{
  "index": "2024-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Consider a circle $\\Omega$ with radius 9 and center at the origin $(0,0)$, and a disc $\\Delta$ with radius 1 and center at $(r,0)$, where $0 \\leq r \\leq 8$. Two points $P$ and $Q$ are chosen independently and uniformly at random on $\\Omega$. Which value(s) of $r$ minimize the probability that the chord $\\overline{PQ}$ intersects $\\Delta$?",
  "solution": "We will show that $r=0$ (and no other value of $r$) minimizes the stated probability.\nNote that $P$ and $Q$ coincide with probability $0$; thus we can assume that $P\\neq Q$.\n\n\\noindent\n\\textbf{First solution.}\nFirst restrict $P,Q$ to points on $\\Omega$ such that the segment $\\overline{PQ}$ makes an angle of $\\theta$ with the $y$ axis, where $\\theta$ is a fixed number with $-\\pi/2<\\theta\\leq\\pi/2$. By rotating the diagram by $-\\theta$ around the origin, we move $\\overline{PQ}$ to be a vertical line and move $\\Delta$ to be centered at $(r\\cos\\theta,-r\\sin\\theta)$. In this rotated picture, $P$ and $Q$ are at $(9\\cos\\phi,\\pm 9\\sin\\phi)$ where $\\phi$ is chosen uniformly at random in $(0,\\pi)$. Now the vertical tangent lines to the boundary of $\\Delta$, $x=r\\cos\\theta\\pm 1$, intersect the $y>0$ semicircle of $\\Omega$ at $(9\\cos\\phi,9\\sin\\phi)$ where $\\phi = \\cos^{-1}\\left(\\frac{r\\cos\\theta\\pm 1}{9}\\right)$. Thus the probability that $\\overline{PQ}$ intersects $\\Delta$ for a specific value of $\\theta$ is\n$\\frac{1}{\\pi} f(r,\\theta)$, where we define \n\\[\nf(r,\\theta) = \\cos^{-1} \\left(\\frac{r\\cos\\theta-1}{9}\\right) - \\cos^{-1}\\left(\\frac{r\\cos\\theta+1}{9}\\right).\n\\]\n\nIf we now allow $\\theta$ to vary (uniformly) in $(-\\pi/2,\\pi/2]$, we find that the overall probability that $\\overline{PQ}$ intersects $\\Delta$ is\n\\[\nP(r) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} f(r,\\theta)\\,d\\theta.\n\\]\nThe function $P(r)$ is differentiable with \n\\[\nP'(r) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} \\frac{\\partial f(r,\\theta)}{\\partial r}\\,d\\theta.\n\\]\nNow\n\\begin{align*}\n\\frac{\\partial f(r,\\theta)}{\\partial r} &= (\\cos t)\\left((80-2r\\cos t-r^2\\cos^2 t)^{-1/2} \\right. \\\\\n&\\qquad \\left. -(80+2r\\cos t-r^2\\cos^2 t)^{-1/2}\\right),\n\\end{align*}\nwhich, for $t\\in (-\\pi/2,\\pi/2)$, is zero for $r=0$ and strictly positive for $r>0$. It follows that $P'(0)=0$ and $P'(r)<0$ for $r\\in (0,8]$, whence $P(r)$ is minimized when $r=0$.\n\n\\noindent\n\\textbf{Second solution.} (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang)\nWe interpret the first paragraph of the first solution as reducing the original problem to the following assertion:\ngiven two parallel lines at distance 2, both of which intersect a circle of radius 9, \nthe length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.\n\nTo see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above.",
  "vars": [
    "P",
    "Q",
    "f",
    "t",
    "\\\\theta",
    "\\\\phi",
    "\\\\Omega",
    "\\\\Delta"
  ],
  "params": [
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "firstpt",
        "Q": "secondpt",
        "f": "arcfunc",
        "t": "angletv",
        "\\theta": "angletheta",
        "\\phi": "anglephi",
        "\\Omega": "maincirc",
        "\\Delta": "smalldisc",
        "r": "centerdis"
      },
      "question": "Consider a circle $maincirc$ with radius 9 and center at the origin $(0,0)$, and a disc $smalldisc$ with radius 1 and center at $(centerdis,0)$, where $0 \\leq centerdis \\leq 8$. Two points $firstpt$ and $secondpt$ are chosen independently and uniformly at random on $maincirc$. Which value(s) of $centerdis$ minimize the probability that the chord $\\overline{firstpt secondpt}$ intersects $smalldisc$?",
      "solution": "We will show that $centerdis=0$ (and no other value of $centerdis$) minimizes the stated probability.\nNote that $firstpt$ and $secondpt$ coincide with probability $0$; thus we can assume that $firstpt\\neq secondpt$.\n\n\\noindent\n\\textbf{First solution.}\nFirst restrict $firstpt,secondpt$ to points on $maincirc$ such that the segment $\\overline{firstpt secondpt}$ makes an angle of $angletheta$ with the $y$ axis, where $angletheta$ is a fixed number with $-\\pi/2<angletheta\\leq\\pi/2$. By rotating the diagram by $-angletheta$ around the origin, we move $\\overline{firstpt secondpt}$ to be a vertical line and move $smalldisc$ to be centered at $(centerdis\\cos angletheta,-centerdis\\sin angletheta)$. In this rotated picture, $firstpt$ and $secondpt$ are at $(9\\cos anglephi,\\pm 9\\sin anglephi)$ where $anglephi$ is chosen uniformly at random in $(0,\\pi)$. Now the vertical tangent lines to the boundary of $smalldisc$, $x=centerdis\\cos angletheta\\pm 1$, intersect the $y>0$ semicircle of $maincirc$ at $(9\\cos anglephi,9\\sin anglephi)$ where $anglephi = \\cos^{-1}\\left(\\frac{centerdis\\cos angletheta\\pm 1}{9}\\right)$. Thus the probability that $\\overline{firstpt secondpt}$ intersects $smalldisc$ for a specific value of $angletheta$ is\n$\\frac{1}{\\pi} \\, arcfunc(centerdis,angletheta)$, where we define \n\\[\narcfunc(centerdis,angletheta) = \\cos^{-1} \\left(\\frac{centerdis\\cos angletheta-1}{9}\\right) - \\cos^{-1}\\left(\\frac{centerdis\\cos angletheta+1}{9}\\right).\n\\]\n\nIf we now allow $angletheta$ to vary (uniformly) in $(-\\pi/2,\\pi/2]$, we find that the overall probability that $\\overline{firstpt secondpt}$ intersects $smalldisc$ is\n\\[\nfirstpt(centerdis) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} arcfunc(centerdis,angletheta)\\,d angletheta.\n\\]\nThe function $firstpt(centerdis)$ is differentiable with \n\\[\nfirstpt'(centerdis) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} \\frac{\\partial arcfunc(centerdis,angletheta)}{\\partial centerdis}\\,d angletheta.\n\\]\nNow\n\\begin{align*}\n\\frac{\\partial arcfunc(centerdis,angletheta)}{\\partial centerdis} &= (\\cos angletv)\\left((80-2 centerdis\\cos angletv-centerdis^2\\cos^2 angletv)^{-1/2} \\right. \\\\\n&\\qquad \\left. -(80+2 centerdis\\cos angletv-centerdis^2\\cos^2 angletv)^{-1/2}\\right),\n\\end{align*}\nwhich, for $angletv\\in (-\\pi/2,\\pi/2)$, is zero for $centerdis=0$ and strictly positive for $centerdis>0$. It follows that $firstpt'(0)=0$ and $firstpt'(centerdis)<0$ for $centerdis\\in (0,8]$, whence $firstpt(centerdis)$ is minimized when $centerdis=0$.\n\n\\noindent\n\\textbf{Second solution.} (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang)\nWe interpret the first paragraph of the first solution as reducing the original problem to the following assertion:\ngiven two parallel lines at distance 2, both of which intersect a circle of radius 9, \nthe length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.\n\nTo see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "almondcup",
        "Q": "bannerdock",
        "f": "mangoaltar",
        "t": "candleveil",
        "\\theta": "orchardkey",
        "\\phi": "lanternmud",
        "\\Omega": "villagewing",
        "\\Delta": "canyonseed",
        "r": "pebblegate"
      },
      "question": "Consider a circle $villagewing$ with radius 9 and center at the origin $(0,0)$, and a disc $canyonseed$ with radius 1 and center at $(pebblegate,0)$, where $0 \\leq pebblegate \\leq 8$. Two points $almondcup$ and $bannerdock$ are chosen independently and uniformly at random on $villagewing$. Which value(s) of $pebblegate$ minimize the probability that the chord $\\overline{almondcupbannerdock}$ intersects $canyonseed$?",
      "solution": "We will show that $pebblegate=0$ (and no other value of $pebblegate$) minimizes the stated probability.\nNote that $almondcup$ and $bannerdock$ coincide with probability $0$; thus we can assume that $almondcup\\neq bannerdock$.\n\n\\noindent\n\\textbf{First solution.}\nFirst restrict $almondcup,bannerdock$ to points on $villagewing$ such that the segment $\\overline{almondcupbannerdock}$ makes an angle of $orchardkey$ with the $y$ axis, where $orchardkey$ is a fixed number with $-\\pi/2<orchardkey\\leq\\pi/2$. By rotating the diagram by $-orchardkey$ around the origin, we move $\\overline{almondcupbannerdock}$ to be a vertical line and move $canyonseed$ to be centered at $(pebblegate\\cos orchardkey,-pebblegate\\sin orchardkey)$. In this rotated picture, $almondcup$ and $bannerdock$ are at $(9\\cos lanternmud,\\pm 9\\sin lanternmud)$ where $lanternmud$ is chosen uniformly at random in $(0,\\pi)$. Now the vertical tangent lines to the boundary of $canyonseed$, $x=pebblegate\\cos orchardkey\\pm 1$, intersect the $y>0$ semicircle of $villagewing$ at $(9\\cos lanternmud,9\\sin lanternmud)$ where $lanternmud = \\cos^{-1}\\left(\\frac{pebblegate\\cos orchardkey\\pm 1}{9}\\right)$. Thus the probability that $\\overline{almondcupbannerdock}$ intersects $canyonseed$ for a specific value of $orchardkey$ is\n$\\frac{1}{\\pi} mangoaltar(pebblegate,orchardkey)$, where we define \n\\[\nmangoaltar(pebblegate,orchardkey) = \\cos^{-1} \\left(\\frac{pebblegate\\cos orchardkey-1}{9}\\right) - \\cos^{-1}\\left(\\frac{pebblegate\\cos orchardkey+1}{9}\\right).\n\\]\n\nIf we now allow $orchardkey$ to vary (uniformly) in $(-\\pi/2,\\pi/2]$, we find that the overall probability that $\\overline{almondcupbannerdock}$ intersects $canyonseed$ is\n\\[\nalmondcup(pebblegate) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} mangoaltar(pebblegate,orchardkey)\\,d orchardkey.\n\\]\nThe function $almondcup(pebblegate)$ is differentiable with \n\\[\nalmondcup'(pebblegate) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} \\frac{\\partial mangoaltar(pebblegate,orchardkey)}{\\partial pebblegate}\\,d orchardkey.\n\\]\nNow\n\\begin{align*}\n\\frac{\\partial mangoaltar(pebblegate,orchardkey)}{\\partial pebblegate} &= (\\cos candleveil)\\left((80-2pebblegate\\cos candleveil-pebblegate^2\\cos^2 candleveil)^{-1/2} \\right. \\\\\n&\\qquad \\left. -(80+2pebblegate\\cos candleveil-pebblegate^2\\cos^2 candleveil)^{-1/2}\\right),\n\\end{align*}\nwhich, for $candleveil\\in (-\\pi/2,\\pi/2)$, is zero for $pebblegate=0$ and strictly positive for $pebblegate>0$. It follows that $almondcup'(0)=0$ and $almondcup'(pebblegate)<0$ for $pebblegate\\in (0,8]$, whence $almondcup(pebblegate)$ is minimized when $pebblegate=0$.\n\n\\noindent\n\\textbf{Second solution.} (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang)\nWe interpret the first paragraph of the first solution as reducing the original problem to the following assertion:\ngiven two parallel lines at distance 2, both of which intersect a circle of radius 9, \nthe length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.\n\nTo see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "voidmass",
        "Q": "emptiness",
        "f": "disfunction",
        "t": "stillness",
        "\\theta": "straighten",
        "\\phi": "linearity",
        "\\Omega": "squarezone",
        "\\Delta": "ringvoid",
        "r": "diameter"
      },
      "question": "Consider a circle $squarezone$ with radius 9 and center at the origin $(0,0)$, and a disc $ringvoid$ with radius 1 and center at $(diameter,0)$, where $0 \\leq diameter \\leq 8$. Two points $voidmass$ and $emptiness$ are chosen independently and uniformly at random on $squarezone$. Which value(s) of $diameter$ minimize the probability that the chord $\\overline{voidmassemptiness}$ intersects $ringvoid$?",
      "solution": "We will show that $diameter=0$ (and no other value of $diameter$) minimizes the stated probability.\nNote that $voidmass$ and $emptiness$ coincide with probability $0$; thus we can assume that $voidmass\\neq emptiness$.\n\n\\noindent\n\\textbf{First solution.}\nFirst restrict $voidmass,emptiness$ to points on $squarezone$ such that the segment $\\overline{voidmassemptiness}$ makes an angle of $straighten$ with the $y$ axis, where $straighten$ is a fixed number with $-\\pi/2<straighten\\leq\\pi/2$. By rotating the diagram by $-straighten$ around the origin, we move $\\overline{voidmassemptiness}$ to be a vertical line and move $ringvoid$ to be centered at $(diameter\\cos straighten,-diameter\\sin straighten)$. In this rotated picture, $voidmass$ and $emptiness$ are at $(9\\cos linearity,\\pm 9\\sin linearity)$ where $linearity$ is chosen uniformly at random in $(0,\\pi)$. Now the vertical tangent lines to the boundary of $ringvoid$, $x=diameter\\cos straighten\\pm 1$, intersect the $y>0$ semicircle of $squarezone$ at $(9\\cos linearity,9\\sin linearity)$ where $linearity = \\cos^{-1}\\left(\\frac{diameter\\cos straighten\\pm 1}{9}\\right)$. Thus the probability that $\\overline{voidmassemptiness}$ intersects $ringvoid$ for a specific value of $straighten$ is\n$\\frac{1}{\\pi} disfunction(diameter,straighten)$, where we define \n\\[\ndisfunction(diameter,straighten) = \\cos^{-1} \\left(\\frac{diameter\\cos straighten-1}{9}\\right) - \\cos^{-1}\\left(\\frac{diameter\\cos straighten+1}{9}\\right).\n\\]\n\nIf we now allow $straighten$ to vary (uniformly) in $(-\\pi/2,\\pi/2]$, we find that the overall probability that $\\overline{voidmassemptiness}$ intersects $ringvoid$ is\n\\[\nvoidmass(diameter) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} disfunction(diameter,straighten)\\,d straighten.\n\\]\nThe function $voidmass(diameter)$ is differentiable with \n\\[\nvoidmass'(diameter) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} \\frac{\\partial disfunction(diameter,straighten)}{\\partial diameter}\\,d straighten.\n\\]\nNow\n\\begin{align*}\n\\frac{\\partial disfunction(diameter,straighten)}{\\partial diameter} &= (\\cos stillness)\\left((80-2diameter\\cos stillness-diameter^2\\cos^2 stillness)^{-1/2} \\right. \\\\\n&\\qquad \\left. -(80+2diameter\\cos stillness-diameter^2\\cos^2 stillness)^{-1/2}\\right),\n\\end{align*}\nwhich, for $stillness\\in (-\\pi/2,\\pi/2)$, is zero for $diameter=0$ and strictly positive for $diameter>0$. It follows that $voidmass'(0)=0$ and $voidmass'(diameter)<0$ for $diameter\\in (0,8]$, whence $voidmass(diameter)$ is minimized when $diameter=0$.\n\n\\noindent\n\\textbf{Second solution.} (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang)\nWe interpret the first paragraph of the first solution as reducing the original problem to the following assertion:\ngiven two parallel lines at distance 2, both of which intersect a circle of radius 9, \nthe length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.\n\nTo see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above."
    },
    "garbled_string": {
      "map": {
        "P": "mnjqztrw",
        "Q": "gvdhcplk",
        "f": "zpeksuxo",
        "t": "rqnvbcya",
        "\\theta": "kfhuzdwp",
        "\\phi": "dbtqsmne",
        "\\Omega": "ycvbazrs",
        "\\Delta": "hwxlojem",
        "r": "xntvgrma"
      },
      "question": "Consider a circle $ycvbazrs$ with radius 9 and center at the origin $(0,0)$, and a disc $hwxlojem$ with radius 1 and center at $(xntvgrma,0)$, where $0 \\leq xntvgrma \\leq 8$. Two points $mnjqztrw$ and $gvdhcplk$ are chosen independently and uniformly at random on $ycvbazrs$. Which value(s) of $xntvgrma$ minimize the probability that the chord $\\overline{mnjqztrwgvdhcplk}$ intersects $hwxlojem$?",
      "solution": "We will show that $xntvgrma=0$ (and no other value of $xntvgrma$) minimizes the stated probability.\nNote that $mnjqztrw$ and $gvdhcplk$ coincide with probability $0$; thus we can assume that $mnjqztrw\\neq gvdhcplk$.\n\n\\noindent\n\\textbf{First solution.}\nFirst restrict $mnjqztrw,gvdhcplk$ to points on $ycvbazrs$ such that the segment $\\overline{mnjqztrwgvdhcplk}$ makes an angle of $kfhuzdwp$ with the $y$ axis, where $kfhuzdwp$ is a fixed number with $-\\pi/2<kfhuzdwp\\leq\\pi/2$. By rotating the diagram by $-kfhuzdwp$ around the origin, we move $\\overline{mnjqztrwgvdhcplk}$ to be a vertical line and move $hwxlojem$ to be centered at $(xntvgrma\\cos kfhuzdwp,-xntvgrma\\sin kfhuzdwp)$. In this rotated picture, $mnjqztrw$ and $gvdhcplk$ are at $(9\\cos dbtqsmne,\\pm 9\\sin dbtqsmne)$ where $dbtqsmne$ is chosen uniformly at random in $(0,\\pi)$. Now the vertical tangent lines to the boundary of $hwxlojem$, $x=xntvgrma\\cos kfhuzdwp\\pm 1$, intersect the $y>0$ semicircle of $ycvbazrs$ at $(9\\cos dbtqsmne,9\\sin dbtqsmne)$ where \n$dbtqsmne = \\cos^{-1}\\left(\\frac{xntvgrma\\cos kfhuzdwp\\pm 1}{9}\\right)$. \nThus the probability that $\\overline{mnjqztrwgvdhcplk}$ intersects $hwxlojem$ for a specific value of $kfhuzdwp$ is\n$\\frac{1}{\\pi} \\, zpeksuxo(xntvgrma,kfhuzdwp)$, where we define \n\\[\nzpeksuxo(xntvgrma,kfhuzdwp) = \\cos^{-1} \\left(\\frac{xntvgrma\\cos kfhuzdwp-1}{9}\\right) - \\cos^{-1}\\left(\\frac{xntvgrma\\cos kfhuzdwp+1}{9}\\right).\n\\]\n\nIf we now allow $kfhuzdwp$ to vary (uniformly) in $(-\\pi/2,\\pi/2]$, we find that the overall probability that $\\overline{mnjqztrwgvdhcplk}$ intersects $hwxlojem$ is\n\\[\nmnjqztrw(xntvgrma) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} zpeksuxo(xntvgrma,kfhuzdwp)\\,d kfhuzdwp.\n\\]\nThe function $mnjqztrw(xntvgrma)$ is differentiable with \n\\[\nmnjqztrw'(xntvgrma) = \\frac{1}{\\pi^2} \\int_{-\\pi/2}^{\\pi/2} \\frac{\\partial zpeksuxo(xntvgrma,kfhuzdwp)}{\\partial xntvgrma}\\,d kfhuzdwp.\n\\]\nNow\n\\begin{align*}\n\\frac{\\partial zpeksuxo(xntvgrma,kfhuzdwp)}{\\partial xntvgrma} &= (\\cos rqnvbcya)\\left((80-2xntvgrma\\cos rqnvbcya-xntvgrma^2\\cos^2 rqnvbcya)^{-1/2} \\right. \\\\\n&\\qquad \\left. -(80+2xntvgrma\\cos rqnvbcya-xntvgrma^2\\cos^2 rqnvbcya)^{-1/2}\\right),\n\\end{align*}\nwhich, for $rqnvbcya\\in (-\\pi/2,\\pi/2)$, is zero for $xntvgrma=0$ and strictly positive for $xntvgrma>0$. It follows that $mnjqztrw'(0)=0$ and $mnjqztrw'(xntvgrma)<0$ for $xntvgrma\\in (0,8]$, whence $mnjqztrw(xntvgrma)$ is minimized when $xntvgrma=0$.\n\n\\noindent\n\\textbf{Second solution.} (based on ideas from Elliott Liu, Bjorn Poonen, Linus Tang, and Allen Wang)\nWe interpret the first paragraph of the first solution as reducing the original problem to the following assertion:\ngiven two parallel lines at distance 2, both of which intersect a circle of radius 9, \nthe length of either of the two congruent arcs of the circle lying between the two lines is minimized when the line halfway between the two parallel lines passes through the center of the circle.\n\nTo see this, note that the length of a minor arc of a circle is a strictly increasing function of the length of the chord connecting the two endpoints of the arc. In this case, the chord connects points on the two given parallel lines, so the distance between these points is minimized by having them be the endpoints of a segment perpendicular to the two lines; this achieves the situation described above."
    },
    "kernel_variant": {
      "question": "Let Ω be the circle of radius 13 centred at the origin O=(0,0).  For a real number s with 0≤s≤11/√2 set c_s=(s,s) and let Δ_s denote the closed disc of radius 2 centred at c_s.  Two points P,Q are chosen independently and uniformly at random on Ω, and the chord PQ is drawn.\n\nFor which value(s) of s is the probability that the chord PQ meets the disc Δ_s minimal?",
      "solution": "Answer.  The probability is strictly increasing in s, hence it is minimal when s=0 and only then.\n\nThroughout we write R=13, r=√2s (so that r is the distance Oc_s) and confine ourselves to 0≤r≤11.\n\n1.  A convenient description of a random chord.\n\nA straight line in the plane is determined by its (signed) distance d from the origin and by the unit normal n=(cosα,sinα), α∈[0,2π).  It can therefore be written as\n      L(d,α):  x·n = d .\n\nFor a chord of the circle Ω we must have |d|≤R.  It is a standard fact (and easy to verify directly) that if P,Q are chosen independently and uniformly on Ω then\n      – the angle α is uniformly distributed on [0,2π), and\n      – conditional on α, the variable d has density\n            g(d) = 1/(π√(R²–d²)),   |d|<R,                       (1)\n        and d is independent of α.\nThus a random chord may be generated by first picking α∼Unif[0,2π) and then d with density (1).  (See e.g. the classical solution of Bertrand’s paradox for chords generated by two random points.)\n\n2.  Reformulating the desired probability.\n\nBecause Ω is rotationally symmetric, the position of the small disc matters only through its distance r from O.  Hence we may rotate the picture and assume from now on that the centre is at c=(r,0).  Let L=L(d,α) be a random chord as above.  The chord meets Δ_r (the disc of radius 2 centred at c) exactly when the distance from c to the line L does not exceed 2:\n      |c·n – d| ≤ 2  ⟺  |r cosα – d| ≤ 2.                       (2)\nThe required probability can therefore be written as\n      P(r) = (1/2π) ∫_{0}^{2π}  ∫_{-R}^{R} 𝟙{|r cosα–d|≤2}\n                                      g(d)  dd dα.             (3)\n\n3.  A more convenient inner integral.\n\nFor a fixed real x with |x|≤R–2 put\n      h(x):=∫_{x–2}^{x+2} g(d) dd.                              (4)\nBecause 0≤r≤11 and |cosα|≤1, the number x=r cosα always satisfies |x|≤11< R–2, so that no endpoint in (4) ever reaches ±R and we may ignore the cut-offs.\nWith this notation (3) becomes\n      P(r) = (1/2π) ∫_{0}^{2π} h(r cosα)\n                            dα.                                (5)\n\n4.  Monotonicity of h.\n\nDifferentiate (4) with respect to x:\n      h′(x) = g(x+2) – g(x–2).                                 (6)\nThe density g in (1) is an even function of d and it is strictly increasing in |d| (the denominator √(R²–d²) decreases with |d|).  Consequently\n      h′(x)   >0  if  x>0,\n               0  if  x=0,                                    (7)\n              <0  if  x<0.\nHence h is strictly increasing on (0,R–2), strictly decreasing on (–R+2,0) and is an odd function.\n\n5.  Derivative of P.\n\nDifferentiate (5) under the integral sign (justified because the integrand is continuous):\n      P′(r) = (1/2π) ∫_{0}^{2π} h′(r cosα) · cosα  dα.         (8)\nFix r>0.  For an angle α with cosα>0 we have r cosα>0, so h′(r cosα)>0 by (7) and the integrand in (8) is positive.  For α with cosα<0 both factors change sign, so the product is again positive.  (Formally, h′ is odd while cosα is even with respect to α↦α+π.)  Except for the negligible set where cosα=0 the integrand is strictly positive; therefore\n      P′(r) > 0   whenever  r>0.                               (9)\nSince P′(0)=0 by symmetry, P is strictly increasing on (0,11] and attains its unique minimum at r=0.\n\n6.  Translating back to s.\n\nRecall r=√2 s, so r=0 corresponds to s=0.  Thus the probability that the random chord meets Δ_s is minimal exactly for\n      s = 0.\n\nRemark.  The analysis also shows that moving the small disc away from the centre (in any direction) always makes it more likely, not less, to be hit by a random chord defined by two uniform points on the circle – a consequence of the fact that such chords are more heavily concentrated near the circumference than near the centre.",
      "_meta": {
        "core_steps": [
          "Exploit rotational symmetry: rotate any fixed–angle chord to a convenient orientation, carrying the small disc’s center to a new point.",
          "For that orientation, the chord meets the disc iff its endpoints lie between two parallel tangents to the disc; the relevant arc-length equals a difference of arccos terms.",
          "Average this arc length over all chord orientations to obtain a probability function P(r).",
          "Differentiate P(r); show P'(0)=0 and P'(r)<0 for r>0, so P(r) is minimized at r=0."
        ],
        "mutable_slots": {
          "slot_outer_radius": {
            "description": "Radius of the large circle (only required to exceed the small disc’s radius)",
            "original": 9
          },
          "slot_inner_radius": {
            "description": "Radius of the smaller disc whose intersection with the chord is tested",
            "original": 1
          },
          "slot_center_path": {
            "description": "Straight line (here the x-axis) along which the small disc’s center moves",
            "original": "(r,0)"
          },
          "slot_offset_range": {
            "description": "Allowed values of the offset r, namely from coincidence with the large circle’s center out to just before the disc would leave the circle",
            "original": "0 ≤ r ≤ 8  ( = outer_radius − inner_radius )"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}