{
  "index": "1983-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "Problem A-2\n\nThe hands of an accurate clock have lengths 3 and 4 . Find the distance between the tips of the hands when that distance is increasing most rapidly.",
  "solution": "A-2.\n\nLet \\( O A \\) be the long hand and \\( O B \\) be the short hand. We can think of \\( O A \\) as fixed and \\( O B \\) as rotating at constant speed. Let \\( v \\) be the vector giving the velocity of point \\( B \\) under this assumption. The rate of change of the distance between \\( A \\) and \\( B \\) is the component of \\( v \\) in the direction of \\( A B \\). Since \\( v \\) is orthogonal to \\( O B \\) and the magnitude of \\( v \\) is constant, this component is maximal when \\( \\measuredangle O B A \\) is a right angle, i.e., when the distance \\( A B \\) is \\( \\sqrt{4^{2}-3^{2}}=\\sqrt{7} \\).\n\nAlternatively, let \\( x \\) be the distance \\( A B \\) and \\( \\theta=\\measuredangle A O B \\). By the Law of Cosines,\n\\[\nx^{2}=3^{2}+4^{2}-2 \\cdot 3 \\cdot 4 \\cos \\theta=25-24 \\cos \\theta\n\\]\n\nSince \\( d \\theta / d t \\) is constant, we may assume units chosen so that \\( \\theta \\) is also time \\( t \\). Now\n\\[\n2 x \\frac{d x}{d \\theta}=24 \\sin \\theta, \\frac{d x}{d \\theta}=\\frac{12 \\sin \\theta}{\\sqrt{25-24 \\cos \\theta}}\n\\]\n\nSince \\( d x / d \\theta \\) is an odd function of \\( \\theta,|d x / d s| \\) is a maximum when \\( d x / d \\theta \\) is a maximum or a minimum. Since \\( d x / d s \\) is a periodic differentiable function of \\( \\theta, d^{2} x / d s^{2}=0 \\) at the extremes for \\( d x / d s \\). For \\( \\operatorname{such} \\theta \\),\n\\[\n12 \\cos \\theta=x \\frac{d^{2} x}{d \\theta^{2}}+\\left(\\frac{d x}{d \\theta}\\right)^{2}=\\left(\\frac{d x}{d \\theta}\\right)^{2}=\\frac{144 \\sin ^{2} \\theta}{x^{2}}\n\\]\n\nThen\n\\[\nx^{2}=\\frac{12 \\sin ^{2} \\theta}{\\cos \\theta}=\\frac{12-12 \\cos ^{2} \\theta}{\\cos \\theta}=25-24 \\cos \\theta\n\\]\nand it follows that\n\\[\n12 \\cos ^{2} \\theta-25 \\cos \\theta+12=0\n\\]\n\nThe only allowable solution for \\( \\cos \\theta \\) is \\( \\cos \\theta=3 / 4 \\) and hence \\( x=\\sqrt{25-24 \\cos \\theta} \\) \\( =\\sqrt{25-18}=\\sqrt{7} \\).",
  "vars": [
    "x",
    "\\\\theta",
    "t",
    "s"
  ],
  "params": [
    "v",
    "O",
    "A",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "distance",
        "\\theta": "angletheta",
        "t": "timevar",
        "s": "arclen",
        "v": "velocity",
        "O": "originpt",
        "A": "longtip",
        "B": "shorttip"
      },
      "question": "Problem A-2\n\nThe hands of an accurate clock have lengths 3 and 4 . Find the distance between the tips of the hands when that distance is increasing most rapidly.",
      "solution": "A-2.\n\nLet \\( originpt longtip \\) be the long hand and \\( originpt shorttip \\) be the short hand. We can think of \\( originpt longtip \\) as fixed and \\( originpt shorttip \\) as rotating at constant speed. Let \\( velocity \\) be the vector giving the velocity of point \\( shorttip \\) under this assumption. The rate of change of the distance between \\( longtip \\) and \\( shorttip \\) is the component of \\( velocity \\) in the direction of \\( longtip shorttip \\). Since \\( velocity \\) is orthogonal to \\( originpt shorttip \\) and the magnitude of \\( velocity \\) is constant, this component is maximal when \\( \\measuredangle originpt shorttip longtip \\) is a right angle, i.e., when the distance \\( longtip shorttip \\) is \\( \\sqrt{4^{2}-3^{2}}=\\sqrt{7} \\).\n\nAlternatively, let \\( distance \\) be the distance \\( longtip shorttip \\) and \\( angletheta=\\measuredangle longtip originpt shorttip \\). By the Law of Cosines,\n\\[\ndistance^{2}=3^{2}+4^{2}-2 \\cdot 3 \\cdot 4 \\cos angletheta=25-24 \\cos angletheta\n\\]\n\nSince \\( d angletheta / d timevar \\) is constant, we may assume units chosen so that \\( angletheta \\) is also time \\( timevar \\). Now\n\\[\n2 distance \\frac{d distance}{d angletheta}=24 \\sin angletheta, \\qquad \\frac{d distance}{d angletheta}=\\frac{12 \\sin angletheta}{\\sqrt{25-24 \\cos angletheta}}\n\\]\n\nSince \\( d distance / d angletheta \\) is an odd function of \\( angletheta,|d distance / d arclen| \\) is a maximum when \\( d distance / d angletheta \\) is a maximum or a minimum. Since \\( d distance / d arclen \\) is a periodic differentiable function of \\( angletheta, d^{2} distance / d arclen^{2}=0 \\) at the extremes for \\( d distance / d arclen \\). For such \\( angletheta \\),\n\\[\n12 \\cos angletheta=distance \\frac{d^{2} distance}{d angletheta^{2}}+\\left(\\frac{d distance}{d angletheta}\\right)^{2}=\\left(\\frac{d distance}{d angletheta}\\right)^{2}=\\frac{144 \\sin ^{2} angletheta}{distance^{2}}\n\\]\n\nThen\n\\[\ndistance^{2}=\\frac{12 \\sin ^{2} angletheta}{\\cos angletheta}=\\frac{12-12 \\cos ^{2} angletheta}{\\cos angletheta}=25-24 \\cos angletheta\n\\]\nand it follows that\n\\[\n12 \\cos ^{2} angletheta-25 \\cos angletheta+12=0\n\\]\n\nThe only allowable solution for \\( \\cos angletheta \\) is \\( \\cos angletheta=3 / 4 \\) and hence \\( distance=\\sqrt{25-24 \\cos angletheta}=\\sqrt{25-18}=\\sqrt{7} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "curvature",
        "\\theta": "humidity",
        "t": "sunshine",
        "s": "lavender",
        "v": "tornadoes",
        "O": "waterfall",
        "A": "backpack",
        "B": "pinecones"
      },
      "question": "Problem A-2\n\nThe hands of an accurate clock have lengths 3 and 4 . Find the distance between the tips of the hands when that distance is increasing most rapidly.",
      "solution": "A-2.\n\nLet \\( waterfall\\ backpack \\) be the long hand and \\( waterfall\\ pinecones \\) be the short hand. We can think of \\( waterfall\\ backpack \\) as fixed and \\( waterfall\\ pinecones \\) as rotating at constant speed. Let \\( tornadoes \\) be the vector giving the velocity of point \\( pinecones \\) under this assumption. The rate of change of the distance between \\( backpack \\) and \\( pinecones \\) is the component of \\( tornadoes \\) in the direction of \\( backpack\\ pinecones \\). Since \\( tornadoes \\) is orthogonal to \\( waterfall\\ pinecones \\) and the magnitude of \\( tornadoes \\) is constant, this component is maximal when \\( \\measuredangle waterfall\\ pinecones\\ backpack \\) is a right angle, i.e., when the distance \\( backpack\\ pinecones \\) is \\( \\sqrt{4^{2}-3^{2}}=\\sqrt{7} \\).\n\nAlternatively, let \\( curvature \\) be the distance \\( backpack\\ pinecones \\) and \\( humidity=\\measuredangle backpack\\ waterfall\\ pinecones \\). By the Law of Cosines,\n\\[\ncurvature^{2}=3^{2}+4^{2}-2 \\cdot 3 \\cdot 4 \\cos humidity=25-24 \\cos humidity\n\\]\n\nSince \\( d\\, humidity / d\\, sunshine \\) is constant, we may assume units chosen so that \\( humidity \\) is also time \\( sunshine \\). Now\n\\[\n2\\, curvature \\frac{d\\, curvature}{d\\, humidity}=24 \\sin humidity,\\quad \\frac{d\\, curvature}{d\\, humidity}=\\frac{12 \\sin humidity}{\\sqrt{25-24 \\cos humidity}}\n\\]\n\nSince \\( d\\, curvature / d\\, humidity \\) is an odd function of \\( humidity,|d\\, curvature / d\\, lavender| \\) is a maximum when \\( d\\, curvature / d\\, humidity \\) is a maximum or a minimum. Since \\( d\\, curvature / d\\, lavender \\) is a periodic differentiable function of \\( humidity, d^{2}\\, curvature / d\\, lavender^{2}=0 \\) at the extremes for \\( d\\, curvature / d\\, lavender \\). For \\( \\operatorname{such} humidity \\),\n\\[\n12 \\cos humidity = curvature \\frac{d^{2}\\, curvature}{d\\, humidity^{2}} + \\left(\\frac{d\\, curvature}{d\\, humidity}\\right)^{2} = \\left(\\frac{d\\, curvature}{d\\, humidity}\\right)^{2} = \\frac{144 \\sin ^{2} humidity}{curvature^{2}}\n\\]\n\nThen\n\\[\ncurvature^{2}= \\frac{12 \\sin ^{2} humidity}{\\cos humidity} = \\frac{12-12 \\cos ^{2} humidity}{\\cos humidity} = 25-24 \\cos humidity\n\\]\nand it follows that\n\\[\n12 \\cos ^{2} humidity - 25 \\cos humidity + 12 = 0\n\\]\n\nThe only allowable solution for \\( \\cos humidity \\) is \\( \\cos humidity = 3 / 4 \\) and hence \\( curvature = \\sqrt{25-24 \\cos humidity} = \\sqrt{25-18} = \\sqrt{7} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "closenessvar",
        "\\theta": "straightness",
        "t": "timelessness",
        "s": "stillnessvar",
        "v": "restvector",
        "O": "infinitypoint",
        "A": "shortnesspoint",
        "B": "longnesspoint"
      },
      "question": "Problem A-2\n\nThe hands of an accurate clock have lengths 3 and 4 . Find the distance between the tips of the hands when that distance is increasing most rapidly.",
      "solution": "A-2.\n\nLet \\( infinitypoint\\ shortnesspoint \\) be the long hand and \\( infinitypoint\\ longnesspoint \\) be the short hand. We can think of \\( infinitypoint\\ shortnesspoint \\) as fixed and \\( infinitypoint\\ longnesspoint \\) as rotating at constant speed. Let \\( restvector \\) be the vector giving the velocity of point \\( longnesspoint \\) under this assumption. The rate of change of the distance between \\( shortnesspoint \\) and \\( longnesspoint \\) is the component of \\( restvector \\) in the direction of \\( shortnesspoint longnesspoint \\). Since \\( restvector \\) is orthogonal to \\( infinitypoint\\ longnesspoint \\) and the magnitude of \\( restvector \\) is constant, this component is maximal when \\( \\measuredangle infinitypoint\\ longnesspoint\\ shortnesspoint \\) is a right angle, i.e., when the distance \\( shortnesspoint\\ longnesspoint \\) is \\( \\sqrt{4^{2}-3^{2}}=\\sqrt{7} \\).\n\nAlternatively, let \\( closenessvar \\) be the distance \\( shortnesspoint\\ longnesspoint \\) and \\( straightness=\\measuredangle shortnesspoint\\ infinitypoint\\ longnesspoint \\). By the Law of Cosines,\n\\[ closenessvar^{2}=3^{2}+4^{2}-2 \\cdot 3 \\cdot 4 \\cos straightness=25-24 \\cos straightness \\]\n\nSince \\( d\\ straightness / d\\ timelessness \\) is constant, we may assume units chosen so that \\( straightness \\) is also time \\( timelessness \\). Now\n\\[ 2\\ closenessvar \\frac{d\\ closenessvar}{d\\ straightness}=24 \\sin straightness, \\frac{d\\ closenessvar}{d\\ straightness}=\\frac{12 \\sin straightness}{\\sqrt{25-24 \\cos straightness}} \\]\n\nSince \\( d\\ closenessvar / d\\ straightness \\) is an odd function of \\( straightness,|d\\ closenessvar / d\\ stillnessvar| \\) is a maximum when \\( d\\ closenessvar / d\\ straightness \\) is a maximum or a minimum. Since \\( d\\ closenessvar / d\\ stillnessvar \\) is a periodic differentiable function of \\( straightness, d^{2}\\ closenessvar / d\\ stillnessvar^{2}=0 \\) at the extremes for \\( d\\ closenessvar / d\\ stillnessvar \\). For such \\( straightness \\),\n\\[ 12 \\cos straightness=closenessvar \\frac{d^{2}\\ closenessvar}{d\\ straightness^{2}}+\\left(\\frac{d\\ closenessvar}{d\\ straightness}\\right)^{2}=\\left(\\frac{d\\ closenessvar}{d\\ straightness}\\right)^{2}=\\frac{144 \\sin ^{2} straightness}{closenessvar^{2}} \\]\n\nThen\n\\[ closenessvar^{2}=\\frac{12 \\sin ^{2} straightness}{\\cos straightness}=\\frac{12-12 \\cos ^{2} straightness}{\\cos straightness}=25-24 \\cos straightness \\]\nand it follows that\n\\[ 12 \\cos ^{2} straightness-25 \\cos straightness+12=0 \\]\n\nThe only allowable solution for \\( \\cos straightness \\) is \\( \\cos straightness=3 / 4 \\) and hence \\( closenessvar=\\sqrt{25-24 \\cos straightness} \\) \\( =\\sqrt{25-18}=\\sqrt{7} \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "\\theta": "hjgrksla",
        "t": "mlfdjhqn",
        "s": "pvrthkcs",
        "v": "rpxdgula",
        "O": "xbcqjlow",
        "A": "gmvhwdnr",
        "B": "szktlepa"
      },
      "question": "Problem A-2\n\nThe hands of an accurate clock have lengths 3 and 4 . Find the distance between the tips of the hands when that distance is increasing most rapidly.",
      "solution": "A-2.\n\nLet \\( xbcqjlow gmvhwdnr \\) be the long hand and \\( xbcqjlow szktlepa \\) be the short hand. We can think of \\( xbcqjlow gmvhwdnr \\) as fixed and \\( xbcqjlow szktlepa \\) as rotating at constant speed. Let \\( rpxdgula \\) be the vector giving the velocity of point \\( szktlepa \\) under this assumption. The rate of change of the distance between \\( gmvhwdnr \\) and \\( szktlepa \\) is the component of \\( rpxdgula \\) in the direction of \\( gmvhwdnr szktlepa \\). Since \\( rpxdgula \\) is orthogonal to \\( xbcqjlow szktlepa \\) and the magnitude of \\( rpxdgula \\) is constant, this component is maximal when \\( \\measuredangle xbcqjlow szktlepa gmvhwdnr \\) is a right angle, i.e., when the distance \\( gmvhwdnr szktlepa \\) is \\( \\sqrt{4^{2}-3^{2}}=\\sqrt{7} \\).\n\nAlternatively, let \\( qzxwvtnp \\) be the distance \\( gmvhwdnr szktlepa \\) and \\( hjgrksla=\\measuredangle gmvhwdnr xbcqjlow szktlepa \\). By the Law of Cosines,\n\\[\nqzxwvtnp^{2}=3^{2}+4^{2}-2 \\cdot 3 \\cdot 4 \\cos hjgrksla=25-24 \\cos hjgrksla\n\\]\n\nSince \\( d hjgrksla / d mlfdjhqn \\) is constant, we may assume units chosen so that \\( hjgrksla \\) is also time \\( mlfdjhqn \\). Now\n\\[\n2 qzxwvtnp \\frac{d qzxwvtnp}{d hjgrksla}=24 \\sin hjgrksla, \\frac{d qzxwvtnp}{d hjgrksla}=\\frac{12 \\sin hjgrksla}{\\sqrt{25-24 \\cos hjgrksla}}\n\\]\n\nSince \\( d qzxwvtnp / d hjgrksla \\) is an odd function of \\( hjgrksla,|d qzxwvtnp / d pvrthkcs| \\) is a maximum when \\( d qzxwvtnp / d hjgrksla \\) is a maximum or a minimum. Since \\( d qzxwvtnp / d pvrthkcs \\) is a periodic differentiable function of \\( hjgrksla, d^{2} qzxwvtnp / d pvrthkcs^{2}=0 \\) at the extremes for \\( d qzxwvtnp / d pvrthkcs \\). For \\( \\operatorname{such} hjgrksla \\),\n\\[\n12 \\cos hjgrksla=qzxwvtnp \\frac{d^{2} qzxwvtnp}{d hjgrksla^{2}}+\\left(\\frac{d qzxwvtnp}{d hjgrksla}\\right)^{2}=\\left(\\frac{d qzxwvtnp}{d hjgrksla}\\right)^{2}=\\frac{144 \\sin ^{2} hjgrksla}{qzxwvtnp^{2}}\n\\]\n\nThen\n\\[\nqzxwvtnp^{2}=\\frac{12 \\sin ^{2} hjgrksla}{\\cos hjgrksla}=\\frac{12-12 \\cos ^{2} hjgrksla}{\\cos hjgrksla}=25-24 \\cos hjgrksla\n\\]\nand it follows that\n\\[\n12 \\cos ^{2} hjgrksla-25 \\cos hjgrksla+12=0\n\\]\n\nThe only allowable solution for \\( \\cos hjgrksla \\) is \\( \\cos hjgrksla=3 / 4 \\) and hence \\( qzxwvtnp=\\sqrt{25-24 \\cos hjgrksla} \\) \\( =\\sqrt{25-18}=\\sqrt{7} \\)."
    },
    "kernel_variant": {
      "question": "“Spatial Clock’’ – final, fully-rigorised variant  \n\nA rigid axle coincides with the \\(x\\)-axis.  \nTwo clock–hands are attached to it so that the planes in which they move are perpendicular.\n\n• Hand \\(A\\) has length \\(5\\rm \\,cm\\) and turns counter-clockwise in the horizontal \\(xy\\)-plane with the constant angular speed  \n \\(\\displaystyle \\omega _1 = 1\\;\\mathrm{rad\\,s^{-1}}\\).\n  At time \\(t=0\\) its tip is on the positive \\(x\\)-axis.\n\n• Hand \\(B\\) has length \\(7\\rm \\,cm\\) and turns clockwise in the vertical \\(xz\\)-plane with the constant angular speed  \n \\(\\displaystyle \\omega _2 = 2\\;\\mathrm{rad\\,s^{-1}}\\).\n  Its tip also starts from the positive \\(x\\)-axis.\n\nFor \\(t\\ge 0\\) denote the tips by \\(A(t)\\) and \\(B(t)\\) and set  \n\\[\nd(t)=|A(t)B(t)|,\\qquad \nv(t)=\\frac{\\mathrm d}{\\mathrm dt}d(t).\n\\]\n\n1.  Prove that  \n\\[\nd^{2}(t)=74-70\\cos t\\cos 2t .\n\\]\n\n2.  Show that  \n\\[\n\\boxed{\\;\nv(t)=\\frac{35\\,\\sin t\\bigl(6\\cos ^{2}t-1\\bigr)}\n           {\\sqrt{\\,74-70\\cos t\\cos 2t\\,}}\\;}\n\\]\n\n3.  Put \\(c=\\cos t\\in[-1,1]\\) and define the sextic  \n\\[\nF(c)=c\\bigl(18c^{2}-13\\bigr)\\bigl(74+70c-140c^{3}\\bigr)\n      -35(1-c^{2})\\bigl(6c^{2}-1\\bigr)^{2}.\n\\]\n\n   (a)  Show that the critical points of \\(v\\) are the real zeros of \\(F\\).\n\n   (b)  Prove that \\(F\\) possesses exactly one zero\n        \\(c_{0}\\in(0,1)\\) and exactly one further zero\n        \\(c_{1}\\in(-1,0)\\).\n\n   (c)  Prove that \\(|v(t_{1})|<|v(t_{0})|\\) where\n        \\(t_{k}=\\arccos c_{k}\\;(k=0,1)\\).\n\n   Deduce  \n   \\[\n   v(t_{0})=\\max_{t\\ge 0}v(t),\\qquad\n   v(2\\pi-t_{0})=\\min_{t\\ge 0}v(t)=-\\,v(t_{0}).\n   \\]\n\n4.  Compute\n\\[\nt_{0}\\approx0.300\\text{ rad}\\;(17.2^{\\circ}),\\qquad\nd(t_{0})\\approx4.35\\text{ cm},\\qquad\nv(t_{0})\\approx10.7\\text{ cm s}^{-1},\\qquad\n|v(t_{1})|\\approx6.3\\text{ cm s}^{-1}.\n\\]\n\nHence the tips separate fastest (about \\(10.7\\rm \\,cm\\,s^{-1}\\))\nwhen they are about \\(4.35\\rm \\,cm\\) apart, and they approach each\nother with the same speed \\(2\\pi-t_{0}\\) seconds later.",
      "solution": "Throughout we write  \n\\[\n\\theta=\\omega _1t=t,\\qquad \\varphi=\\omega _2t=2t,\\qquad\nc=\\cos t,\\qquad s=\\sin t .\n\\]\n\n--------------------------------------------------------------------\n1.  Coordinates and the squared distance  \n\n\\[\nA(t)=(5\\cos\\theta,\\;5\\sin\\theta,\\;0),\\qquad\nB(t)=(7\\cos\\varphi,\\;0,\\;-7\\sin\\varphi).\n\\]\n\nHence  \n\\[\n\\overrightarrow{AB}=(7\\cos2t-5\\cos t,\\; -5\\sin t,\\; -7\\sin2t).\n\\]\nA direct calculation gives  \n\\[\nd^{2}(t)=\n(7\\cos2t-5\\cos t)^{2}+(5\\sin t)^{2}+(7\\sin2t)^{2}\n       =74-70\\cos t\\cos 2t .\n\\]\n\n--------------------------------------------------------------------\n2.  A compact formula for \\(v(t)\\)  \n\nDifferentiate \\(d^{2}(t)\\):\n\\[\n\\frac{\\mathrm d}{\\mathrm dt}d^{2}(t)=\n70\\!\\left(\\sin t\\cos 2t+2\\cos t\\sin 2t\\right).\n\\]\nPut  \n\\[\nN(t)=35\\bigl(\\sin t\\cos 2t+2\\cos t\\sin 2t\\bigr),\\qquad\nD(t)=d(t)=\\sqrt{74-70\\cos t\\cos 2t}.\n\\]\n\nBecause \\((D^{2})'=2DD'=2N\\) we have \\(DD'=N\\).  Consequently\n\\[\nv(t)=D'(t)=\\frac{N(t)}{D(t)}\n      =\\frac{35\\sin t\\,(6\\cos ^2t-1)}{D(t)},\n\\]\nwhich is exactly the expression stated in Part 2.\n\n--------------------------------------------------------------------\n3.  The critical–point equation  \n\nWrite \\(v(t)=N(t)/D(t)\\) with \\(DD'=N\\) as just established.  Then\n\\[\nv'(t)=\\frac{N'D-ND'}{D^{2}}\n     =\\frac{N'}{D}-\\frac{N^{2}}{D^{3}}.\n\\]\nHence \\(v'(t)=0\\iff N'D^{2}=N^{2}\\).\n\nWith \\(c=\\cos t\\) one finds\n\\[\nN(c)=35\\sqrt{1-c^{2}}\\,(6c^{2}-1),\\qquad\nN'(t)=35c\\,(18c^{2}-13),\\qquad\nD^{2}(c)=74+70c-140c^{3}.\n\\]\nSubstituting these three expressions in \\(N'D^{2}=N^{2}\\) yields\nprecisely the sextic\n\\[\nF(c)=c(18c^{2}-13)(74+70c-140c^{3})\n     -35(1-c^{2})(6c^{2}-1)^{2},\n\\]\nso the critical points of \\(v\\) are the real zeros of \\(F\\).  \nThis completes 3 (a).\n\n--------------------------------------------------------------------\n4.  Uniqueness of the two real roots  \n\nAfter expansion\n\\[\nF(c)=-35-962c-455c^{2}+1332c^{3}+1400c^{4}-1260c^{6}.\n\\tag{1}\n\\]\nA (checked) Sturm sequence for \\(F\\) is\n\n\\[\n\\begin{aligned}\nS_{0}&=F,\\\\\nS_{1}&=F'=-962-910c+3996c^{2}+5600c^{3}-7560c^{5},\\\\\nS_{2}&=-1400c^{4}-1998c^{3}+910c^{2}+2405c+105,\\\\\nS_{3}&=-\\bigl(53946c^{4}+3430c^{3}-44955c^{2}-7385c-4810\\bigr),\\\\\nS_{4}&=-\\bigl(258496c^{3}+119070c^{2}-329721c-81785\\bigr),\\\\\nS_{5}&= 2417712345c^{2}+598378088c+350518480,\\\\\nS_{6}&=-98863809851872c-26467140819200,\\\\\nS_{7}&= 939553966571110838665216\\; (>0).\n\\end{aligned}\n\\]\n\nSigns of \\(S_{k}(-1),S_{k}(0),S_{k}(1)\\) are\n\n\\[\n\\begin{array}{c|ccccccc}\nc & S_{0} & S_{1} & S_{2} & S_{3} & S_{4} & S_{5} & S_{6} \\\\ \\hline\n-1 & - & + & - & - & + & - & +\\\\\n0  & - & - & + & + & - & + & -\\\\\n1  & + & + & + & - & - & + & -\n\\end{array}\n\\]\n\nso  \n\\(\n\\operatorname{Var}(-1)=2,\\;\n\\operatorname{Var}(0)=1,\\;\n\\operatorname{Var}(1)=0.\n\\)\n\nTherefore \\(F\\) has exactly one root in \\((-1,0)\\) and exactly one in \\((0,1)\\), establishing 3 (b).\n\n--------------------------------------------------------------------\n5.  Localisation of the two roots  \n\n\\[\nF\\Bigl(\\tfrac1{\\sqrt2}\\Bigr)\\approx-2.80\\times10^{2}<0,\\qquad\nF(1)=20>0,\n\\]\nso the positive root satisfies \\(c_{0}\\in(\\tfrac1{\\sqrt2},1)\\).\n\n\\[\nF\\Bigl(-\\tfrac1{\\sqrt2}\\Bigr)\\approx+1.40\\times10^{2}>0,\\qquad\nF(-1)=-720<0,\n\\]\nso the negative root satisfies \\(c_{1}\\in(-1,-\\tfrac1{\\sqrt2})\\).\n\n--------------------------------------------------------------------\n6.  Comparison of the extreme speeds  \n\nWrite\n\\[\n\\Phi(c)=\\frac{35\\sqrt{1-c^{2}}\\;|6c^{2}-1|}\n               {\\sqrt{74+70c-140c^{3}}}\n       =|v(t)|,\\qquad c=\\cos t .\n\\]\nThe numerator of \\(\\Phi\\) is even in \\(c\\), whereas\n\\[\nD^{2}(c)=74+70c-140c^{3}\n\\quad\\Longrightarrow\\quad\nD^{2}(-x)-D^{2}(x)=140x(2x^{2}-1).\n\\]\nFor \\(x>\\tfrac1{\\sqrt2}\\) the right–hand side is positive, hence\n\\(D(-x)>D(x)\\).  Because \\(|c_{0}|>\\tfrac1{\\sqrt2}\\) and\n\\(|c_{1}|>\\tfrac1{\\sqrt2}\\) (previous step) we obtain\n\\(D(c_{1})>D(c_{0})\\) and consequently\n\\(|v(t_{1})|<|v(t_{0})|\\).  This proves 3 (c).\n\nSince \\(d^{2}(t)\\) is \\(2\\pi\\)-periodic and \\(v(2\\pi-t)=-v(t)\\),\nthe global maximum and minimum of \\(v\\) are\n\\(v(t_{0})\\) and \\(-v(t_{0})\\) as asserted.\n\n--------------------------------------------------------------------\n7.  Numerical evaluation  \n\nA few Newton iterations for the positive root of \\(F\\):\n\n\\[\nc_0^{(0)}=0.95\\;\\longrightarrow\\;0.9545\\;\\longrightarrow\\;\n0.954930\\;(\\text{stable to }10^{-6}).\n\\]\n\nThus  \n\\[\nc_{0}\\approx0.95493,\\qquad\nt_{0}=\\arccos c_{0}\\approx0.3001\\text{ rad }(17.2^{\\circ}).\n\\]\nCorrespondingly  \n\\[\n\\begin{aligned}\nd^{2}(t_{0})&=74-70c_{0}(2c_{0}^{2}-1)\\approx18.94,\\\\\nd(t_{0})    &\\approx4.35\\text{ cm},\\\\\nv(t_{0})    &\\approx10.7\\text{ cm\\,s}^{-1}.\n\\end{aligned}\n\\]\n\nFor the negative root \\(c_{1}\\approx-0.803\\;\n(t_{1}\\approx2.517\\text{ rad})\\) the same formula yields  \n\\(|v(t_{1})|\\approx6.3\\text{ cm\\,s}^{-1}\\).\n\nHence the tips separate most rapidly\n(\\(10.7\\rm \\,cm\\,s^{-1}\\)) when they are \\(4.35\\rm \\,cm\\) apart\nand re-approach with the same speed \\(2\\pi-t_{0}\\approx5.98\\) s later.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.668414",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The original 2-D clock problem is replaced by one in three dimensions; the two hands now move in perpendicular planes, so their tips’ trajectories are space curves.  \n\n• Additional variables.  Two independently evolving angles (θ, φ) enter, introducing a non-trivial coupling term cos θ cos 2θ in d²(t).  \n\n• Non-elementary optimisation.  Maximising v(t) is no longer a single-variable trigonometric extremum but involves a rational function whose derivative leads to a transcendental equation (†) with large coefficients and no obvious “nice’’ root.  \n\n• Vector-calculus viewpoint.  A coordinate-free derivation via the projection of the relative-velocity vector onto the A–B line must be combined with explicit component computations.  \n\n• Multiple interacting concepts.  The solution requires trigonometric identities, differentiation under a square-root, rational optimisation, sign analysis, and careful numerical estimation.  \n\nHence the enhanced variant is substantially harder than both the original problem and the simpler kernel variant: it demands work in 3-space, copes with two distinct angular velocities, and culminates in solving (†), an equation inaccessible to inspection or elementary “right-angle’’ arguments."
      }
    },
    "original_kernel_variant": {
      "question": "“Spatial Clock’’ – final, fully-rigorised variant  \n\nA rigid axle coincides with the \\(x\\)-axis.  \nTwo clock–hands are attached to it so that the planes in which they move are perpendicular.\n\n• Hand \\(A\\) has length \\(5\\rm \\,cm\\) and turns counter-clockwise in the horizontal \\(xy\\)-plane with the constant angular speed  \n \\(\\displaystyle \\omega _1 = 1\\;\\mathrm{rad\\,s^{-1}}\\).\n  At time \\(t=0\\) its tip is on the positive \\(x\\)-axis.\n\n• Hand \\(B\\) has length \\(7\\rm \\,cm\\) and turns clockwise in the vertical \\(xz\\)-plane with the constant angular speed  \n \\(\\displaystyle \\omega _2 = 2\\;\\mathrm{rad\\,s^{-1}}\\).\n  Its tip also starts from the positive \\(x\\)-axis.\n\nFor \\(t\\ge 0\\) denote the tips by \\(A(t)\\) and \\(B(t)\\) and set  \n\\[\nd(t)=|A(t)B(t)|,\\qquad \nv(t)=\\frac{\\mathrm d}{\\mathrm dt}d(t).\n\\]\n\n1.  Prove that  \n\\[\nd^{2}(t)=74-70\\cos t\\cos 2t .\n\\]\n\n2.  Show that  \n\\[\n\\boxed{\\;\nv(t)=\\frac{35\\,\\sin t\\bigl(6\\cos ^{2}t-1\\bigr)}\n           {\\sqrt{\\,74-70\\cos t\\cos 2t\\,}}\\;}\n\\]\n\n3.  Put \\(c=\\cos t\\in[-1,1]\\) and define the sextic  \n\\[\nF(c)=c\\bigl(18c^{2}-13\\bigr)\\bigl(74+70c-140c^{3}\\bigr)\n      -35(1-c^{2})\\bigl(6c^{2}-1\\bigr)^{2}.\n\\]\n\n   (a)  Show that the critical points of \\(v\\) are the real zeros of \\(F\\).\n\n   (b)  Prove that \\(F\\) possesses exactly one zero\n        \\(c_{0}\\in(0,1)\\) and exactly one further zero\n        \\(c_{1}\\in(-1,0)\\).\n\n   (c)  Prove that \\(|v(t_{1})|<|v(t_{0})|\\) where\n        \\(t_{k}=\\arccos c_{k}\\;(k=0,1)\\).\n\n   Deduce  \n   \\[\n   v(t_{0})=\\max_{t\\ge 0}v(t),\\qquad\n   v(2\\pi-t_{0})=\\min_{t\\ge 0}v(t)=-\\,v(t_{0}).\n   \\]\n\n4.  Compute\n\\[\nt_{0}\\approx0.300\\text{ rad}\\;(17.2^{\\circ}),\\qquad\nd(t_{0})\\approx4.35\\text{ cm},\\qquad\nv(t_{0})\\approx10.7\\text{ cm s}^{-1},\\qquad\n|v(t_{1})|\\approx6.3\\text{ cm s}^{-1}.\n\\]\n\nHence the tips separate fastest (about \\(10.7\\rm \\,cm\\,s^{-1}\\))\nwhen they are about \\(4.35\\rm \\,cm\\) apart, and they approach each\nother with the same speed \\(2\\pi-t_{0}\\) seconds later.",
      "solution": "Throughout we write  \n\\[\n\\theta=\\omega _1t=t,\\qquad \\varphi=\\omega _2t=2t,\\qquad\nc=\\cos t,\\qquad s=\\sin t .\n\\]\n\n--------------------------------------------------------------------\n1.  Coordinates and the squared distance  \n\n\\[\nA(t)=(5\\cos\\theta,\\;5\\sin\\theta,\\;0),\\qquad\nB(t)=(7\\cos\\varphi,\\;0,\\;-7\\sin\\varphi).\n\\]\n\nHence  \n\\[\n\\overrightarrow{AB}=(7\\cos2t-5\\cos t,\\; -5\\sin t,\\; -7\\sin2t).\n\\]\nA direct calculation gives  \n\\[\nd^{2}(t)=\n(7\\cos2t-5\\cos t)^{2}+(5\\sin t)^{2}+(7\\sin2t)^{2}\n       =74-70\\cos t\\cos 2t .\n\\]\n\n--------------------------------------------------------------------\n2.  A compact formula for \\(v(t)\\)  \n\nDifferentiate \\(d^{2}(t)\\):\n\\[\n\\frac{\\mathrm d}{\\mathrm dt}d^{2}(t)=\n70\\!\\left(\\sin t\\cos 2t+2\\cos t\\sin 2t\\right).\n\\]\nPut  \n\\[\nN(t)=35\\bigl(\\sin t\\cos 2t+2\\cos t\\sin 2t\\bigr),\\qquad\nD(t)=d(t)=\\sqrt{74-70\\cos t\\cos 2t}.\n\\]\n\nBecause \\((D^{2})'=2DD'=2N\\) we have \\(DD'=N\\).  Consequently\n\\[\nv(t)=D'(t)=\\frac{N(t)}{D(t)}\n      =\\frac{35\\sin t\\,(6\\cos ^2t-1)}{D(t)},\n\\]\nwhich is exactly the expression stated in Part 2.\n\n--------------------------------------------------------------------\n3.  The critical–point equation  \n\nWrite \\(v(t)=N(t)/D(t)\\) with \\(DD'=N\\) as just established.  Then\n\\[\nv'(t)=\\frac{N'D-ND'}{D^{2}}\n     =\\frac{N'}{D}-\\frac{N^{2}}{D^{3}}.\n\\]\nHence \\(v'(t)=0\\iff N'D^{2}=N^{2}\\).\n\nWith \\(c=\\cos t\\) one finds\n\\[\nN(c)=35\\sqrt{1-c^{2}}\\,(6c^{2}-1),\\qquad\nN'(t)=35c\\,(18c^{2}-13),\\qquad\nD^{2}(c)=74+70c-140c^{3}.\n\\]\nSubstituting these three expressions in \\(N'D^{2}=N^{2}\\) yields\nprecisely the sextic\n\\[\nF(c)=c(18c^{2}-13)(74+70c-140c^{3})\n     -35(1-c^{2})(6c^{2}-1)^{2},\n\\]\nso the critical points of \\(v\\) are the real zeros of \\(F\\).  \nThis completes 3 (a).\n\n--------------------------------------------------------------------\n4.  Uniqueness of the two real roots  \n\nAfter expansion\n\\[\nF(c)=-35-962c-455c^{2}+1332c^{3}+1400c^{4}-1260c^{6}.\n\\tag{1}\n\\]\nA (checked) Sturm sequence for \\(F\\) is\n\n\\[\n\\begin{aligned}\nS_{0}&=F,\\\\\nS_{1}&=F'=-962-910c+3996c^{2}+5600c^{3}-7560c^{5},\\\\\nS_{2}&=-1400c^{4}-1998c^{3}+910c^{2}+2405c+105,\\\\\nS_{3}&=-\\bigl(53946c^{4}+3430c^{3}-44955c^{2}-7385c-4810\\bigr),\\\\\nS_{4}&=-\\bigl(258496c^{3}+119070c^{2}-329721c-81785\\bigr),\\\\\nS_{5}&= 2417712345c^{2}+598378088c+350518480,\\\\\nS_{6}&=-98863809851872c-26467140819200,\\\\\nS_{7}&= 939553966571110838665216\\; (>0).\n\\end{aligned}\n\\]\n\nSigns of \\(S_{k}(-1),S_{k}(0),S_{k}(1)\\) are\n\n\\[\n\\begin{array}{c|ccccccc}\nc & S_{0} & S_{1} & S_{2} & S_{3} & S_{4} & S_{5} & S_{6} \\\\ \\hline\n-1 & - & + & - & - & + & - & +\\\\\n0  & - & - & + & + & - & + & -\\\\\n1  & + & + & + & - & - & + & -\n\\end{array}\n\\]\n\nso  \n\\(\n\\operatorname{Var}(-1)=2,\\;\n\\operatorname{Var}(0)=1,\\;\n\\operatorname{Var}(1)=0.\n\\)\n\nTherefore \\(F\\) has exactly one root in \\((-1,0)\\) and exactly one in \\((0,1)\\), establishing 3 (b).\n\n--------------------------------------------------------------------\n5.  Localisation of the two roots  \n\n\\[\nF\\Bigl(\\tfrac1{\\sqrt2}\\Bigr)\\approx-2.80\\times10^{2}<0,\\qquad\nF(1)=20>0,\n\\]\nso the positive root satisfies \\(c_{0}\\in(\\tfrac1{\\sqrt2},1)\\).\n\n\\[\nF\\Bigl(-\\tfrac1{\\sqrt2}\\Bigr)\\approx+1.40\\times10^{2}>0,\\qquad\nF(-1)=-720<0,\n\\]\nso the negative root satisfies \\(c_{1}\\in(-1,-\\tfrac1{\\sqrt2})\\).\n\n--------------------------------------------------------------------\n6.  Comparison of the extreme speeds  \n\nWrite\n\\[\n\\Phi(c)=\\frac{35\\sqrt{1-c^{2}}\\;|6c^{2}-1|}\n               {\\sqrt{74+70c-140c^{3}}}\n       =|v(t)|,\\qquad c=\\cos t .\n\\]\nThe numerator of \\(\\Phi\\) is even in \\(c\\), whereas\n\\[\nD^{2}(c)=74+70c-140c^{3}\n\\quad\\Longrightarrow\\quad\nD^{2}(-x)-D^{2}(x)=140x(2x^{2}-1).\n\\]\nFor \\(x>\\tfrac1{\\sqrt2}\\) the right–hand side is positive, hence\n\\(D(-x)>D(x)\\).  Because \\(|c_{0}|>\\tfrac1{\\sqrt2}\\) and\n\\(|c_{1}|>\\tfrac1{\\sqrt2}\\) (previous step) we obtain\n\\(D(c_{1})>D(c_{0})\\) and consequently\n\\(|v(t_{1})|<|v(t_{0})|\\).  This proves 3 (c).\n\nSince \\(d^{2}(t)\\) is \\(2\\pi\\)-periodic and \\(v(2\\pi-t)=-v(t)\\),\nthe global maximum and minimum of \\(v\\) are\n\\(v(t_{0})\\) and \\(-v(t_{0})\\) as asserted.\n\n--------------------------------------------------------------------\n7.  Numerical evaluation  \n\nA few Newton iterations for the positive root of \\(F\\):\n\n\\[\nc_0^{(0)}=0.95\\;\\longrightarrow\\;0.9545\\;\\longrightarrow\\;\n0.954930\\;(\\text{stable to }10^{-6}).\n\\]\n\nThus  \n\\[\nc_{0}\\approx0.95493,\\qquad\nt_{0}=\\arccos c_{0}\\approx0.3001\\text{ rad }(17.2^{\\circ}).\n\\]\nCorrespondingly  \n\\[\n\\begin{aligned}\nd^{2}(t_{0})&=74-70c_{0}(2c_{0}^{2}-1)\\approx18.94,\\\\\nd(t_{0})    &\\approx4.35\\text{ cm},\\\\\nv(t_{0})    &\\approx10.7\\text{ cm\\,s}^{-1}.\n\\end{aligned}\n\\]\n\nFor the negative root \\(c_{1}\\approx-0.803\\;\n(t_{1}\\approx2.517\\text{ rad})\\) the same formula yields  \n\\(|v(t_{1})|\\approx6.3\\text{ cm\\,s}^{-1}\\).\n\nHence the tips separate most rapidly\n(\\(10.7\\rm \\,cm\\,s^{-1}\\)) when they are \\(4.35\\rm \\,cm\\) apart\nand re-approach with the same speed \\(2\\pi-t_{0}\\approx5.98\\) s later.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.524168",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The original 2-D clock problem is replaced by one in three dimensions; the two hands now move in perpendicular planes, so their tips’ trajectories are space curves.  \n\n• Additional variables.  Two independently evolving angles (θ, φ) enter, introducing a non-trivial coupling term cos θ cos 2θ in d²(t).  \n\n• Non-elementary optimisation.  Maximising v(t) is no longer a single-variable trigonometric extremum but involves a rational function whose derivative leads to a transcendental equation (†) with large coefficients and no obvious “nice’’ root.  \n\n• Vector-calculus viewpoint.  A coordinate-free derivation via the projection of the relative-velocity vector onto the A–B line must be combined with explicit component computations.  \n\n• Multiple interacting concepts.  The solution requires trigonometric identities, differentiation under a square-root, rational optimisation, sign analysis, and careful numerical estimation.  \n\nHence the enhanced variant is substantially harder than both the original problem and the simpler kernel variant: it demands work in 3-space, copes with two distinct angular velocities, and culminates in solving (†), an equation inaccessible to inspection or elementary “right-angle’’ arguments."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}