{
  "index": "1961-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "COMB"
  ],
  "difficulty": "",
  "question": "2. Let \\( \\alpha \\) and \\( \\beta \\) be given positive real numbers, with \\( \\alpha<\\beta \\). If two points are selected at random from a straight line segment of length \\( \\beta \\), what is the probability that the distance between them is at least \\( \\alpha \\) ?",
  "solution": "Solution. We interpret \"at random\" to mean that the pair of points \\( x, y \\) is chosen so that the probability that \\( \\langle x, y\\rangle \\) falls in any region in the square \\( [0, \\beta] \\times[0, \\beta] \\) is proportional to the area of that region. Then the\nfavorable region is evidently the union of the two triangular regions shown and the probability of a favorable outcome is\n\\[\n\\frac{(\\beta-\\alpha)^{2}}{\\beta^{2}}=\\left(1-\\frac{\\alpha}{\\beta}\\right)^{2}\n\\]",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "\\\\alpha",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "firstpoint",
        "y": "secondpoint",
        "\\alpha": "mindist",
        "\\beta": "segmentlen"
      },
      "question": "2. Let \\( mindist \\) and \\( segmentlen \\) be given positive real numbers, with \\( mindist<segmentlen \\). If two points are selected at random from a straight line segment of length \\( segmentlen \\), what is the probability that the distance between them is at least \\( mindist \\)?",
      "solution": "Solution. We interpret \"at random\" to mean that the pair of points \\( firstpoint, secondpoint \\) is chosen so that the probability that \\( \\langle firstpoint, secondpoint\\rangle \\) falls in any region in the square \\( [0, segmentlen] \\times[0, segmentlen] \\) is proportional to the area of that region. Then the favorable region is evidently the union of the two triangular regions shown and the probability of a favorable outcome is\n\\[\n\\frac{(segmentlen-mindist)^{2}}{segmentlen^{2}}=\\left(1-\\frac{mindist}{segmentlen}\\right)^{2}\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orangepulp",
        "y": "violetink",
        "\\alpha": "silvercloud",
        "\\beta": "goldenleaf"
      },
      "question": "2. Let \\( silvercloud \\) and \\( goldenleaf \\) be given positive real numbers, with \\( silvercloud<goldenleaf \\). If two points are selected at random from a straight line segment of length \\( goldenleaf \\), what is the probability that the distance between them is at least \\( silvercloud \\) ?",
      "solution": "Solution. We interpret \"at random\" to mean that the pair of points \\( orangepulp, violetink \\) is chosen so that the probability that \\( \\langle orangepulp, violetink\\rangle \\) falls in any region in the square \\( [0, goldenleaf] \\times[0, goldenleaf] \\) is proportional to the area of that region. Then the favorable region is evidently the union of the two triangular regions shown and the probability of a favorable outcome is\n\\[\n\\frac{(goldenleaf-silvercloud)^{2}}{goldenleaf^{2}}=\\left(1-\\frac{silvercloud}{goldenleaf}\\right)^{2}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stationary",
        "y": "immobile",
        "\\alpha": "variableone",
        "\\beta": "variabletwo"
      },
      "question": "Let \\( variableone \\) and \\( variabletwo \\) be given positive real numbers, with \\( variableone<variabletwo \\). If two points are selected at random from a straight line segment of length \\( variabletwo \\), what is the probability that the distance between them is at least \\( variableone \\) ?",
      "solution": "We interpret \"at random\" to mean that the pair of points \\( stationary, immobile \\) is chosen so that the probability that \\( \\langle stationary, immobile\\rangle \\) falls in any region in the square \\( [0, variabletwo] \\times[0, variabletwo] \\) is proportional to the area of that region. Then the\nfavorable region is evidently the union of the two triangular regions shown and the probability of a favorable outcome is\n\\[\n\\frac{(variabletwo-variableone)^{2}}{variabletwo^{2}}=\\left(1-\\frac{variableone}{variabletwo}\\right)^{2}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "\\alpha": "mgztrfpl",
        "\\beta": "jkdvhsqm"
      },
      "question": "2. Let \\( mgztrfpl \\) and \\( jkdvhsqm \\) be given positive real numbers, with \\( mgztrfpl<jkdvhsqm \\). If two points are selected at random from a straight line segment of length \\( jkdvhsqm \\), what is the probability that the distance between them is at least \\( mgztrfpl \\) ?",
      "solution": "Solution. We interpret \"at random\" to mean that the pair of points \\( qzxwvtnp, hjgrksla \\) is chosen so that the probability that \\( \\langle qzxwvtnp, hjgrksla\\rangle \\) falls in any region in the square \\( [0, jkdvhsqm] \\times[0, jkdvhsqm] \\) is proportional to the area of that region. Then the favorable region is evidently the union of the two triangular regions shown and the probability of a favorable outcome is\n\\[\n\\frac{(jkdvhsqm-mgztrfpl)^{2}}{jkdvhsqm^{2}}=\\left(1-\\frac{mgztrfpl}{jkdvhsqm}\\right)^{2}\n\\]"
    },
    "kernel_variant": {
      "question": "Let $\\ell,\\delta,\\varepsilon,\\eta$ be positive real numbers that satisfy  \n\\[\n\\delta+2\\varepsilon+2\\eta\\le \\ell .\n\\]\nThree points $X_{1},X_{2},X_{3}$ are chosen independently and uniformly at random on the segment $[0,\\ell]$.  \nDetermine, in closed form, the probability that the following three conditions are fulfilled simultaneously  \n\n(i)\\, the diameter of the triple $\\{X_{1},X_{2},X_{3}\\}$ does not exceed $\\delta$, that is  \n\\[\n\\max\\{|X_{i}-X_{j}|:i,j\\in\\{1,2,3\\}\\}\\le\\delta;\n\\]\n\n(ii)\\, the midpoint $M=\\dfrac{X_{1}+X_{2}}{2}$ of the first two points satisfies  \n\\[\n\\bigl|M-\\tfrac{\\ell}{2}\\bigr|\\le\\varepsilon;\n\\]\n\n(iii)\\, the third point satisfies  \n\\[\n|X_{3}-M|\\le\\eta .\n\\]\n\nYour answer must be an explicit (possibly piece-wise) formula in $\\ell,\\delta,\\varepsilon$ and $\\eta$.",
      "solution": "We work throughout with the joint density of independent, uniformly distributed points on $[0,\\ell]$.\n\nStep 0.  Change of variables for $(X_{1},X_{2})$.  \nDefine\n\\[\nu:=\\frac{X_{1}+X_{2}}{2}\\quad\\text{(midpoint)},\\qquad\ns:=\\frac{X_{2}-X_{1}}{2}\\quad\\text{(half–separation, possibly negative)} .\n\\]\nThe inverse map is $X_{1}=u-s$, $X_{2}=u+s$, whose Jacobian has absolute value $|\\partial(X_{1},X_{2})/\\partial(u,s)|=2$.  \nBecause $0\\le X_{1},X_{2}\\le\\ell$, the pair $(u,s)$ ranges over  \n\\[\n-\\frac{\\ell}{2}\\le s\\le\\frac{\\ell}{2},\\qquad |s|\\le u\\le\\ell-|s|. \\tag{1}\n\\]\nHence the joint density of $(u,s)$ equals\n\\[\nf_{(u,s)}(u,s)=\\frac{1}{\\ell^{2}}\\cdot 2=\\frac{2}{\\ell^{2}} .\n\\]\n\nStep 1.  Translating the three requirements.\n\n(i)  Diameter condition: $|s|\\le\\delta/2$.  \nAdditionally, $X_{3}$ must lie within $\\delta$ of both $X_{1}$ and $X_{2}$; this is equivalent to  \n\\[\n|X_{3}-u|\\le\\delta-|s|. \\tag{2}\n\\]\n\n(ii)  Midpoint window:  \n\\[\n|u-\\tfrac{\\ell}{2}|\\le\\varepsilon. \\tag{3}\n\\]\n\n(iii)  Third–point window:  \n\\[\n|X_{3}-u|\\le\\eta. \\tag{4}\n\\]\n\nConditions (2) and (4) together give  \n\\[\n|X_{3}-u|\\le\\min\\bigl(\\eta,\\delta-|s|\\bigr). \\tag{5}\n\\]\n\nStep 2.  Using $\\delta+2\\varepsilon+2\\eta\\le\\ell$.  \nThe global inequality implies  \n\\[\n\\frac{\\delta}{2}+\\varepsilon\\le\\frac{\\ell}{2}-\\eta<\\frac{\\ell}{2}. \\tag{6}\n\\]\nThus, for every $s$ with $|s|\\le\\delta/2$, the $\\varepsilon$–window (3) lies fully inside the interval for $u$ described in (1).  Consequently  \n\\[\n|s|\\le\\frac{\\delta}{2},\\qquad |u-\\tfrac{\\ell}{2}|\\le\\varepsilon \\tag{7}\n\\]\nparametrise a complete rectangle in the $(u,s)$–plane:\n\\[\nu\\in\\bigl[\\tfrac{\\ell}{2}-\\varepsilon,\\tfrac{\\ell}{2}+\\varepsilon\\bigr],\\qquad\ns\\in\\bigl[-\\tfrac{\\delta}{2},\\tfrac{\\delta}{2}\\bigr]. \\tag{8}\n\\]\n\nStep 3.  Conditional probability for $X_{3}$.  \nFor fixed $(u,s)$ satisfying (7), inequality (6) guarantees that the interval (5) lies inside $[0,\\ell]$.  Therefore\n\\[\n\\Pr\\bigl(|X_{3}-u|\\le\\min(\\eta,\\delta-|s|)\\,\\bigl|\\,u,s\\bigr)=\n\\frac{2\\min(\\eta,\\delta-|s|)}{\\ell}. \\tag{9}\n\\]\n\nStep 4.  Integrating over $(u,s)$.  \nWith (8) and (9),\n\\[\n\\begin{aligned}\n\\Pr(\\text{all three events})&=\n\\iint_{(8)}\\frac{2}{\\ell^{2}}\\cdot\\frac{2\\min(\\eta,\\delta-|s|)}{\\ell}\\,du\\,ds\\\\\n&=\\frac{8\\varepsilon}{\\ell^{3}}\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds. \\tag{10}\n\\end{aligned}\n\\]\nBecause the integrand is even in $s$,\n\\[\n\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds\n=2\\int_{0}^{\\delta/2}\\min(\\eta,\\delta-s)\\,ds.  \\tag{11}\n\\]\n\nStep 5.  Evaluating the $s$–integral.  \nSet $s_{0}:=\\max(0,\\delta-\\eta)$; note that $s_{0}\\in[0,\\delta/2]$ exactly when $\\eta\\in[\\delta/2,\\delta]$.\n\nCase A: $0<\\eta\\le\\delta/2$ (hence $s_{0}=\\delta-\\eta\\ge\\delta/2$).  \nHere $\\delta-s\\ge\\eta$ for all $s\\in[0,\\delta/2]$, so  \n\\[\nI:=\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds\n=2\\eta\\cdot\\frac{\\delta}{2}=\\delta\\eta. \\tag{12A}\n\\]\n\nCase B: $\\delta/2<\\eta\\le\\delta$ (hence $s_{0}=\\delta-\\eta\\in(0,\\delta/2)$).  \nSplit the integral at $s_{0}$:\n\\[\n\\begin{aligned}\nI&=\n2\\!\\left[\\int_{0}^{s_{0}}\\eta\\,ds+\\int_{s_{0}}^{\\delta/2}(\\delta-s)\\,ds\\right]\\\\\n&=2\\Bigl[\\eta s_{0}+\\bigl(\\delta s-\\tfrac{s^{2}}{2}\\bigr)_{s_{0}}^{\\delta/2}\\Bigr]\\\\\n&=2\\Bigl[\\eta s_{0}+\\frac{3\\delta^{2}}{8}-\\delta s_{0}+\\frac{s_{0}^{2}}{2}\\Bigr]\\\\\n&=2\\Bigl[-s_{0}^{2}+\\frac{3\\delta^{2}}{8}+\\frac{s_{0}^{2}}{2}\\Bigr]\n=\\frac{3\\delta^{2}}{4}-(\\delta-\\eta)^{2}. \\tag{12B}\n\\end{aligned}\n\\]\n\nCase C: $\\eta\\ge\\delta$ ($s_{0}=0$).  Then $\\min(\\eta,\\delta-s)=\\delta-s$ and\n\\[\nI=2\\int_{0}^{\\delta/2}(\\delta-s)\\,ds\n=2\\Bigl[\\delta s-\\tfrac{s^{2}}{2}\\Bigr]_{0}^{\\delta/2}\n=\\frac{3\\delta^{2}}{4}. \\tag{12C}\n\\]\n\nStep 6.  Final probability.  \nInsert (12A)–(12C) into (10):\n\\[\n\\boxed{\n\\Pr(\\ell,\\delta,\\varepsilon,\\eta)=\n\\begin{cases}\n\\dfrac{8\\varepsilon\\,\\delta\\,\\eta}{\\ell^{3}}, & 0<\\eta\\le\\delta/2,\\\\[6pt]\n\\dfrac{8\\varepsilon}{\\ell^{3}}\\Bigl[\\dfrac{3\\delta^{2}}{4}-(\\delta-\\eta)^{2}\\Bigr],\n     & \\delta/2<\\eta\\le\\delta,\\\\[6pt]\n\\dfrac{6\\varepsilon\\,\\delta^{2}}{\\ell^{3}}, & \\eta\\ge\\delta .\n\\end{cases}}\n\\]\n\nStep 7.  Sanity checks.  \n(i)  $\\eta\\to0^{+}$ gives probability $0$, as expected.  \n(ii)  For $\\eta\\ge\\delta$ the diameter restriction alone governs the event, so the result is constant in $\\eta$; the formula indeed becomes independent of $\\eta$.  \n(iii)  $\\varepsilon\\to0^{+}$ or $\\delta\\to0^{+}$ forces the probability to $0$, which the expression reflects.  \n(iv)  Continuity at $\\eta=\\delta/2$ and $\\eta=\\delta$ follows from direct substitution.\n\nThe probability above is therefore correct and complete.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.531623",
        "was_fixed": false,
        "difficulty_analysis": "• More random points: The problem involves three independent points instead of two, increasing dimensionality from 2-D to 3-D in the sampling space.\n\n• Coupled geometric constraints:  \n  – a global diameter bound,  \n  – a constraint on the midpoint of the first two points,  \n  – and a proximity requirement between that midpoint and the third point.  \n  These interact non-trivially; one must recognise that some inequalities (the diameter bound) are automatically satisfied once the others are enforced.\n\n• Change of variables and Jacobian: Solving requires a non-obvious affine transformation (from (X₁ ,X₂) to (u ,s)) and careful handling of its Jacobian.\n\n• Region geometry in two parameters: One must determine the exact shape (here a rectangle) of the feasible (u ,s)-region after simultaneously imposing disparate linear constraints; this demands thoughtful use of the given inequality δ + 2ε + 2η ≤ ℓ.\n\n• Conditional probability inside a moving window: The probability for X₃ depends on u; showing that it becomes a constant 2η/ℓ requires proving that the window never touches the boundary, which in turn hinges on the interplay of all four parameters.\n\n• Multi-step integration: The final computation involves integrating a constant conditional probability over a two-dimensional region found via geometric reasoning and variable transformation.\n\nOverall, the enhanced variant raises the problem’s dimension, introduces multiple interacting inequalities, and obliges the solver to use advanced techniques from multivariable calculus and geometric probability, making it decidedly harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\ell,\\delta,\\varepsilon,\\eta$ be positive real numbers that satisfy  \n\\[\n\\delta+2\\varepsilon+2\\eta\\le \\ell .\n\\]\nThree points $X_{1},X_{2},X_{3}$ are chosen independently and uniformly at random on the segment $[0,\\ell]$.  \nDetermine, in closed form, the probability that the following three conditions are fulfilled simultaneously  \n\n(i)\\, the diameter of the triple $\\{X_{1},X_{2},X_{3}\\}$ does not exceed $\\delta$, that is  \n\\[\n\\max\\{|X_{i}-X_{j}|:i,j\\in\\{1,2,3\\}\\}\\le\\delta;\n\\]\n\n(ii)\\, the midpoint $M=\\dfrac{X_{1}+X_{2}}{2}$ of the first two points satisfies  \n\\[\n\\bigl|M-\\tfrac{\\ell}{2}\\bigr|\\le\\varepsilon;\n\\]\n\n(iii)\\, the third point satisfies  \n\\[\n|X_{3}-M|\\le\\eta .\n\\]\n\nYour answer must be an explicit (possibly piece-wise) formula in $\\ell,\\delta,\\varepsilon$ and $\\eta$.",
      "solution": "We work throughout with the joint density of independent, uniformly distributed points on $[0,\\ell]$.\n\nStep 0.  Change of variables for $(X_{1},X_{2})$.  \nDefine\n\\[\nu:=\\frac{X_{1}+X_{2}}{2}\\quad\\text{(midpoint)},\\qquad\ns:=\\frac{X_{2}-X_{1}}{2}\\quad\\text{(half–separation, possibly negative)} .\n\\]\nThe inverse map is $X_{1}=u-s$, $X_{2}=u+s$, whose Jacobian has absolute value $|\\partial(X_{1},X_{2})/\\partial(u,s)|=2$.  \nBecause $0\\le X_{1},X_{2}\\le\\ell$, the pair $(u,s)$ ranges over  \n\\[\n-\\frac{\\ell}{2}\\le s\\le\\frac{\\ell}{2},\\qquad |s|\\le u\\le\\ell-|s|. \\tag{1}\n\\]\nHence the joint density of $(u,s)$ equals\n\\[\nf_{(u,s)}(u,s)=\\frac{1}{\\ell^{2}}\\cdot 2=\\frac{2}{\\ell^{2}} .\n\\]\n\nStep 1.  Translating the three requirements.\n\n(i)  Diameter condition: $|s|\\le\\delta/2$.  \nAdditionally, $X_{3}$ must lie within $\\delta$ of both $X_{1}$ and $X_{2}$; this is equivalent to  \n\\[\n|X_{3}-u|\\le\\delta-|s|. \\tag{2}\n\\]\n\n(ii)  Midpoint window:  \n\\[\n|u-\\tfrac{\\ell}{2}|\\le\\varepsilon. \\tag{3}\n\\]\n\n(iii)  Third–point window:  \n\\[\n|X_{3}-u|\\le\\eta. \\tag{4}\n\\]\n\nConditions (2) and (4) together give  \n\\[\n|X_{3}-u|\\le\\min\\bigl(\\eta,\\delta-|s|\\bigr). \\tag{5}\n\\]\n\nStep 2.  Using $\\delta+2\\varepsilon+2\\eta\\le\\ell$.  \nThe global inequality implies  \n\\[\n\\frac{\\delta}{2}+\\varepsilon\\le\\frac{\\ell}{2}-\\eta<\\frac{\\ell}{2}. \\tag{6}\n\\]\nThus, for every $s$ with $|s|\\le\\delta/2$, the $\\varepsilon$–window (3) lies fully inside the interval for $u$ described in (1).  Consequently  \n\\[\n|s|\\le\\frac{\\delta}{2},\\qquad |u-\\tfrac{\\ell}{2}|\\le\\varepsilon \\tag{7}\n\\]\nparametrise a complete rectangle in the $(u,s)$–plane:\n\\[\nu\\in\\bigl[\\tfrac{\\ell}{2}-\\varepsilon,\\tfrac{\\ell}{2}+\\varepsilon\\bigr],\\qquad\ns\\in\\bigl[-\\tfrac{\\delta}{2},\\tfrac{\\delta}{2}\\bigr]. \\tag{8}\n\\]\n\nStep 3.  Conditional probability for $X_{3}$.  \nFor fixed $(u,s)$ satisfying (7), inequality (6) guarantees that the interval (5) lies inside $[0,\\ell]$.  Therefore\n\\[\n\\Pr\\bigl(|X_{3}-u|\\le\\min(\\eta,\\delta-|s|)\\,\\bigl|\\,u,s\\bigr)=\n\\frac{2\\min(\\eta,\\delta-|s|)}{\\ell}. \\tag{9}\n\\]\n\nStep 4.  Integrating over $(u,s)$.  \nWith (8) and (9),\n\\[\n\\begin{aligned}\n\\Pr(\\text{all three events})&=\n\\iint_{(8)}\\frac{2}{\\ell^{2}}\\cdot\\frac{2\\min(\\eta,\\delta-|s|)}{\\ell}\\,du\\,ds\\\\\n&=\\frac{8\\varepsilon}{\\ell^{3}}\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds. \\tag{10}\n\\end{aligned}\n\\]\nBecause the integrand is even in $s$,\n\\[\n\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds\n=2\\int_{0}^{\\delta/2}\\min(\\eta,\\delta-s)\\,ds.  \\tag{11}\n\\]\n\nStep 5.  Evaluating the $s$–integral.  \nSet $s_{0}:=\\max(0,\\delta-\\eta)$; note that $s_{0}\\in[0,\\delta/2]$ exactly when $\\eta\\in[\\delta/2,\\delta]$.\n\nCase A: $0<\\eta\\le\\delta/2$ (hence $s_{0}=\\delta-\\eta\\ge\\delta/2$).  \nHere $\\delta-s\\ge\\eta$ for all $s\\in[0,\\delta/2]$, so  \n\\[\nI:=\\int_{-\\delta/2}^{\\delta/2}\\min(\\eta,\\delta-|s|)\\,ds\n=2\\eta\\cdot\\frac{\\delta}{2}=\\delta\\eta. \\tag{12A}\n\\]\n\nCase B: $\\delta/2<\\eta\\le\\delta$ (hence $s_{0}=\\delta-\\eta\\in(0,\\delta/2)$).  \nSplit the integral at $s_{0}$:\n\\[\n\\begin{aligned}\nI&=\n2\\!\\left[\\int_{0}^{s_{0}}\\eta\\,ds+\\int_{s_{0}}^{\\delta/2}(\\delta-s)\\,ds\\right]\\\\\n&=2\\Bigl[\\eta s_{0}+\\bigl(\\delta s-\\tfrac{s^{2}}{2}\\bigr)_{s_{0}}^{\\delta/2}\\Bigr]\\\\\n&=2\\Bigl[\\eta s_{0}+\\frac{3\\delta^{2}}{8}-\\delta s_{0}+\\frac{s_{0}^{2}}{2}\\Bigr]\\\\\n&=2\\Bigl[-s_{0}^{2}+\\frac{3\\delta^{2}}{8}+\\frac{s_{0}^{2}}{2}\\Bigr]\n=\\frac{3\\delta^{2}}{4}-(\\delta-\\eta)^{2}. \\tag{12B}\n\\end{aligned}\n\\]\n\nCase C: $\\eta\\ge\\delta$ ($s_{0}=0$).  Then $\\min(\\eta,\\delta-s)=\\delta-s$ and\n\\[\nI=2\\int_{0}^{\\delta/2}(\\delta-s)\\,ds\n=2\\Bigl[\\delta s-\\tfrac{s^{2}}{2}\\Bigr]_{0}^{\\delta/2}\n=\\frac{3\\delta^{2}}{4}. \\tag{12C}\n\\]\n\nStep 6.  Final probability.  \nInsert (12A)–(12C) into (10):\n\\[\n\\boxed{\n\\Pr(\\ell,\\delta,\\varepsilon,\\eta)=\n\\begin{cases}\n\\dfrac{8\\varepsilon\\,\\delta\\,\\eta}{\\ell^{3}}, & 0<\\eta\\le\\delta/2,\\\\[6pt]\n\\dfrac{8\\varepsilon}{\\ell^{3}}\\Bigl[\\dfrac{3\\delta^{2}}{4}-(\\delta-\\eta)^{2}\\Bigr],\n     & \\delta/2<\\eta\\le\\delta,\\\\[6pt]\n\\dfrac{6\\varepsilon\\,\\delta^{2}}{\\ell^{3}}, & \\eta\\ge\\delta .\n\\end{cases}}\n\\]\n\nStep 7.  Sanity checks.  \n(i)  $\\eta\\to0^{+}$ gives probability $0$, as expected.  \n(ii)  For $\\eta\\ge\\delta$ the diameter restriction alone governs the event, so the result is constant in $\\eta$; the formula indeed becomes independent of $\\eta$.  \n(iii)  $\\varepsilon\\to0^{+}$ or $\\delta\\to0^{+}$ forces the probability to $0$, which the expression reflects.  \n(iv)  Continuity at $\\eta=\\delta/2$ and $\\eta=\\delta$ follows from direct substitution.\n\nThe probability above is therefore correct and complete.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.443451",
        "was_fixed": false,
        "difficulty_analysis": "• More random points: The problem involves three independent points instead of two, increasing dimensionality from 2-D to 3-D in the sampling space.\n\n• Coupled geometric constraints:  \n  – a global diameter bound,  \n  – a constraint on the midpoint of the first two points,  \n  – and a proximity requirement between that midpoint and the third point.  \n  These interact non-trivially; one must recognise that some inequalities (the diameter bound) are automatically satisfied once the others are enforced.\n\n• Change of variables and Jacobian: Solving requires a non-obvious affine transformation (from (X₁ ,X₂) to (u ,s)) and careful handling of its Jacobian.\n\n• Region geometry in two parameters: One must determine the exact shape (here a rectangle) of the feasible (u ,s)-region after simultaneously imposing disparate linear constraints; this demands thoughtful use of the given inequality δ + 2ε + 2η ≤ ℓ.\n\n• Conditional probability inside a moving window: The probability for X₃ depends on u; showing that it becomes a constant 2η/ℓ requires proving that the window never touches the boundary, which in turn hinges on the interplay of all four parameters.\n\n• Multi-step integration: The final computation involves integrating a constant conditional probability over a two-dimensional region found via geometric reasoning and variable transformation.\n\nOverall, the enhanced variant raises the problem’s dimension, introduces multiple interacting inequalities, and obliges the solver to use advanced techniques from multivariable calculus and geometric probability, making it decidedly harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}