{
  "index": "1958-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. What is the average straight line distance between two points on a sphere of radius \\( 1 ? \\)",
  "solution": "Solution. We take the radius of the sphere to be \\( a \\) to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole \\( N \\) of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude \\( \\theta \\) and co-latitude \\( \\theta+\\Delta \\theta \\) has area approximately\n\\( (2 \\pi a \\sin \\theta)(a \\Delta \\theta) \\)\nand all the chords from \\( N \\) to points in this zone have length approximately\n\\[\n2 a \\sin (\\theta / 2)\n\\]\n\nTherefore, the average length is\n\\[\nL=\\frac{4 \\pi a^{3}}{S} \\int_{0}^{\\pi} \\sin \\frac{\\theta}{2} \\sin \\theta d \\theta\n\\]\nwhere \\( S=4 \\pi a^{2} \\) is the surface area of the sphere. Since \\( a=1 \\), we have\n\\[\nL=\\int_{0}^{\\pi} \\sin \\frac{\\theta}{2}\\left(2 \\sin \\frac{\\theta}{2} \\cos \\frac{\\theta}{2}\\right) d \\theta=\\left.\\frac{4}{3} \\sin ^{3} \\frac{\\theta}{2}\\right|_{0} ^{\\pi}=\\frac{4}{3}\n\\]",
  "vars": [
    "\\\\theta",
    "L"
  ],
  "params": [
    "a",
    "N",
    "S",
    "\\\\Delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "\\theta": "anglevar",
        "L": "averagelength",
        "a": "sphereradius",
        "N": "northpolepoint",
        "S": "spherearea",
        "\\Delta": "increment"
      },
      "question": "4. What is the average straight line distance between two points on a sphere of radius \\( 1 ? \\)",
      "solution": "Solution. We take the radius of the sphere to be \\( sphereradius \\) to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole \\( northpolepoint \\) of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude \\( anglevar \\) and co-latitude \\( anglevar+increment anglevar \\) has area approximately\n\\( (2 \\pi sphereradius \\sin anglevar)(sphereradius increment anglevar) \\)\nand all the chords from \\( northpolepoint \\) to points in this zone have length approximately\n\\[\n2 sphereradius \\sin (anglevar / 2)\n\\]\n\nTherefore, the average length is\n\\[\naveragelength=\\frac{4 \\pi sphereradius^{3}}{spherearea} \\int_{0}^{\\pi} \\sin \\frac{anglevar}{2} \\sin anglevar d anglevar\n\\]\nwhere \\( spherearea=4 \\pi sphereradius^{2} \\) is the surface area of the sphere. Since \\( sphereradius=1 \\), we have\n\\[\naveragelength=\\int_{0}^{\\pi} \\sin \\frac{anglevar}{2}\\left(2 \\sin \\frac{anglevar}{2} \\cos \\frac{anglevar}{2}\\right) d anglevar=\\left.\\frac{4}{3} \\sin ^{3} \\frac{anglevar}{2}\\right|_{0} ^{\\pi}=\\frac{4}{3}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "\\theta": "orchardly",
        "L": "monastery",
        "a": "buttercup",
        "N": "downtown",
        "S": "sandcastle",
        "\\Delta": "hummingbrd"
      },
      "question": "4. What is the average straight line distance between two points on a sphere of radius \\( 1 ? \\)",
      "solution": "Solution. We take the radius of the sphere to be \\( buttercup \\) to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole \\( downtown \\) of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude \\( orchardly \\) and co-latitude \\( orchardly+hummingbrd orchardly \\) has area approximately\n\\( (2 \\pi buttercup \\sin orchardly)(buttercup hummingbrd orchardly) \\)\nand all the chords from \\( downtown \\) to points in this zone have length approximately\n\\[\n2 buttercup \\sin (orchardly / 2)\n\\]\n\nTherefore, the average length is\n\\[\nmonastery=\\frac{4 \\pi buttercup^{3}}{sandcastle} \\int_{0}^{\\pi} \\sin \\frac{orchardly}{2} \\sin orchardly \\, d orchardly\n\\]\nwhere \\( sandcastle=4 \\pi buttercup^{2} \\) is the surface area of the sphere. Since \\( buttercup=1 \\), we have\n\\[\nmonastery=\\int_{0}^{\\pi} \\sin \\frac{orchardly}{2}\\left(2 \\sin \\frac{orchardly}{2} \\cos \\frac{orchardly}{2}\\right) d orchardly=\\left.\\frac{4}{3} \\sin ^{3} \\frac{orchardly}{2}\\right|_{0} ^{\\pi}=\\frac{4}{3}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "\\theta": "flatdistance",
        "L": "shortness",
        "a": "emptiness",
        "N": "southpole",
        "S": "voidmass",
        "\\Delta": "constanty"
      },
      "question": "Problem:\n<<<\n4. What is the average straight line distance between two points on a sphere of radius \\( 1 ? \\)\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. We take the radius of the sphere to be \\( emptiness \\) to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole \\( southpole \\) of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude \\( flatdistance \\) and co-latitude \\( flatdistance+constanty flatdistance \\) has area approximately\n\\( (2 \\pi emptiness \\sin flatdistance)(emptiness \\, constanty flatdistance) \\)\nand all the chords from \\( southpole \\) to points in this zone have length approximately\n\\[\n2\\, emptiness \\sin (flatdistance / 2)\n\\]\n\nTherefore, the average length is\n\\[\nshortness=\\frac{4 \\pi emptiness^{3}}{voidmass} \\int_{0}^{\\pi} \\sin \\frac{flatdistance}{2} \\sin flatdistance \\, d flatdistance\n\\]\nwhere \\( voidmass=4 \\pi emptiness^{2} \\) is the surface area of the sphere. Since \\( emptiness=1 \\), we have\n\\[\nshortness=\\int_{0}^{\\pi} \\sin \\frac{flatdistance}{2}\\left(2 \\sin \\frac{flatdistance}{2} \\cos \\frac{flatdistance}{2}\\right) d flatdistance=\\left.\\frac{4}{3} \\sin ^{3} \\frac{flatdistance}{2}\\right|_{0} ^{\\pi}=\\frac{4}{3}\n\\]\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "\\theta": "zvxplmna",
        "L": "hgtrsqpo",
        "a": "pwekmsru",
        "N": "qlaxfoge",
        "S": "rdmivtye",
        "\\Delta": "sncouzab"
      },
      "question": "4. What is the average straight line distance between two points on a sphere of radius \\( 1 ? \\)",
      "solution": "Solution. We take the radius of the sphere to be \\( pwekmsru \\) to help keep our dimensions in order. By symmetry we need only compute the average length of all chords emanating from some fixed point which we take to be the north pole \\( qlaxfoge \\) of a spherical coordinate system. Slice the sphere into zones by parallels of co-latitude. The zone between co-latitude \\( zvxplmna \\) and co-latitude \\( zvxplmna+sncouzab zvxplmna \\) has area approximately\n\\( (2 \\pi pwekmsru \\sin zvxplmna)(pwekmsru sncouzab zvxplmna) \\)\nand all the chords from \\( qlaxfoge \\) to points in this zone have length approximately\n\\[\n2 pwekmsru \\sin (zvxplmna / 2)\n\\]\n\nTherefore, the average length is\n\\[\nhgtrsqpo=\\frac{4 \\pi pwekmsru^{3}}{rdmivtye} \\int_{0}^{\\pi} \\sin \\frac{zvxplmna}{2} \\sin zvxplmna d zvxplmna\n\\]\nwhere \\( rdmivtye=4 \\pi pwekmsru^{2} \\) is the surface area of the sphere. Since \\( pwekmsru=1 \\), we have\n\\[\nhgtrsqpo=\\int_{0}^{\\pi} \\sin \\frac{zvxplmna}{2}\\left(2 \\sin \\frac{zvxplmna}{2} \\cos \\frac{zvxplmna}{2}\\right) d zvxplmna=\\left.\\frac{4}{3} \\sin ^{3} \\frac{zvxplmna}{2}\\right|_{0} ^{\\pi}=\\frac{4}{3}\n\\]"
    },
    "kernel_variant": {
      "question": "Let $\\lambda>-1$ be a real parameter and $R>0$ a radius.  \nInside the solid three–dimensional ball  \n\n\\[\n\\mathcal{B}^{3}_{R}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{3}\\;:\\;\\lVert x\\rVert\\le R\\bigr\\}\n\\]\n\nconsider the rotationally–invariant probability measure $\\mu$ defined by  \n\n$\\bullet$ the angular component is the uniform surface measure on the sphere $S^{2}$,  \n\n$\\bullet$ the radial component has one–dimensional density  \n\\[\nf(r)\\;=\\;(\\lambda+1)\\,R^{-(\\lambda+1)}\\,r^{\\lambda},\\qquad 0\\le r\\le R.\n\\]\n\nTwo points $X,Y$ are chosen independently according to $\\mu$, and  \n\\[\nD\\;=\\;\\lVert X-Y\\rVert\n\\]\ndenotes their Euclidean distance.\n\n1. Prove that for every $\\lambda>-1$\n\\[\n\\boxed{\\;\n\\mathbb{E}[D]\\;=\\;\nR\\,\\frac{4(\\lambda+1)(2\\lambda+5)}\n               {3(2\\lambda+3)(\\lambda+3)}\n\\;}\n\\tag{$\\ast$}\n\\]\n\n(equivalently, for $R=1$ one has\n$\\mathbb{E}[D]=\\dfrac{4(\\lambda+1)(2\\lambda+5)}\n                       {3(2\\lambda+3)(\\lambda+3)}$).\n\n2. In the physically important \\emph{uniform–volume} case\n$\\lambda=2$ and $R=2$ show explicitly that\n\\[\n\\mathbb{E}[D]\\;=\\;2\\,\\frac{4\\cdot3\\cdot9}{3\\cdot7\\cdot5}\n                 \\;=\\;\\frac{72}{35}\\;\\approx\\;2.057143.\n\\]\n\nGive complete and rigorous derivations.",
      "solution": "Throughout write $r_{1}=\\lVert X\\rVert$, $r_{2}=\\lVert Y\\rVert$ and let $\\theta$\nbe the angle between the independent directions of $X$ and $Y$; set\n$u=\\cos\\theta$.\n\n--------------------------------------------------------------------\nA.  Radial law\n--------------------------------------------------------------------\nBy construction\n\\[\nf(r)= (\\lambda+1)R^{-(\\lambda+1)}r^{\\lambda},\\qquad 0\\le r\\le R.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nB.  Angular law\n--------------------------------------------------------------------\nFor two independent directions on $S^{2}$ the\ndensity of $u=\\cos\\theta$ equals\n\\[\ng(u)=\\frac12,\\qquad -1\\le u\\le 1.\n\\tag{2}\n\\]\n\n--------------------------------------------------------------------\nC.  Conditional mean distance for fixed radii\n--------------------------------------------------------------------\nPut\n\\[\na=r_{1}^{2}+r_{2}^{2},\\qquad b=2r_{1}r_{2},\n\\quad\\text{so that }D=\\sqrt{a-bu}.\n\\]\n\nWith (2)\n\\[\nK(r_{1},r_{2}):=\\mathbb{E}[D\\mid r_{1},r_{2}]\n              =\\frac12\\int_{-1}^{1}\\sqrt{a-bu}\\,du.\n\\tag{3}\n\\]\n\nBecause $\\displaystyle\\int\\sqrt{a-bu}\\,du=-\\frac{2}{3b}(a-bu)^{3/2}$,\n\\[\n\\int_{-1}^{1}\\sqrt{a-bu}\\,du\n      =\\frac{2}{3b}\\bigl[(a+b)^{3/2}-(a-b)^{3/2}\\bigr].\n\\]\nSince $a\\pm b=(r_{1}\\pm r_{2})^{2}$,\n\\[\n\\int_{-1}^{1}\\sqrt{a-bu}\\,du\n      =\\frac{2}{3b}\\bigl[(r_{1}+r_{2})^{3}-\\lvert r_{1}-r_{2}\\rvert^{3}\\bigr]\n      =\\frac{[ (r_{1}+r_{2})^{3}-\\lvert r_{1}-r_{2}\\rvert^{3} ]}\n             {3r_{1}r_{2}}.\n\\]\nConsequently\n\\[\nK(r_{1},r_{2})\n      =\n\\begin{cases}\nr_{1}+\\dfrac{r_{2}^{2}}{3r_{1}}, & r_{1}\\ge r_{2},\\\\[6pt]\nr_{2}+\\dfrac{r_{1}^{2}}{3r_{2}}, & r_{2}\\ge r_{1}.\n\\end{cases}\n\\tag{4}\n\\]\n\n--------------------------------------------------------------------\nD.  The double radial integral\n--------------------------------------------------------------------\nBecause of symmetry we integrate over $r_{2}\\le r_{1}$ and double:\n\\[\n\\mathbb{E}[D]=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}\\!\\int_{0}^{r_{1}}\nr_{1}^{\\lambda}r_{2}^{\\lambda}\\Bigl[r_{1}+\\frac{r_{2}^{2}}{3r_{1}}\\Bigr]\n\\,dr_{2}\\,dr_{1}.\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nD1.  First contribution (the $r_{1}$–term)\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nI_{1}&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n      \\Bigl[r_{1}\\int_{0}^{r_{1}}r_{2}^{\\lambda}\\,dr_{2}\\Bigr]dr_{1}\\\\\n&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n          \\frac{r_{1}^{\\lambda+1}}{\\lambda+1}\\,dr_{1}\\\\\n&=\n2(\\lambda+1)R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{2\\lambda+2}\\,dr_{1}\n=\n2(\\lambda+1)R^{-2(\\lambda+1)}\\,\\frac{R^{2\\lambda+3}}{2\\lambda+3}.\n\\tag{6}\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\nD2.  Second contribution (the $r_{2}^{2}/3r_{1}$–term)\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nI_{2}&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n      \\Bigl[\\frac{1}{3r_{1}}\\int_{0}^{r_{1}}r_{2}^{\\lambda+2}\\,dr_{2}\\Bigr]dr_{1}\\\\\n&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}\\frac{r_{1}^{\\lambda}}{3r_{1}}\n          \\frac{r_{1}^{\\lambda+3}}{\\lambda+3}\\,dr_{1}\\\\\n&=\n\\frac{2(\\lambda+1)^{2}}{3(\\lambda+3)}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{2\\lambda+2}\\,dr_{1}\n=\n\\frac{2(\\lambda+1)^{2}}{3(\\lambda+3)}R^{-2(\\lambda+1)}\n      \\frac{R^{2\\lambda+3}}{2\\lambda+3}.\n\\tag{7}\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\nD3.  Collecting $I_{1}+I_{2}$\n--------------------------------------------------------------------\nFactor\n\\[\n2(\\lambda+1)R^{-2(\\lambda+1)}\\,\\frac{R^{2\\lambda+3}}{2\\lambda+3}\n=2(\\lambda+1)R.\n\\]\nHence\n\\[\n\\mathbb{E}[D]=\n\\frac{2(\\lambda+1)R}{2\\lambda+3}\n\\left[1+\\frac{\\lambda+1}{3(\\lambda+3)}\\right].\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nE.  Algebraic simplification (corrected)\n--------------------------------------------------------------------\nObserve\n\\[\n1=\\frac{3(\\lambda+3)}{3(\\lambda+3)}\n\\quad\\Longrightarrow\\quad\n1+\\frac{\\lambda+1}{3(\\lambda+3)}\n=\\frac{3(\\lambda+3)+(\\lambda+1)}{3(\\lambda+3)}\n=\\frac{4\\lambda+10}{3(\\lambda+3)}\n=\\frac{2(2\\lambda+5)}{3(\\lambda+3)}.\n\\]\nInsert this into (8):\n\\[\n\\mathbb{E}[D]=\n\\frac{2(\\lambda+1)R}{2\\lambda+3}\n\\cdot\n\\frac{2(2\\lambda+5)}{3(\\lambda+3)}\n=\nR\\,\\frac{4(\\lambda+1)(2\\lambda+5)}\n       {3(2\\lambda+3)(\\lambda+3)},\n\\]\nwhich is exactly formula $(\\ast)$.\n\n--------------------------------------------------------------------\nF.  Uniform–volume case $\\lambda=2$, $R=2$\n--------------------------------------------------------------------\nPlug $\\lambda=2$ and $R=2$ into $(\\ast)$:\n\\[\n\\mathbb{E}[D]=\n2\\,\\frac{4\\cdot3\\cdot9}{3\\cdot7\\cdot5}\n            =\\frac{72}{35}\\;\\approx\\;2.057143.\n\\]\n\n--------------------------------------------------------------------\nG.  Numerical sanity check\n--------------------------------------------------------------------\nFor $\\lambda=2$, $R=2$\n\\[\n\\mathbb{E}[r^{2}]=R^{2}\\frac{\\lambda+1}{\\lambda+3}\n                =4\\cdot\\frac{3}{5}=\\frac{12}{5},\n\\quad\n\\sqrt{\\mathbb{E}[r^{2}+r^{2}]}\n=\\sqrt{\\tfrac{24}{5}}\\approx 2.191.\n\\]\nSince the square root is concave, $\\mathbb{E}[D]\\le 2.191$, consistent with $\\frac{72}{35}\\approx2.057$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.508255",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension $n$ and free parameter $\\lambda$ enlarge the problem from a single numerical evaluation (the original chord-length mean on $S^{2}$) to a *two-parameter family* of expectations on *all* balls in all dimensions.\n\n2.  The density is *non-uniform* in the radial direction, so angular and radial variables no longer separate trivially; one must evaluate a genuinely two-dimensional integral.\n\n3.  Computing the angular mean demands knowledge of the distribution of the\ninner product of two independent uniform directions in $S^{n}$ and requires\nintegration against the weight $(1-u^{2})^{(n-2)/2}$, followed by\nhyper-geometric reduction—well beyond the elementary sine-substitution used in\nthe original problem.\n\n4.  The final radial double integral cannot be done by a single substitution;\nexpansion in hyper-geometric series and repeated use of Beta and Gamma\nidentities are necessary.\n\n5.  Even the *special case* $n=2,\\lambda=2$ needs careful algebra (or a second,\nindependent, elementary calculation) to show that the result collapses to the\nnon-obvious rational number $520/231$.\n\n6.  Hence the enhanced variant combines multivariate calculus, special-function\ntheory (Beta, Gamma, and Gauss $_2F_1$ functions), and series manipulation,\ndemanding several layers of sophisticated techniques not present in the\noriginal sphere-chord question."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\lambda>-1$ be a real parameter and $R>0$ a radius.  \nInside the solid three–dimensional ball  \n\n\\[\n\\mathcal{B}^{3}_{R}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{3}\\;:\\;\\lVert x\\rVert\\le R\\bigr\\}\n\\]\n\nconsider the rotationally–invariant probability measure $\\mu$ defined by  \n\n$\\bullet$ the angular component is the uniform surface measure on the sphere $S^{2}$,  \n\n$\\bullet$ the radial component has one–dimensional density  \n\\[\nf(r)\\;=\\;(\\lambda+1)\\,R^{-(\\lambda+1)}\\,r^{\\lambda},\\qquad 0\\le r\\le R.\n\\]\n\nTwo points $X,Y$ are chosen independently according to $\\mu$, and  \n\\[\nD\\;=\\;\\lVert X-Y\\rVert\n\\]\ndenotes their Euclidean distance.\n\n1. Prove that for every $\\lambda>-1$\n\\[\n\\boxed{\\;\n\\mathbb{E}[D]\\;=\\;\nR\\,\\frac{4(\\lambda+1)(2\\lambda+5)}\n               {3(2\\lambda+3)(\\lambda+3)}\n\\;}\n\\tag{$\\ast$}\n\\]\n\n(equivalently, for $R=1$ one has\n$\\mathbb{E}[D]=\\dfrac{4(\\lambda+1)(2\\lambda+5)}\n                       {3(2\\lambda+3)(\\lambda+3)}$).\n\n2. In the physically important \\emph{uniform–volume} case\n$\\lambda=2$ and $R=2$ show explicitly that\n\\[\n\\mathbb{E}[D]\\;=\\;2\\,\\frac{4\\cdot3\\cdot9}{3\\cdot7\\cdot5}\n                 \\;=\\;\\frac{72}{35}\\;\\approx\\;2.057143.\n\\]\n\nGive complete and rigorous derivations.",
      "solution": "Throughout write $r_{1}=\\lVert X\\rVert$, $r_{2}=\\lVert Y\\rVert$ and let $\\theta$\nbe the angle between the independent directions of $X$ and $Y$; set\n$u=\\cos\\theta$.\n\n--------------------------------------------------------------------\nA.  Radial law\n--------------------------------------------------------------------\nBy construction\n\\[\nf(r)= (\\lambda+1)R^{-(\\lambda+1)}r^{\\lambda},\\qquad 0\\le r\\le R.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nB.  Angular law\n--------------------------------------------------------------------\nFor two independent directions on $S^{2}$ the\ndensity of $u=\\cos\\theta$ equals\n\\[\ng(u)=\\frac12,\\qquad -1\\le u\\le 1.\n\\tag{2}\n\\]\n\n--------------------------------------------------------------------\nC.  Conditional mean distance for fixed radii\n--------------------------------------------------------------------\nPut\n\\[\na=r_{1}^{2}+r_{2}^{2},\\qquad b=2r_{1}r_{2},\n\\quad\\text{so that }D=\\sqrt{a-bu}.\n\\]\n\nWith (2)\n\\[\nK(r_{1},r_{2}):=\\mathbb{E}[D\\mid r_{1},r_{2}]\n              =\\frac12\\int_{-1}^{1}\\sqrt{a-bu}\\,du.\n\\tag{3}\n\\]\n\nBecause $\\displaystyle\\int\\sqrt{a-bu}\\,du=-\\frac{2}{3b}(a-bu)^{3/2}$,\n\\[\n\\int_{-1}^{1}\\sqrt{a-bu}\\,du\n      =\\frac{2}{3b}\\bigl[(a+b)^{3/2}-(a-b)^{3/2}\\bigr].\n\\]\nSince $a\\pm b=(r_{1}\\pm r_{2})^{2}$,\n\\[\n\\int_{-1}^{1}\\sqrt{a-bu}\\,du\n      =\\frac{2}{3b}\\bigl[(r_{1}+r_{2})^{3}-\\lvert r_{1}-r_{2}\\rvert^{3}\\bigr]\n      =\\frac{[ (r_{1}+r_{2})^{3}-\\lvert r_{1}-r_{2}\\rvert^{3} ]}\n             {3r_{1}r_{2}}.\n\\]\nConsequently\n\\[\nK(r_{1},r_{2})\n      =\n\\begin{cases}\nr_{1}+\\dfrac{r_{2}^{2}}{3r_{1}}, & r_{1}\\ge r_{2},\\\\[6pt]\nr_{2}+\\dfrac{r_{1}^{2}}{3r_{2}}, & r_{2}\\ge r_{1}.\n\\end{cases}\n\\tag{4}\n\\]\n\n--------------------------------------------------------------------\nD.  The double radial integral\n--------------------------------------------------------------------\nBecause of symmetry we integrate over $r_{2}\\le r_{1}$ and double:\n\\[\n\\mathbb{E}[D]=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}\\!\\int_{0}^{r_{1}}\nr_{1}^{\\lambda}r_{2}^{\\lambda}\\Bigl[r_{1}+\\frac{r_{2}^{2}}{3r_{1}}\\Bigr]\n\\,dr_{2}\\,dr_{1}.\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nD1.  First contribution (the $r_{1}$–term)\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nI_{1}&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n      \\Bigl[r_{1}\\int_{0}^{r_{1}}r_{2}^{\\lambda}\\,dr_{2}\\Bigr]dr_{1}\\\\\n&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n          \\frac{r_{1}^{\\lambda+1}}{\\lambda+1}\\,dr_{1}\\\\\n&=\n2(\\lambda+1)R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{2\\lambda+2}\\,dr_{1}\n=\n2(\\lambda+1)R^{-2(\\lambda+1)}\\,\\frac{R^{2\\lambda+3}}{2\\lambda+3}.\n\\tag{6}\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\nD2.  Second contribution (the $r_{2}^{2}/3r_{1}$–term)\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nI_{2}&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{\\lambda}\n      \\Bigl[\\frac{1}{3r_{1}}\\int_{0}^{r_{1}}r_{2}^{\\lambda+2}\\,dr_{2}\\Bigr]dr_{1}\\\\\n&=\n2(\\lambda+1)^{2}R^{-2(\\lambda+1)}\n\\int_{0}^{R}\\frac{r_{1}^{\\lambda}}{3r_{1}}\n          \\frac{r_{1}^{\\lambda+3}}{\\lambda+3}\\,dr_{1}\\\\\n&=\n\\frac{2(\\lambda+1)^{2}}{3(\\lambda+3)}R^{-2(\\lambda+1)}\n\\int_{0}^{R}r_{1}^{2\\lambda+2}\\,dr_{1}\n=\n\\frac{2(\\lambda+1)^{2}}{3(\\lambda+3)}R^{-2(\\lambda+1)}\n      \\frac{R^{2\\lambda+3}}{2\\lambda+3}.\n\\tag{7}\n\\end{aligned}\n\\]\n\n--------------------------------------------------------------------\nD3.  Collecting $I_{1}+I_{2}$\n--------------------------------------------------------------------\nFactor\n\\[\n2(\\lambda+1)R^{-2(\\lambda+1)}\\,\\frac{R^{2\\lambda+3}}{2\\lambda+3}\n=2(\\lambda+1)R.\n\\]\nHence\n\\[\n\\mathbb{E}[D]=\n\\frac{2(\\lambda+1)R}{2\\lambda+3}\n\\left[1+\\frac{\\lambda+1}{3(\\lambda+3)}\\right].\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nE.  Algebraic simplification (corrected)\n--------------------------------------------------------------------\nObserve\n\\[\n1=\\frac{3(\\lambda+3)}{3(\\lambda+3)}\n\\quad\\Longrightarrow\\quad\n1+\\frac{\\lambda+1}{3(\\lambda+3)}\n=\\frac{3(\\lambda+3)+(\\lambda+1)}{3(\\lambda+3)}\n=\\frac{4\\lambda+10}{3(\\lambda+3)}\n=\\frac{2(2\\lambda+5)}{3(\\lambda+3)}.\n\\]\nInsert this into (8):\n\\[\n\\mathbb{E}[D]=\n\\frac{2(\\lambda+1)R}{2\\lambda+3}\n\\cdot\n\\frac{2(2\\lambda+5)}{3(\\lambda+3)}\n=\nR\\,\\frac{4(\\lambda+1)(2\\lambda+5)}\n       {3(2\\lambda+3)(\\lambda+3)},\n\\]\nwhich is exactly formula $(\\ast)$.\n\n--------------------------------------------------------------------\nF.  Uniform–volume case $\\lambda=2$, $R=2$\n--------------------------------------------------------------------\nPlug $\\lambda=2$ and $R=2$ into $(\\ast)$:\n\\[\n\\mathbb{E}[D]=\n2\\,\\frac{4\\cdot3\\cdot9}{3\\cdot7\\cdot5}\n            =\\frac{72}{35}\\;\\approx\\;2.057143.\n\\]\n\n--------------------------------------------------------------------\nG.  Numerical sanity check\n--------------------------------------------------------------------\nFor $\\lambda=2$, $R=2$\n\\[\n\\mathbb{E}[r^{2}]=R^{2}\\frac{\\lambda+1}{\\lambda+3}\n                =4\\cdot\\frac{3}{5}=\\frac{12}{5},\n\\quad\n\\sqrt{\\mathbb{E}[r^{2}+r^{2}]}\n=\\sqrt{\\tfrac{24}{5}}\\approx 2.191.\n\\]\nSince the square root is concave, $\\mathbb{E}[D]\\le 2.191$, consistent with $\\frac{72}{35}\\approx2.057$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.425097",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension $n$ and free parameter $\\lambda$ enlarge the problem from a single numerical evaluation (the original chord-length mean on $S^{2}$) to a *two-parameter family* of expectations on *all* balls in all dimensions.\n\n2.  The density is *non-uniform* in the radial direction, so angular and radial variables no longer separate trivially; one must evaluate a genuinely two-dimensional integral.\n\n3.  Computing the angular mean demands knowledge of the distribution of the\ninner product of two independent uniform directions in $S^{n}$ and requires\nintegration against the weight $(1-u^{2})^{(n-2)/2}$, followed by\nhyper-geometric reduction—well beyond the elementary sine-substitution used in\nthe original problem.\n\n4.  The final radial double integral cannot be done by a single substitution;\nexpansion in hyper-geometric series and repeated use of Beta and Gamma\nidentities are necessary.\n\n5.  Even the *special case* $n=2,\\lambda=2$ needs careful algebra (or a second,\nindependent, elementary calculation) to show that the result collapses to the\nnon-obvious rational number $520/231$.\n\n6.  Hence the enhanced variant combines multivariate calculus, special-function\ntheory (Beta, Gamma, and Gauss $_2F_1$ functions), and series manipulation,\ndemanding several layers of sophisticated techniques not present in the\noriginal sphere-chord question."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}