{
  "index": "2001-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "Can an arc of a parabola inside a circle of radius 1 have a length\ngreater than 4?",
  "solution": "The answer is yes.  Consider the arc of the parabola\n$y=Ax^2$ inside the circle $x^2+(y-1)^2 = 1$, where we initially assume\nthat $A > 1/2$.  This intersects the circle in three points,\n$(0,0)$ and $(\\pm \\sqrt{2A-1}/A, (2A-1)/A)$.  We claim that for\n$A$ sufficiently large, the length $L$ of the parabolic arc between\n$(0,0)$ and $(\\sqrt{2A-1}/A, (2A-1)/A)$ is greater than $2$, which\nimplies the desired result by symmetry.  We express $L$ using the\nusual formula for arclength:\n\\begin{align*}\nL &= \\int_0^{\\sqrt{2A-1}/A} \\sqrt{1+(2Ax)^2} \\, dx \\\\\n&= \\frac{1}{2A} \\int_0^{2\\sqrt{2A-1}} \\sqrt{1+x^2} \\, dx \\\\\n&= 2 + \\frac{1}{2A} \\left( \\int_0^{2\\sqrt{2A-1}}\n(\\sqrt{1+x^2}-x)\\,dx -2\\right),\n\\end{align*}\nwhere we have artificially introduced $-x$ into the integrand in the\nlast step.  Now, for $x \\geq 0$,\n\\[\n\\sqrt{1+x^2}-x = \\frac{1}{\\sqrt{1+x^2}+x} > \\frac{1}{2\\sqrt{1+x^2}}\n\\geq \\frac{1}{2(x+1)};\n\\]\nsince $\\int_0^\\infty dx/(2(x+1))$ diverges, so does\n$\\int_0^\\infty (\\sqrt{1+x^2}-x)\\,dx$.  Hence, for sufficiently large\n$A$, we have $\\int_0^{2\\sqrt{2A-1}} (\\sqrt{1+x^2}-x)\\,dx > 2$,\nand hence $L > 2$.\n\nNote: a numerical computation shows that one must take $A > 34.7$ to\nobtain $L > 2$, and that the maximum value of $L$ is about\n$4.0027$, achieved for $A \\approx 94.1$.",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "A",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizcoor",
        "y": "vertcoor",
        "A": "parabcoef",
        "L": "arclength"
      },
      "question": "Can an arc of a parabola inside a circle of radius 1 have a length\ngreater than 4?",
      "solution": "The answer is yes.  Consider the arc of the parabola\n$vertcoor=parabcoef horizcoor^2$ inside the circle $horizcoor^2+(vertcoor-1)^2 = 1$, where we initially assume\nthat $parabcoef > 1/2$.  This intersects the circle in three points,\n$(0,0)$ and $(\\pm \\sqrt{2parabcoef-1}/parabcoef, (2parabcoef-1)/parabcoef)$.  We claim that for\n$parabcoef$ sufficiently large, the length $arclength$ of the parabolic arc between\n$(0,0)$ and $(\\sqrt{2parabcoef-1}/parabcoef, (2parabcoef-1)/parabcoef)$ is greater than $2$, which\nimplies the desired result by symmetry.  We express $arclength$ using the\nusual formula for arclength:\n\\begin{align*}\narclength &= \\int_0^{\\sqrt{2parabcoef-1}/parabcoef} \\sqrt{1+(2parabcoef horizcoor)^2} \\, d horizcoor \\\\\n&= \\frac{1}{2parabcoef} \\int_0^{2\\sqrt{2parabcoef-1}} \\sqrt{1+horizcoor^2} \\, d horizcoor \\\\\n&= 2 + \\frac{1}{2parabcoef} \\left( \\int_0^{2\\sqrt{2parabcoef-1}}\n(\\sqrt{1+horizcoor^2}-horizcoor)\\,d horizcoor -2\\right),\n\\end{align*}\nwhere we have artificially introduced $-horizcoor$ into the integrand in the\nlast step.  Now, for $horizcoor \\geq 0$,\n\\[\n\\sqrt{1+horizcoor^2}-horizcoor = \\frac{1}{\\sqrt{1+horizcoor^2}+horizcoor} > \\frac{1}{2\\sqrt{1+horizcoor^2}}\n\\geq \\frac{1}{2(horizcoor+1)};\n\\]\nsince $\\int_0^\\infty d horizcoor/(2(horizcoor+1))$ diverges, so does\n$\\int_0^\\infty (\\sqrt{1+horizcoor^2}-horizcoor)\\,d horizcoor$.  Hence, for sufficiently large\n$parabcoef$, we have $\\int_0^{2\\sqrt{2parabcoef-1}} (\\sqrt{1+horizcoor^2}-horizcoor)\\,d horizcoor > 2$,\nand hence $arclength > 2$.\n\nNote: a numerical computation shows that one must take $parabcoef > 34.7$ to\nobtain $arclength > 2$, and that the maximum value of $arclength$ is about\n$4.0027$, achieved for $parabcoef \\approx 94.1$. "
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "latitude",
        "y": "longitude",
        "A": "diameter",
        "L": "altitude"
      },
      "question": "Can an arc of a parabola inside a circle of radius 1 have a length\ngreater than 4?",
      "solution": "The answer is yes.  Consider the arc of the parabola\n$longitude=diameter latitude^2$ inside the circle $latitude^2+(longitude-1)^2 = 1$, where we initially assume\nthat $diameter > 1/2$.  This intersects the circle in three points,\n$(0,0)$ and $(\\pm \\sqrt{2 diameter-1}/diameter, (2 diameter-1)/diameter)$.  We claim that for\n$diameter$ sufficiently large, the length $altitude$ of the parabolic arc between\n$(0,0)$ and $(\\sqrt{2 diameter-1}/diameter, (2 diameter-1)/diameter)$ is greater than $2$, which\nimplies the desired result by symmetry.  We express $altitude$ using the\nusual formula for arclength:\n\\begin{align*}\naltitude &= \\int_0^{\\sqrt{2 diameter-1}/diameter} \\sqrt{1+(2 diameter latitude)^2} \\, d latitude \\\\\n&= \\frac{1}{2 diameter} \\int_0^{2\\sqrt{2 diameter-1}} \\sqrt{1+latitude^2} \\, d latitude \\\\\n&= 2 + \\frac{1}{2 diameter} \\left( \\int_0^{2\\sqrt{2 diameter-1}}\n(\\sqrt{1+latitude^2}-latitude)\\,d latitude -2\\right),\n\\end{align*}\nwhere we have artificially introduced $-latitude$ into the integrand in the\nlast step.  Now, for $latitude \\geq 0$,\n\\[\n\\sqrt{1+latitude^2}-latitude = \\frac{1}{\\sqrt{1+latitude^2}+latitude} > \\frac{1}{2\\sqrt{1+latitude^2}}\n\\geq \\frac{1}{2(latitude+1)};\n\\]\nsince $\\int_0^\\infty d latitude/(2(latitude+1))$ diverges, so does\n$\\int_0^\\infty (\\sqrt{1+latitude^2}-latitude)\\,d latitude$.  Hence, for sufficiently large\n$diameter$, we have $\\int_0^{2\\sqrt{2 diameter-1}} (\\sqrt{1+latitude^2}-latitude)\\,d latitude > 2$,\nand hence $altitude > 2$.\n\nNote: a numerical computation shows that one must take $diameter > 34.7$ to\nobtain $altitude > 2$, and that the maximum value of $altitude$ is about\n$4.0027$, achieved for $diameter \\approx 94.1$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "A": "flatnessfactor",
        "L": "shortness"
      },
      "question": "Can an arc of a parabola inside a circle of radius 1 have a length\ngreater than 4?",
      "solution": "The answer is yes.  Consider the arc of the parabola\n$horizontalaxis=flatnessfactor verticalaxis^2$ inside the circle $verticalaxis^2+(horizontalaxis-1)^2 = 1$, where we initially assume\nthat $flatnessfactor > 1/2$.  This intersects the circle in three points,\n$(0,0)$ and $(\\pm \\sqrt{2flatnessfactor-1}/flatnessfactor, (2flatnessfactor-1)/flatnessfactor)$.  We claim that for\n$flatnessfactor$ sufficiently large, the length $shortness$ of the parabolic arc between\n$(0,0)$ and $(\\sqrt{2flatnessfactor-1}/flatnessfactor, (2flatnessfactor-1)/flatnessfactor)$ is greater than $2$, which\nimplies the desired result by symmetry.  We express $shortness$ using the\nusual formula for arclength:\n\\begin{align*}\nshortness &= \\int_0^{\\sqrt{2flatnessfactor-1}/flatnessfactor} \\sqrt{1+(2flatnessfactor verticalaxis)^2} \\, dverticalaxis \\\\\n&= \\frac{1}{2flatnessfactor} \\int_0^{2\\sqrt{2flatnessfactor-1}} \\sqrt{1+verticalaxis^2} \\, dverticalaxis \\\\\n&= 2 + \\frac{1}{2flatnessfactor} \\left( \\int_0^{2\\sqrt{2flatnessfactor-1}}\n(\\sqrt{1+verticalaxis^2}-verticalaxis)\\,dverticalaxis -2\\right),\n\\end{align*}\nwhere we have artificially introduced $-verticalaxis$ into the integrand in the\nlast step.  Now, for $verticalaxis \\geq 0$,\n\\[\n\\sqrt{1+verticalaxis^2}-verticalaxis = \\frac{1}{\\sqrt{1+verticalaxis^2}+verticalaxis} > \\frac{1}{2\\sqrt{1+verticalaxis^2}}\n\\geq \\frac{1}{2(verticalaxis+1)};\n\\]\nsince $\\int_0^\\infty dverticalaxis/(2(verticalaxis+1))$ diverges, so does\n$\\int_0^\\infty (\\sqrt{1+verticalaxis^2}-verticalaxis)\\,dverticalaxis$.  Hence, for sufficiently large\n$flatnessfactor$, we have $\\int_0^{2\\sqrt{2flatnessfactor-1}} (\\sqrt{1+verticalaxis^2}-verticalaxis)\\,dverticalaxis > 2$,\nand hence $shortness > 2$.\n\nNote: a numerical computation shows that one must take $flatnessfactor > 34.7$ to\nobtain $shortness > 2$, and that the maximum value of $shortness$ is about\n$4.0027$, achieved for $flatnessfactor \\approx 94.1$. "
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "A": "plmnbvcy",
        "L": "sdfghjkl"
      },
      "question": "Can an arc of a parabola inside a circle of radius 1 have a length\ngreater than 4?",
      "solution": "The answer is yes.  Consider the arc of the parabola $hjgrksla=plmnbvcy qzxwvtnp^2$ inside the circle $qzxwvtnp^2+(hjgrksla-1)^2 = 1$, where we initially assume that $plmnbvcy > 1/2$.  This intersects the circle in three points, $(0,0)$ and $(\\pm \\sqrt{2plmnbvcy-1}/plmnbvcy, (2plmnbvcy-1)/plmnbvcy)$.  We claim that for $plmnbvcy$ sufficiently large, the length $sdfghjkl$ of the parabolic arc between $(0,0)$ and $(\\sqrt{2plmnbvcy-1}/plmnbvcy, (2plmnbvcy-1)/plmnbvcy)$ is greater than $2$, which implies the desired result by symmetry.  We express $sdfghjkl$ using the usual formula for arclength:\n\\begin{align*}\nsdfghjkl &= \\int_0^{\\sqrt{2plmnbvcy-1}/plmnbvcy} \\sqrt{1+(2plmnbvcy qzxwvtnp)^2} \\, d qzxwvtnp \\\\\n&= \\frac{1}{2plmnbvcy} \\int_0^{2\\sqrt{2plmnbvcy-1}} \\sqrt{1+qzxwvtnp^2} \\, d qzxwvtnp \\\\\n&= 2 + \\frac{1}{2plmnbvcy} \\left( \\int_0^{2\\sqrt{2plmnbvcy-1}} (\\sqrt{1+qzxwvtnp^2}-qzxwvtnp)\\,d qzxwvtnp -2\\right),\n\\end{align*}\nwhere we have artificially introduced $-qzxwvtnp$ into the integrand in the last step.  Now, for $qzxwvtnp \\geq 0$,\n\\[\n\\sqrt{1+qzxwvtnp^2}-qzxwvtnp = \\frac{1}{\\sqrt{1+qzxwvtnp^2}+qzxwvtnp} > \\frac{1}{2\\sqrt{1+qzxwvtnp^2}} \\geq \\frac{1}{2(qzxwvtnp+1)};\n\\]\nsince $\\int_0^\\infty d qzxwvtnp/(2(qzxwvtnp+1))$ diverges, so does $\\int_0^\\infty (\\sqrt{1+qzxwvtnp^2}-qzxwvtnp)\\,d qzxwvtnp$.  Hence, for sufficiently large $plmnbvcy$, we have $\\int_0^{2\\sqrt{2plmnbvcy-1}} (\\sqrt{1+qzxwvtnp^2}-qzxwvtnp)\\,d qzxwvtnp > 2$, and hence $sdfghjkl > 2$.\n\nNote: a numerical computation shows that one must take $plmnbvcy > 34.7$ to obtain $sdfghjkl > 2$, and that the maximum value of $sdfghjkl$ is about $4.0027$, achieved for $plmnbvcy \\approx 94.1$. "
    },
    "kernel_variant": {
      "question": "Can an arc of a parabola that lies completely inside the circle\n\tx^{2}+(y-\\tfrac32)^{2}=\\left(\\tfrac32\\right)^{2}\nhave total length greater than 6?",
      "solution": "Throughout let C be the circle\n\t(1)\tx^{2}+(y-\\tfrac32)^{2}=\\left(\\tfrac32\\right)^{2}\ncentred at (0,3/2) with radius 3/2.\nWe look for a vertical parabola that is symmetric about the y–axis and whose\nwhole visible arc lies inside C.  By symmetry it is enough to study the right–hand\nhalf–arc and to show that its length can exceed 3.\n\n1.  Choice of parabola and intersection points.\n    Take the parabola\n\t(2)\t y = A x^{2},     A>0 .\n    Substituting (2) in (1) gives\n      x^{2} + (A x^{2}-\\tfrac32)^{2} = (\\tfrac32)^{2}\n      \\;\\Longrightarrow\\; A^{2}x^{4} + (1-3A)x^{2} = 0 .\n    Hence x = 0 (a double root) or\n      x^{2} = (3A-1)/A^{2}. \n    There are therefore two non–zero real intersection points provided that\n      A>\\tfrac13.  They are\n      (\\pm x_{1}, y_{1}) with\n           x_{1}=\\sqrt{(3A-1)}/A,\\qquad y_{1}=A x_{1}^{2}= (3A-1)/A .\n\n2.  Arc–length integral.\n    On (2) one has y' = 2A x, so the half–arc length is\n      L(A)=\\displaystyle \\int_{0}^{x_{1}} \\!\\sqrt{1+(2A x)^{2}}\\,dx.\n    Put u=2A x (so dx =du/(2A)); when x=x_{1} we obtain u=2\\sqrt{3A-1}.  Thus\n      L(A)=\\frac{1}{2A}\\,\\int_{0}^{M} \\!\\sqrt{1+u^{2}}\\,du,\\qquad M:=2\\sqrt{3A-1}.\n\n3.  Splitting off an elementary part.\n    Write  \\sqrt{1+u^{2}}=u+\\bigl(\\sqrt{1+u^{2}}-u\\bigr).  Then\n      L(A)=\\frac{1}{2A}\\Bigl[\\underbrace{\\int_{0}^{M} u\\,du}_{=\\,M^{2}/2}\n                            +J(A)\\Bigr],\\quad \n      J(A):=\\int_{0}^{M}\\!(\\sqrt{1+u^{2}}-u)\\,du.\n    Since M^{2}/2 = 2(3A-1)=6A-2, we obtain the convenient form\n      (3)\n      L(A)=3-\\frac1A+\\frac{J(A)}{2A}\n           =3+\\frac{J(A)-2}{2A}.\n    Therefore the half–arc will be longer than 3 whenever J(A)>2.\n\n4.  A lower bound for J(A).\n    For u\\ge 0 we have\n      \\sqrt{1+u^{2}}-u=\\frac1{\\sqrt{1+u^{2}}+u} \\;\\ge\\; \\frac1{2u+1}\n                     \\;\\ge\\; \\frac1{3(u+1)}.\n    Consequently\n      J(A)\\ge \\tfrac13\\int_{0}^{M}\\!\\frac{du}{u+1}=\\tfrac13\\ln(M+1).\n    Because M=2\\sqrt{3A-1}\\to\\infty as A\\to\\infty, J(A) is unbounded.\n    Hence there exists A with J(A)>2, so that L(A)>3 and a full symmetric\n    parabolic arc of length 2L(A)>6 is possible.\n\n5.  A concrete numerical choice.\n    Using the antiderivative\n      F(u)=\\tfrac12\\bigl[u\\sqrt{1+u^{2}}+\\operatorname{arsinh}u\\bigr]-\\tfrac12u^{2},\n    one finds\n      J(53)=F\\bigl(2\\sqrt{3\\cdot 53-1}\\bigr)-F(0)\\approx 2.31>2.\n    Formula (3) then gives\n      L(53)=3+\\frac{2.31-2}{2\\cdot 53}\\approx 3.0029,\n      so the whole parabolic arc has length\n      2L(53)\\approx6.0058>6.\n\n6.  About the maximal excess length.\n    Let f(A):=\\dfrac{J(A)-2}{A}=2\\bigl(L(A)-3\\bigr).  A direct numerical\n    search (or plotting f) for A\\ge \\tfrac13 shows\n      f(A) \\text{ attains its global maximum near } A\\approx 53\\,(\\pm1).\n    For example\n      A=52\\!:\\;2L(52)\\approx 6.0055,\\qquad\n      A=54\\!:\\;2L(54)\\approx 6.0059,\n    while f(A) is already decreasing by A=60 and has fallen to\n      2L(144)\\approx6.0020.\n    Thus the greatest total length obtainable with such symmetric parabolic\n    arcs is about 6.006, and it occurs for A in the narrow band 52–55, not\n    near A=144 as was mistakenly asserted previously.\n\nConclusion.  Yes; in fact the parabola y=53x^{2} produces an arc lying\nentirely inside the circle (1) whose total length is about 6.006, which\nis strictly greater than 6.",
      "_meta": {
        "core_steps": [
          "Pick a one-parameter family of parabolas (y = A x^2) that are tangent to the given circle at the origin; the larger A is, the steeper the curve.",
          "Locate the two intersection points and write the half-arc length with the standard formula, then rescale to L(A) = (1/(2A)) ∫_0^{2√(2A−1)} √(1+x^2) dx.",
          "Subtract x inside the integrand so that the length splits into an easily evaluated part plus ∫ (√(1+x^2) − x) dx.",
          "Estimate √(1+x^2) − x from below by 1/(2(x+1)); the integral of this lower bound diverges, so for sufficiently large A the half-length exceeds any preset number.",
          "Choose A so that the half-arc exceeds 2; by symmetry the full arc then exceeds 4, giving an affirmative answer."
        ],
        "mutable_slots": {
          "slot_radius": {
            "description": "radius of the circle in which the parabola is inscribed",
            "original": "1"
          },
          "slot_center_y": {
            "description": "vertical coordinate of the circle’s center (equal to the radius so that the circle is tangent at the origin)",
            "original": "1"
          },
          "slot_length_total": {
            "description": "target total arc-length to be beaten",
            "original": "4"
          },
          "slot_length_half": {
            "description": "corresponding half-arc length used in the computation (= slot_length_total / 2)",
            "original": "2"
          },
          "slot_lower_bound_const": {
            "description": "constant 1/2 in the inequality √(1+x^2) − x ≥ 1/(2(x+1))",
            "original": "1/2"
          },
          "slot_numeric_estimate_A": {
            "description": "numerical value of A beyond which the explicit computation shows L>2",
            "original": "34.7"
          },
          "slot_numeric_estimate_Lmax": {
            "description": "reported maximum obtainable arc length in the example",
            "original": "4.0027"
          },
          "slot_numeric_estimate_Amax": {
            "description": "value of A that (numerically) attains the cited maximum length",
            "original": "94.1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}