{
  "index": "1946-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "6. A particle moves on a circle with center \\( O \\), starting from rest at a point \\( P \\) and coming to rest again at a point \\( Q \\), without coming to rest at any intermediate point. Prove that the acceleration vector of the particle does not vanish at any point between \\( P \\) and \\( Q \\) and that, at some point \\( R \\) between \\( P \\) and \\( Q \\), the acceleration vector points in along the radius \\( R O \\).",
  "solution": "Solution. Suppose the circle has radius \\( r \\). If Cartesian coordinates are chosen with origin at the center of the circle, then the coordinates of the particle are \\( (r \\cos \\theta, r \\sin \\theta) \\), where \\( \\theta \\) is a function of the time \\( t \\). Differentiating this twice, we see that the acceleration vector is\n\\[\nr \\frac{d \\omega}{d t}(-\\sin \\theta, \\cos \\theta)+r \\omega^{2}(-\\cos \\theta,-\\sin \\theta)\n\\]\nwhere \\( \\omega=d \\theta / d t \\) is the angular velocity. Since \\( (-\\sin \\theta, \\cos \\theta) \\) and \\( (-\\cos \\theta \\), \\( -\\sin \\theta \\) ) are orthogonal unit vectors in the directions of the tangent and the inward normal, we see that the two terms of (1) are, respectively, the tangential and normal components of the acceleration. Since \\( \\omega \\neq 0 \\) at any time during the motion, the normal component of acceleration, and hence the acceleration vector itself, is never zero. Since \\( \\omega=0 \\) at the start and at the finish, by Rolle's theorem there is an intermediate time at which \\( d \\omega / d t=0 \\). At that time the acceleration vector points inward along the radius because \\( r \\omega^{2}>0 \\).\n\nRemark. We have interpreted \"coming to rest\" to mean \"having velocity zero.\" If \"coming to rest\" means \"remaining stationary through some time interval,\" then the first statement is false, for it is certainly possible that \\( \\omega \\) and \\( d \\omega / d t \\) vanish simultaneously but not on an interval. The second statement is true, however, because the usual proof of Rolle's theorem shows that \\( d \\omega / d t=0 \\) at some point where \\( \\omega \\neq 0 \\), unless \\( \\omega=0 \\) identically, which is ruled out.",
  "vars": [
    "O",
    "P",
    "Q",
    "R",
    "\\\\theta",
    "t",
    "\\\\omega"
  ],
  "params": [
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "O": "centerpo",
        "P": "startpoint",
        "Q": "endpoint",
        "R": "radialpoi",
        "\\theta": "anglevar",
        "t": "timevar",
        "\\omega": "angveloc",
        "r": "circradiu"
      },
      "question": "6. A particle moves on a circle with center \\( centerpo \\), starting from rest at a point \\( startpoint \\) and coming to rest again at a point \\( endpoint \\), without coming to rest at any intermediate point. Prove that the acceleration vector of the particle does not vanish at any point between \\( startpoint \\) and \\( endpoint \\) and that, at some point \\( radialpoi \\) between \\( startpoint \\) and \\( endpoint \\), the acceleration vector points in along the radius \\( radialpoi centerpo \\).",
      "solution": "Suppose the circle has radius \\( circradiu \\). If Cartesian coordinates are chosen with origin at the center of the circle, then the coordinates of the particle are \\( (circradiu \\cos anglevar,\\; circradiu \\sin anglevar) \\), where \\( anglevar \\) is a function of the time \\( timevar \\). Differentiating this twice, we see that the acceleration vector is\n\\[\ncircradiu \\frac{d angveloc}{d timevar}(-\\sin anglevar,\\; \\cos anglevar)+circradiu \nangveloc^{2}(-\\cos anglevar,-\\sin anglevar)\n\\]\nwhere \\( angveloc=d anglevar / d timevar \\) is the angular velocity. Since \\( (-\\sin anglevar,\\; \\cos anglevar) \\) and \\( (-\\cos anglevar, -\\sin anglevar) \\) are orthogonal unit vectors in the directions of the tangent and the inward normal, we see that the two terms of (1) are, respectively, the tangential and normal components of the acceleration. Since \\( angveloc \\neq 0 \\) at any time during the motion, the normal component of acceleration, and hence the acceleration vector itself, is never zero. Since \\( angveloc=0 \\) at the start and at the finish, by Rolle's theorem there is an intermediate time at which \\( d angveloc / d timevar=0 \\). At that time the acceleration vector points inward along the radius because \\( circradiu angveloc^{2}>0 \\).\n\nRemark. We have interpreted \"coming to rest\" to mean \"having velocity zero.\" If \"coming to rest\" means \"remaining stationary through some time interval,\" then the first statement is false, for it is certainly possible that \\( angveloc \\) and \\( d angveloc / d timevar \\) vanish simultaneously but not on an interval. The second statement is true, however, because the usual proof of Rolle's theorem shows that \\( d angveloc / d timevar=0 \\) at some point where \\( angveloc \\neq 0 \\), unless \\( angveloc=0 \\) identically, which is ruled out."
    },
    "descriptive_long_confusing": {
      "map": {
        "O": "waterfall",
        "P": "hedgehog",
        "Q": "labyrinth",
        "R": "carpenter",
        "\\theta": "pineapple",
        "t": "molecule",
        "\\omega": "lighthouse",
        "r": "avalanche"
      },
      "question": "6. A particle moves on a circle with center \\( waterfall \\), starting from rest at a point \\( hedgehog \\) and coming to rest again at a point \\( labyrinth \\), without coming to rest at any intermediate point. Prove that the acceleration vector of the particle does not vanish at any point between \\( hedgehog \\) and \\( labyrinth \\) and that, at some point \\( carpenter \\) between \\( hedgehog \\) and \\( labyrinth \\), the acceleration vector points in along the radius \\( carpenter waterfall \\).",
      "solution": "Solution. Suppose the circle has radius \\( avalanche \\). If Cartesian coordinates are chosen with origin at the center of the circle, then the coordinates of the particle are \\( (avalanche \\cos pineapple, avalanche \\sin pineapple) \\), where \\( pineapple \\) is a function of the time \\( molecule \\). Differentiating this twice, we see that the acceleration vector is\n\\[\navalanche \\frac{d lighthouse}{d molecule}(-\\sin pineapple, \\cos pineapple)+avalanche lighthouse^{2}(-\\cos pineapple,-\\sin pineapple)\n\\]\nwhere \\( lighthouse=d pineapple / d molecule \\) is the angular velocity. Since \\( (-\\sin pineapple, \\cos pineapple) \\) and \\( (-\\cos pineapple \\), \\( -\\sin pineapple \\) ) are orthogonal unit vectors in the directions of the tangent and the inward normal, we see that the two terms of (1) are, respectively, the tangential and normal components of the acceleration. Since \\( lighthouse \\neq 0 \\) at any time during the motion, the normal component of acceleration, and hence the acceleration vector itself, is never zero. Since \\( lighthouse=0 \\) at the start and at the finish, by Rolle's theorem there is an intermediate time at which \\( d lighthouse / d molecule=0 \\). At that time the acceleration vector points inward along the radius because \\( avalanche lighthouse^{2}>0 \\).\n\nRemark. We have interpreted \"coming to rest\" to mean \"having velocity zero.\" If \"coming to rest\" means \"remaining stationary through some time interval,\" then the first statement is false, for it is certainly possible that \\( lighthouse \\) and \\( d lighthouse / d molecule \\) vanish simultaneously but not on an interval. The second statement is true, however, because the usual proof of Rolle's theorem shows that \\( d lighthouse / d molecule=0 \\) at some point where \\( lighthouse \\neq 0 \\), unless \\( lighthouse=0 \\) identically, which is ruled out."
    },
    "descriptive_long_misleading": {
      "map": {
        "O": "boundarypoint",
        "P": "terminalspot",
        "Q": "initialspot",
        "R": "extremespot",
        "\\theta": "distanceval",
        "t": "spacevalue",
        "\\omega": "stillness",
        "r": "infiniteval"
      },
      "question": "A particle moves on a circle with center \\( boundarypoint \\), starting from rest at a point \\( terminalspot \\) and coming to rest again at a point \\( initialspot \\), without coming to rest at any intermediate point. Prove that the acceleration vector of the particle does not vanish at any point between \\( terminalspot \\) and \\( initialspot \\) and that, at some point \\( extremespot \\) between \\( terminalspot \\) and \\( initialspot \\), the acceleration vector points in along the radius \\( extremespot boundarypoint \\).",
      "solution": "Solution. Suppose the circle has radius \\( infiniteval \\). If Cartesian coordinates are chosen with origin at the center of the circle, then the coordinates of the particle are \\( (infiniteval \\cos distanceval, infiniteval \\sin distanceval) \\), where \\( distanceval \\) is a function of the spacevalue \\( spacevalue \\). Differentiating this twice, we see that the acceleration vector is\n\\[\ninfiniteval \\frac{d stillness}{d spacevalue}(-\\sin distanceval, \\cos distanceval)+infiniteval stillness^{2}(-\\cos distanceval,-\\sin distanceval)\n\\]\nwhere \\( stillness=d distanceval / d spacevalue \\) is the angular velocity. Since \\( (-\\sin distanceval, \\cos distanceval) \\) and \\( (-\\cos distanceval , -\\sin distanceval ) \\) are orthogonal unit vectors in the directions of the tangent and the inward normal, we see that the two terms of (1) are, respectively, the tangential and normal components of the acceleration. Since \\( stillness \\neq 0 \\) at any time during the motion, the normal component of acceleration, and hence the acceleration vector itself, is never zero. Since \\( stillness=0 \\) at the start and at the finish, by Rolle's theorem there is an intermediate time at which \\( d stillness / d spacevalue=0 \\). At that time the acceleration vector points inward along the radius because \\( infiniteval stillness^{2}>0 \\).\n\nRemark. We have interpreted \"coming to rest\" to mean \"having velocity zero.\" If \"coming to rest\" means \"remaining stationary through some time interval,\" then the first statement is false, for it is certainly possible that \\( stillness \\) and \\( d stillness / d spacevalue \\) vanish simultaneously but not on an interval. The second statement is true, however, because the usual proof of Rolle's theorem shows that \\( d stillness / d spacevalue=0 \\) at some point where \\( stillness \\neq 0 \\), unless \\( stillness=0 \\) identically, which is ruled out."
    },
    "garbled_string": {
      "map": {
        "O": "lmskdfjo",
        "P": "gqtrdoxz",
        "Q": "cnvbikas",
        "R": "juyplase",
        "\\theta": "kzjndpqr",
        "t": "plmqrdst",
        "\\omega": "vhczmqpe",
        "r": "xtrsapow"
      },
      "question": "6. A particle moves on a circle with center \\( lmskdfjo \\), starting from rest at a point \\( gqtrdoxz \\) and coming to rest again at a point \\( cnvbikas \\), without coming to rest at any intermediate point. Prove that the acceleration vector of the particle does not vanish at any point between \\( gqtrdoxz \\) and \\( cnvbikas \\) and that, at some point \\( juyplase \\) between \\( gqtrdoxz \\) and \\( cnvbikas \\), the acceleration vector points in along the radius \\( juyplase lmskdfjo \\).",
      "solution": "Solution. Suppose the circle has radius \\( xtrsapow \\). If Cartesian coordinates are chosen with origin at the center of the circle, then the coordinates of the particle are \\( (xtrsapow \\cos kzjndpqr, xtrsapow \\sin kzjndpqr) \\), where \\( kzjndpqr \\) is a function of the time \\( plmqrdst \\). Differentiating this twice, we see that the acceleration vector is\n\\[\nxtrsapow \\frac{d vhczmqpe}{d plmqrdst}(-\\sin kzjndpqr, \\cos kzjndpqr)+xtrsapow vhczmqpe^{2}(-\\cos kzjndpqr,-\\sin kzjndpqr)\n\\]\nwhere \\( vhczmqpe=d kzjndpqr / d plmqrdst \\) is the angular velocity. Since \\( (-\\sin kzjndpqr, \\cos kzjndpqr) \\) and \\( (-\\cos kzjndpqr, -\\sin kzjndpqr ) \\) are orthogonal unit vectors in the directions of the tangent and the inward normal, we see that the two terms of (1) are, respectively, the tangential and normal components of the acceleration. Since \\( vhczmqpe \\neq 0 \\) at any time during the motion, the normal component of acceleration, and hence the acceleration vector itself, is never zero. Since \\( vhczmqpe=0 \\) at the start and at the finish, by Rolle's theorem there is an intermediate time at which \\( d vhczmqpe / d plmqrdst=0 \\). At that time the acceleration vector points inward along the radius because \\( xtrsapow vhczmqpe^{2}>0 \\).\n\nRemark. We have interpreted \"coming to rest\" to mean \"having velocity zero.\" If \"coming to rest\" means \"remaining stationary through some time interval,\" then the first statement is false, for it is certainly possible that \\( vhczmqpe \\) and \\( d vhczmqpe / d plmqrdst \\) vanish simultaneously but not on an interval. The second statement is true, however, because the usual proof of Rolle's theorem shows that \\( d vhczmqpe / d plmqrdst=0 \\) at some point where \\( vhczmqpe \\neq 0 \\), unless \\( vhczmqpe=0 \\) identically, which is ruled out."
    },
    "kernel_variant": {
      "question": "Two identical point–particles \\(P_{1},P_{2}\\) of unit mass move on the same fixed circle  \n\\[\n\\mathcal C=\\{X\\in\\mathbb R^{2}\\mid\\|X-H\\|=a\\},\\qquad a>0 ,\n\\]\nwhose centre is the point \\(H\\) in the plane.  \n\nThe particles are linked by a rigid, mass–less rod of length  \n\\[\nd,\\qquad 0<d<2a ,\n\\]\nso that  \n\\[\n\\|P_{1}(t)-P_{2}(t)\\|=d\\qquad\\text{for all }t .\n\\]\n\nAt time \\(t=0\\) the system is at rest in the configuration  \n\\[\nP_{1}(0)=A,\\;P_{2}(0)=A^{*},\\qquad A^{*}\\neq A ,\n\\]\nand at a later time \\(T>0\\) it is again at rest in the (different) configuration  \n\\[\nP_{1}(T)=B,\\;P_{2}(T)=B^{*},\\qquad B^{*}\\neq B .\n\\]\nNeither particle is momentarily at rest at any instant \\(0<t<T\\).\n\n1.  Prove that there is no instant \\(t\\in(0,T)\\) at which the two acceleration vectors vanish simultaneously.\n\n2.  Show that there exists at least one instant \\(t_{1}\\in(0,T)\\) for which both acceleration vectors are directed exactly toward the centre \\(H\\).\n\n3.  Denote  \n\\[\ng_{1}(t)=\\bigl(P_{2}(t)-P_{1}(t)\\bigr)\\cdot\\ddot P_{1}(t),\\qquad  \ng_{2}(t)=\\bigl(P_{2}(t)-P_{1}(t)\\bigr)\\cdot\\ddot P_{2}(t).\n\\]\nProve that each of the two continuous functions \\(g_{1},g_{2}\\) possesses at least one zero in the open interval \\((0,T)\\); that is, there exist (possibly different) instants  \n\\[\nt_{2},t_{3}\\in(0,T)\n\\]\nsuch that\n\\[\ng_{1}(t_{2})=0,\\qquad g_{2}(t_{3})=0 .\n\\]\nShow moreover that the coincidence \\(t_{2}=t_{3}\\) cannot occur unless the two particles are simultaneously at rest, a situation ruled out by the hypotheses.",
      "solution": "Preliminaries and notation.  \nBecause both particles always lie on the same circle and are connected by a rigid rod of fixed length, every admissible configuration is described by a single angle.  \nFix a positively–oriented polar frame centred at \\(H\\) and denote by\n\\[\n\\theta(t)\\in\\mathbb R\n\\]\nthe polar angle of \\(P_{1}\\).  Put  \n\\[\n\\beta\\in(0,\\pi),\\qquad 2a\\sin\\frac{\\beta}{2}=d ,\n\\]\nso that  \n\\[\nP_{1}(t)=H+a(\\cos\\theta,\\sin\\theta),\\quad\nP_{2}(t)=H+a\\bigl(\\cos(\\theta+\\beta),\\sin(\\theta+\\beta)\\bigr).\n\\]\n\nWrite  \n\\[\n\\omega(t)=\\dot\\theta(t),\\qquad\\alpha(t)=\\dot\\omega(t)=\\ddot\\theta(t).\n\\]\nBecause the system is at rest at the endpoints and never stops in the open interval,\n\\[\n\\omega(0)=\\omega(T)=0,\\qquad \\omega(t)\\neq 0\\quad(0<t<T),\n\\]\nand \\(\\omega\\in C^{1}[0,T]\\).\n\nAcceleration decomposition.  \nLet \\(\\mathbf t_{j},\\mathbf n_{j}\\;(j=1,2)\\) be, respectively, the unit tangent (direction of increasing \\(\\theta\\)) and the inward unit normal at \\(P_{j}\\).  One readily computes\n\\[\n\\boxed{\\;\\ddot P_{j}=a\\alpha\\,\\mathbf t_{j}+a\\omega^{2}\\,\\mathbf n_{j}\\quad(j=1,2).}\n\\]\n\n--------------------------------------------------------------------\n1.  Simultaneous cancellation of both accelerations is impossible.  \nAssume that \\(\\ddot P_{1}(t_{0})=\\ddot P_{2}(t_{0})=\\mathbf0\\) for some \\(t_{0}\\in(0,T)\\).  \nThe boxed formula forces  \n\\[\n\\alpha(t_{0})=0,\\qquad\\omega(t_{0})=0,\n\\]\ncontradicting the strict non–vanishing of \\(\\omega\\) on \\((0,T)\\).  Hence part 1 is proved.\n\n--------------------------------------------------------------------\n2.  Existence of an instant where both accelerations are purely radial.  \nSince \\(\\omega\\) vanishes only at the endpoints, it keeps a constant sign on \\((0,T)\\).  Assume \\(\\omega>0\\) (the case \\(\\omega<0\\) is identical).  \nBecause \\(\\omega(0)=\\omega(T)=0\\) while \\(\\omega\\not\\equiv 0\\), Rolle’s theorem yields \\(t_{1}\\in(0,T)\\) with  \n\\[\n\\alpha(t_{1})=\\dot\\omega(t_{1})=0.\n\\]\nAt that instant\n\\[\n\\ddot P_{j}(t_{1})=a\\omega(t_{1})^{2}\\,\\mathbf n_{j}(t_{1})\\qquad(j=1,2),\n\\]\na positive multiple of the inward normal; therefore each acceleration vector is directed exactly toward \\(H\\).  Part 2 is settled.\n\n--------------------------------------------------------------------\n3.  Orthogonality of each acceleration to the rod at (possibly different) instants.  \n\n3 (a)  Closed expressions for \\(g_{1},g_{2}\\).  \nA direct trigonometric calculation (best performed in the \\(\\{\\mathbf t_{1},\\mathbf n_{1}\\}\\) frame) yields  \n\\[\n\\boxed{\\;g_{1}(t)=a^{2}\\Bigl[\\alpha(t)\\sin\\beta\n      +2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)\\,\\omega(t)^{2}\\Bigr]},\\tag{1}\n\\]\n\\[\n\\boxed{\\;g_{2}(t)=a^{2}\\Bigl[\\alpha(t)\\sin\\beta\n      -2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)\\,\\omega(t)^{2}\\Bigr]}. \\tag{2}\n\\]\nNote that the sign of both terms in (2) is the opposite of what appeared in the earlier (incorrect) version.\n\n3 (b)  Producing opposite signs for \\(g_{1}\\).  \nSet  \n\\[\n\\kappa:=2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)>0,\\qquad\n\\sigma:=\\sin\\beta>0 .\n\\]\nBecause \\(\\omega(0)=\\omega(T)=0\\) and \\(\\omega\\not\\equiv0\\), its derivative \\(\\alpha\\) attains both positive and negative values on \\((0,T)\\).  Define  \n\\[\n\\mathcal P:=\\{t\\in(0,T)\\mid\\alpha(t)>0\\},\\qquad\n\\mathcal N:=\\{t\\in(0,T)\\mid\\alpha(t)<0\\},\n\\]\nboth non–empty.\n\nPick \\(t^{+}\\in\\mathcal P\\) so close to \\(0\\) that \\(\\omega(t^{+})^{2}<\\dfrac{\\sigma}{2\\kappa}\\alpha(t^{+})\\).  \nThen\n\\[\ng_{1}(t^{+})=a^{2}\\bigl[\\alpha(t^{+})\\sigma+\\kappa\\omega(t^{+})^{2}\\bigr]\n           >\\tfrac12a^{2}\\alpha(t^{+})\\sigma>0. \\tag{3}\n\\]\n\nSimilarly, choose \\(t^{-}\\in\\mathcal N\\) near \\(T\\) so that  \n\\(\\omega(t^{-})^{2}<\\dfrac{\\sigma}{2\\kappa}\\,|\\alpha(t^{-})|\\).  Then  \n\\[\ng_{1}(t^{-})=a^{2}\\bigl[\\alpha(t^{-})\\sigma+\\kappa\\omega(t^{-})^{2}\\bigr]\n           <-\\tfrac12a^{2}|\\alpha(t^{-})|\\sigma<0. \\tag{4}\n\\]\n\nBecause \\(g_{1}\\) is continuous, the Intermediate Value Theorem furnishes  \n\\[\nt_{2}\\in(t^{+},t^{-})\\subset(0,T)\\quad\\text{with}\\quad g_{1}(t_{2})=0.\n\\]\n\n3 (c)  Producing opposite signs for \\(g_{2}\\).  \nUsing the same two instants \\(t^{+},t^{-}\\) we now have, from (2),\n\\[\ng_{2}(t^{+})=a^{2}\\bigl[\\alpha(t^{+})\\sigma-\\kappa\\omega(t^{+})^{2}\\bigr]\n           >\\tfrac12a^{2}\\alpha(t^{+})\\sigma>0,\n\\]\n\\[\ng_{2}(t^{-})=a^{2}\\bigl[\\alpha(t^{-})\\sigma-\\kappa\\omega(t^{-})^{2}\\bigr]\n           <-\\tfrac12a^{2}|\\alpha(t^{-})|\\sigma<0.\n\\]\nHence a second application of the Intermediate Value Theorem yields  \n\\[\nt_{3}\\in(t^{+},t^{-})\\subset(0,T)\\quad\\text{with}\\quad g_{2}(t_{3})=0.\n\\]\n\n3 (d)  The zeros cannot coincide.  \nAssume \\(t_{2}=t_{3}\\).  Then \\(g_{1}(t_{2})=g_{2}(t_{2})=0\\).  Adding and subtracting (1)–(2) give  \n\\[\ng_{1}+g_{2}=2a^{2}\\alpha(t_{2})\\sin\\beta=0,\\qquad\ng_{1}-g_{2}=2a^{2}\\kappa\\omega(t_{2})^{2}=0.\n\\]\nHence \\(\\alpha(t_{2})=\\omega(t_{2})=0\\), contradicting the hypothesis that \\(\\omega(t)\\neq0\\) for \\(0<t<T\\).  Therefore \\(t_{2}\\neq t_{3}\\).\n\n--------------------------------------------------------------------\nAll three statements have now been proved with complete rigour.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.407992",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-particle circle problem, the enhanced variant is markedly harder for several reasons.\n\n• Two bodies and an additional holonomic constraint (fixed chord length) raise the configuration space from a 1- to a 2-particle system with an internal geometric restriction, so one must keep track of three different vector quantities: the common centre, the individual positions and the rigid rod.\n\n• The proof requires introducing a single angular coordinate that simultaneously describes both trajectories, then translating kinematic data (velocity, acceleration) of {\\em two} particles in terms of that coordinate.  This leads to coupled identities that are absent in the original problem.\n\n• Part 1 demands exclusion of {\\em simultaneous} vanishing of two vector functions, not just one, forcing a careful examination of how the vanishing of the common scalar functions \\(\\omega\\) and \\(\\alpha\\) would propagate to both accelerations.\n\n• Part 3 is genuinely new: showing orthogonality to the rod entails constructing and analysing a non–trivial scalar product function \\(g(t)\\) that mixes the geometry of the constraint (the chord) with the dynamics (the acceleration), then applying the Intermediate Value Theorem after working out its sign at the end–points by means of higher–order Taylor expansions.\n\n• Throughout, one must manage multiple interacting concepts (polar coordinates, rigid constraints, vector decomposition, Rolle’s theorem, continuity arguments) rather than the single mean-value argument of the original.\n\nHence the enhanced variant demands a substantially broader toolkit and a longer chain of reasoning than either the original problem or the basic kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Two identical point–particles \\(P_{1},P_{2}\\) of unit mass move on the same fixed circle  \n\\[\n\\mathcal C=\\{X\\in\\mathbb R^{2}\\mid\\|X-H\\|=a\\},\\qquad a>0 ,\n\\]\nwhose centre is the point \\(H\\) in the plane.  \n\nThe particles are linked by a rigid, mass–less rod of length  \n\\[\nd,\\qquad 0<d<2a ,\n\\]\nso that  \n\\[\n\\|P_{1}(t)-P_{2}(t)\\|=d\\qquad\\text{for all }t .\n\\]\n\nAt time \\(t=0\\) the system is at rest in the configuration  \n\\[\nP_{1}(0)=A,\\;P_{2}(0)=A^{*},\\qquad A^{*}\\neq A ,\n\\]\nand at a later time \\(T>0\\) it is again at rest in the (different) configuration  \n\\[\nP_{1}(T)=B,\\;P_{2}(T)=B^{*},\\qquad B^{*}\\neq B .\n\\]\nNeither particle is momentarily at rest at any instant \\(0<t<T\\).\n\n1.  Prove that there is no instant \\(t\\in(0,T)\\) at which the two acceleration vectors vanish simultaneously.\n\n2.  Show that there exists at least one instant \\(t_{1}\\in(0,T)\\) for which both acceleration vectors are directed exactly toward the centre \\(H\\).\n\n3.  Denote  \n\\[\ng_{1}(t)=\\bigl(P_{2}(t)-P_{1}(t)\\bigr)\\cdot\\ddot P_{1}(t),\\qquad  \ng_{2}(t)=\\bigl(P_{2}(t)-P_{1}(t)\\bigr)\\cdot\\ddot P_{2}(t).\n\\]\nProve that each of the two continuous functions \\(g_{1},g_{2}\\) possesses at least one zero in the open interval \\((0,T)\\); that is, there exist (possibly different) instants  \n\\[\nt_{2},t_{3}\\in(0,T)\n\\]\nsuch that\n\\[\ng_{1}(t_{2})=0,\\qquad g_{2}(t_{3})=0 .\n\\]\nShow moreover that the coincidence \\(t_{2}=t_{3}\\) cannot occur unless the two particles are simultaneously at rest, a situation ruled out by the hypotheses.",
      "solution": "Preliminaries and notation.  \nBecause both particles always lie on the same circle and are connected by a rigid rod of fixed length, every admissible configuration is described by a single angle.  \nFix a positively–oriented polar frame centred at \\(H\\) and denote by\n\\[\n\\theta(t)\\in\\mathbb R\n\\]\nthe polar angle of \\(P_{1}\\).  Put  \n\\[\n\\beta\\in(0,\\pi),\\qquad 2a\\sin\\frac{\\beta}{2}=d ,\n\\]\nso that  \n\\[\nP_{1}(t)=H+a(\\cos\\theta,\\sin\\theta),\\quad\nP_{2}(t)=H+a\\bigl(\\cos(\\theta+\\beta),\\sin(\\theta+\\beta)\\bigr).\n\\]\n\nWrite  \n\\[\n\\omega(t)=\\dot\\theta(t),\\qquad\\alpha(t)=\\dot\\omega(t)=\\ddot\\theta(t).\n\\]\nBecause the system is at rest at the endpoints and never stops in the open interval,\n\\[\n\\omega(0)=\\omega(T)=0,\\qquad \\omega(t)\\neq 0\\quad(0<t<T),\n\\]\nand \\(\\omega\\in C^{1}[0,T]\\).\n\nAcceleration decomposition.  \nLet \\(\\mathbf t_{j},\\mathbf n_{j}\\;(j=1,2)\\) be, respectively, the unit tangent (direction of increasing \\(\\theta\\)) and the inward unit normal at \\(P_{j}\\).  One readily computes\n\\[\n\\boxed{\\;\\ddot P_{j}=a\\alpha\\,\\mathbf t_{j}+a\\omega^{2}\\,\\mathbf n_{j}\\quad(j=1,2).}\n\\]\n\n--------------------------------------------------------------------\n1.  Simultaneous cancellation of both accelerations is impossible.  \nAssume that \\(\\ddot P_{1}(t_{0})=\\ddot P_{2}(t_{0})=\\mathbf0\\) for some \\(t_{0}\\in(0,T)\\).  \nThe boxed formula forces  \n\\[\n\\alpha(t_{0})=0,\\qquad\\omega(t_{0})=0,\n\\]\ncontradicting the strict non–vanishing of \\(\\omega\\) on \\((0,T)\\).  Hence part 1 is proved.\n\n--------------------------------------------------------------------\n2.  Existence of an instant where both accelerations are purely radial.  \nSince \\(\\omega\\) vanishes only at the endpoints, it keeps a constant sign on \\((0,T)\\).  Assume \\(\\omega>0\\) (the case \\(\\omega<0\\) is identical).  \nBecause \\(\\omega(0)=\\omega(T)=0\\) while \\(\\omega\\not\\equiv 0\\), Rolle’s theorem yields \\(t_{1}\\in(0,T)\\) with  \n\\[\n\\alpha(t_{1})=\\dot\\omega(t_{1})=0.\n\\]\nAt that instant\n\\[\n\\ddot P_{j}(t_{1})=a\\omega(t_{1})^{2}\\,\\mathbf n_{j}(t_{1})\\qquad(j=1,2),\n\\]\na positive multiple of the inward normal; therefore each acceleration vector is directed exactly toward \\(H\\).  Part 2 is settled.\n\n--------------------------------------------------------------------\n3.  Orthogonality of each acceleration to the rod at (possibly different) instants.  \n\n3 (a)  Closed expressions for \\(g_{1},g_{2}\\).  \nA direct trigonometric calculation (best performed in the \\(\\{\\mathbf t_{1},\\mathbf n_{1}\\}\\) frame) yields  \n\\[\n\\boxed{\\;g_{1}(t)=a^{2}\\Bigl[\\alpha(t)\\sin\\beta\n      +2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)\\,\\omega(t)^{2}\\Bigr]},\\tag{1}\n\\]\n\\[\n\\boxed{\\;g_{2}(t)=a^{2}\\Bigl[\\alpha(t)\\sin\\beta\n      -2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)\\,\\omega(t)^{2}\\Bigr]}. \\tag{2}\n\\]\nNote that the sign of both terms in (2) is the opposite of what appeared in the earlier (incorrect) version.\n\n3 (b)  Producing opposite signs for \\(g_{1}\\).  \nSet  \n\\[\n\\kappa:=2\\sin^{2}\\!\\Bigl(\\tfrac{\\beta}{2}\\Bigr)>0,\\qquad\n\\sigma:=\\sin\\beta>0 .\n\\]\nBecause \\(\\omega(0)=\\omega(T)=0\\) and \\(\\omega\\not\\equiv0\\), its derivative \\(\\alpha\\) attains both positive and negative values on \\((0,T)\\).  Define  \n\\[\n\\mathcal P:=\\{t\\in(0,T)\\mid\\alpha(t)>0\\},\\qquad\n\\mathcal N:=\\{t\\in(0,T)\\mid\\alpha(t)<0\\},\n\\]\nboth non–empty.\n\nPick \\(t^{+}\\in\\mathcal P\\) so close to \\(0\\) that \\(\\omega(t^{+})^{2}<\\dfrac{\\sigma}{2\\kappa}\\alpha(t^{+})\\).  \nThen\n\\[\ng_{1}(t^{+})=a^{2}\\bigl[\\alpha(t^{+})\\sigma+\\kappa\\omega(t^{+})^{2}\\bigr]\n           >\\tfrac12a^{2}\\alpha(t^{+})\\sigma>0. \\tag{3}\n\\]\n\nSimilarly, choose \\(t^{-}\\in\\mathcal N\\) near \\(T\\) so that  \n\\(\\omega(t^{-})^{2}<\\dfrac{\\sigma}{2\\kappa}\\,|\\alpha(t^{-})|\\).  Then  \n\\[\ng_{1}(t^{-})=a^{2}\\bigl[\\alpha(t^{-})\\sigma+\\kappa\\omega(t^{-})^{2}\\bigr]\n           <-\\tfrac12a^{2}|\\alpha(t^{-})|\\sigma<0. \\tag{4}\n\\]\n\nBecause \\(g_{1}\\) is continuous, the Intermediate Value Theorem furnishes  \n\\[\nt_{2}\\in(t^{+},t^{-})\\subset(0,T)\\quad\\text{with}\\quad g_{1}(t_{2})=0.\n\\]\n\n3 (c)  Producing opposite signs for \\(g_{2}\\).  \nUsing the same two instants \\(t^{+},t^{-}\\) we now have, from (2),\n\\[\ng_{2}(t^{+})=a^{2}\\bigl[\\alpha(t^{+})\\sigma-\\kappa\\omega(t^{+})^{2}\\bigr]\n           >\\tfrac12a^{2}\\alpha(t^{+})\\sigma>0,\n\\]\n\\[\ng_{2}(t^{-})=a^{2}\\bigl[\\alpha(t^{-})\\sigma-\\kappa\\omega(t^{-})^{2}\\bigr]\n           <-\\tfrac12a^{2}|\\alpha(t^{-})|\\sigma<0.\n\\]\nHence a second application of the Intermediate Value Theorem yields  \n\\[\nt_{3}\\in(t^{+},t^{-})\\subset(0,T)\\quad\\text{with}\\quad g_{2}(t_{3})=0.\n\\]\n\n3 (d)  The zeros cannot coincide.  \nAssume \\(t_{2}=t_{3}\\).  Then \\(g_{1}(t_{2})=g_{2}(t_{2})=0\\).  Adding and subtracting (1)–(2) give  \n\\[\ng_{1}+g_{2}=2a^{2}\\alpha(t_{2})\\sin\\beta=0,\\qquad\ng_{1}-g_{2}=2a^{2}\\kappa\\omega(t_{2})^{2}=0.\n\\]\nHence \\(\\alpha(t_{2})=\\omega(t_{2})=0\\), contradicting the hypothesis that \\(\\omega(t)\\neq0\\) for \\(0<t<T\\).  Therefore \\(t_{2}\\neq t_{3}\\).\n\n--------------------------------------------------------------------\nAll three statements have now been proved with complete rigour.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.349858",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-particle circle problem, the enhanced variant is markedly harder for several reasons.\n\n• Two bodies and an additional holonomic constraint (fixed chord length) raise the configuration space from a 1- to a 2-particle system with an internal geometric restriction, so one must keep track of three different vector quantities: the common centre, the individual positions and the rigid rod.\n\n• The proof requires introducing a single angular coordinate that simultaneously describes both trajectories, then translating kinematic data (velocity, acceleration) of {\\em two} particles in terms of that coordinate.  This leads to coupled identities that are absent in the original problem.\n\n• Part 1 demands exclusion of {\\em simultaneous} vanishing of two vector functions, not just one, forcing a careful examination of how the vanishing of the common scalar functions \\(\\omega\\) and \\(\\alpha\\) would propagate to both accelerations.\n\n• Part 3 is genuinely new: showing orthogonality to the rod entails constructing and analysing a non–trivial scalar product function \\(g(t)\\) that mixes the geometry of the constraint (the chord) with the dynamics (the acceleration), then applying the Intermediate Value Theorem after working out its sign at the end–points by means of higher–order Taylor expansions.\n\n• Throughout, one must manage multiple interacting concepts (polar coordinates, rigid constraints, vector decomposition, Rolle’s theorem, continuity arguments) rather than the single mean-value argument of the original.\n\nHence the enhanced variant demands a substantially broader toolkit and a longer chain of reasoning than either the original problem or the basic kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}