{
  "index": "1995-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "rolls without slipping on the curve $y = c \\sin \\left( \\frac{x}{a}\n\\right)$. How are $a,b,c$ related, given that the ellipse completes\none revolution when it traverses one period of the curve?",
  "solution": "For those who haven't taken enough physics, ``rolling without\nslipping'' means that the perimeter of the ellipse and the curve pass\nat the same rate, so all we're saying is that the perimeter of the\nellipse equals the length of one period of the sine curve. So set up\nthe integrals:\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{(-a \\sin \\theta)^{2} + (b \\cos \\theta)^{2}}\\,\nd\\theta\\\\\n = \\int_{0}^{2\\pi a} \\sqrt{1 + (c/a \\cos x/a)^{2}}\\,dx.\n\\end{multline*}\nLet $\\theta = x/a$ in the second integral and write 1 as $\\sin^{2}\n\\theta + \\cos^{2} \\theta$ and you get\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{a^{2} \\sin^{2} \\theta + b^{2} \\cos^{2}\n\\theta}\\,d\\theta\\\\\n = \\int_{0}^{2\\pi} \\sqrt{a^{2} \\sin^{2} \\theta +\n(a^{2} + c^{2}) \\cos^{2} \\theta}\\,d\\theta.\n\\end{multline*}\nSince the left side is increasing as a function of $b$, we have\nequality if and only if $b^{2} = a^{2} + c^{2}$.",
  "vars": [
    "x",
    "y",
    "\\\\theta"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizpos",
        "y": "vertpos",
        "\\theta": "anglepar",
        "a": "horizscal",
        "b": "vertirad",
        "c": "curveamp"
      },
      "question": "rolls without slipping on the curve $vertpos = curveamp \\sin \\left( \\frac{horizpos}{horizscal}\\right)$. How are $horizscal,vertirad,curveamp$ related, given that the ellipse completes one revolution when it traverses one period of the curve?",
      "solution": "For those who haven't taken enough physics, ``rolling without\nslipping'' means that the perimeter of the ellipse and the curve pass\nat the same rate, so all we're saying is that the perimeter of the\nellipse equals the length of one period of the sine curve. So set up\nthe integrals:\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{(-horizscal \\sin anglepar)^{2} + (vertirad \\cos anglepar)^{2}}\\, danglepar\\\\\n = \\int_{0}^{2\\pi horizscal} \\sqrt{1 + (curveamp/horizscal \\cos horizpos/horizscal)^{2}}\\,dhorizpos.\n\\end{multline*}\nLet $anglepar = horizpos/horizscal$ in the second integral and write 1 as $\\sin^{2}\nanglepar + \\cos^{2} anglepar$ and you get\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{horizscal^{2} \\sin^{2} anglepar + vertirad^{2} \\cos^{2}\nanglepar}\\,danglepar\\\\\n = \\int_{0}^{2\\pi} \\sqrt{horizscal^{2} \\sin^{2} anglepar +\n(horizscal^{2} + curveamp^{2}) \\cos^{2} anglepar}\\,danglepar.\n\\end{multline*}\nSince the left side is increasing as a function of $vertirad$, we have\nequality if and only if $vertirad^{2} = horizscal^{2} + curveamp^{2}$. "
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "hazelnuts",
        "y": "sailormoon",
        "\\\\theta": "toothbrush",
        "a": "lemonade",
        "b": "marshmallow",
        "c": "rhinoceros"
      },
      "question": "rolls without slipping on the curve $sailormoon = rhinoceros \\sin \\left( \\frac{hazelnuts}{lemonade}\n\\right)$. How are $lemonade,marshmallow,rhinoceros$ related, given that the ellipse completes\none revolution when it traverses one period of the curve?",
      "solution": "For those who haven't taken enough physics, ``rolling without\nslipping'' means that the perimeter of the ellipse and the curve pass\nat the same rate, so all we're saying is that the perimeter of the\nellipse equals the length of one period of the sine curve. So set up\nthe integrals:\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{(-lemonade \\sin toothbrush)^{2} + (marshmallow \\cos toothbrush)^{2}}\\,\n d\\text{toothbrush}\\\\\n = \\int_{0}^{2\\pi lemonade} \\sqrt{1 + (rhinoceros/lemonade \\cos hazelnuts/lemonade)^{2}}\\,d\\text{hazelnuts}.\n\\end{multline*}\nLet $toothbrush = hazelnuts/lemonade$ in the second integral and write 1 as $\\sin^{2}\n tothbrush + \\cos^{2} toothbrush$ and you get\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{lemonade^{2} \\sin^{2} toothbrush + marshmallow^{2} \\cos^{2}\n toothbrush}\\,d\\text{toothbrush}\\\\\n = \\int_{0}^{2\\pi} \\sqrt{lemonade^{2} \\sin^{2} toothbrush +\n(lemonade^{2} + rhinoceros^{2}) \\cos^{2} toothbrush}\\,d\\text{toothbrush}.\n\\end{multline*}\nSince the left side is increasing as a function of $marshmallow$, we have\nequality if and only if $marshmallow^{2} = lemonade^{2} + rhinoceros^{2}$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "\\\\theta": "straightangle",
        "a": "constricted",
        "b": "elongated",
        "c": "flatline"
      },
      "question": "rolls without slipping on the curve $horizontalaxis = flatline \\sin \\left( \\frac{verticalaxis}{constricted}\n\\right)$. How are $constricted,elongated,flatline$ related, given that the ellipse completes\none revolution when it traverses one period of the curve?",
      "solution": "For those who haven't taken enough physics, ``rolling without\nslipping'' means that the perimeter of the ellipse and the curve pass\nat the same rate, so all we're saying is that the perimeter of the\nellipse equals the length of one period of the sine curve. So set up\nthe integrals:\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{(-constricted \\sin straightangle)^{2} + (elongated \\cos straightangle)^{2}}\\,\ndstraightangle\\\\\n = \\int_{0}^{2\\pi constricted} \\sqrt{1 + (flatline/constricted \\cos verticalaxis/constricted)^{2}}\\,dverticalaxis.\n\\end{multline*}\nLet $straightangle = verticalaxis/constricted$ in the second integral and write 1 as $\\sin^{2}\nstraightangle + \\cos^{2} straightangle$ and you get\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{constricted^{2} \\sin^{2} straightangle + elongated^{2} \\cos^{2}\nstraightangle}\\,dstraightangle\\\\\n = \\int_{0}^{2\\pi} \\sqrt{constricted^{2} \\sin^{2} straightangle +\n(constricted^{2} + flatline^{2}) \\cos^{2} straightangle}\\,dstraightangle.\n\\end{multline*}\nSince the left side is increasing as a function of elongated, we have\nequality if and only if $elongated^{2} = constricted^{2} + flatline^{2}$.}",
      "confidence": "0.18"
    },
    "garbled_string": {
      "map": {
        "x": "qazwsxed",
        "y": "rfvtgbyh",
        "\\\\theta": "ujmkolpq",
        "a": "plokmijn",
        "b": "qlazwsed",
        "c": "wsxedcrf"
      },
      "question": "rolls without slipping on the curve $rfvtgbyh = wsxedcrf \\sin \\left( \\frac{qazwsxed}{plokmijn}\n\\right)$. How are $plokmijn,qlazwsed,wsxedcrf$ related, given that the ellipse completes\none revolution when it traverses one period of the curve?",
      "solution": "For those who haven't taken enough physics, ``rolling without\nslipping'' means that the perimeter of the ellipse and the curve pass\nat the same rate, so all we're saying is that the perimeter of the\nellipse equals the length of one period of the sine curve. So set up\nthe integrals:\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{(-plokmijn \\sin ujmkolpq)^{2} + (qlazwsed \\cos ujmkolpq)^{2}}\\,\ndujmkolpq\\\\\n = \\int_{0}^{2\\pi plokmijn} \\sqrt{1 + (wsxedcrf/plokmijn \\cos qazwsxed/plokmijn)^{2}}\\,dqazwsxed.\n\\end{multline*}\nLet $ujmkolpq = qazwsxed/plokmijn$ in the second integral and write 1 as $\\sin^{2}\nujmkolpq + \\cos^{2} ujmkolpq$ and you get\n\\begin{multline*}\n\\int_{0}^{2\\pi} \\sqrt{plokmijn^{2} \\sin^{2} ujmkolpq + qlazwsed^{2} \\cos^{2}\nujmkolpq}\\,dujmkolpq\\\\\n = \\int_{0}^{2\\pi} \\sqrt{plokmijn^{2} \\sin^{2} ujmkolpq +\n(plokmijn^{2} + wsxedcrf^{2}) \\cos^{2} ujmkolpq}\\,dujmkolpq.\n\\end{multline*}\nSince the left side is increasing as a function of $qlazwsed$, we have\nequality if and only if $qlazwsed^{2} = plokmijn^{2} + wsxedcrf^{2}$.",
      "error": null
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nA,B,C,R,S,H>0\n\\]\nbe fixed real parameters.  \nDefine the $2\\pi$–periodic parametrisation of the triaxial ellipsoid  \n\\[\nE(\\theta,\\varphi)\\;=\\;\\bigl(A\\sin\\theta\\cos\\varphi,\\;\n                              B\\sin\\theta\\sin\\varphi,\\;\n                              C\\cos\\theta\\bigr),\\qquad (\\theta,\\varphi)\\in\\mathbb R^{2},\n\\]\nwhose image is  \n\\[\n\\widehat E:\\qquad\\frac{x^{2}}{A^{2}}+\\frac{y^{2}}{B^{2}}+\\frac{z^{2}}{C^{2}}=1.\n\\tag{0}\n\\]\n\nIn $\\mathbb R^{3}$ fix the “egg–crate’’ surface  \n\\[\n\\Sigma:\\qquad z \\;=\\;H\n          +R\\cos\\!\\Bigl(\\frac{x}{A}\\Bigr)\n          +S\\cos\\!\\Bigl(\\frac{y}{B}\\Bigr).\n\\tag{1}\n\\]\nAll data are of class $C^{\\infty}$.\n\nThroughout the motion the ellipsoid will be described by a one–parameter family  \n\\[\ng(t)=\\bigl(Q(t),c(t)\\bigr)\\in SE(3),\\qquad 0\\le t\\le T,\n\\]\nwhere $Q(t)\\in SO(3)$ is the time–dependent body attitude and\n$c(t)\\in\\mathbb R^{3}$ the centre–of–mass position.  \nLet\n\\[\n(\\theta(t),\\varphi(t))\\in\\mathbb R^{2},\\qquad 0\\le t\\le T,\n\\]\nbe $C^{1}$ functions whose derivatives are allowed to vanish and that can be\nre–parametrised in time at will (in particular we may require\n$\\dot\\theta$ and $\\dot\\varphi$ to attain the value $0$ at prescribed instants).\n\nThe triple $\\bigl(g(t),\\theta(t),\\varphi(t)\\bigr)$ is said to perform a\n\\emph{rolling of $\\widehat E$ on $\\Sigma$} provided\n\n(i) (contact point)\n\\[\nP(t):=c(t)+Q(t)\\,E\\bigl(\\theta(t),\\varphi(t)\\bigr)\\;\\in\\;\\Sigma;\n\\tag{2}\n\\]\n\n(ii) (no–slip) the velocity of the material point of the ellipsoid that\nis in contact with $\\Sigma$ vanishes:\n\\[\n\\dot c(t)+\\dot Q(t)\\,E\\bigl(\\theta(t),\\varphi(t)\\bigr)=0;\n\\tag{3}\n\\]\n\n(iii) (no–twist) the angular velocity\n\\[\n\\omega(t)=\\bigl(\\dot Q(t)Q(t)^{-1}\\bigr)^{\\vee}\\in\\mathbb R^{3}\n\\]\nis orthogonal to the common unit normal\n\\[\nn(t):=\\frac{Q(t)\\,n_{E}\\bigl(\\theta(t),\\varphi(t)\\bigr)}\n           {\\lvert n_{E}(\\theta(t),\\varphi(t))\\rvert}\n      \\;=\\;\\pm n_{\\Sigma}\\bigl(x(t),y(t)\\bigr)\\quad \\text{at }P(t),\n\\tag{4}\n\\]\nwhere\n\\[\nn_{E}(\\theta,\\varphi)=\n\\Bigl(\\tfrac{\\sin\\theta\\cos\\varphi}{A^{2}},\n      \\tfrac{\\sin\\theta\\sin\\varphi}{B^{2}},\n      \\tfrac{\\cos\\theta}{C^{2}}\\Bigr),\\qquad\nn_{\\Sigma}(x,y)=\n\\frac{\\bigl(-f_{x},-f_{y},1\\bigr)}\n     {\\sqrt{1+f_{x}^{2}+f_{y}^{2}}},\n\\]\nwith $f(x,y)=H+R\\cos(x/A)+S\\cos(y/B)$.\n\nDuring the motion the contact point makes two successive lattice steps\non $\\Sigma$.\n\nStage I ($0\\le t\\le T_{1}$).  \nFix $\\theta(t)\\equiv\\pi/2$ and let $\\varphi(t)$ increase monotonically\nfrom $-\\pi$ to $\\pi$, while both derivatives are required to satisfy\n$\\dot\\varphi(0)=\\dot\\varphi(T_{1})=0$.  \nIn space the contact point stays in the plane $y=y_{0}$ and moves in $x$:\n\\[\nx(0)=x_{0},\\qquad x(T_{1})=x_{0}+2\\pi A.\n\\tag{S$_{\\!I}$}\n\\]\n\nStage II ($T_{1}\\le t\\le T$).  \nFix $\\varphi(t)\\equiv\\pi$ and let $\\theta(t)$ increase monotonically\nfrom $\\pi/2$ to $3\\pi/2$, again with $\\dot\\theta(T_{1})=\\dot\\theta(T)=0$.\nThe $x$–coordinate is frozen at\n$x(t)\\equiv x_{1}:=x_{0}+2\\pi A$ and\n\\[\ny(T_{1})=y_{0},\\qquad y(T)=y_{0}+2\\pi B.\n\\tag{S$_{\\!II}$}\n\\]\n(The starting ordinates $x_{0},y_{0}$ are arbitrary.  Taking\n$\\varphi\\equiv\\pi$ instead of $0$ is immaterial for the local geometry\nbut simplifies the $C^{1}$ sewing of the two stages.)\n\na) Kinematics along the prescribed curves.  \n\n a$_1$) Show that on Stage I there exists  \n\\[\n\\lambda_{I}(\\varphi,x)=\n\\frac{\\lvert E_{\\varphi}(\\tfrac{\\pi}{2},\\varphi)\\rvert}\n     {\\sqrt{\\,1+f_{x}(x,y_{0})^{2}}}\n\\tag{5}\n\\]\nsuch that $\\dot x(t)=\\lambda_{I}\\bigl(\\varphi(t),x(t)\\bigr)\\,\\dot\\varphi(t)$.\n\n a$_2$) Prove the analogous statement for Stage II:\n\\[\n\\lambda_{II}(\\theta,y)=\n\\frac{\\lvert E_{\\theta}(\\theta,\\pi)\\rvert}\n     {\\sqrt{\\,1+f_{y}(x_{1},y)^{2}}},\n\\tag{6}\n\\]\nso that $\\dot y(t)=\\lambda_{II}\\bigl(\\theta(t),y(t)\\bigr)\\,\\dot\\theta(t)$.\n\nb) Using the no–slip condition point–wise together with\n$(S_{\\!I})$–$(S_{\\!II})$, prove the\n\\emph{metric identities}\n\\[\nB^{2}=A^{2}+R^{2}\\qquad\\text{and}\\qquad\nC^{2}=A^{2}+S^{2}.\n\\tag{7}\n\\]\n\nc) Conversely, assume that (7) holds.\n\n(i) Stage I.  \nInsert (7) into (5) to obtain  \n\\[\n\\frac{dx}{d\\varphi}\n  =\\Lambda_{I}(\\varphi,x):=\n   \\frac{\\sqrt{A^{2}+R^{2}\\cos^{2}\\varphi}}\n        {\\sqrt{\\,1+(R^{2}/A^{2})\\sin^{2}(x/A)}}.\n\\tag{8}\n\\]\nShow that for every $x_{0}$ equation (8) admits a unique $C^{1}$ solution\non $[-\\pi,\\pi]$ satisfying\n\\[\nx(\\pi)-x(-\\pi)=2\\pi A.\n\\tag{9}\n\\]\n\n(ii) Stage II.  \nAnalogously, prove that the solution of  \n\\[\n\\frac{dy}{d\\theta}\n  =\\Lambda_{II}(\\theta,y):=\n   \\frac{\\sqrt{A^{2}+S^{2}\\sin^{2}\\theta}}\n        {\\sqrt{\\,1+(S^{2}/B^{2})\\sin^{2}(y/B)}}\n\\tag{10}\n\\]\nsatisfies\n\\[\ny(3\\pi/2)-y(\\pi/2)=2\\pi B.\n\\tag{11}\n\\]\n\n(iii) Construct a rolling realising Stages I–II and prove that the\nresulting rigid motion $g(t)$ is of class $C^{1}$ on $[0,T]$.  In\nparticular, justify carefully that the body configurations at $t=T_{1}$\nmatch up to a (possibly trivial) \\emph{body–fixed} rotation about the\ncommon normal and a $(\\theta,\\varphi)$ shift leaving the contact point\ninvariant.  \n(You may use that the rolling constraints define a smooth Ehresmann\nconnection on the bundle $SE(3)\\times\\widehat E\\longrightarrow\\widehat\nE$ and that the time parametrisation of Stage I can always be slowed\ndown so that $\\dot\\varphi(T_{1})=0$, likewise for Stage II.)\n\n(Hints.  \n– For (c-i) evaluate the integral\n$\\displaystyle\\int\\sqrt{1+k^{2}\\sin^{2}u}\\,du$ explicitly and exploit\nits $\\pi$–periodicity.  \n– For (c-iii) choose the initial orientation of Stage II equal to the\nfinal orientation of Stage I in order to guarantee $C^{1}$–matching.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we write  \n\\[\nr(\\theta,\\varphi):=E(\\theta,\\varphi),\\qquad\nf(x,y)=H+R\\cos\\!\\Bigl(\\tfrac{x}{A}\\Bigr)+S\\cos\\!\\Bigl(\\tfrac{y}{B}\\Bigr),\\qquad\nP=c+Qr.\n\\]\nAll quantities are $C^{\\infty}$.\n\n--------------------------------------------------------------------\n\\textbf{a) Kinematics.}\n\nDifferentiating $P=c+Qr$ gives\n\\[\n\\dot P=\\dot c+\\dot Q\\,r+Q\\,\\dot r.\n\\tag{12}\n\\]\nBy the no–slip condition $(\\dot c+\\dot Q\\,r)=0$, hence\n\\[\n\\dot P=Q\\,\\dot r.\n\\tag{13}\n\\]\nThus the motion of the contact point along $\\Sigma$ is governed\n\\emph{solely} by the variation of $(\\theta,\\varphi)$.\n\n------------------------------------------------\na$_1$) Stage I $(\\theta\\equiv\\pi/2,\\;y\\equiv y_{0})$.  \nHere\n\\[\nr_{\\varphi}\\bigl(\\tfrac{\\pi}{2},\\varphi\\bigr)=\n(-A\\sin\\varphi,\\;B\\cos\\varphi,\\;0),\\qquad\nv_{E}:=r_{\\varphi}\\,\\dot\\varphi.\n\\]\nWith (13)\n\\[\n\\dot P=Qv_{E}=\\dot x\\,(1,\\,0,\\,f_{x}(x,y_{0})).\n\\]\nBoth vectors lie in the one–dimensional subspace $T_{P}\\Sigma$, so\nthere exists $\\lambda_{I}$ with\n\\[\nQv_{E}=\\lambda_{I}(\\varphi,x)\\bigl(1,\\,0,\\,f_{x}(x,y_{0})\\bigr).\n\\]\nTaking Euclidean norms gives (5); equating first components yields\n$\\dot x=\\lambda_{I}\\dot\\varphi$.\n\n------------------------------------------------\na$_2$) Stage II $(\\varphi\\equiv\\pi,\\;x\\equiv x_{1})$.  \nNow\n\\[\nr_{\\theta}(\\theta,\\pi)=(-A\\cos\\theta,\\;0,\\;-C\\sin\\theta),\\qquad\nv_{E}:=r_{\\theta}\\,\\dot\\theta.\n\\]\nProceeding as above we obtain (6) and\n$\\dot y=\\lambda_{II}\\dot\\theta$.\n\n--------------------------------------------------------------------\n\\textbf{b) Metric identities.}\n\nStage I: by (13) and a$_1$\n\\[\n\\lvert r_{\\varphi}\\rvert\\,\\dot\\varphi\n   =\\sqrt{1+f_{x}^{2}}\\;\\dot x.\n\\tag{14}\n\\]\nSince $\\dot\\varphi,\\dot x\\ge 0$ no absolute values are needed.\nIntegrating over $[0,T_{1}]$ and using $(S_{\\!I})$ gives\n\\[\n\\int_{-\\pi}^{\\pi}\n   \\lvert r_{\\varphi}(\\tfrac{\\pi}{2},\\varphi)\\rvert\\,d\\varphi\n=\\int_{x_{0}}^{x_{0}+2\\pi A}\n   \\sqrt{1+f_{x}(x,y_{0})^{2}}\\,dx.\n\\tag{15}\n\\]\nThe left–hand side equals\n\\[\n\\int_{-\\pi}^{\\pi}\n   \\sqrt{A^{2}\\sin^{2}\\varphi+B^{2}\\cos^{2}\\varphi}\\,d\\varphi,\n\\]\nwhile the right–hand side equals\n\\[\n\\int_{0}^{2\\pi}\\sqrt{A^{2}+R^{2}\\sin^{2}u}\\,du\n\\quad\\text{with }u:=x/A.\n\\]\n\nStage II (interchanging $(x,y),(R,S),(B,C)$) yields\n\\[\n\\int_{0}^{2\\pi}\\sqrt{A^{2}+S^{2}\\sin^{2}u}\\,du\n=\\int_{0}^{2\\pi}\\sqrt{A^{2}\\sin^{2}\\theta+C^{2}\\cos^{2}\\theta}\\,d\\theta.\n\\]\n\nSet\n\\[\nF(b):=\\int_{0}^{2\\pi}\\sqrt{A^{2}\\sin^{2}\\vartheta+b^{2}\\cos^{2}\\vartheta}\\,d\\vartheta,\n\\quad b>0.\n\\]\nDifferentiating under the integral sign gives\n\\[\nF'(b)=\\int_{0}^{2\\pi}\\frac{b\\cos^{2}\\vartheta}\n              {\\sqrt{A^{2}\\sin^{2}\\vartheta+b^{2}\\cos^{2}\\vartheta}}\n       \\,d\\vartheta>0,\n\\]\nso $F$ is strictly increasing.  Equation (15) therefore forces\n$B^{2}=A^{2}+R^{2}$, and Stage II yields $C^{2}=A^{2}+S^{2}$, i.e.~(7).\n\n--------------------------------------------------------------------\n\\textbf{c) Inverse construction under assumption (7).}\n\n(i) Stage I.  \nWith $u:=x/A$ and $k:=R/A$ equation (8) becomes\n\\[\n\\frac{du}{d\\varphi}\n   =\\Phi(\\varphi,u):=\n    \\frac{\\sqrt{1+k^{2}\\cos^{2}\\varphi}}\n         {\\sqrt{1+k^{2}\\sin^{2}u}}.\n\\tag{16}\n\\]\n$\\Phi$ is $C^{1}$ and strictly positive on the cylinder\n$(\\varphi,u)\\in[-\\pi,\\pi]\\times\\mathbb R$, hence the ODE has a unique\n$C^{1}$ solution for all $\\varphi\\in[-\\pi,\\pi]$.  Separation of\nvariables gives\n\\[\n\\int_{u_{0}}^{u(\\varphi)}\\sqrt{1+k^{2}\\sin^{2}s}\\,ds\n   =\\int_{-\\pi}^{\\varphi}\\sqrt{1+k^{2}\\cos^{2}\\tau}\\,d\\tau.\n\\tag{17}\n\\]\nThe integrands are $\\pi$–periodic and even, so evaluating at\n$\\varphi=\\pi$ yields $u(\\pi)=u_{0}+2\\pi$, i.e.~(9).\n\n(ii) Stage II.  \nSubstituting $v:=y/B$ and $\\ell:=S/B$ converts (10) into the same model\nequation, proving (11).\n\n(iii) \\emph{Existence of a $C^{1}$ rolling realising both stages.}\n\n\\emph{Step 1: horizontal lift along Stage I.}  \nChoose an arbitrary initial configuration $g(0)=(Q_{0},c_{0})$.\nBecause the base curve $\\bigl(\\theta(t),\\varphi(t)\\bigr)$ and the\ncontact point on $\\Sigma$ are prescribed, the rolling connection\nprovides a unique horizontal lift $g(t)=(Q(t),c(t))$ on $[0,T_{1}]$.\nBy construction $\\dot\\varphi(0)=0$ and we have arranged $\\dot\\varphi(T_{1})=0$,\nso\n\\[\n\\dot Q(T_{1})=\\dot c(T_{1})=0.\n\\tag{18}\n\\]\n\n\\emph{Step 2: initial data for Stage II.}  \nAt $t=T_{1}$ the contact point satisfies\n$\\theta=\\pi/2,\\varphi=\\pi$ and the common normal is\n$n(T_{1})=\\tfrac{Q(T_{1})\\,n_{E}(\\pi/2,\\pi)}{\\lvert n_{E}(\\pi/2,\\pi)\\rvert}$.\nBecause for a fixed contact point and normal the admissible body\norientations form a one–parameter family obtained by rotations around\nthe normal, we \\emph{choose}\n\\[\ng(T_{1}^{+})=(Q(T_{1}),c(T_{1})).\n\\tag{19}\n\\]\nThis choice suppresses any jump in $Q$ or $c$.\n\n\\emph{Step 3: horizontal lift along Stage II.}  \nFrom the initial data (19) the rolling connection yields a unique\nhorizontal lift on $[T_{1},T]$.  Because $\\dot\\theta(T_{1})=0$ the\nright–hand sides of the differential equations for $Q$ and $c$ vanish\nat $t=T_{1}$, so with (18)\n\\[\n\\dot Q(T_{1}^{+})=\\dot Q(T_{1}^{-}),\\qquad\n\\dot c(T_{1}^{+})=\\dot c(T_{1}^{-}),\n\\]\nand the overall map $g:[0,T]\\to SE(3)$ is $C^{1}$.\n\n\\emph{Step 4: summary.}  \nThe curve\n$t\\mapsto\\bigl(g(t),\\theta(t),\\varphi(t)\\bigr)$ meets all rolling\nconstraints, realises the required lattice steps (thanks to (9) and\n(11)), and is of class $C^{1}$ on $[0,T]$.  Consequently the rolling\nexists if and only if the metric identities (7) hold.\n\n--------------------------------------------------------------------\n\\textbf{Conclusion.}  \nA $C^{1}$ rolling of the triaxial ellipsoid \\eqref{0} on the egg–crate\nsurface \\eqref{1} performing the two prescribed lattice steps exists\n\\emph{iff}\n\\[\n\\boxed{\\;B^{2}=A^{2}+R^{2},\\qquad C^{2}=A^{2}+S^{2}\\;}.\n\\qquad\\qedhere\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.743293",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension — The original problem is planar; the enhanced\nvariant takes place in ℝ³ with a full ellipsoid rolling on a doubly\nperiodic surface, so one deals with two-parameter families of contact\npoints instead of a single curve.\n\n2. Additional constraints — Besides equality of arc lengths, the problem\ndemands second-order tangency (“no twist’’) and *simultaneous* matching\nof two independent periodic motions (one about each principal axis of\nthe ellipsoid).\n\n3. Sophisticated structures — The solution needs first fundamental\nforms, differential geometry of surfaces, and monotonicity properties of\ncomplete elliptic integrals rather than elementary arc-length calculus\nalone.\n\n4. Deeper theory — Showing that the coincidence of the metrics along two\nindependent directions suffices for full metric coincidence, and that\nhorizontal motion of the centre follows from the common normal argument,\nrequires familiarity with rigid surface rolling theory.\n\n5. Multiple interacting concepts — The argument couples surface geometry\n(elliptic integrals, metric tensors) with rigid-body kinematics\n(synchronous rotations about two axes and horizontal centre-of-mass\nmotion).\n\nAll these layers make the enhanced kernel variant substantially more\ntechnically demanding than both the original Olympiad-style question and\nthe current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nA,B,C,R,S,H>0\n\\]\nbe fixed real parameters.  \nDefine the $2\\pi$–periodic parametrisation of the triaxial ellipsoid  \n\\[\nE(\\theta,\\varphi)\\;=\\;\\bigl(A\\sin\\theta\\cos\\varphi,\\;\n                              B\\sin\\theta\\sin\\varphi,\\;\n                              C\\cos\\theta\\bigr),\\qquad (\\theta,\\varphi)\\in\\mathbb R^{2},\n\\]\nwhose image is  \n\\[\n\\widehat E:\\qquad\\frac{x^{2}}{A^{2}}+\\frac{y^{2}}{B^{2}}+\\frac{z^{2}}{C^{2}}=1.\n\\tag{0}\n\\]\n\nIn $\\mathbb R^{3}$ fix the “egg–crate’’ surface  \n\\[\n\\Sigma:\\qquad z \\;=\\;H\n          +R\\cos\\!\\Bigl(\\frac{x}{A}\\Bigr)\n          +S\\cos\\!\\Bigl(\\frac{y}{B}\\Bigr).\n\\tag{1}\n\\]\nAll data are of class $C^{\\infty}$.\n\nThroughout the motion the ellipsoid will be described by a one–parameter family  \n\\[\ng(t)=\\bigl(Q(t),c(t)\\bigr)\\in SE(3),\\qquad 0\\le t\\le T,\n\\]\nwhere $Q(t)\\in SO(3)$ is the time–dependent body attitude and\n$c(t)\\in\\mathbb R^{3}$ the centre–of–mass position.  \nLet\n\\[\n(\\theta(t),\\varphi(t))\\in\\mathbb R^{2},\\qquad 0\\le t\\le T,\n\\]\nbe $C^{1}$ functions whose derivatives are allowed to vanish and that can be\nre–parametrised in time at will (in particular we may require\n$\\dot\\theta$ and $\\dot\\varphi$ to attain the value $0$ at prescribed instants).\n\nThe triple $\\bigl(g(t),\\theta(t),\\varphi(t)\\bigr)$ is said to perform a\n\\emph{rolling of $\\widehat E$ on $\\Sigma$} provided\n\n(i) (contact point)\n\\[\nP(t):=c(t)+Q(t)\\,E\\bigl(\\theta(t),\\varphi(t)\\bigr)\\;\\in\\;\\Sigma;\n\\tag{2}\n\\]\n\n(ii) (no–slip) the velocity of the material point of the ellipsoid that\nis in contact with $\\Sigma$ vanishes:\n\\[\n\\dot c(t)+\\dot Q(t)\\,E\\bigl(\\theta(t),\\varphi(t)\\bigr)=0;\n\\tag{3}\n\\]\n\n(iii) (no–twist) the angular velocity\n\\[\n\\omega(t)=\\bigl(\\dot Q(t)Q(t)^{-1}\\bigr)^{\\vee}\\in\\mathbb R^{3}\n\\]\nis orthogonal to the common unit normal\n\\[\nn(t):=\\frac{Q(t)\\,n_{E}\\bigl(\\theta(t),\\varphi(t)\\bigr)}\n           {\\lvert n_{E}(\\theta(t),\\varphi(t))\\rvert}\n      \\;=\\;\\pm n_{\\Sigma}\\bigl(x(t),y(t)\\bigr)\\quad \\text{at }P(t),\n\\tag{4}\n\\]\nwhere\n\\[\nn_{E}(\\theta,\\varphi)=\n\\Bigl(\\tfrac{\\sin\\theta\\cos\\varphi}{A^{2}},\n      \\tfrac{\\sin\\theta\\sin\\varphi}{B^{2}},\n      \\tfrac{\\cos\\theta}{C^{2}}\\Bigr),\\qquad\nn_{\\Sigma}(x,y)=\n\\frac{\\bigl(-f_{x},-f_{y},1\\bigr)}\n     {\\sqrt{1+f_{x}^{2}+f_{y}^{2}}},\n\\]\nwith $f(x,y)=H+R\\cos(x/A)+S\\cos(y/B)$.\n\nDuring the motion the contact point makes two successive lattice steps\non $\\Sigma$.\n\nStage I ($0\\le t\\le T_{1}$).  \nFix $\\theta(t)\\equiv\\pi/2$ and let $\\varphi(t)$ increase monotonically\nfrom $-\\pi$ to $\\pi$, while both derivatives are required to satisfy\n$\\dot\\varphi(0)=\\dot\\varphi(T_{1})=0$.  \nIn space the contact point stays in the plane $y=y_{0}$ and moves in $x$:\n\\[\nx(0)=x_{0},\\qquad x(T_{1})=x_{0}+2\\pi A.\n\\tag{S$_{\\!I}$}\n\\]\n\nStage II ($T_{1}\\le t\\le T$).  \nFix $\\varphi(t)\\equiv\\pi$ and let $\\theta(t)$ increase monotonically\nfrom $\\pi/2$ to $3\\pi/2$, again with $\\dot\\theta(T_{1})=\\dot\\theta(T)=0$.\nThe $x$–coordinate is frozen at\n$x(t)\\equiv x_{1}:=x_{0}+2\\pi A$ and\n\\[\ny(T_{1})=y_{0},\\qquad y(T)=y_{0}+2\\pi B.\n\\tag{S$_{\\!II}$}\n\\]\n(The starting ordinates $x_{0},y_{0}$ are arbitrary.  Taking\n$\\varphi\\equiv\\pi$ instead of $0$ is immaterial for the local geometry\nbut simplifies the $C^{1}$ sewing of the two stages.)\n\na) Kinematics along the prescribed curves.  \n\n a$_1$) Show that on Stage I there exists  \n\\[\n\\lambda_{I}(\\varphi,x)=\n\\frac{\\lvert E_{\\varphi}(\\tfrac{\\pi}{2},\\varphi)\\rvert}\n     {\\sqrt{\\,1+f_{x}(x,y_{0})^{2}}}\n\\tag{5}\n\\]\nsuch that $\\dot x(t)=\\lambda_{I}\\bigl(\\varphi(t),x(t)\\bigr)\\,\\dot\\varphi(t)$.\n\n a$_2$) Prove the analogous statement for Stage II:\n\\[\n\\lambda_{II}(\\theta,y)=\n\\frac{\\lvert E_{\\theta}(\\theta,\\pi)\\rvert}\n     {\\sqrt{\\,1+f_{y}(x_{1},y)^{2}}},\n\\tag{6}\n\\]\nso that $\\dot y(t)=\\lambda_{II}\\bigl(\\theta(t),y(t)\\bigr)\\,\\dot\\theta(t)$.\n\nb) Using the no–slip condition point–wise together with\n$(S_{\\!I})$–$(S_{\\!II})$, prove the\n\\emph{metric identities}\n\\[\nB^{2}=A^{2}+R^{2}\\qquad\\text{and}\\qquad\nC^{2}=A^{2}+S^{2}.\n\\tag{7}\n\\]\n\nc) Conversely, assume that (7) holds.\n\n(i) Stage I.  \nInsert (7) into (5) to obtain  \n\\[\n\\frac{dx}{d\\varphi}\n  =\\Lambda_{I}(\\varphi,x):=\n   \\frac{\\sqrt{A^{2}+R^{2}\\cos^{2}\\varphi}}\n        {\\sqrt{\\,1+(R^{2}/A^{2})\\sin^{2}(x/A)}}.\n\\tag{8}\n\\]\nShow that for every $x_{0}$ equation (8) admits a unique $C^{1}$ solution\non $[-\\pi,\\pi]$ satisfying\n\\[\nx(\\pi)-x(-\\pi)=2\\pi A.\n\\tag{9}\n\\]\n\n(ii) Stage II.  \nAnalogously, prove that the solution of  \n\\[\n\\frac{dy}{d\\theta}\n  =\\Lambda_{II}(\\theta,y):=\n   \\frac{\\sqrt{A^{2}+S^{2}\\sin^{2}\\theta}}\n        {\\sqrt{\\,1+(S^{2}/B^{2})\\sin^{2}(y/B)}}\n\\tag{10}\n\\]\nsatisfies\n\\[\ny(3\\pi/2)-y(\\pi/2)=2\\pi B.\n\\tag{11}\n\\]\n\n(iii) Construct a rolling realising Stages I–II and prove that the\nresulting rigid motion $g(t)$ is of class $C^{1}$ on $[0,T]$.  In\nparticular, justify carefully that the body configurations at $t=T_{1}$\nmatch up to a (possibly trivial) \\emph{body–fixed} rotation about the\ncommon normal and a $(\\theta,\\varphi)$ shift leaving the contact point\ninvariant.  \n(You may use that the rolling constraints define a smooth Ehresmann\nconnection on the bundle $SE(3)\\times\\widehat E\\longrightarrow\\widehat\nE$ and that the time parametrisation of Stage I can always be slowed\ndown so that $\\dot\\varphi(T_{1})=0$, likewise for Stage II.)\n\n(Hints.  \n– For (c-i) evaluate the integral\n$\\displaystyle\\int\\sqrt{1+k^{2}\\sin^{2}u}\\,du$ explicitly and exploit\nits $\\pi$–periodicity.  \n– For (c-iii) choose the initial orientation of Stage II equal to the\nfinal orientation of Stage I in order to guarantee $C^{1}$–matching.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we write  \n\\[\nr(\\theta,\\varphi):=E(\\theta,\\varphi),\\qquad\nf(x,y)=H+R\\cos\\!\\Bigl(\\tfrac{x}{A}\\Bigr)+S\\cos\\!\\Bigl(\\tfrac{y}{B}\\Bigr),\\qquad\nP=c+Qr.\n\\]\nAll quantities are $C^{\\infty}$.\n\n--------------------------------------------------------------------\n\\textbf{a) Kinematics.}\n\nDifferentiating $P=c+Qr$ gives\n\\[\n\\dot P=\\dot c+\\dot Q\\,r+Q\\,\\dot r.\n\\tag{12}\n\\]\nBy the no–slip condition $(\\dot c+\\dot Q\\,r)=0$, hence\n\\[\n\\dot P=Q\\,\\dot r.\n\\tag{13}\n\\]\nThus the motion of the contact point along $\\Sigma$ is governed\n\\emph{solely} by the variation of $(\\theta,\\varphi)$.\n\n------------------------------------------------\na$_1$) Stage I $(\\theta\\equiv\\pi/2,\\;y\\equiv y_{0})$.  \nHere\n\\[\nr_{\\varphi}\\bigl(\\tfrac{\\pi}{2},\\varphi\\bigr)=\n(-A\\sin\\varphi,\\;B\\cos\\varphi,\\;0),\\qquad\nv_{E}:=r_{\\varphi}\\,\\dot\\varphi.\n\\]\nWith (13)\n\\[\n\\dot P=Qv_{E}=\\dot x\\,(1,\\,0,\\,f_{x}(x,y_{0})).\n\\]\nBoth vectors lie in the one–dimensional subspace $T_{P}\\Sigma$, so\nthere exists $\\lambda_{I}$ with\n\\[\nQv_{E}=\\lambda_{I}(\\varphi,x)\\bigl(1,\\,0,\\,f_{x}(x,y_{0})\\bigr).\n\\]\nTaking Euclidean norms gives (5); equating first components yields\n$\\dot x=\\lambda_{I}\\dot\\varphi$.\n\n------------------------------------------------\na$_2$) Stage II $(\\varphi\\equiv\\pi,\\;x\\equiv x_{1})$.  \nNow\n\\[\nr_{\\theta}(\\theta,\\pi)=(-A\\cos\\theta,\\;0,\\;-C\\sin\\theta),\\qquad\nv_{E}:=r_{\\theta}\\,\\dot\\theta.\n\\]\nProceeding as above we obtain (6) and\n$\\dot y=\\lambda_{II}\\dot\\theta$.\n\n--------------------------------------------------------------------\n\\textbf{b) Metric identities.}\n\nStage I: by (13) and a$_1$\n\\[\n\\lvert r_{\\varphi}\\rvert\\,\\dot\\varphi\n   =\\sqrt{1+f_{x}^{2}}\\;\\dot x.\n\\tag{14}\n\\]\nSince $\\dot\\varphi,\\dot x\\ge 0$ no absolute values are needed.\nIntegrating over $[0,T_{1}]$ and using $(S_{\\!I})$ gives\n\\[\n\\int_{-\\pi}^{\\pi}\n   \\lvert r_{\\varphi}(\\tfrac{\\pi}{2},\\varphi)\\rvert\\,d\\varphi\n=\\int_{x_{0}}^{x_{0}+2\\pi A}\n   \\sqrt{1+f_{x}(x,y_{0})^{2}}\\,dx.\n\\tag{15}\n\\]\nThe left–hand side equals\n\\[\n\\int_{-\\pi}^{\\pi}\n   \\sqrt{A^{2}\\sin^{2}\\varphi+B^{2}\\cos^{2}\\varphi}\\,d\\varphi,\n\\]\nwhile the right–hand side equals\n\\[\n\\int_{0}^{2\\pi}\\sqrt{A^{2}+R^{2}\\sin^{2}u}\\,du\n\\quad\\text{with }u:=x/A.\n\\]\n\nStage II (interchanging $(x,y),(R,S),(B,C)$) yields\n\\[\n\\int_{0}^{2\\pi}\\sqrt{A^{2}+S^{2}\\sin^{2}u}\\,du\n=\\int_{0}^{2\\pi}\\sqrt{A^{2}\\sin^{2}\\theta+C^{2}\\cos^{2}\\theta}\\,d\\theta.\n\\]\n\nSet\n\\[\nF(b):=\\int_{0}^{2\\pi}\\sqrt{A^{2}\\sin^{2}\\vartheta+b^{2}\\cos^{2}\\vartheta}\\,d\\vartheta,\n\\quad b>0.\n\\]\nDifferentiating under the integral sign gives\n\\[\nF'(b)=\\int_{0}^{2\\pi}\\frac{b\\cos^{2}\\vartheta}\n              {\\sqrt{A^{2}\\sin^{2}\\vartheta+b^{2}\\cos^{2}\\vartheta}}\n       \\,d\\vartheta>0,\n\\]\nso $F$ is strictly increasing.  Equation (15) therefore forces\n$B^{2}=A^{2}+R^{2}$, and Stage II yields $C^{2}=A^{2}+S^{2}$, i.e.~(7).\n\n--------------------------------------------------------------------\n\\textbf{c) Inverse construction under assumption (7).}\n\n(i) Stage I.  \nWith $u:=x/A$ and $k:=R/A$ equation (8) becomes\n\\[\n\\frac{du}{d\\varphi}\n   =\\Phi(\\varphi,u):=\n    \\frac{\\sqrt{1+k^{2}\\cos^{2}\\varphi}}\n         {\\sqrt{1+k^{2}\\sin^{2}u}}.\n\\tag{16}\n\\]\n$\\Phi$ is $C^{1}$ and strictly positive on the cylinder\n$(\\varphi,u)\\in[-\\pi,\\pi]\\times\\mathbb R$, hence the ODE has a unique\n$C^{1}$ solution for all $\\varphi\\in[-\\pi,\\pi]$.  Separation of\nvariables gives\n\\[\n\\int_{u_{0}}^{u(\\varphi)}\\sqrt{1+k^{2}\\sin^{2}s}\\,ds\n   =\\int_{-\\pi}^{\\varphi}\\sqrt{1+k^{2}\\cos^{2}\\tau}\\,d\\tau.\n\\tag{17}\n\\]\nThe integrands are $\\pi$–periodic and even, so evaluating at\n$\\varphi=\\pi$ yields $u(\\pi)=u_{0}+2\\pi$, i.e.~(9).\n\n(ii) Stage II.  \nSubstituting $v:=y/B$ and $\\ell:=S/B$ converts (10) into the same model\nequation, proving (11).\n\n(iii) \\emph{Existence of a $C^{1}$ rolling realising both stages.}\n\n\\emph{Step 1: horizontal lift along Stage I.}  \nChoose an arbitrary initial configuration $g(0)=(Q_{0},c_{0})$.\nBecause the base curve $\\bigl(\\theta(t),\\varphi(t)\\bigr)$ and the\ncontact point on $\\Sigma$ are prescribed, the rolling connection\nprovides a unique horizontal lift $g(t)=(Q(t),c(t))$ on $[0,T_{1}]$.\nBy construction $\\dot\\varphi(0)=0$ and we have arranged $\\dot\\varphi(T_{1})=0$,\nso\n\\[\n\\dot Q(T_{1})=\\dot c(T_{1})=0.\n\\tag{18}\n\\]\n\n\\emph{Step 2: initial data for Stage II.}  \nAt $t=T_{1}$ the contact point satisfies\n$\\theta=\\pi/2,\\varphi=\\pi$ and the common normal is\n$n(T_{1})=\\tfrac{Q(T_{1})\\,n_{E}(\\pi/2,\\pi)}{\\lvert n_{E}(\\pi/2,\\pi)\\rvert}$.\nBecause for a fixed contact point and normal the admissible body\norientations form a one–parameter family obtained by rotations around\nthe normal, we \\emph{choose}\n\\[\ng(T_{1}^{+})=(Q(T_{1}),c(T_{1})).\n\\tag{19}\n\\]\nThis choice suppresses any jump in $Q$ or $c$.\n\n\\emph{Step 3: horizontal lift along Stage II.}  \nFrom the initial data (19) the rolling connection yields a unique\nhorizontal lift on $[T_{1},T]$.  Because $\\dot\\theta(T_{1})=0$ the\nright–hand sides of the differential equations for $Q$ and $c$ vanish\nat $t=T_{1}$, so with (18)\n\\[\n\\dot Q(T_{1}^{+})=\\dot Q(T_{1}^{-}),\\qquad\n\\dot c(T_{1}^{+})=\\dot c(T_{1}^{-}),\n\\]\nand the overall map $g:[0,T]\\to SE(3)$ is $C^{1}$.\n\n\\emph{Step 4: summary.}  \nThe curve\n$t\\mapsto\\bigl(g(t),\\theta(t),\\varphi(t)\\bigr)$ meets all rolling\nconstraints, realises the required lattice steps (thanks to (9) and\n(11)), and is of class $C^{1}$ on $[0,T]$.  Consequently the rolling\nexists if and only if the metric identities (7) hold.\n\n--------------------------------------------------------------------\n\\textbf{Conclusion.}  \nA $C^{1}$ rolling of the triaxial ellipsoid \\eqref{0} on the egg–crate\nsurface \\eqref{1} performing the two prescribed lattice steps exists\n\\emph{iff}\n\\[\n\\boxed{\\;B^{2}=A^{2}+R^{2},\\qquad C^{2}=A^{2}+S^{2}\\;}.\n\\qquad\\qedhere\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.574401",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension — The original problem is planar; the enhanced\nvariant takes place in ℝ³ with a full ellipsoid rolling on a doubly\nperiodic surface, so one deals with two-parameter families of contact\npoints instead of a single curve.\n\n2. Additional constraints — Besides equality of arc lengths, the problem\ndemands second-order tangency (“no twist’’) and *simultaneous* matching\nof two independent periodic motions (one about each principal axis of\nthe ellipsoid).\n\n3. Sophisticated structures — The solution needs first fundamental\nforms, differential geometry of surfaces, and monotonicity properties of\ncomplete elliptic integrals rather than elementary arc-length calculus\nalone.\n\n4. Deeper theory — Showing that the coincidence of the metrics along two\nindependent directions suffices for full metric coincidence, and that\nhorizontal motion of the centre follows from the common normal argument,\nrequires familiarity with rigid surface rolling theory.\n\n5. Multiple interacting concepts — The argument couples surface geometry\n(elliptic integrals, metric tensors) with rigid-body kinematics\n(synchronous rotations about two axes and horizontal centre-of-mass\nmotion).\n\nAll these layers make the enhanced kernel variant substantially more\ntechnically demanding than both the original Olympiad-style question and\nthe current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}