axisdist)^{2}} \\, radial \\, d radial \\, d angletheta\n\\]\nwhich contributes\n\\[\n2 \\, radial \\sin angletheta \\, \\sqrt{radius^{2}-(radial-axisdist)^{2}} \\, radial \\, d radial \\, d angletheta\n\\]\nto the moment \\( momenty \\) of the solid in the \\( ycoord \\) direction. We have \\( \\bar{ycoord}=momenty / volume \\) where \\( volume \\) is the volume of the semi-toroid.\n\\[\n\\begin{aligned}\nvolume & = \\int_{0}^{\\pi} \\int_{axisdist-radius}^{axisdist+radius} 2 \\sqrt{radius^{2}}-\\overline{(radial-axisdist)^{2}} \\, radial \\, d radial \\, d angletheta \\\\\n       & = 2\\pi \\int_{axisdist-radius}^{axisdist+radius} \\sqrt{radius^{2}-(radial-\\bar{axisdist})^{2}} \\, radial \\, d radial \\\\\n       & = 2\\pi \\, radius^{2} \\int_{-\\pi/2}^{\\pi/2} \\cos anglephi \\, (axisdist+radius \\sin anglephi) \\cos anglephi \\, d anglephi \\\\\n       & = \\pi^{2} \\, radius^{2} \\, axisdist . \\\\\nmomenty & = \\int_{0}^{\\pi} \\int_{axisdist-radius}^{axisdist+radius} 2 \\sqrt{radius^{2}-(radial-axisdist)^{2}} \\, radial^{2} \\, d radial \\sin angletheta \\, d angletheta \\\\\n        & = 4 \\int_{axisdist-radius}^{axisdist+radius} \\sqrt{radius^{2}-\\frac{1}{(radial-axisdist)^{2}}} \\, radial^{2} \\, d radial \\\\\n        & = 4 \\, radius^{2} \\int_{-\\pi/2}^{\\pi/2} \\cos anglephi \\, (axisdist+radius \\sin anglephi)^{2} \\cos anglephi \\, d anglephi \\\\\n        & = 2\\pi \\, radius^{2} \\, axisdist^{2} + \\frac{\\pi}{2} radius^{4} .\n\\end{aligned}\n\\]\n\nIn both integrals we used the substitution \\( radial = axisdist + radius \\sin anglephi \\).\nHence\n\\[\n\\bar{ycoord}=\\frac{momenty}{volume}=\\frac{radius^{2}+4 \\, axisdist^{2}}{2\\pi \\, axisdist} .\n\\]\nBut we require \\( \\bar{ycoord}=axisdist-radius \\), so\n\\[\n2\\pi \\, axisdist^{2} - 2\\pi \\, radius \\, axisdist = radius^{2} + 4 \\, axisdist^{2} .\n\\]\nIf \\( ratiofactor = axisdist / radius \\), then\n\\[\n(2\\pi-4) \\, ratiofactor^{2} - 2\\pi \\, ratiofactor - 1 = 0\n\\]\nand\n\\[\nratiofactor = \\frac{\\pi + \\sqrt{\\pi^{2}+2\\pi-4}}{2\\pi-4} .\n\\]\nWe choose the positive sign since \\( \\boldsymbol{ratiofactor} \\) must be positive.\n\nRemarks. The volume of the semi-toroid could have been obtained from Pappus's theorem that the volume of a solid of revolution is the product of the area of the generating region and the length of the circle traversed by the centroid of the area.\n\nThe critical ratio is about 2.91 ."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lemonade",
        "z": "pancakes",
        "y": "orchestra",
        "r": "treasure",
        "\\theta": "butterfly",
        "\\phi": "harmony",
        "M_y": "strawberry",
        "V": "pineapple",
        "c": "dandelion",
        "a": "waterlily",
        "b": "rainstorm"
      },
      "question": "A circle of radius \\( waterlily \\) is revolved through \\( 180^{\\circ} \\) about a line in its plane, distant \\( rainstorm \\) from the center of the circle, where \\( rainstorm>waterlily \\). For what value of the ratio \\( rainstorm / waterlily \\) does the center of gravity of the solid thus generated lie on the surface of the solid?",
      "solution": "Solution. We choose axes so that the generating circle starts in the \\( lemonade\\text{-}pancakes \\) plane and is revolved about the \\( pancakes \\)-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).\n\nIt is clear from symmetry that the centroid lies at a point \\( (0, \\bar{orchestra}, 0) \\) on the y-axis, and the requirement of the problem is that \\( \\bar{orchestra}=rainstorm-waterlily \\).\n\nTo find the centroid we introduce polar coordinates in the \\( lemonade\\text{-}orchestra \\) plane. Corresponding to the element of area \\( treasure d treasure d butterfly \\) in the plane there is the element of volume\n\\[\n2 \\sqrt{waterlily^{2}-(treasure-rainstorm)^{2}}\\, treasure \\, d treasure \\, d butterfly\n\\]\nwhich contributes\n\\[\n2\\, treasure \\sin butterfly \\sqrt{waterlily^{2}-(treasure-rainstorm)^{2}}\\, treasure \\, d treasure \\, d butterfly\n\\]\nto the moment \\( strawberry \\) of the solid in the y direction: We have \\( \\bar{orchestra}=strawberry/pineapple \\) where \\( pineapple \\) is the volume of the semi-toroid.\n\\[\n\\begin{aligned}\npineapple &= \\int_{0}^{\\pi} \\int_{rainstorm-waterlily}^{rainstorm+waterlily} 2 \\sqrt{waterlily^{2}}-\\overline{(treasure-rainstorm)^{2}}\\, treasure \\, d treasure \\, d butterfly \\\\ & = 2\\pi \\int_{rainstorm-waterlily}^{rainstorm+waterlily} \\sqrt{waterlily^{2}-(treasure-\\bar{rainstorm})^{2}}\\, treasure \\, d treasure \\\\ & = 2\\pi waterlily^{2} \\int_{-\\pi/2}^{\\pi/2} \\cos harmony\\,(rainstorm+waterlily \\sin harmony) \\cos harmony \\, d harmony \\\\ & = \\pi^{2} waterlily^{2} rainstorm . \\\\[6pt]\nstrawberry &= \\int_{0}^{\\pi} \\int_{rainstorm-waterlily}^{rainstorm+waterlily} 2 \\sqrt{waterlily^{2}-(treasure-rainstorm)^{2}}\\, treasure^{2} \\, d treasure \\sin butterfly \\, d butterfly \\\\ & = 4 \\int_{rainstorm-waterlily}^{rainstorm+waterlily} \\sqrt{waterlily^{2}-\\frac{1}{(treasure-rainstorm)^{2}}}\\, treasure^{2} \\, d treasure \\\\ & = 4 waterlily^{2} \\int_{-\\pi/2}^{\\pi/2} \\cos harmony\\,(rainstorm+waterlily \\sin harmony)^{2} \\cos harmony \\, d harmony \\\\ & = 2\\pi waterlily^{2} rainstorm^{2}+\\frac{\\pi}{2} waterlily^{4} .\n\\end{aligned}\n\\]\n\nIn both integrals we used the substitution \\( treasure = rainstorm + waterlily \\sin harmony \\). Hence\n\\[\n\\bar{orchestra}=\\frac{strawberry}{pineapple}=\\frac{waterlily^{2}+4\\, rainstorm^{2}}{2\\pi rainstorm} .\n\\]\n\nBut we require \\( \\bar{orchestra}=rainstorm-waterlily \\), so\n\\[\n2\\pi rainstorm^{2}-2\\pi waterlily\\, rainstorm = waterlily^{2}+4\\, rainstorm^{2} .\n\\]\n\nIf \\( dandelion = rainstorm / waterlily \\), then\n\\[\n(2\\pi-4) dandelion^{2}-2\\pi dandelion-1 = 0\n\\]\nand\n\\[\ndandelion = \\frac{\\pi+\\sqrt{\\pi^{2}+2\\pi-4}}{2\\pi-4} .\n\\]\n\nWe chose the positive sign since \\( \\boldsymbol{dandelion} \\) must be positive.\n\nRemarks. The volume of the semi-toroid could have been obtained from Pappus's Theorem that the volume of a solid of revolution is the product of the area of the generating region times the length of the circle traversed by the centroid of the area.\n\nThe critical ratio is about 2.91 ."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "z": "horizontalaxis",
        "y": "depthcoordinate",
        "r": "angularmeasure",
        "\\theta": "linearoffset",
        "\\phi": "axialoffset",
        "M_y": "restenergy",
        "V": "hollowshell",
        "c": "summationfactor",
        "a": "centerpoint",
        "b": "nearshift"
      },
      "question": "Problem:\n<<<\n5. A circle of radius \\( centerpoint \\) is revolved through \\( 180^{\\circ} \\) about a line in its plane, distant \\( nearshift \\) from the center of the circle, where \\( nearshift>centerpoint \\). For what value of the ratio \\( nearshift / centerpoint \\) does the center of gravity of the solid thus generated lie on the surface of the solid?\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. We choose axes so that the generating circle starts in the \\( verticalaxis-horizontalaxis \\) plane and is revolved about the \\( horizontalaxis \\)-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).\n\nIt is clear from symmetry that the centroid lies at a point \\( (0, \\bar{depthcoordinate}, 0) \\) on the \\( depthcoordinate \\)-axis, and the requirement of the problem is that \\( \\bar{depthcoordinate}=nearshift-centerpoint \\).\n\nTo find the centroid we introduce polar coordinates in the \\( verticalaxis-depthcoordinate \\) plane. Corresponding to the element of area \\( angularmeasure d angularmeasure d linearoffset \\) in the plane there is the element of volume\n\\[\n2 \\sqrt{centerpoint^{2}-(angularmeasure-nearshift)^{2}} angularmeasure d angularmeasure d linearoffset\n\\]\nwhich contributes\n\\[\n2 angularmeasure \\sin linearoffset \\sqrt{centerpoint^{2}-(angularmeasure-nearshift)^{2}} angularmeasure d angularmeasure d linearoffset\n\\]\nto the moment \\( restenergy \\) of the solid in the \\( depthcoordinate \\) direction: We have \\( \\bar{depthcoordinate}=restenergy / hollowshell \\) where \\( hollowshell \\) is the volume of the semi-toroid.\n\\[\n\\begin{aligned}\nhollowshell & =\\int_{0}^{\\pi} \\int_{nearshift-centerpoint}^{nearshift+centerpoint} 2 \\sqrt{centerpoint^{2}}-\\overline{(angularmeasure-nearshift)^{2}} angularmeasure d angularmeasure d linearoffset \\\\\n& =2 \\pi \\int_{nearshift-centerpoint}^{nearshift+centerpoint} \\sqrt{centerpoint^{2}-(angularmeasure-\\bar{nearshift})^{2}} angularmeasure d angularmeasure \\\\\n& =2 \\pi centerpoint^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos axialoffset(nearshift+centerpoint \\sin axialoffset) \\cos axialoffset d axialoffset \\\\\n& =\\pi^{2} centerpoint^{2} nearshift . \\\\\nrestenergy & =\\int_{0}^{\\pi} \\int_{nearshift-centerpoint}^{nearshift+centerpoint} 2 \\sqrt{centerpoint^{2}-(angularmeasure-nearshift)^{2}} angularmeasure^{2} d angularmeasure \\sin linearoffset d linearoffset \\\\\n& =4 \\int_{nearshift-centerpoint}^{nearshift+centerpoint} \\sqrt{centerpoint^{2}-\\frac{1}{(angularmeasure-nearshift)^{2}} angularmeasure^{2} d angularmeasure} \\\\\n& =4 centerpoint^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos axialoffset(nearshift+centerpoint \\sin axialoffset)^{2} \\cos axialoffset d axialoffset \\\\\n& =2 \\pi centerpoint^{2} nearshift^{2}+\\frac{\\pi}{2} centerpoint^{4} .\n\\end{aligned}\n\\]\n\nIn both integrals we used the substitution \\( angularmeasure=nearshift+centerpoint \\sin axialoffset \\).\nHence\n\\[\n\\bar{depthcoordinate}=\\frac{restenergy}{hollowshell}=\\frac{centerpoint^{2}+4 nearshift^{2}}{2 \\pi nearshift} .\n\\]\n\nBut we require \\( \\bar{depthcoordinate}=nearshift-centerpoint \\), so\n\\[\n2 \\pi nearshift^{2}-2 \\pi centerpoint nearshift=centerpoint^{2}+4 nearshift^{2} .\n\\]\n\nIf \\( summationfactor=nearshift / centerpoint \\), then\n\\[\n(2 \\pi-4) summationfactor^{2}-2 \\pi summationfactor-1=0\n\\]\nand\n\\[\nsummationfactor=\\frac{\\pi+\\sqrt{\\pi^{2}+2 \\pi-4}}{2 \\pi-4} .\n\\]\n\nWe chose the positive sign since \\( \\boldsymbol{summationfactor} \\) must be positive.\nRemarks. The volume of the semi-toroid could have been obtained from Pappus's Theorem that the volume of a solid of revolution is the product of the area of the generating region times the length of the circle traversed by the centroid of the area.\n\nThe critical ratio is about 2.91 .\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "z": "nbvcyqwe",
        "y": "hjgrksla",
        "r": "ldkfjsie",
        "\\\\theta": "pqowieur",
        "\\\\phi": "mxnbvzhr",
        "M_y": "ghrtyuio",
        "V": "asdfghjk",
        "c": "zxcvbnml",
        "a": "klmnbvcx",
        "b": "sdfghjkl"
      },
      "question": "5. A circle of radius \\( klmnbvcx \\) is revolved through \\( 180^{\\circ} \\) about a line in its plane, distant \\( sdfghjkl \\) from the center of the circle, where \\( sdfghjkl>klmnbvcx \\). For what value of the ratio \\( sdfghjkl / klmnbvcx \\) does the center of gravity of the solid thus generated lie on the surface of the solid?",
      "solution": "Solution. We choose axes so that the generating circle starts in the \\( qzxwvtnp-nbvcyqwe \\) plane and is revolved about the \\( nbvcyqwe \\)-axis. The generated solid is half of a toroid (i.e., a solid bounded by a torus).\n\nIt is clear from symmetry that the centroid lies at a point \\( (0, \\bar{hjgrksla}, 0) \\) on the \\( hjgrksla \\)-axis, and the requirement of the problem is that \\( \\bar{hjgrksla}=sdfghjkl-klmnbvcx \\).\n\nTo find the centroid we introduce polar coordinates in the \\( qzxwvtnp-hjgrksla \\) plane. Corresponding to the element of area \\( ldkfjsie d ldkfjsie d pqowieur \\) in the plane there is the element of volume\n\\[\n2 \\sqrt{klmnbvcx^{2}-(ldkfjsie-sdfghjkl)^{2}} ldkfjsie d ldkfjsie d pqowieur\n\\]\nwhich contributes\n\\[\n2 ldkfjsie \\sin pqowieur \\sqrt{klmnbvcx^{2}-(ldkfjsie-sdfghjkl)^{2}} ldkfjsie d ldkfjsie d pqowieur\n\\]\nto the moment \\( ghrtyuio \\) of the solid in the \\( hjgrksla \\) direction: We have \\( \\bar{hjgrksla}=ghrtyuio / asdfghjk \\) where \\( asdfghjk \\) is the volume of the semi-toroid.\n\\[\n\\begin{aligned}\nasdfghjk & =\\int_{0}^{\\pi} \\int_{sdfghjkl-klmnbvcx}^{sdfghjkl+klmnbvcx} 2 \\sqrt{klmnbvcx^{2}}-\\overline{(ldkfjsie-sdfghjkl)^{2}} ldkfjsie d ldkfjsie d pqowieur \\\\\n& =2 \\pi \\int_{sdfghjkl-klmnbvcx}^{sdfghjkl+klmnbvcx} \\sqrt{klmnbvcx^{2}-(ldkfjsie-\\bar{sdfghjkl})^{2}} ldkfjsie d ldkfjsie \\\\\n& =2 \\pi klmnbvcx^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos mxnbvzhr(sdfghjkl+klmnbvcx \\sin mxnbvzhr) \\cos mxnbvzhr d mxnbvzhr \\\\\n& =\\pi^{2} klmnbvcx^{2} sdfghjkl . \\\\\nghrtyuio & =\\int_{0}^{\\pi} \\int_{sdfghjkl-klmnbvcx}^{sdfghjkl+klmnbvcx} 2 \\sqrt{klmnbvcx^{2}-(ldkfjsie-sdfghjkl)^{2}} ldkfjsie^{2} d ldkfjsie \\sin pqowieur d pqowieur \\\\\n& =4 \\int_{sdfghjkl-klmnbvcx}^{sdfghjkl+klmnbvcx} \\sqrt{klmnbvcx^{2}-\\frac{1}{(ldkfjsie-sdfghjkl)^{2}} ldkfjsie^{2} d ldkfjsie} \\\\\n& =4 klmnbvcx^{2} \\int_{-\\pi / 2}^{\\pi / 2} \\cos mxnbvzhr(sdfghjkl+klmnbvcx \\sin mxnbvzhr)^{2} \\cos mxnbvzhr d mxnbvzhr \\\\\n& =2 \\pi klmnbvcx^{2} sdfghjkl^{2}+\\frac{\\pi}{2} klmnbvcx^{4} .\n\\end{aligned}\n\\]\n\nIn both integrals we used the substitution \\( ldkfjsie=sdfghjkl+klmnbvcx \\sin mxnbvzhr \\).\nHence\n\\[\n\\bar{hjgrksla}=\\frac{ghrtyuio}{asdfghjk}=\\frac{klmnbvcx^{2}+4 sdfghjkl^{2}}{2 \\pi sdfghjkl} .\n\\]\n\nBut we require \\( \\bar{hjgrksla}=sdfghjkl-klmnbvcx \\), so\n\\[\n2 \\pi sdfghjkl^{2}-2 \\pi klmnbvcx sdfghjkl=klmnbvcx^{2}+4 sdfghjkl^{2} .\n\\]\n\nIf \\( zxcvbnml=sdfghjkl / klmnbvcx \\), then\n\\[\n(2 \\pi-4) zxcvbnml^{2}-2 \\pi zxcvbnml-1=0\n\\]\nand\n\\[\nzxcvbnml=\\frac{\\pi+\\sqrt{\\pi^{2}+2 \\pi-4}}{2 \\pi-4} .\n\\]\n\nWe chose the positive sign since \\( \\boldsymbol{zxcvbnml} \\) must be positive.\nRemarks. The volume of the semi-toroid could have been obtained from Pappus's Theorem that the volume of a solid of revolution is the product of the area of the generating region times the length of the circle traversed by the centroid of the area.\n\nThe critical ratio is about 2.91 ."
    },
    "kernel_variant": {
      "question": "Fix a radius $a>0$.  \nIn cylindrical coordinates $(\\rho,\\theta ,z)$ consider the circle  \n\\[\n(\\rho-d)^{2}+z^{2}=a^{2},\\qquad d>a .\n\\]\nRevolving this circle about the $z$–axis, but only through the angle $3\\pi/2$,\n\\[\n0\\le\\theta\\le\\frac{3\\pi}{2},\n\\]\ngenerates the three–quarter solid torus $T$ with\nmajor (central) radius $d$ and tube radius $a$.\n\nInside $T$ the material is not homogeneous; its mass–density is prescribed by the power law  \n\\[\n\\varrho(\\rho,\\theta ,z)=\\varrho_{0}\\,\\rho^{m},\\qquad m\\in\\mathbb{N}\\cup\\{0\\},\\;\n\\varrho_{0}>0 .\n\\]\n\nFor which value(s) of the ratio  \n\\[\nc:=\\frac{d}{a}>1\n\\]\ndoes the centre of mass $\\,( \\bar{\\rho},\\bar{\\theta},\\bar{z})\\,$ of $T$ lie on the *outer* cylindrical surface\n\\[\n\\bar{\\rho}=d+a=a(c+1)\\;?\n\\]\n(For symmetry reasons $\\bar{\\theta}=3\\pi/4$ and $\\bar{z}=0$, so only the radial coordinate $\\bar{\\rho}$ is unknown.)",
      "solution": "Throughout let $c=d/a>1$ and $\\varrho_{0}=1$ (it cancels in all ratios).\n\n1.  Mass and first moment  \nAn infinitesimal mass element is\n\\[\ndm=\\rho^{m}\\rho\\,d\\rho\\,d\\theta\\,dz .\n\\]\nFor a fixed $\\rho$ the circular cross–section satisfies\n$|z|\\le\\sqrt{a^{2}-(\\rho-d)^{2}}$; hence the corresponding slice of volume is\n\\[\n2\\sqrt{a^{2}-(\\rho-d)^{2}}\\;\\rho\\,d\\rho\\,d\\theta .\n\\]\nIntroduce the family of integrals\n\\[\nI_{k}(c):=\\int_{d-a}^{d+a}\\rho^{\\,k}\\;2\\sqrt{a^{2}-(\\rho-d)^{2}}\\;d\\rho ,\n\\qquad k\\in\\mathbb{N}.\n\\]\nBecause the density is independent of $\\theta$ and $\\theta$ ranges over\n$3\\pi/2$, we obtain\n\\[\nM   =\\frac{3\\pi}{2}\\,I_{m+1}(c),\\qquad\nM_{\\rho}=\\frac{3\\pi}{2}\\,I_{m+2}(c),\n\\]\nso that\n\\[\n\\bar{\\rho}=\\frac{M_{\\rho}}{M}=\\frac{I_{m+2}(c)}{I_{m+1}(c)} .\n\\tag{1}\n\\]\n\n2.  Trigonometric substitution  \nPut $\\rho=d+a\\sin\\varphi$ with $-\\pi/2\\le\\varphi\\le\\pi/2$; then\n$d\\rho=a\\cos\\varphi\\,d\\varphi$ and\n$\\sqrt{a^{2}-(\\rho-d)^{2}}=a\\cos\\varphi$.  Hence\n\\[\nI_{k}(c)=2a^{k+2}\\int_{-\\pi/2}^{\\pi/2}(c+\\sin\\varphi)^{k}(\\cos\\varphi)^{2}\\,d\\varphi .\n\\]\nWrite\n\\[\nJ_{k}(c):=\\int_{-\\pi/2}^{\\pi/2}(c+\\sin\\varphi)^{k}(\\cos\\varphi)^{2}\\,d\\varphi ,\n\\qquad k\\in\\mathbb{N},\n\\]\nso that by (1)\n\\[\n\\frac{\\bar{\\rho}}{a}=\\frac{J_{m+2}(c)}{J_{m+1}(c)} .\n\\tag{2}\n\\]\n\n3.  A combinatorial expansion  \nExpand\n$(c+\\sin\\varphi)^{k}$ via the binomial theorem and exploit evenness of $\\sin\\varphi$:\n\\[\n(c+\\sin\\varphi)^{k}\n=\\sum_{j=0}^{\\lfloor k/2\\rfloor}\\binom{k}{2j}c^{\\,k-2j}\\sin^{2j}\\varphi .\n\\]\nConsequently\n\\[\nJ_{k}(c)=\\sum_{j=0}^{\\lfloor k/2\\rfloor}\\binom{k}{2j}c^{\\,k-2j}B_{j},\n\\qquad\nB_{j}:=\\int_{-\\pi/2}^{\\pi/2}\\sin^{2j}\\varphi(\\cos\\varphi)^{2}\\,d\\varphi .\n\\]\nA standard Beta–function evaluation yields\n\\[\nB_{j}=\\frac{\\pi(2j)!}{2^{2j+1}\\,j!\\,(j+1)!},\\qquad j=0,1,2,\\dots .\n\\tag{3}\n\\]\n\n4.  Centroid condition  \nRequiring $\\bar{\\rho}=a(c+1)$ is equivalent, via (2), to\n\\[\nP_{m}(c):=J_{m+2}(c)-(c+1)J_{m+1}(c)=0 .\n\\tag{4}\n\\]\nInsertion of the series for $J_{k}$ shows that $P_{m}(c)$ is a *polynomial*\nof degree $m+2$ whose coefficients have mixed signs (already for $m=0$ one\nfinds $P_{0}(c)=\\tfrac{\\pi}{8}(1-4c)$).  \nOnly its *sign* for $c>1$ will matter, not the individual coefficients.\n\n5.  An illuminating integral representation  \nReturning to the defining integral one obtains directly\n\\[\nP_{m}(c)=\\int_{-\\pi/2}^{\\pi/2}\n(c+\\sin\\varphi)^{\\,m+1}\\bigl(\\sin\\varphi-1\\bigr)(\\cos\\varphi)^{2}\\,d\\varphi .\n\\tag{5}\n\\]\nFor $-\\pi/2<\\varphi<\\pi/2$ we have $\\cos\\varphi>0$ and\n$\\sin\\varphi<1$, so the factor $\\sin\\varphi-1$ is *strictly negative* on\na set of positive measure.  \nMoreover $(c+\\sin\\varphi)^{\\,m+1}>0$ for every $c>1$ and all $\\varphi$ in the\ndomain.  Therefore\n\\[\nP_{m}(c)<0\\qquad\\text{for every }c>1\\text{ and every }m\\in\\mathbb{N}\\cup\\{0\\}.\n\\tag{6}\n\\]\n\n6.  Non–existence of admissible ratios  \nEquation (4) can be satisfied only if $P_{m}(c)=0$, but (6) shows that\n$P_{m}(c)$ is **strictly negative** throughout the physically relevant range\n$c>1$.  Hence $P_{m}(c)$ possesses *no* root in this interval.  Consequently\n\\[\n\\boxed{\\text{There is no ratio }c=d/a>1\\text{ for which the centroid of the\nthree–quarter torus, endowed with the density }\\rho^{m},\\text{ lies on its\nouter cylindrical surface.}}\n\\]\n\n7.  Low–order checks (consistency)  \n(i) $m=0$ (homogeneous material):\n$P_{0}(c)=\\tfrac{\\pi}{8}(1-4c)$ has the single root $c=\\tfrac14<1$.\n\n(ii) $m=1$ (density $\\propto\\rho$):\n$P_{1}(c)=\\tfrac{\\pi}{8}(-4c^{2}+2c-1)$ has negative discriminant, hence no real\nroot at all.\n\n(iii) $m=2$ (density $\\propto\\rho^{2}$):\n$P_{2}(c)=\\tfrac{\\pi}{16}(-8c^{3}+6c^{2}-6c+1)$ has the unique real root\n$c\\approx0.18<1$.\n\nAll examples confirm the general result.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.393924",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher technical level: the density is no longer constant but grows like ρ^m; the integrals therefore depend on an extra (unbounded) parameter m and lead to a whole *family* of algebraic equations instead of one quadratic.  \n•  Algebra–analytic interplay: solving condition (5) required transforming volume-type integrals to Beta–function form, then converting the centroid condition into the polynomial (6).  \n•  Advanced existence arguments: one has to analyse monotonicity of S_k and f_m to show uniqueness (or non-existence) of solutions for general m, not merely solve a single equation.  \n•  Multiple interacting concepts: Beta/Gamma functions, binomial expansions, polynomial algebra, and qualitative root analysis all enter.  \n•  Original solution was a single quadratic; the enhanced variant demands manipulating an infinite *sequence* of increasingly high-degree polynomials and understanding how their solvability depends on the parameter m."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a radius $a>0$.  \nIn cylindrical coordinates $(\\rho,\\theta ,z)$ consider the circle  \n\\[\n(\\rho-d)^{2}+z^{2}=a^{2},\\qquad d>a .\n\\]\nRevolving this circle about the $z$–axis, but only through the angle $3\\pi/2$,\n\\[\n0\\le\\theta\\le\\frac{3\\pi}{2},\n\\]\ngenerates the three–quarter solid torus $T$ with\nmajor (central) radius $d$ and tube radius $a$.\n\nInside $T$ the material is not homogeneous; its mass–density is prescribed by the power law  \n\\[\n\\varrho(\\rho,\\theta ,z)=\\varrho_{0}\\,\\rho^{m},\\qquad m\\in\\mathbb{N}\\cup\\{0\\},\\;\n\\varrho_{0}>0 .\n\\]\n\nFor which value(s) of the ratio  \n\\[\nc:=\\frac{d}{a}>1\n\\]\ndoes the centre of mass $\\,( \\bar{\\rho},\\bar{\\theta},\\bar{z})\\,$ of $T$ lie on the *outer* cylindrical surface\n\\[\n\\bar{\\rho}=d+a=a(c+1)\\;?\n\\]\n(For symmetry reasons $\\bar{\\theta}=3\\pi/4$ and $\\bar{z}=0$, so only the radial coordinate $\\bar{\\rho}$ is unknown.)",
      "solution": "Throughout let $c=d/a>1$ and $\\varrho_{0}=1$ (it cancels in all ratios).\n\n1.  Mass and first moment  \nAn infinitesimal mass element is\n\\[\ndm=\\rho^{m}\\rho\\,d\\rho\\,d\\theta\\,dz .\n\\]\nFor a fixed $\\rho$ the circular cross–section satisfies\n$|z|\\le\\sqrt{a^{2}-(\\rho-d)^{2}}$; hence the corresponding slice of volume is\n\\[\n2\\sqrt{a^{2}-(\\rho-d)^{2}}\\;\\rho\\,d\\rho\\,d\\theta .\n\\]\nIntroduce the family of integrals\n\\[\nI_{k}(c):=\\int_{d-a}^{d+a}\\rho^{\\,k}\\;2\\sqrt{a^{2}-(\\rho-d)^{2}}\\;d\\rho ,\n\\qquad k\\in\\mathbb{N}.\n\\]\nBecause the density is independent of $\\theta$ and $\\theta$ ranges over\n$3\\pi/2$, we obtain\n\\[\nM   =\\frac{3\\pi}{2}\\,I_{m+1}(c),\\qquad\nM_{\\rho}=\\frac{3\\pi}{2}\\,I_{m+2}(c),\n\\]\nso that\n\\[\n\\bar{\\rho}=\\frac{M_{\\rho}}{M}=\\frac{I_{m+2}(c)}{I_{m+1}(c)} .\n\\tag{1}\n\\]\n\n2.  Trigonometric substitution  \nPut $\\rho=d+a\\sin\\varphi$ with $-\\pi/2\\le\\varphi\\le\\pi/2$; then\n$d\\rho=a\\cos\\varphi\\,d\\varphi$ and\n$\\sqrt{a^{2}-(\\rho-d)^{2}}=a\\cos\\varphi$.  Hence\n\\[\nI_{k}(c)=2a^{k+2}\\int_{-\\pi/2}^{\\pi/2}(c+\\sin\\varphi)^{k}(\\cos\\varphi)^{2}\\,d\\varphi .\n\\]\nWrite\n\\[\nJ_{k}(c):=\\int_{-\\pi/2}^{\\pi/2}(c+\\sin\\varphi)^{k}(\\cos\\varphi)^{2}\\,d\\varphi ,\n\\qquad k\\in\\mathbb{N},\n\\]\nso that by (1)\n\\[\n\\frac{\\bar{\\rho}}{a}=\\frac{J_{m+2}(c)}{J_{m+1}(c)} .\n\\tag{2}\n\\]\n\n3.  A combinatorial expansion  \nExpand\n$(c+\\sin\\varphi)^{k}$ via the binomial theorem and exploit evenness of $\\sin\\varphi$:\n\\[\n(c+\\sin\\varphi)^{k}\n=\\sum_{j=0}^{\\lfloor k/2\\rfloor}\\binom{k}{2j}c^{\\,k-2j}\\sin^{2j}\\varphi .\n\\]\nConsequently\n\\[\nJ_{k}(c)=\\sum_{j=0}^{\\lfloor k/2\\rfloor}\\binom{k}{2j}c^{\\,k-2j}B_{j},\n\\qquad\nB_{j}:=\\int_{-\\pi/2}^{\\pi/2}\\sin^{2j}\\varphi(\\cos\\varphi)^{2}\\,d\\varphi .\n\\]\nA standard Beta–function evaluation yields\n\\[\nB_{j}=\\frac{\\pi(2j)!}{2^{2j+1}\\,j!\\,(j+1)!},\\qquad j=0,1,2,\\dots .\n\\tag{3}\n\\]\n\n4.  Centroid condition  \nRequiring $\\bar{\\rho}=a(c+1)$ is equivalent, via (2), to\n\\[\nP_{m}(c):=J_{m+2}(c)-(c+1)J_{m+1}(c)=0 .\n\\tag{4}\n\\]\nInsertion of the series for $J_{k}$ shows that $P_{m}(c)$ is a *polynomial*\nof degree $m+2$ whose coefficients have mixed signs (already for $m=0$ one\nfinds $P_{0}(c)=\\tfrac{\\pi}{8}(1-4c)$).  \nOnly its *sign* for $c>1$ will matter, not the individual coefficients.\n\n5.  An illuminating integral representation  \nReturning to the defining integral one obtains directly\n\\[\nP_{m}(c)=\\int_{-\\pi/2}^{\\pi/2}\n(c+\\sin\\varphi)^{\\,m+1}\\bigl(\\sin\\varphi-1\\bigr)(\\cos\\varphi)^{2}\\,d\\varphi .\n\\tag{5}\n\\]\nFor $-\\pi/2<\\varphi<\\pi/2$ we have $\\cos\\varphi>0$ and\n$\\sin\\varphi<1$, so the factor $\\sin\\varphi-1$ is *strictly negative* on\na set of positive measure.  \nMoreover $(c+\\sin\\varphi)^{\\,m+1}>0$ for every $c>1$ and all $\\varphi$ in the\ndomain.  Therefore\n\\[\nP_{m}(c)<0\\qquad\\text{for every }c>1\\text{ and every }m\\in\\mathbb{N}\\cup\\{0\\}.\n\\tag{6}\n\\]\n\n6.  Non–existence of admissible ratios  \nEquation (4) can be satisfied only if $P_{m}(c)=0$, but (6) shows that\n$P_{m}(c)$ is **strictly negative** throughout the physically relevant range\n$c>1$.  Hence $P_{m}(c)$ possesses *no* root in this interval.  Consequently\n\\[\n\\boxed{\\text{There is no ratio }c=d/a>1\\text{ for which the centroid of the\nthree–quarter torus, endowed with the density }\\rho^{m},\\text{ lies on its\nouter cylindrical surface.}}\n\\]\n\n7.  Low–order checks (consistency)  \n(i) $m=0$ (homogeneous material):\n$P_{0}(c)=\\tfrac{\\pi}{8}(1-4c)$ has the single root $c=\\tfrac14<1$.\n\n(ii) $m=1$ (density $\\propto\\rho$):\n$P_{1}(c)=\\tfrac{\\pi}{8}(-4c^{2}+2c-1)$ has negative discriminant, hence no real\nroot at all.\n\n(iii) $m=2$ (density $\\propto\\rho^{2}$):\n$P_{2}(c)=\\tfrac{\\pi}{16}(-8c^{3}+6c^{2}-6c+1)$ has the unique real root\n$c\\approx0.18<1$.\n\nAll examples confirm the general result.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.338819",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher technical level: the density is no longer constant but grows like ρ^m; the integrals therefore depend on an extra (unbounded) parameter m and lead to a whole *family* of algebraic equations instead of one quadratic.  \n•  Algebra–analytic interplay: solving condition (5) required transforming volume-type integrals to Beta–function form, then converting the centroid condition into the polynomial (6).  \n•  Advanced existence arguments: one has to analyse monotonicity of S_k and f_m to show uniqueness (or non-existence) of solutions for general m, not merely solve a single equation.  \n•  Multiple interacting concepts: Beta/Gamma functions, binomial expansions, polynomial algebra, and qualitative root analysis all enter.  \n•  Original solution was a single quadratic; the enhanced variant demands manipulating an infinite *sequence* of increasingly high-degree polynomials and understanding how their solvability depends on the parameter m."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}