{
  "index": "1941-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. If the \\( x \\)-coordinate \\( \\bar{x} \\) of the center of mass of the area lying between the \\( x \\)-axis and the curve \\( y=f(x),(f(x)>0) \\), and between the lines \\( x=0 \\) and \\( x \\) \\( =a \\) is given by\n\\[\n\\bar{x}=g(a)\n\\]\nshow that\n\\[\nf(x)=A \\frac{g^{\\prime}(x)}{[x-g(x)]^{2}} e^{\\int d x /(x-g(x))}\n\\]\nwhere \\( A \\) is a positive constant.",
  "solution": "Solution. By the definition of centroid,\n\\[\ng(x)=\\frac{\\int_{0}^{x} t f(t) d t}{\\int_{0}^{x} f(t) d t}, \\quad x \\neq 0 .\n\\]\n\nWe confine our attention to positive values of \\( x \\). Put\n\\[\nF(x)=\\int_{0}^{x} f(t) d t ; \\quad \\text { then } F^{\\prime}=f\n\\]\n\nWrite (1) in the form\n\\[\nF(x) g(x)=\\int_{0}^{x} t f(t) d t,\n\\]\nand differentiate to get\n\\[\nF^{\\prime}(x) g(x)+F(x) g^{\\prime}(x)=x f(x)=x F^{\\prime}(x)\n\\]\n\nThus \\( \\boldsymbol{F} \\) satisfies the linear differential equation\n\\[\nF^{\\prime}(x)=\\frac{g^{\\prime}(x)}{x-g(x)} F(x)\n\\]\n\nHence\n\\[\nF(x)=A e^{\\psi(x)}\n\\]\nwhere \\( A \\) is a positive constant and\n\\[\n\\psi(x)=\\int^{x} \\frac{g^{\\prime}(t)}{t-g(t)} d t=\\int^{x} \\frac{d t}{t-g(t)}-\\log (x-g(x))\n\\]\n\nThus\n\\[\nF(x)=\\frac{A}{x-g(x)} \\exp \\int^{x} \\frac{d t}{t-g(t)}\n\\]\n\nSubstituting (5) into (2) gives us\n\\[\nf(x)=A \\frac{g^{\\prime}(x)}{(x-g(x))^{2}} \\exp \\int^{x} \\frac{d t}{t-g(t)}\n\\]\nas required.\nUnder the conditions of the problem, \\( x>g(x) \\) for all positive \\( x \\), so the denominator \\( x-g(x) \\) causes no singularities for positive \\( x \\). For \\( x<0 \\), we have \\( g(x)>x \\), so we must replace \\( \\log (x-g(x)) \\) by \\( \\log (g(x)-x) \\) in (4). Then (6) becomes\n\\[\nf(x)=\\frac{-A g^{\\prime}(x)}{(x-g(x))^{2}} \\exp \\int^{x} \\frac{d t}{t-g(t)}\n\\]\n\nHowever, in this case the constant \\( A \\) in (3) must be negative so we again have the required form.",
  "vars": [
    "x",
    "t",
    "g",
    "f",
    "F",
    "\\\\psi"
  ],
  "params": [
    "a",
    "A"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "t": "paramtr",
        "g": "centroid",
        "f": "curvefn",
        "F": "antider",
        "\\psi": "auxfunc",
        "a": "upperbd",
        "A": "constan"
      },
      "question": "6. If the \\( abscissa \\)-coordinate \\( \\bar{abscissa} \\) of the center of mass of the area lying between the \\( abscissa \\)-axis and the curve \\( y=curvefn(abscissa),\\,(curvefn(abscissa)>0) \\), and between the lines \\( abscissa=0 \\) and \\( abscissa =upperbd \\) is given by\n\\[\n\\bar{abscissa}=centroid(upperbd)\n\\]\nshow that\n\\[\ncurvefn(abscissa)=constan \\frac{centroid^{\\prime}(abscissa)}{[abscissa-centroid(abscissa)]^{2}} e^{\\int d abscissa /(abscissa-centroid(abscissa))}\n\\]\nwhere \\( constan \\) is a positive constant.",
      "solution": "Solution. By the definition of centroid,\n\\[\ncentroid(abscissa)=\\frac{\\int_{0}^{abscissa} paramtr\\, curvefn(paramtr)\\, d paramtr}{\\int_{0}^{abscissa} curvefn(paramtr)\\, d paramtr}, \\quad abscissa \\neq 0 .\n\\]\n\nWe confine our attention to positive values of \\( abscissa \\). Put\n\\[\nantider(abscissa)=\\int_{0}^{abscissa} curvefn(paramtr)\\, d paramtr ; \\quad \\text { then } antider^{\\prime}=curvefn\n\\]\n\nWrite (1) in the form\n\\[\nantider(abscissa)\\, centroid(abscissa)=\\int_{0}^{abscissa} paramtr\\, curvefn(paramtr)\\, d paramtr,\n\\]\nand differentiate to get\n\\[\nantider^{\\prime}(abscissa)\\, centroid(abscissa)+antider(abscissa)\\, centroid^{\\prime}(abscissa)=abscissa\\, curvefn(abscissa)=abscissa\\, antider^{\\prime}(abscissa)\n\\]\n\nThus \\( \\mathbf{antider} \\) satisfies the linear differential equation\n\\[\nantider^{\\prime}(abscissa)=\\frac{centroid^{\\prime}(abscissa)}{abscissa-centroid(abscissa)}\\, antider(abscissa)\n\\]\n\nHence\n\\[\nantider(abscissa)=constan\\, e^{auxfunc(abscissa)}\n\\]\nwhere \\( constan \\) is a positive constant and\n\\[\nauxfunc(abscissa)=\\int^{abscissa} \\frac{centroid^{\\prime}(paramtr)}{paramtr-centroid(paramtr)}\\, d paramtr=\\int^{abscissa} \\frac{d paramtr}{paramtr-centroid(paramtr)}-\\log (abscissa-centroid(abscissa))\n\\]\n\nThus\n\\[\nantider(abscissa)=\\frac{constan}{abscissa-centroid(abscissa)} \\exp \\int^{abscissa} \\frac{d paramtr}{paramtr-centroid(paramtr)}\n\\]\n\nSubstituting (5) into (2) gives us\n\\[\ncurvefn(abscissa)=constan \\frac{centroid^{\\prime}(abscissa)}{(abscissa-centroid(abscissa))^{2}} \\exp \\int^{abscissa} \\frac{d paramtr}{paramtr-centroid(paramtr)}\n\\]\nas required.\n\nUnder the conditions of the problem, \\( abscissa>centroid(abscissa) \\) for all positive \\( abscissa \\), so the denominator \\( abscissa-centroid(abscissa) \\) causes no singularities for positive \\( abscissa \\). For \\( abscissa<0 \\), we have \\( centroid(abscissa)>abscissa \\), so we must replace \\( \\log (abscissa-centroid(abscissa)) \\) by \\( \\log (centroid(abscissa)-abscissa) \\) in (4). Then (6) becomes\n\\[\ncurvefn(abscissa)=\\frac{-constan\\, centroid^{\\prime}(abscissa)}{(abscissa-centroid(abscissa))^{2}} \\exp \\int^{abscissa} \\frac{d paramtr}{paramtr-centroid(paramtr)}\n\\]\n\nHowever, in this case the constant \\( constan \\) in (3) must be negative so we again have the required form."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pebblestone",
        "t": "marigold",
        "g": "lighthouse",
        "f": "windchimes",
        "F": "buttercup",
        "\\psi": "tumbleweed",
        "a": "raincloud",
        "A": "moonflower"
      },
      "question": "6. If the \\( pebblestone \\)-coordinate \\( \\bar{pebblestone} \\) of the center of mass of the area lying between the x-axis and the curve \\( y=windchimes(pebblestone),(windchimes(pebblestone)>0) \\), and between the lines \\( pebblestone=0 \\) and \\( pebblestone \\) \\( =raincloud \\) is given by\n\\[\n\\bar{pebblestone}=lighthouse(raincloud)\n\\]\nshow that\n\\[\nwindchimes(pebblestone)=moonflower \\frac{lighthouse^{\\prime}(pebblestone)}{[pebblestone-lighthouse(pebblestone)]^{2}} e^{\\int d pebblestone /(pebblestone-lighthouse(pebblestone))}\n\\]\nwhere \\( moonflower \\) is a positive constant.",
      "solution": "Solution. By the definition of centroid,\n\\[\nlighthouse(pebblestone)=\\frac{\\int_{0}^{pebblestone} marigold\\, windchimes(marigold) d marigold}{\\int_{0}^{pebblestone} windchimes(marigold) d marigold}, \\quad pebblestone \\neq 0 .\n\\]\n\nWe confine our attention to positive values of \\( pebblestone \\). Put\n\\[\nbuttercup(pebblestone)=\\int_{0}^{pebblestone} windchimes(marigold) d marigold ; \\quad \\text { then } buttercup^{\\prime}=windchimes\n\\]\n\nWrite (1) in the form\n\\[\nbuttercup(pebblestone) \\, lighthouse(pebblestone)=\\int_{0}^{pebblestone} marigold \\, windchimes(marigold) d marigold,\n\\]\nand differentiate to get\n\\[\nbuttercup^{\\prime}(pebblestone) \\, lighthouse(pebblestone)+buttercup(pebblestone) \\, lighthouse^{\\prime}(pebblestone)=pebblestone \\, windchimes(pebblestone)=pebblestone \\, buttercup^{\\prime}(pebblestone)\n\\]\n\nThus \\( \\boldsymbol{buttercup} \\) satisfies the linear differential equation\n\\[\nbuttercup^{\\prime}(pebblestone)=\\frac{lighthouse^{\\prime}(pebblestone)}{pebblestone-lighthouse(pebblestone)} \\, buttercup(pebblestone)\n\\]\n\nHence\n\\[\nbuttercup(pebblestone)=moonflower \\, e^{tumbleweed(pebblestone)}\n\\]\nwhere \\( moonflower \\) is a positive constant and\n\\[\ntumbleweed(pebblestone)=\\int^{pebblestone} \\frac{lighthouse^{\\prime}(marigold)}{marigold-lighthouse(marigold)} d marigold=\\int^{pebblestone} \\frac{d marigold}{marigold-lighthouse(marigold)}-\\log (pebblestone-lighthouse(pebblestone))\n\\]\n\nThus\n\\[\nbuttercup(pebblestone)=\\frac{moonflower}{pebblestone-lighthouse(pebblestone)} \\exp \\int^{pebblestone} \\frac{d marigold}{marigold-lighthouse(marigold)}\n\\]\n\nSubstituting (5) into (2) gives us\n\\[\nwindchimes(pebblestone)=moonflower \\frac{lighthouse^{\\prime}(pebblestone)}{(pebblestone-lighthouse(pebblestone))^{2}} \\exp \\int^{pebblestone} \\frac{d marigold}{marigold-lighthouse(marigold)}\n\\]\nas required.\nUnder the conditions of the problem, \\( pebblestone>lighthouse(pebblestone) \\) for all positive \\( pebblestone \\), so the denominator \\( pebblestone-lighthouse(pebblestone) \\) causes no singularities for positive \\( pebblestone \\). For \\( pebblestone<0 \\), we have \\( lighthouse(pebblestone)>pebblestone \\), so we must replace \\( \\log (pebblestone-lighthouse(pebblestone)) \\) by \\( \\log (lighthouse(pebblestone)-pebblestone) \\) in (4). Then (6) becomes\n\\[\nwindchimes(pebblestone)=\\frac{-moonflower \\, lighthouse^{\\prime}(pebblestone)}{(pebblestone-lighthouse(pebblestone))^{2}} \\exp \\int^{pebblestone} \\frac{d marigold}{marigold-lighthouse(marigold)}\n\\]\n\nHowever, in this case the constant \\( moonflower \\) in (3) must be negative so we again have the required form."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "timeless",
        "g": "randomness",
        "f": "constant",
        "F": "derivative",
        "\\psi": "ignorance",
        "a": "variable",
        "A": "negative"
      },
      "question": "6. If the \\( fixedvalue \\)-coordinate \\( \\bar{fixedvalue} \\) of the center of mass of the area lying between the \\( fixedvalue \\)-axis and the curve \\( y=constant(fixedvalue),(constant(fixedvalue)>0) \\), and between the lines \\( fixedvalue=0 \\) and \\( fixedvalue \\) \\( =variable \\) is given by\n\\[\n\\bar{fixedvalue}=randomness(variable)\n\\]\nshow that\n\\[\nconstant(fixedvalue)=negative \\frac{randomness^{\\prime}(fixedvalue)}{[fixedvalue-randomness(fixedvalue)]^{2}} e^{\\int d fixedvalue /(fixedvalue-randomness(fixedvalue))}\n\\]\nwhere \\( negative \\) is a positive constant.",
      "solution": "Solution. By the definition of centroid,\n\\[\nrandomness(fixedvalue)=\\frac{\\int_{0}^{fixedvalue} timeless constant(timeless) d timeless}{\\int_{0}^{fixedvalue} constant(timeless) d timeless}, \\quad fixedvalue \\neq 0 .\n\\]\n\nWe confine our attention to positive values of \\( fixedvalue \\). Put\n\\[\nderivative(fixedvalue)=\\int_{0}^{fixedvalue} constant(timeless) d timeless ; \\quad \\text { then } derivative^{\\prime}=constant\n\\]\n\nWrite (1) in the form\n\\[\nderivative(fixedvalue) randomness(fixedvalue)=\\int_{0}^{fixedvalue} timeless constant(timeless) d timeless,\n\\]\nand differentiate to get\n\\[\nderivative^{\\prime}(fixedvalue) randomness(fixedvalue)+derivative(fixedvalue) randomness^{\\prime}(fixedvalue)=fixedvalue constant(fixedvalue)=fixedvalue derivative^{\\prime}(fixedvalue)\n\\]\n\nThus \\( \\boldsymbol{derivative} \\) satisfies the linear differential equation\n\\[\nderivative^{\\prime}(fixedvalue)=\\frac{randomness^{\\prime}(fixedvalue)}{fixedvalue-randomness(fixedvalue)} derivative(fixedvalue)\n\\]\n\nHence\n\\[\nderivative(fixedvalue)=negative e^{ignorance(fixedvalue)}\n\\]\nwhere \\( negative \\) is a positive constant and\n\\[\nignorance(fixedvalue)=\\int^{fixedvalue} \\frac{randomness^{\\prime}(timeless)}{timeless-randomness(timeless)} d timeless=\\int^{fixedvalue} \\frac{d timeless}{timeless-randomness(timeless)}-\\log (fixedvalue-randomness(fixedvalue))\n\\]\n\nThus\n\\[\nderivative(fixedvalue)=\\frac{negative}{fixedvalue-randomness(fixedvalue)} \\exp \\int^{fixedvalue} \\frac{d timeless}{timeless-randomness(timeless)}\n\\]\n\nSubstituting (5) into (2) gives us\n\\[\nconstant(fixedvalue)=negative \\frac{randomness^{\\prime}(fixedvalue)}{(fixedvalue-randomness(fixedvalue))^{2}} \\exp \\int^{fixedvalue} \\frac{d timeless}{timeless-randomness(timeless)}\n\\]\nas required.\nUnder the conditions of the problem, \\( fixedvalue>randomness(fixedvalue) \\) for all positive \\( fixedvalue \\), so the denominator \\( fixedvalue-randomness(fixedvalue) \\) causes no singularities for positive \\( fixedvalue \\). For \\( fixedvalue<0 \\), we have \\( randomness(fixedvalue)>fixedvalue \\), so we must replace \\( \\log (fixedvalue-randomness(fixedvalue)) \\) by \\( \\log (randomness(fixedvalue)-fixedvalue) \\) in (4). Then (6) becomes\n\\[\nconstant(fixedvalue)=\\frac{-negative randomness^{\\prime}(fixedvalue)}{(fixedvalue-randomness(fixedvalue))^{2}} \\exp \\int^{fixedvalue} \\frac{d timeless}{timeless-randomness(timeless)}\n\\]\n\nHowever, in this case the constant \\( negative \\) in (3) must be negative so we again have the required form."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "g": "plmnxbcq",
        "f": "rtyhqsld",
        "F": "bdmvxpla",
        "\\psi": "zpkhmytr",
        "a": "scvmbnld",
        "A": "wqlkzjtr"
      },
      "question": "6. If the \\( qzxwvtnp \\)-coordinate \\( \\bar{qzxwvtnp} \\) of the center of mass of the area lying between the \\( qzxwvtnp \\)-axis and the curve \\( y=rtyhqsld(qzxwvtnp),(rtyhqsld(qzxwvtnp)>0) \\), and between the lines \\( qzxwvtnp=0 \\) and \\( qzxwvtnp \\) \\( =scvmbnld \\) is given by\n\\[\n\\bar{qzxwvtnp}=plmnxbcq(scvmbnld)\n\\]\nshow that\n\\[\nrtyhqsld(qzxwvtnp)=wqlkzjtr \\frac{plmnxbcq^{\\prime}(qzxwvtnp)}{[qzxwvtnp-plmnxbcq(qzxwvtnp)]^{2}} e^{\\int d qzxwvtnp /(qzxwvtnp-plmnxbcq(qzxwvtnp))}\n\\]\nwhere \\( wqlkzjtr \\) is a positive constant.",
      "solution": "Solution. By the definition of centroid,\n\\[\nplmnxbcq(qzxwvtnp)=\\frac{\\int_{0}^{qzxwvtnp} hjgrksla \\, rtyhqsld(hjgrksla) d hjgrksla}{\\int_{0}^{qzxwvtnp} rtyhqsld(hjgrksla) d hjgrksla}, \\quad qzxwvtnp \\neq 0 .\n\\]\n\nWe confine our attention to positive values of \\( qzxwvtnp \\). Put\n\\[\nbdmvxpla(qzxwvtnp)=\\int_{0}^{qzxwvtnp} rtyhqsld(hjgrksla) d hjgrksla ; \\quad \\text { then } bdmvxpla^{\\prime}=rtyhqsld\n\\]\n\nWrite (1) in the form\n\\[\nbdmvxpla(qzxwvtnp) plmnxbcq(qzxwvtnp)=\\int_{0}^{qzxwvtnp} hjgrksla \\, rtyhqsld(hjgrksla) d hjgrksla,\n\\]\nand differentiate to get\n\\[\nbdmvxpla^{\\prime}(qzxwvtnp) plmnxbcq(qzxwvtnp)+bdmvxpla(qzxwvtnp) plmnxbcq^{\\prime}(qzxwvtnp)=qzxwvtnp \\, rtyhqsld(qzxwvtnp)=qzxwvtnp \\, bdmvxpla^{\\prime}(qzxwvtnp)\n\\]\n\nThus \\( \\boldsymbol{bdmvxpla} \\) satisfies the linear differential equation\n\\[\nbdmvxpla^{\\prime}(qzxwvtnp)=\\frac{plmnxbcq^{\\prime}(qzxwvtnp)}{qzxwvtnp - plmnxbcq(qzxwvtnp)} bdmvxpla(qzxwvtnp)\n\\]\n\nHence\n\\[\nbdmvxpla(qzxwvtnp)=wqlkzjtr e^{zpkhmytr(qzxwvtnp)}\n\\]\nwhere \\( wqlkzjtr \\) is a positive constant and\n\\[\nzpkhmytr(qzxwvtnp)=\\int^{qzxwvtnp} \\frac{plmnxbcq^{\\prime}(hjgrksla)}{hjgrksla-plmnxbcq(hjgrksla)} d hjgrksla=\\int^{qzxwvtnp} \\frac{d hjgrksla}{hjgrksla-plmnxbcq(hjgrksla)}-\\log (qzxwvtnp-plmnxbcq(qzxwvtnp))\n\\]\n\nThus\n\\[\nbdmvxpla(qzxwvtnp)=\\frac{wqlkzjtr}{qzxwvtnp-plmnxbcq(qzxwvtnp)} \\exp \\int^{qzxwvtnp} \\frac{d hjgrksla}{hjgrksla-plmnxbcq(hjgrksla)}\n\\]\n\nSubstituting (5) into (2) gives us\n\\[\nrtyhqsld(qzxwvtnp)=wqlkzjtr \\frac{plmnxbcq^{\\prime}(qzxwvtnp)}{(qzxwvtnp-plmnxbcq(qzxwvtnp))^{2}} \\exp \\int^{qzxwvtnp} \\frac{d hjgrksla}{hjgrksla-plmnxbcq(hjgrksla)}\n\\]\nas required.\n\nUnder the conditions of the problem, \\( qzxwvtnp>plmnxbcq(qzxwvtnp) \\) for all positive \\( qzxwvtnp \\), so the denominator \\( qzxwvtnp-plmnxbcq(qzxwvtnp) \\) causes no singularities for positive \\( qzxwvtnp \\). For \\( qzxwvtnp<0 \\), we have \\( plmnxbcq(qzxwvtnp)>qzxwvtnp \\), so we must replace \\( \\log (qzxwvtnp-plmnxbcq(qzxwvtnp)) \\) by \\( \\log (plmnxbcq(qzxwvtnp)-qzxwvtnp) \\) in (4). Then (6) becomes\n\\[\nrtyhqsld(qzxwvtnp)=\\frac{-wqlkzjtr \\, plmnxbcq^{\\prime}(qzxwvtnp)}{(qzxwvtnp-plmnxbcq(qzxwvtnp))^{2}} \\exp \\int^{qzxwvtnp} \\frac{d hjgrksla}{hjgrksla-plmnxbcq(hjgrksla)}\n\\]\n\nHowever, in this case the constant \\( wqlkzjtr \\) in (3) must be negative so we again have the required form."
    },
    "kernel_variant": {
      "question": "Fix an integer $n\\ge 3$ and a real number $b$.  \nLet  \n\n\\[\nu,\\;v:(b,\\infty)\\longrightarrow(0,\\infty)\n\\]\n\nbe $C^{1}$–functions satisfying  \n\n\\[\n0<v(x)<u(x)\\qquad\\forall x>b .\n\\]\n\nFor every $a>b$ form the hollow $n$–dimensional body  \n\n\\[\nS(a)=\\Bigl\\{(x,y)\\in\\mathbb R\\times\\mathbb R^{\\,n-1}:\n               \\;b\\le x\\le a,\\;\n               v(x)\\le |y|\\le u(x)\\Bigr\\},\n\\]\n\nobtained by revolving, about the $x$–axis, the graphs $r=u(x)$ (outer boundary) and  \n$r=v(x)$ (inner cavity).  \nThe material has uniform density $\\rho\\equiv1$.\n\nIntroduce  \n\n\\[\n\\begin{aligned}\n&V(a)=\\text{volume of }S(a),\\\\[2pt]\n&g(a)=\\text{$x$–coordinate of the centroid of }S(a),\\\\[2pt]\n&J(a)=\\int_{S(a)}|y|^{2}\\,dV\\quad\\text{(polar moment of inertia about the $x$–axis)}.\n\\end{aligned}\n\\]\n\nAssume that, for every $x>b$,  \n\n\\[\nx>g(x),\\qquad g'(x)>0 .\n\\]\n\nFurthermore suppose  \n\n\\[\n\\begin{cases}\n\\text{\\rm(i)} & g\\in C^{1}(b,\\infty),\\\\[4pt]\n\\text{\\rm(ii)}& \\displaystyle\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\quad\\text{converges for every }x>b,\\\\[10pt]\n\\text{\\rm(iii)}& h(a):=\\dfrac{J(a)}{V(a)}\\in C^{1}(b,\\infty)\\ \\text{and } h(a)\\neq0,\\\\[10pt]\n\\text{\\rm(iv)} & V'(x)>0,\\;J'(x)>0\\quad\\forall x>b .\n\\end{cases}\n\\]\n\nThroughout set $N:=n-1$ and denote by  \n\n\\[\n\\omega_{N}:=\\frac{\\pi^{N/2}}{\\Gamma\\!\\bigl(1+N/2\\bigr)}\n\\]\n\nthe volume of the $N$–dimensional unit ball.\n\nA.  (Reconstruction of the cross–sectional area)  \nShow that a positive constant  \n\n\\[\nK=K(n,b,u,v)>0\n\\]\n\nexists such that, for every $x>b$,\n\n\\[\n\\boxed{\nu(x)^{N}-v(x)^{N}=K\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\Bigr)\n}\\tag{$\\star$}\n\\]\n\nB.  (First and second sectional moments)  \nDefine  \n\n\\[\n\\Delta(x):=u(x)^{N}-v(x)^{N},\\qquad\n\\Sigma(x):=u(x)^{N+2}-v(x)^{N+2}.\n\\]\n\nProve that  \n\n\\[\n\\boxed{\\Delta(x)=\\frac{V'(x)}{\\omega_{N}}}\\tag{B$_1$}\\qquad\\text{and}\\qquad\n\\boxed{\\Sigma(x)=\\frac{N+2}{N\\,\\omega_{N}}\\,J'(x)}\\tag{B$_2$}.\n\\]\n\nUsing $J(x)=h(x)\\,V(x)$ deduce  \n\n\\[\n\\boxed{\n\\Sigma(x)=\\frac{N+2}{N}\\,\n           \\Bigl[\\,h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\\Bigr]\\,\n           \\Delta(x)\n}\\tag{$\\star\\star$}\n\\]\n\nC.  (Explicit formulae when $n=3$)  \nFix $n=3$ (hence $N=2$).  Show that  \n\n\\[\n\\begin{aligned}\nu(x)^{2}+v(x)^{2}&=\\frac{\\Sigma(x)}{\\Delta(x)},\\tag{C$_1$}\\\\\nu(x)^{2}-v(x)^{2}&=\\Delta(x).\\tag{C$_2$}\n\\end{aligned}\n\\]\n\nHence prove  \n\n\\[\n\\boxed{\nu(x)=\\sqrt{\\dfrac{\\Sigma(x)+\\Delta(x)^{2}}{2\\Delta(x)}},\\qquad\nv(x)=\\sqrt{\\dfrac{\\Sigma(x)-\\Delta(x)^{2}}{2\\Delta(x)}}\n}\\tag{C$_3$}\n\\]\n\nwhere  \n\n\\[\n\\begin{aligned}\n\\Delta(x)&=K\\,\n           g'(x)\\,[x-g(x)]^{-2}\\,\n           \\exp\\!\\Bigl(\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\Bigr),\\tag{C$_4$}\\\\[6pt]\n\\Sigma(x)&=2\\Bigl[\n              h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\n           \\Bigr]\\,\n           \\Delta(x).\\tag{C$_5$}\n\\end{aligned}\n\\]\n\nFinally verify that $\\Sigma(x)\\pm\\Delta(x)^{2}>0$ for all $x>b$, so that $u(x)>v(x)>0$.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "\\textbf{Step 0.  Cross–sectional geometry.}  \nLet $N:=n-1$.  For each $x>b$ define the annular cross–section  \n\n\\[\nA(x)=\\{\\,y\\in\\mathbb R^{N}: v(x)\\le|y|\\le u(x)\\}.\n\\]\n\nIn polar coordinates  \n\n\\[\n\\lvert A(x)\\rvert=\\omega_{N}\\bigl(u(x)^{N}-v(x)^{N}\\bigr),\\tag{S0.1}\n\\]\n\n\\[\nI_{2}(x):=\\int_{A(x)}|y|^{2}\\,dA=\\frac{N\\,\\omega_{N}}{N+2}\\bigl(u(x)^{N+2}-v(x)^{N+2}\\bigr).\\tag{S0.2}\n\\]\n\nSet  \n\n\\[\nV(x):=\\int_{b}^{x}\\lvert A(t)\\rvert\\,dt,\\qquad  \nJ(x):=\\int_{b}^{x}I_{2}(t)\\,dt,\n\\]\n\nso that $V'(x)=\\lvert A(x)\\rvert$ and $J'(x)=I_{2}(x)$.\n\n\\medskip\n\\textbf{Step 1.  A differential equation for $V$.}  \nBy definition of the centroid,\n\n\\[\ng(x)\\,V(x)=\\int_{b}^{x}t\\,\\lvert A(t)\\rvert\\,dt.\\tag{S1.1}\n\\]\n\nDifferentiating,\n\n\\[\n\\lvert A(x)\\rvert\\,g(x)+V(x)\\,g'(x)=x\\,\\lvert A(x)\\rvert.\\tag{S1.2}\n\\]\n\nBecause $\\lvert A\\rvert=V'$, division by $V$ gives  \n\n\\[\n\\frac{V'(x)}{V(x)}=\\frac{g'(x)}{x-g(x)}.\\tag{S1.3}\n\\]\n\nAssumption (ii) guarantees that the integrating factor is well–defined, yielding  \n\n\\[\nV(x)=A_{0}\\,\n      \\frac{\\exp\\!\\bigl(\\displaystyle\\int_{b}^{x}\\frac{dt}{t-g(t)}\\bigr)}\n           {x-g(x)},\\qquad A_{0}>0.\\tag{S1.4}\n\\]\n\n\\textbf{Strict positivity of $A_{0}$.}  \nBecause $x>g(x)$ and $V'(x)=\\lvert A(x)\\rvert>0$, formula (S1.3) implies $V>0$ on $(b,\\infty)$, hence $A_{0}>0$.\n\nDifferentiating (S1.4),\n\n\\[\n\\lvert A(x)\\rvert=V'(x)=A_{0}\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr).\\tag{S1.5}\n\\]\n\nInsert (S0.1) into (S1.5) and set $K:=A_{0}/\\omega_{N}>0$; then  \n\n\\[\nu(x)^{N}-v(x)^{N}=K\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr),\n\\]\n\nwhich is exactly ($\\star$).\n\n\\medskip\n\\textbf{Step 2.  Proof of (B$_1$) and (B$_2$).}  \nFormula (S0.1) gives $\\Delta(x)=\\lvert A(x)\\rvert/\\omega_{N}=V'(x)/\\omega_{N}$, proving (B$_1$).  \nSimilarly, (S0.2) implies  \n\n\\[\nJ'(x)=I_{2}(x)=\\frac{N\\,\\omega_{N}}{N+2}\\,\\Sigma(x)\n\\quad\\Longrightarrow\\quad\n\\Sigma(x)=\\frac{N+2}{N\\,\\omega_{N}}\\,J'(x),\n\\]\n\nwhich is (B$_2$).\n\n\\medskip\n\\textbf{Step 3.  Eliminating $J$ and $V$.}  \nWith $J(x)=h(x)\\,V(x)$,\n\n\\[\nJ'(x)=h'(x)\\,V(x)+h(x)\\,V'(x).\\tag{S3.1}\n\\]\n\nInsert (S3.1) into (B$_2$) and use (B$_1$):\n\n\\[\n\\Sigma(x)=\\frac{N+2}{N}\\Bigl[h'(x)\\,\\frac{V(x)}{V'(x)}+h(x)\\Bigr]\\Delta(x).\n\\]\n\nFinally, (S1.3) gives $V(x)/V'(x)=(x-g(x))/g'(x)$; hence ($\\star\\star$) follows.\n\n\\medskip\n\\textbf{Step 4.  Specialisation to $n=3$ ($N=2$).}  \nWith $N=2$, $\\omega_{2}=\\pi$ and $(N+2)/N=2$.  \nEquation ($\\star$) becomes  \n\n\\[\n\\Delta(x)=K\\,\n          g'(x)\\,[x-g(x)]^{-2}\\,\n          \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr),\\tag{C4}\n\\]\n\nwhile ($\\star\\star$) reads  \n\n\\[\n\\Sigma(x)=2\\Bigl[\n              h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\n           \\Bigr]\\,\\Delta(x).\\tag{C5}\n\\]\n\n\\medskip\n\\textbf{Step 5.  Extracting $u$ and $v$.}  \nFor $N=2$ the algebraic identity  \n\n\\[\nu^{4}-v^{4}=(u^{2}-v^{2})(u^{2}+v^{2})\\tag{S5.1}\n\\]\n\nholds, i.e.\\ $\\Sigma(x)=\\Delta(x)\\bigl(u(x)^{2}+v(x)^{2}\\bigr)$.  Consequently\n\n\\[\nu(x)^{2}+v(x)^{2}=\\frac{\\Sigma(x)}{\\Delta(x)},\\qquad\nu(x)^{2}-v(x)^{2}=\\Delta(x),\\tag{S5.2}\n\\]\n\nwhich are (C$_1$)–(C$_2$).  Solving the linear system (add and subtract) yields  \n\n\\[\nu(x)^{2}=\\frac{\\Sigma(x)+\\Delta(x)^{2}}{2\\Delta(x)},\\qquad\nv(x)^{2}=\\frac{\\Sigma(x)-\\Delta(x)^{2}}{2\\Delta(x)},\\tag{S5.3}\n\\]\n\nnamely (C$_3$).\n\n\\smallskip\n\\textbf{Positivity of the radicands.}  \nBecause $u>v>0$ we have $\\Delta>0$.  Moreover, $\\Sigma=u^{4}-v^{4}>\\Delta^{2}$, hence  \n$\\Sigma\\pm\\Delta^{2} >0$, so the square roots in (C$_3$) are real and positive. With $\\Sigma>\\Delta^{2}$ the second square root is smaller, giving $u(x)>v(x)>0$ for all $x>b$.\n\n\\smallskip\n\\textbf{Dependence on prescribed data.}  \nThe right-hand sides of (C$_3$)–(C$_5$) involve only $g,h$, the constant $K$, and universal constants.  Thus the radial profiles $u$ and $v$ are uniquely determined by the one–dimensional data $g$ and $h$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.384067",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $n\\ge 3$ and a real number $b$.  \nLet  \n\n\\[\nu,\\;v:(b,\\infty)\\longrightarrow(0,\\infty)\n\\]\n\nbe $C^{1}$–functions satisfying  \n\n\\[\n0<v(x)<u(x)\\qquad\\forall x>b .\n\\]\n\nFor every $a>b$ form the hollow $n$–dimensional body  \n\n\\[\nS(a)=\\Bigl\\{(x,y)\\in\\mathbb R\\times\\mathbb R^{\\,n-1}:\n               \\;b\\le x\\le a,\\;\n               v(x)\\le |y|\\le u(x)\\Bigr\\},\n\\]\n\nobtained by revolving, about the $x$–axis, the graphs $r=u(x)$ (outer boundary) and  \n$r=v(x)$ (inner cavity).  \nThe material has uniform density $\\rho\\equiv1$.\n\nIntroduce  \n\n\\[\n\\begin{aligned}\n&V(a)=\\text{volume of }S(a),\\\\[2pt]\n&g(a)=\\text{$x$–coordinate of the centroid of }S(a),\\\\[2pt]\n&J(a)=\\int_{S(a)}|y|^{2}\\,dV\\quad\\text{(polar moment of inertia about the $x$–axis)}.\n\\end{aligned}\n\\]\n\nAssume that, for every $x>b$,  \n\n\\[\nx>g(x),\\qquad g'(x)>0 .\n\\]\n\nFurthermore suppose  \n\n\\[\n\\begin{cases}\n\\text{\\rm(i)} & g\\in C^{1}(b,\\infty),\\\\[4pt]\n\\text{\\rm(ii)}& \\displaystyle\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\quad\\text{converges for every }x>b,\\\\[10pt]\n\\text{\\rm(iii)}& h(a):=\\dfrac{J(a)}{V(a)}\\in C^{1}(b,\\infty)\\ \\text{and } h(a)\\neq0,\\\\[10pt]\n\\text{\\rm(iv)} & V'(x)>0,\\;J'(x)>0\\quad\\forall x>b .\n\\end{cases}\n\\]\n\nThroughout set $N:=n-1$ and denote by  \n\n\\[\n\\omega_{N}:=\\frac{\\pi^{N/2}}{\\Gamma\\!\\bigl(1+N/2\\bigr)}\n\\]\n\nthe volume of the $N$–dimensional unit ball.\n\nA.  (Reconstruction of the cross–sectional area)  \nShow that a positive constant  \n\n\\[\nK=K(n,b,u,v)>0\n\\]\n\nexists such that, for every $x>b$,\n\n\\[\n\\boxed{\nu(x)^{N}-v(x)^{N}=K\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\Bigr)\n}\\tag{$\\star$}\n\\]\n\nB.  (First and second sectional moments)  \nDefine  \n\n\\[\n\\Delta(x):=u(x)^{N}-v(x)^{N},\\qquad\n\\Sigma(x):=u(x)^{N+2}-v(x)^{N+2}.\n\\]\n\nProve that  \n\n\\[\n\\boxed{\\Delta(x)=\\frac{V'(x)}{\\omega_{N}}}\\tag{B$_1$}\\qquad\\text{and}\\qquad\n\\boxed{\\Sigma(x)=\\frac{N+2}{N\\,\\omega_{N}}\\,J'(x)}\\tag{B$_2$}.\n\\]\n\nUsing $J(x)=h(x)\\,V(x)$ deduce  \n\n\\[\n\\boxed{\n\\Sigma(x)=\\frac{N+2}{N}\\,\n           \\Bigl[\\,h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\\Bigr]\\,\n           \\Delta(x)\n}\\tag{$\\star\\star$}\n\\]\n\nC.  (Explicit formulae when $n=3$)  \nFix $n=3$ (hence $N=2$).  Show that  \n\n\\[\n\\begin{aligned}\nu(x)^{2}+v(x)^{2}&=\\frac{\\Sigma(x)}{\\Delta(x)},\\tag{C$_1$}\\\\\nu(x)^{2}-v(x)^{2}&=\\Delta(x).\\tag{C$_2$}\n\\end{aligned}\n\\]\n\nHence prove  \n\n\\[\n\\boxed{\nu(x)=\\sqrt{\\dfrac{\\Sigma(x)+\\Delta(x)^{2}}{2\\Delta(x)}},\\qquad\nv(x)=\\sqrt{\\dfrac{\\Sigma(x)-\\Delta(x)^{2}}{2\\Delta(x)}}\n}\\tag{C$_3$}\n\\]\n\nwhere  \n\n\\[\n\\begin{aligned}\n\\Delta(x)&=K\\,\n           g'(x)\\,[x-g(x)]^{-2}\\,\n           \\exp\\!\\Bigl(\\int_{b}^{x}\\frac{dt}{\\,t-g(t)\\,}\\Bigr),\\tag{C$_4$}\\\\[6pt]\n\\Sigma(x)&=2\\Bigl[\n              h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\n           \\Bigr]\\,\n           \\Delta(x).\\tag{C$_5$}\n\\end{aligned}\n\\]\n\nFinally verify that $\\Sigma(x)\\pm\\Delta(x)^{2}>0$ for all $x>b$, so that $u(x)>v(x)>0$.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "\\textbf{Step 0.  Cross–sectional geometry.}  \nLet $N:=n-1$.  For each $x>b$ define the annular cross–section  \n\n\\[\nA(x)=\\{\\,y\\in\\mathbb R^{N}: v(x)\\le|y|\\le u(x)\\}.\n\\]\n\nIn polar coordinates  \n\n\\[\n\\lvert A(x)\\rvert=\\omega_{N}\\bigl(u(x)^{N}-v(x)^{N}\\bigr),\\tag{S0.1}\n\\]\n\n\\[\nI_{2}(x):=\\int_{A(x)}|y|^{2}\\,dA=\\frac{N\\,\\omega_{N}}{N+2}\\bigl(u(x)^{N+2}-v(x)^{N+2}\\bigr).\\tag{S0.2}\n\\]\n\nSet  \n\n\\[\nV(x):=\\int_{b}^{x}\\lvert A(t)\\rvert\\,dt,\\qquad  \nJ(x):=\\int_{b}^{x}I_{2}(t)\\,dt,\n\\]\n\nso that $V'(x)=\\lvert A(x)\\rvert$ and $J'(x)=I_{2}(x)$.\n\n\\medskip\n\\textbf{Step 1.  A differential equation for $V$.}  \nBy definition of the centroid,\n\n\\[\ng(x)\\,V(x)=\\int_{b}^{x}t\\,\\lvert A(t)\\rvert\\,dt.\\tag{S1.1}\n\\]\n\nDifferentiating,\n\n\\[\n\\lvert A(x)\\rvert\\,g(x)+V(x)\\,g'(x)=x\\,\\lvert A(x)\\rvert.\\tag{S1.2}\n\\]\n\nBecause $\\lvert A\\rvert=V'$, division by $V$ gives  \n\n\\[\n\\frac{V'(x)}{V(x)}=\\frac{g'(x)}{x-g(x)}.\\tag{S1.3}\n\\]\n\nAssumption (ii) guarantees that the integrating factor is well–defined, yielding  \n\n\\[\nV(x)=A_{0}\\,\n      \\frac{\\exp\\!\\bigl(\\displaystyle\\int_{b}^{x}\\frac{dt}{t-g(t)}\\bigr)}\n           {x-g(x)},\\qquad A_{0}>0.\\tag{S1.4}\n\\]\n\n\\textbf{Strict positivity of $A_{0}$.}  \nBecause $x>g(x)$ and $V'(x)=\\lvert A(x)\\rvert>0$, formula (S1.3) implies $V>0$ on $(b,\\infty)$, hence $A_{0}>0$.\n\nDifferentiating (S1.4),\n\n\\[\n\\lvert A(x)\\rvert=V'(x)=A_{0}\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr).\\tag{S1.5}\n\\]\n\nInsert (S0.1) into (S1.5) and set $K:=A_{0}/\\omega_{N}>0$; then  \n\n\\[\nu(x)^{N}-v(x)^{N}=K\\,\n      g'(x)\\,[x-g(x)]^{-2}\\,\n      \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr),\n\\]\n\nwhich is exactly ($\\star$).\n\n\\medskip\n\\textbf{Step 2.  Proof of (B$_1$) and (B$_2$).}  \nFormula (S0.1) gives $\\Delta(x)=\\lvert A(x)\\rvert/\\omega_{N}=V'(x)/\\omega_{N}$, proving (B$_1$).  \nSimilarly, (S0.2) implies  \n\n\\[\nJ'(x)=I_{2}(x)=\\frac{N\\,\\omega_{N}}{N+2}\\,\\Sigma(x)\n\\quad\\Longrightarrow\\quad\n\\Sigma(x)=\\frac{N+2}{N\\,\\omega_{N}}\\,J'(x),\n\\]\n\nwhich is (B$_2$).\n\n\\medskip\n\\textbf{Step 3.  Eliminating $J$ and $V$.}  \nWith $J(x)=h(x)\\,V(x)$,\n\n\\[\nJ'(x)=h'(x)\\,V(x)+h(x)\\,V'(x).\\tag{S3.1}\n\\]\n\nInsert (S3.1) into (B$_2$) and use (B$_1$):\n\n\\[\n\\Sigma(x)=\\frac{N+2}{N}\\Bigl[h'(x)\\,\\frac{V(x)}{V'(x)}+h(x)\\Bigr]\\Delta(x).\n\\]\n\nFinally, (S1.3) gives $V(x)/V'(x)=(x-g(x))/g'(x)$; hence ($\\star\\star$) follows.\n\n\\medskip\n\\textbf{Step 4.  Specialisation to $n=3$ ($N=2$).}  \nWith $N=2$, $\\omega_{2}=\\pi$ and $(N+2)/N=2$.  \nEquation ($\\star$) becomes  \n\n\\[\n\\Delta(x)=K\\,\n          g'(x)\\,[x-g(x)]^{-2}\\,\n          \\exp\\!\\Bigl(\\int_{b}^{x}\\tfrac{dt}{t-g(t)}\\Bigr),\\tag{C4}\n\\]\n\nwhile ($\\star\\star$) reads  \n\n\\[\n\\Sigma(x)=2\\Bigl[\n              h'(x)\\,\\frac{x-g(x)}{g'(x)}+h(x)\n           \\Bigr]\\,\\Delta(x).\\tag{C5}\n\\]\n\n\\medskip\n\\textbf{Step 5.  Extracting $u$ and $v$.}  \nFor $N=2$ the algebraic identity  \n\n\\[\nu^{4}-v^{4}=(u^{2}-v^{2})(u^{2}+v^{2})\\tag{S5.1}\n\\]\n\nholds, i.e.\\ $\\Sigma(x)=\\Delta(x)\\bigl(u(x)^{2}+v(x)^{2}\\bigr)$.  Consequently\n\n\\[\nu(x)^{2}+v(x)^{2}=\\frac{\\Sigma(x)}{\\Delta(x)},\\qquad\nu(x)^{2}-v(x)^{2}=\\Delta(x),\\tag{S5.2}\n\\]\n\nwhich are (C$_1$)–(C$_2$).  Solving the linear system (add and subtract) yields  \n\n\\[\nu(x)^{2}=\\frac{\\Sigma(x)+\\Delta(x)^{2}}{2\\Delta(x)},\\qquad\nv(x)^{2}=\\frac{\\Sigma(x)-\\Delta(x)^{2}}{2\\Delta(x)},\\tag{S5.3}\n\\]\n\nnamely (C$_3$).\n\n\\smallskip\n\\textbf{Positivity of the radicands.}  \nBecause $u>v>0$ we have $\\Delta>0$.  Moreover, $\\Sigma=u^{4}-v^{4}>\\Delta^{2}$, hence  \n$\\Sigma\\pm\\Delta^{2} >0$, so the square roots in (C$_3$) are real and positive. With $\\Sigma>\\Delta^{2}$ the second square root is smaller, giving $u(x)>v(x)>0$ for all $x>b$.\n\n\\smallskip\n\\textbf{Dependence on prescribed data.}  \nThe right-hand sides of (C$_3$)–(C$_5$) involve only $g,h$, the constant $K$, and universal constants.  Thus the radial profiles $u$ and $v$ are uniquely determined by the one–dimensional data $g$ and $h$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.331455",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}