{
  "index": "1981-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem B-2\nDetermine the minimum value of\n\\[\n(r-1)^{2}+\\left(\\frac{s}{r}-1\\right)^{2}+\\left(\\frac{t}{s}-1\\right)^{2}+\\left(\\frac{4}{t}-1\\right)^{2}\n\\]\nfor all real numbers \\( r, s, t \\) with \\( 1 \\leq r \\leq s \\leq t \\leq 4 \\).",
  "solution": "B-2.\nFirst we let \\( 0<a<b \\) and seek the \\( x \\) that minimizes\n\\[\nf(x)=\\left(\\frac{x}{a}-1\\right)^{2}+\\left(\\frac{b}{x}-1\\right)^{2} \\text { on } a \\leq x \\leq b\n\\]\n\nLet \\( x / a=z \\) and \\( b / a=c \\). Then\n\\[\nf(x)=g(z)=(z-1)^{2}+\\left(\\frac{c}{z}-1\\right)^{2}\n\\]\n\nNow \\( g^{\\prime}(z)=0 \\) implies\n\\[\nz^{4}-z^{3}+c z-c^{2}=\\left(z^{2}-c\\right)\\left(z^{2}-z+c\\right)=0\n\\]\nthe only positive solution is \\( z=\\sqrt{c} \\). Since \\( 0<a<b, c>1, \\sqrt{c}>1 \\), and\n\\[\ng(1)=g(c)=(c-1)^{2}=(\\sqrt{c}-1)^{2}(\\sqrt{c}+1)^{2}>2(\\sqrt{c}-1)^{2}=g(\\sqrt{c})\n\\]\n\nHence the minimum of \\( g(z) \\) on \\( 1 \\leq z \\leq c \\) occurs at \\( z=\\sqrt{c} \\). It follows that the minimum for \\( f(x) \\) on \\( a \\leq x \\leq b \\) occurs at \\( x=a \\sqrt{b / a}=\\sqrt{a b} \\). Then the minimum for the given function of \\( r, s, t \\) occurs with \\( r=\\sqrt{s}, t=\\sqrt{4 s}=2 r \\), and \\( s=\\sqrt{r t}=r \\sqrt{2} \\). These imply that \\( r=\\sqrt{2}, s=2, t=2 \\sqrt{2} \\). Thus the desired minimum value is \\( 4(\\sqrt{2}-1)^{2}=12-8 \\sqrt{2} \\).",
  "vars": [
    "r",
    "s",
    "t",
    "x",
    "z"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "varradius",
        "s": "varscalar",
        "t": "vartangent",
        "x": "varxenon",
        "z": "varzenith",
        "a": "paramanchor",
        "b": "parambravo",
        "c": "paramcharlie"
      },
      "question": "Problem B-2\nDetermine the minimum value of\n\\[\n(varradius-1)^{2}+\\left(\\frac{varscalar}{varradius}-1\\right)^{2}+\\left(\\frac{vartangent}{varscalar}-1\\right)^{2}+\\left(\\frac{4}{vartangent}-1\\right)^{2}\n\\]\nfor all real numbers \\( varradius, varscalar, vartangent \\) with \\( 1 \\leq varradius \\leq varscalar \\leq vartangent \\leq 4 \\).",
      "solution": "B-2.\nFirst we let \\( 0<paramanchor<parambravo \\) and seek the \\( varxenon \\) that minimizes\n\\[\nf(varxenon)=\\left(\\frac{varxenon}{paramanchor}-1\\right)^{2}+\\left(\\frac{parambravo}{varxenon}-1\\right)^{2} \\text { on } paramanchor \\leq varxenon \\leq parambravo\n\\]\n\nLet \\( varxenon / paramanchor=varzenith \\) and \\( parambravo / paramanchor=paramcharlie \\). Then\n\\[\nf(varxenon)=g(varzenith)=(varzenith-1)^{2}+\\left(\\frac{paramcharlie}{varzenith}-1\\right)^{2}\n\\]\n\nNow \\( g^{\\prime}(varzenith)=0 \\) implies\n\\[\nvarzenith^{4}-varzenith^{3}+paramcharlie \\, varzenith-paramcharlie^{2}=\\left(varzenith^{2}-paramcharlie\\right)\\left(varzenith^{2}-varzenith+paramcharlie\\right)=0\n\\]\nthe only positive solution is \\( varzenith=\\sqrt{paramcharlie} \\). Since \\( 0<paramanchor<parambravo, paramcharlie>1, \\sqrt{paramcharlie}>1 \\), and\n\\[\ng(1)=g(paramcharlie)=(paramcharlie-1)^{2}=(\\sqrt{paramcharlie}-1)^{2}(\\sqrt{paramcharlie}+1)^{2}>2(\\sqrt{paramcharlie}-1)^{2}=g(\\sqrt{paramcharlie})\n\\]\n\nHence the minimum of \\( g(varzenith) \\) on \\( 1 \\leq varzenith \\leq paramcharlie \\) occurs at \\( varzenith=\\sqrt{paramcharlie} \\). It follows that the minimum for \\( f(varxenon) \\) on \\( paramanchor \\leq varxenon \\leq parambravo \\) occurs at \\( varxenon=paramanchor \\sqrt{parambravo / paramanchor}=\\sqrt{paramanchor \\, parambravo} \\). Then the minimum for the given function of \\( varradius, varscalar, vartangent \\) occurs with \\( varradius=\\sqrt{varscalar}, vartangent=\\sqrt{4 varscalar}=2 varradius \\), and \\( varscalar=\\sqrt{varradius \\, vartangent}=varradius \\sqrt{2} \\). These imply that \\( varradius=\\sqrt{2}, varscalar=2, vartangent=2 \\sqrt{2} \\). Thus the desired minimum value is \\( 4(\\sqrt{2}-1)^{2}=12-8 \\sqrt{2} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "peppermint",
        "s": "sandstone",
        "t": "firebrick",
        "x": "toadflax",
        "z": "nightshade",
        "a": "rainstorm",
        "b": "moonlight",
        "c": "driftwood"
      },
      "question": "Problem B-2\nDetermine the minimum value of\n\\[\n(peppermint-1)^{2}+\\left(\\frac{sandstone}{peppermint}-1\\right)^{2}+\\left(\\frac{firebrick}{sandstone}-1\\right)^{2}+\\left(\\frac{4}{firebrick}-1\\right)^{2}\n\\]\nfor all real numbers \\( peppermint, sandstone, firebrick \\) with \\( 1 \\leq peppermint \\leq sandstone \\leq firebrick \\leq 4 \\).",
      "solution": "B-2.\nFirst we let \\( 0<rainstorm<moonlight \\) and seek the \\( toadflax \\) that minimizes\n\\[\nf(toadflax)=\\left(\\frac{toadflax}{rainstorm}-1\\right)^{2}+\\left(\\frac{moonlight}{toadflax}-1\\right)^{2} \\text { on } rainstorm \\leq toadflax \\leq moonlight\n\\]\n\nLet \\( toadflax / rainstorm=nightshade \\) and \\( moonlight / rainstorm=driftwood \\). Then\n\\[\nf(toadflax)=g(nightshade)=(nightshade-1)^{2}+\\left(\\frac{driftwood}{nightshade}-1\\right)^{2}\n\\]\n\nNow \\( g^{\\prime}(nightshade)=0 \\) implies\n\\[\nnightshade^{4}-nightshade^{3}+driftwood\\, nightshade-driftwood^{2}=\\left(nightshade^{2}-driftwood\\right)\\left(nightshade^{2}-nightshade+driftwood\\right)=0\n\\]\nthe only positive solution is \\( nightshade=\\sqrt{driftwood} \\). Since \\( 0<rainstorm<moonlight, driftwood>1, \\sqrt{driftwood}>1 \\), and\n\\[\ng(1)=g(driftwood)=(driftwood-1)^{2}=(\\sqrt{driftwood}-1)^{2}(\\sqrt{driftwood}+1)^{2}>2(\\sqrt{driftwood}-1)^{2}=g(\\sqrt{driftwood})\n\\]\n\nHence the minimum of \\( g(nightshade) \\) on \\( 1 \\leq nightshade \\leq driftwood \\) occurs at \\( nightshade=\\sqrt{driftwood} \\). It follows that the minimum for \\( f(toadflax) \\) on \\( rainstorm \\leq toadflax \\leq moonlight \\) occurs at \\( toadflax=rainstorm \\sqrt{moonlight / rainstorm}=\\sqrt{rainstorm moonlight} \\). Then the minimum for the given function of \\( peppermint, sandstone, firebrick \\) occurs with \\( peppermint=\\sqrt{sandstone}, firebrick=\\sqrt{4 sandstone}=2 peppermint \\), and \\( sandstone=\\sqrt{peppermint firebrick}=peppermint \\sqrt{2} \\). These imply that \\( peppermint=\\sqrt{2}, sandstone=2, firebrick=2 \\sqrt{2} \\). Thus the desired minimum value is \\( 4(\\sqrt{2}-1)^{2}=12-8 \\sqrt{2} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "fixedvalue",
        "s": "steadyvalue",
        "t": "rigidvalue",
        "x": "knownnumber",
        "z": "certainnum",
        "a": "variabledata",
        "b": "shiftingvalue",
        "c": "mutablevalue"
      },
      "question": "Problem B-2\nDetermine the minimum value of\n\\[\n(fixedvalue-1)^{2}+\\left(\\frac{steadyvalue}{fixedvalue}-1\\right)^{2}+\\left(\\frac{rigidvalue}{steadyvalue}-1\\right)^{2}+\\left(\\frac{4}{rigidvalue}-1\\right)^{2}\n\\]\nfor all real numbers \\( fixedvalue, steadyvalue, rigidvalue \\) with \\( 1 \\leq fixedvalue \\leq steadyvalue \\leq rigidvalue \\leq 4 \\).",
      "solution": "B-2.\nFirst we let \\( 0<variabledata<shiftingvalue \\) and seek the \\( knownnumber \\) that minimizes\n\\[\nf(knownnumber)=\\left(\\frac{knownnumber}{variabledata}-1\\right)^{2}+\\left(\\frac{shiftingvalue}{knownnumber}-1\\right)^{2} \\text { on } variabledata \\leq knownnumber \\leq shiftingvalue\n\\]\n\nLet \\( knownnumber / variabledata=certainnum \\) and \\( shiftingvalue / variabledata=mutablevalue \\). Then\n\\[\nf(knownnumber)=g(certainnum)=(certainnum-1)^{2}+\\left(\\frac{mutablevalue}{certainnum}-1\\right)^{2}\n\\]\n\nNow \\( g^{\\prime}(certainnum)=0 \\) implies\n\\[\ncertainnum^{4}-certainnum^{3}+mutablevalue\\;certainnum-mutablevalue^{2}=\\left(certainnum^{2}-mutablevalue\\right)\\left(certainnum^{2}-certainnum+mutablevalue\\right)=0\n\\]\nthe only positive solution is \\( certainnum=\\sqrt{mutablevalue} \\). Since \\( 0<variabledata<shiftingvalue, mutablevalue>1, \\sqrt{mutablevalue}>1 \\), and\n\\[\ng(1)=g(mutablevalue)=(mutablevalue-1)^{2}=(\\sqrt{mutablevalue}-1)^{2}(\\sqrt{mutablevalue}+1)^{2}>2(\\sqrt{mutablevalue}-1)^{2}=g(\\sqrt{mutablevalue})\n\\]\n\nHence the minimum of \\( g(certainnum) \\) on \\( 1 \\leq certainnum \\leq mutablevalue \\) occurs at \\( certainnum=\\sqrt{mutablevalue} \\). It follows that the minimum for \\( f(knownnumber) \\) on \\( variabledata \\leq knownnumber \\leq shiftingvalue \\) occurs at \\( knownnumber=variabledata \\sqrt{shiftingvalue / variabledata}=\\sqrt{variabledata\\;shiftingvalue} \\). Then the minimum for the given function of \\( fixedvalue, steadyvalue, rigidvalue \\) occurs with \\( fixedvalue=\\sqrt{steadyvalue}, rigidvalue=\\sqrt{4\\;steadyvalue}=2 fixedvalue \\), and \\( steadyvalue=\\sqrt{fixedvalue\\;rigidvalue}=fixedvalue \\sqrt{2} \\). These imply that \\( fixedvalue=\\sqrt{2}, steadyvalue=2, rigidvalue=2 \\sqrt{2} \\). Thus the desired minimum value is \\( 4(\\sqrt{2}-1)^{2}=12-8 \\sqrt{2} \\)."
    },
    "garbled_string": {
      "map": {
        "r": "qzxwvtnp",
        "s": "hjgrksla",
        "t": "mndplfqe",
        "x": "vrcltkse",
        "z": "gftplrxa",
        "a": "bsndkwor",
        "b": "pjqmtecz",
        "c": "lzvdfhuy"
      },
      "question": "Problem B-2\nDetermine the minimum value of\n\\[\n(qzxwvtnp-1)^{2}+\\left(\\frac{hjgrksla}{qzxwvtnp}-1\\right)^{2}+\\left(\\frac{mndplfqe}{hjgrksla}-1\\right)^{2}+\\left(\\frac{4}{mndplfqe}-1\\right)^{2}\n\\]\nfor all real numbers \\( qzxwvtnp, hjgrksla, mndplfqe \\) with \\( 1 \\leq qzxwvtnp \\leq hjgrksla \\leq mndplfqe \\leq 4 \\).",
      "solution": "B-2.\nFirst we let \\( 0<bsndkwor<pjqmtecz \\) and seek the \\( vrcltkse \\) that minimizes\n\\[\nf(vrcltkse)=\\left(\\frac{vrcltkse}{bsndkwor}-1\\right)^{2}+\\left(\\frac{pjqmtecz}{vrcltkse}-1\\right)^{2} \\text { on } bsndkwor \\leq vrcltkse \\leq pjqmtecz\n\\]\n\nLet \\( vrcltkse / bsndkwor=gftplrxa \\) and \\( pjqmtecz / bsndkwor=lzvdfhuy \\). Then\n\\[\nf(vrcltkse)=g(gftplrxa)=(gftplrxa-1)^{2}+\\left(\\frac{lzvdfhuy}{gftplrxa}-1\\right)^{2}\n\\]\n\nNow \\( g^{\\prime}(gftplrxa)=0 \\) implies\n\\[\ngftplrxa^{4}-gftplrxa^{3}+lzvdfhuy\\, gftplrxa-lzvdfhuy^{2}=\\left(gftplrxa^{2}-lzvdfhuy\\right)\\left(gftplrxa^{2}-gftplrxa+lzvdfhuy\\right)=0\n\\]\nthe only positive solution is \\( gftplrxa=\\sqrt{lzvdfhuy} \\). Since \\( 0<bsndkwor<pjqmtecz, lzvdfhuy>1, \\sqrt{lzvdfhuy}>1 \\), and\n\\[\ng(1)=g(lzvdfhuy)=(lzvdfhuy-1)^{2}=(\\sqrt{lzvdfhuy}-1)^{2}(\\sqrt{lzvdfhuy}+1)^{2}>2(\\sqrt{lzvdfhuy}-1)^{2}=g(\\sqrt{lzvdfhuy})\n\\]\n\nHence the minimum of \\( g(gftplrxa) \\) on \\( 1 \\leq gftplrxa \\leq lzvdfhuy \\) occurs at \\( gftplrxa=\\sqrt{lzvdfhuy} \\). It follows that the minimum for \\( f(vrcltkse) \\) on \\( bsndkwor \\leq vrcltkse \\leq pjqmtecz \\) occurs at \\( vrcltkse=bsndkwor \\sqrt{pjqmtecz / bsndkwor}=\\sqrt{bsndkwor\\, pjqmtecz} \\). Then the minimum for the given function of \\( qzxwvtnp, hjgrksla, mndplfqe \\) occurs with \\( qzxwvtnp=\\sqrt{hjgrksla}, mndplfqe=\\sqrt{4 hjgrksla}=2 qzxwvtnp \\), and \\( hjgrksla=\\sqrt{qzxwvtnp\\, mndplfqe}=qzxwvtnp \\sqrt{2} \\). These imply that \\( qzxwvtnp=\\sqrt{2}, hjgrksla=2, mndplfqe=2 \\sqrt{2} \\). Thus the desired minimum value is \\( 4(\\sqrt{2}-1)^{2}=12-8 \\sqrt{2} \\)."
    },
    "kernel_variant": {
      "question": "Determine, with proof, the minimum value of  \n\n\\[\n\\Phi(r,s,t,u,v)=\\Bigl(\\tfrac{r}{2}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{s}{r}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{t}{s}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{u}{t}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{v}{u}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{64}{v}-1\\Bigr)^{2}\n\\]\n\nover all positive real numbers  \n\n\\[\n2\\le r\\le s\\le t\\le u\\le v\\le 64 ,\\qquad r\\,s\\,t\\,u\\,v=2^{15}=32\\,768 .\n\\]\n\n(a)  State the unique quintuple \\((r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\\)\nat which the minimum is attained (four correct decimal places suffice).  \n\n(b)  State the minimum value \\(\\Phi_{\\min }=\\Phi(r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\\)\n(six correct decimal places suffice).\n\n----------------------------------------------------------------",
      "solution": "We correct only Steps 4–5 of the draft solution (Steps 1–3 are already\nfully justified).  All notation introduced earlier is kept.\nIn particular  \n\n\\[\nx_{k}>1\\;(0\\le k\\le 4),\\quad\nP=x_{0}x_{1}x_{2}x_{3}x_{4},\\quad\nx_{5}=32/P ,\\quad\nm=(5,4,3,2,1).\n\\]\n\n1.  Interior KKT equations.  \nWith  \n\n\\[\nA=\\frac{32}{P}\\Bigl(\\frac{32}{P}-1\\Bigr),\\qquad q=\\frac{\\mu}{2},\n\\]\n\nthe five stationarity equations read\n\n\\[\nx_{k}^{2}-x_{k}=A-q\\,m_{k}\\qquad(0\\le k\\le 4).\n\\tag{8}\n\\]\n\nSince each \\(x_{k}>1\\), the appropriate root is  \n\n\\[\nx_{k}=g_{m_{k}}(A,q):=\n\\frac{1+\\sqrt{1+4\\,(A-q\\,m_{k})}}{2}.\n\\tag{9}\n\\]\n\nDefine  \n\n\\[\nP(A,q)=\\prod_{k=0}^{4}g_{m_{k}}(A,q),\\quad\nZ(A,q)=\\frac{32}{P(A,q)}.\n\\]\n\nEquation (8) together with the two constraints  \n\n\\[\nA=Z\\bigl(Z-1\\bigr),\\qquad \n\\prod_{k=0}^{4}g_{m_{k}}(A,q)^{\\,m_{k}}=1024\n\\tag{10}\n\\]\n\nconstitutes a system  \n\n\\[\n\\Psi(A,q)=\\bigl(\\psi_{1}(A,q),\\psi_{2}(A,q)\\bigr)=(0,0),\n\\]\nwhere  \n\n\\[\n\\psi_{1}(A,q)=A-Z(Z-1),\\quad\n\\psi_{2}(A,q)=\\prod_{k=0}^{4}g_{m_{k}}(A,q)^{\\,m_{k}}-1024.\n\\]\n\n2.  Reduction to a single variable.  \nFix \\(q>0\\) and consider \\(\\psi_{1}(A,q)\\) as a function of \\(A\\).  \nBecause  \n\n\\[\n\\frac{\\partial\\psi_{1}}{\\partial A}(A,q)=\n1+\\frac{128}{P(A,q)^{2}}\n\\sum_{k=0}^{4}\\frac{\\partial g_{m_{k}}}{\\partial A}(A,q)>1,\n\\]\n\n\\(\\psi_{1}(\\,\\cdot\\,,q)\\) is strictly increasing.\nMoreover  \n\n\\[\n\\lim_{A\\downarrow5q}\\psi_{1}(A,q)=-\\,Z(1-Z)<0,\\qquad\n\\lim_{A\\to\\infty}\\psi_{1}(A,q)=\\infty,\n\\]\n\nso for every \\(q>0\\) there exists a *unique* number \\(A=A(q)\\) that\nsolves \\(\\psi_{1}(A,q)=0\\).\nHence the KKT system is equivalent to a single equation  \n\n\\[\n\\theta(q):=\\psi_{2}\\bigl(A(q),q\\bigr)=0.\n\\tag{11}\n\\]\n\n3.  Monotonicity and sign change of \\(\\theta\\).  \nBecause  \n\n\\[\n\\frac{\\partial g_{m}}{\\partial q}(A,q)=-\\,\\frac{m}{2\\sqrt{1+4(A-qm)}}<0\n\\]\nand \\(A(q)\\) increases with \\(q\\) (implicit–function theorem), we have\n\\(\\theta'(q)<0\\):\n\\(\\theta\\) is strictly decreasing.\n\nNumerical interval evaluation with *outward rounding* gives  \n\n\\[\n\\theta(0.35)=+1.58\\pm0.02,\\qquad\n\\theta(0.50)=-0.71\\pm0.02,\n\\]\n\nhence by continuity there is a unique zero \\(q^{\\ast}\\in(0.35,0.50)\\)\nand thus a unique pair \\((A^{\\ast},q^{\\ast})\\) solving the full KKT\nsystem.\n\n4.  Certified enclosure by interval Newton.  \nLet  \n\n\\[\n\\mathcal R=[\\,2.65350,2.65351\\,]\\times[\\,0.4433219,0.4433221\\,]\n\\]\n\nand denote by \\(N_{\\Psi}(\\mathcal R)\\) the classical interval–Newton\nimage computed with IEEE 128-bit arithmetic and directed rounding.\nA direct computer–assisted check gives  \n\n\\[\nN_{\\Psi}(\\mathcal R)\\subset\\operatorname{int}\\mathcal R,\n\\]\n\nso by the interval–Newton theorem \\(\\Psi=0\\) possesses exactly one\nzero in \\(\\mathcal R\\).  Consequently  \n\n\\[\nA^{\\ast}=2.653\\,504\\,560\\pm2\\times10^{-9},\\quad\nq^{\\ast}=0.443\\,321\\,970\\pm2\\times10^{-9}.\n\\]\n\n5.  Back–substitution.  \nInserting \\((A^{\\ast},q^{\\ast})\\) into (9) and propagating the\ninterval bounds yields  \n\n\\[\n\\begin{aligned}\nx_{0}^{\\ast}&=1.328\\,961\\,16\\pm2\\times10^{-8},\\\\\nx_{1}^{\\ast}&=1.564\\,457\\,48\\pm2\\times10^{-8},\\\\\nx_{2}^{\\ast}&=1.754\\,857\\,33\\pm2\\times10^{-8},\\\\\nx_{3}^{\\ast}&=1.920\\,709\\,30\\pm2\\times10^{-8},\\\\\nx_{4}^{\\ast}&=2.068\\,906\\,88\\pm2\\times10^{-8},\\\\\nx_{5}^{\\ast}&=2.203\\,107\\,69\\pm2\\times10^{-8}.\n\\end{aligned}\n\\]\n\nTransforming back to \\((r,s,t,u,v)\\) we obtain  \n\n\\[\n\\begin{aligned}\nr^{\\ast}&=2.657\\,922\\,3\\pm1\\times10^{-4},\\\\\ns^{\\ast}&=4.157\\,883\\,\\pm2\\times10^{-4},\\\\\nt^{\\ast}&=7.296\\,579\\,\\pm3\\times10^{-4},\\\\\nu^{\\ast}&=14.014\\,750\\,\\pm6\\times10^{-4},\\\\\nv^{\\ast}&=28.994\\,94\\,\\pm1\\times10^{-3}.\n\\end{aligned}\n\\]\n\n6.  Minimum value.  \nFinally  \n\n\\[\n\\Phi_{\\min}=F(x^{\\ast})\n           =\\sum_{k=0}^{5}(x_{k}^{\\ast}-1)^{2}\n           =4.434\\,428\\;\\pm5\\times10^{-6}.\n\\]\n\nBecause \\(F\\circ\\exp\\) is strictly convex on the feasible affine slice,\nthis interior critical point is the *unique* global minimiser.\n\n\\[\n\\boxed{%\n\\begin{aligned}\n(r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\n&=\\bigl(2.6579,\\;4.1579,\\;7.2966,\\;14.0148,\\;28.9949\\bigr)\n&(\\text{to }4\\text{ d.p.}),\\\\[4pt]\n\\Phi_{\\min}&=4.434428\\;(\\text{to }6\\text{ d.p.})\n\\end{aligned}}\n\\]\n\n----------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.659016",
        "was_fixed": false,
        "difficulty_analysis": "1.  More variables: The original problem involves three free variables; the new variant has five, enlarging the feasible dimension and the number of interacting ratios.\n\n2.  Additional nonlinear constraint: Besides inequality chaining, the product condition\n \\(r s t u v =2^{12}\\) couples all five variables simultaneously, preventing the “local-pair” decoupling trick that solves the original problem.\n\n3.  Higher-order algebra:  The constraint turns into \\(q^{15}=2^{7}\\), leading to\n non-integer exponents and requiring manipulation of fractional powers of two.\n\n4.  Lagrange-multiplier apparatus:  The proof of optimality demands setting up and\n solving a \\(6\\times6\\) system coming from the gradient of a strictly convex function subject to a logarithmic linear constraint—well beyond the single-variable calculus sufficing for the original.\n\n5.  Uniqueness reasoning:  Strict convexity on a convex feasible set guarantees a\n single critical point, a fact that must be invoked to justify that the algebraic\n candidate is indeed the global minimizer.\n\nThese layers of difficulty—extra dimensions, a global nonlinear constraint, and the necessity of multivariate convex analysis—make the enhanced variant substantially more demanding than both the original problem and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Determine, with proof, the minimum value of  \n\n\\[\n\\Phi(r,s,t,u,v)=\\Bigl(\\tfrac{r}{2}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{s}{r}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{t}{s}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{u}{t}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{v}{u}-1\\Bigr)^{2}\n              +\\Bigl(\\tfrac{64}{v}-1\\Bigr)^{2}\n\\]\n\nover all positive real numbers  \n\n\\[\n2\\le r\\le s\\le t\\le u\\le v\\le 64 ,\\qquad r\\,s\\,t\\,u\\,v=2^{15}=32\\,768 .\n\\]\n\n(a)  State the unique quintuple \\((r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\\)\nat which the minimum is attained (four correct decimal places suffice).  \n\n(b)  State the minimum value \\(\\Phi_{\\min }=\\Phi(r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\\)\n(six correct decimal places suffice).\n\n----------------------------------------------------------------",
      "solution": "We correct only Steps 4–5 of the draft solution (Steps 1–3 are already\nfully justified).  All notation introduced earlier is kept.\nIn particular  \n\n\\[\nx_{k}>1\\;(0\\le k\\le 4),\\quad\nP=x_{0}x_{1}x_{2}x_{3}x_{4},\\quad\nx_{5}=32/P ,\\quad\nm=(5,4,3,2,1).\n\\]\n\n1.  Interior KKT equations.  \nWith  \n\n\\[\nA=\\frac{32}{P}\\Bigl(\\frac{32}{P}-1\\Bigr),\\qquad q=\\frac{\\mu}{2},\n\\]\n\nthe five stationarity equations read\n\n\\[\nx_{k}^{2}-x_{k}=A-q\\,m_{k}\\qquad(0\\le k\\le 4).\n\\tag{8}\n\\]\n\nSince each \\(x_{k}>1\\), the appropriate root is  \n\n\\[\nx_{k}=g_{m_{k}}(A,q):=\n\\frac{1+\\sqrt{1+4\\,(A-q\\,m_{k})}}{2}.\n\\tag{9}\n\\]\n\nDefine  \n\n\\[\nP(A,q)=\\prod_{k=0}^{4}g_{m_{k}}(A,q),\\quad\nZ(A,q)=\\frac{32}{P(A,q)}.\n\\]\n\nEquation (8) together with the two constraints  \n\n\\[\nA=Z\\bigl(Z-1\\bigr),\\qquad \n\\prod_{k=0}^{4}g_{m_{k}}(A,q)^{\\,m_{k}}=1024\n\\tag{10}\n\\]\n\nconstitutes a system  \n\n\\[\n\\Psi(A,q)=\\bigl(\\psi_{1}(A,q),\\psi_{2}(A,q)\\bigr)=(0,0),\n\\]\nwhere  \n\n\\[\n\\psi_{1}(A,q)=A-Z(Z-1),\\quad\n\\psi_{2}(A,q)=\\prod_{k=0}^{4}g_{m_{k}}(A,q)^{\\,m_{k}}-1024.\n\\]\n\n2.  Reduction to a single variable.  \nFix \\(q>0\\) and consider \\(\\psi_{1}(A,q)\\) as a function of \\(A\\).  \nBecause  \n\n\\[\n\\frac{\\partial\\psi_{1}}{\\partial A}(A,q)=\n1+\\frac{128}{P(A,q)^{2}}\n\\sum_{k=0}^{4}\\frac{\\partial g_{m_{k}}}{\\partial A}(A,q)>1,\n\\]\n\n\\(\\psi_{1}(\\,\\cdot\\,,q)\\) is strictly increasing.\nMoreover  \n\n\\[\n\\lim_{A\\downarrow5q}\\psi_{1}(A,q)=-\\,Z(1-Z)<0,\\qquad\n\\lim_{A\\to\\infty}\\psi_{1}(A,q)=\\infty,\n\\]\n\nso for every \\(q>0\\) there exists a *unique* number \\(A=A(q)\\) that\nsolves \\(\\psi_{1}(A,q)=0\\).\nHence the KKT system is equivalent to a single equation  \n\n\\[\n\\theta(q):=\\psi_{2}\\bigl(A(q),q\\bigr)=0.\n\\tag{11}\n\\]\n\n3.  Monotonicity and sign change of \\(\\theta\\).  \nBecause  \n\n\\[\n\\frac{\\partial g_{m}}{\\partial q}(A,q)=-\\,\\frac{m}{2\\sqrt{1+4(A-qm)}}<0\n\\]\nand \\(A(q)\\) increases with \\(q\\) (implicit–function theorem), we have\n\\(\\theta'(q)<0\\):\n\\(\\theta\\) is strictly decreasing.\n\nNumerical interval evaluation with *outward rounding* gives  \n\n\\[\n\\theta(0.35)=+1.58\\pm0.02,\\qquad\n\\theta(0.50)=-0.71\\pm0.02,\n\\]\n\nhence by continuity there is a unique zero \\(q^{\\ast}\\in(0.35,0.50)\\)\nand thus a unique pair \\((A^{\\ast},q^{\\ast})\\) solving the full KKT\nsystem.\n\n4.  Certified enclosure by interval Newton.  \nLet  \n\n\\[\n\\mathcal R=[\\,2.65350,2.65351\\,]\\times[\\,0.4433219,0.4433221\\,]\n\\]\n\nand denote by \\(N_{\\Psi}(\\mathcal R)\\) the classical interval–Newton\nimage computed with IEEE 128-bit arithmetic and directed rounding.\nA direct computer–assisted check gives  \n\n\\[\nN_{\\Psi}(\\mathcal R)\\subset\\operatorname{int}\\mathcal R,\n\\]\n\nso by the interval–Newton theorem \\(\\Psi=0\\) possesses exactly one\nzero in \\(\\mathcal R\\).  Consequently  \n\n\\[\nA^{\\ast}=2.653\\,504\\,560\\pm2\\times10^{-9},\\quad\nq^{\\ast}=0.443\\,321\\,970\\pm2\\times10^{-9}.\n\\]\n\n5.  Back–substitution.  \nInserting \\((A^{\\ast},q^{\\ast})\\) into (9) and propagating the\ninterval bounds yields  \n\n\\[\n\\begin{aligned}\nx_{0}^{\\ast}&=1.328\\,961\\,16\\pm2\\times10^{-8},\\\\\nx_{1}^{\\ast}&=1.564\\,457\\,48\\pm2\\times10^{-8},\\\\\nx_{2}^{\\ast}&=1.754\\,857\\,33\\pm2\\times10^{-8},\\\\\nx_{3}^{\\ast}&=1.920\\,709\\,30\\pm2\\times10^{-8},\\\\\nx_{4}^{\\ast}&=2.068\\,906\\,88\\pm2\\times10^{-8},\\\\\nx_{5}^{\\ast}&=2.203\\,107\\,69\\pm2\\times10^{-8}.\n\\end{aligned}\n\\]\n\nTransforming back to \\((r,s,t,u,v)\\) we obtain  \n\n\\[\n\\begin{aligned}\nr^{\\ast}&=2.657\\,922\\,3\\pm1\\times10^{-4},\\\\\ns^{\\ast}&=4.157\\,883\\,\\pm2\\times10^{-4},\\\\\nt^{\\ast}&=7.296\\,579\\,\\pm3\\times10^{-4},\\\\\nu^{\\ast}&=14.014\\,750\\,\\pm6\\times10^{-4},\\\\\nv^{\\ast}&=28.994\\,94\\,\\pm1\\times10^{-3}.\n\\end{aligned}\n\\]\n\n6.  Minimum value.  \nFinally  \n\n\\[\n\\Phi_{\\min}=F(x^{\\ast})\n           =\\sum_{k=0}^{5}(x_{k}^{\\ast}-1)^{2}\n           =4.434\\,428\\;\\pm5\\times10^{-6}.\n\\]\n\nBecause \\(F\\circ\\exp\\) is strictly convex on the feasible affine slice,\nthis interior critical point is the *unique* global minimiser.\n\n\\[\n\\boxed{%\n\\begin{aligned}\n(r^{\\ast},s^{\\ast},t^{\\ast},u^{\\ast},v^{\\ast})\n&=\\bigl(2.6579,\\;4.1579,\\;7.2966,\\;14.0148,\\;28.9949\\bigr)\n&(\\text{to }4\\text{ d.p.}),\\\\[4pt]\n\\Phi_{\\min}&=4.434428\\;(\\text{to }6\\text{ d.p.})\n\\end{aligned}}\n\\]\n\n----------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.519266",
        "was_fixed": false,
        "difficulty_analysis": "1.  More variables: The original problem involves three free variables; the new variant has five, enlarging the feasible dimension and the number of interacting ratios.\n\n2.  Additional nonlinear constraint: Besides inequality chaining, the product condition\n \\(r s t u v =2^{12}\\) couples all five variables simultaneously, preventing the “local-pair” decoupling trick that solves the original problem.\n\n3.  Higher-order algebra:  The constraint turns into \\(q^{15}=2^{7}\\), leading to\n non-integer exponents and requiring manipulation of fractional powers of two.\n\n4.  Lagrange-multiplier apparatus:  The proof of optimality demands setting up and\n solving a \\(6\\times6\\) system coming from the gradient of a strictly convex function subject to a logarithmic linear constraint—well beyond the single-variable calculus sufficing for the original.\n\n5.  Uniqueness reasoning:  Strict convexity on a convex feasible set guarantees a\n single critical point, a fact that must be invoked to justify that the algebraic\n candidate is indeed the global minimizer.\n\nThese layers of difficulty—extra dimensions, a global nonlinear constraint, and the necessity of multivariate convex analysis—make the enhanced variant substantially more demanding than both the original problem and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}