{
  "index": "1984-B-6",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem B-6\nA sequence of convex polygons \\( \\left\\{P_{n}\\right\\}, n \\geqslant 0 \\), is defined inductively as follows. \\( P_{0} \\) is an equilateral triangle with sides of length 1 . Once \\( P_{n} \\) has been determined, its sides are trisected; the vertices of \\( P_{n+1} \\) are the interior trisection points of the sides of \\( P_{n} \\). Thus, \\( P_{n+1} \\) is obtained by cutting corners off \\( P_{n} \\), and \\( P_{n} \\) has \\( 3 \\cdot 2^{n} \\) sides. ( \\( P_{1} \\) is a regular hexagon with sides of length \\( 1 / 3 \\).)\n\nExpress \\( \\lim _{n \\rightarrow \\infty} \\operatorname{Area}\\left(P_{n}\\right) \\) in the form \\( \\sqrt{a} / b \\), where \\( a \\) and \\( b \\) are positive integers.",
  "solution": "B-6.\nSuppose that \\( \\vec{u} \\) and \\( \\vec{v} \\) are consecutive edges in \\( P_{n} \\). Then \\( \\vec{u} / 3,(\\vec{u}+\\vec{v}) / 3 \\), and \\( \\vec{v} / 3 \\) are consecutive edges in \\( P_{n+1} \\). Further,\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{u}}{3} \\times \\frac{\\vec{v}}{3}\\right\\|=\\frac{1}{18}\\|\\vec{u} \\times \\vec{v}\\|\n\\]\nis removed at this corner in making \\( P_{n+1} \\). But at the next step, the amount from these three consecutive edges is\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{u}}{9} \\times \\frac{\\vec{u}+\\vec{v}}{9}\\right\\|+\\frac{1}{2}\\left\\|\\frac{\\vec{u}+\\vec{v}}{9} \\times \\frac{\\vec{v}}{9}\\right\\|=\\frac{1}{81}\\|\\vec{u} \\times \\vec{v}\\| .\n\\]\n\nThus, the amount removed in the \\( (k+1) \\) st snip is \\( 2 / 9 \\) times the amount removed in the \\( k \\) th.\nNote that one-third of the original area is removed at the first step. Thus, the amount removed altogether is\n\\[\n\\frac{1}{3}\\left[1+(2 / 9)+(2 / 9)^{2}+\\cdots\\right]=\\frac{1}{3} \\cdot \\frac{9}{7}=\\frac{3}{7}\n\\]\nof the original area. Since the original area is \\( \\sqrt{3} / 4 \\), we have\n\\[\n\\lim _{n \\rightarrow \\infty} \\text { Area } P_{n}=\\frac{4}{7} \\cdot \\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{7} .\n\\]\n\nThe curve in this problem has been studied extensively by Georges de Rham. (See \"Un peu de mathématiques à propos d'une courbe plane,\" Elem. Math., 2 (1947), 73-76, 89-97; \"Sur une courbe plane,\" J. Math. Pures Appl., 35 (1956), 25-42; and \"Sur les courbes limites de polygones obtenus par trisection,\" Enseign. Math., 5 (1959), 29-43.) Among de Rham's results are the following. The limiting curve is \\( C^{1} \\) with zero curvature almost everywhere, but every subarc contains points where the curvature is infinite. Consequently, the curve is nowhere analytic. De Rham parametrizes pieces of the curve so that the tangent vector is intimately related to the Minkowski ?-function. If the construction is repeated, but with each edge divided in the ratio \\( (1 / 4,1 / 2,1 / 4) \\) rather than \\( (1 / 3,1 / 3,1 / 3) \\), then the resulting limit curve is analytic, consisting of piecewise parabolic arcs.",
  "vars": [
    "P_0",
    "P_1",
    "P_n",
    "P_n+1",
    "k",
    "n",
    "u",
    "v"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P_0": "initpoly",
        "P_1": "firstpoly",
        "P_n": "nthpoly",
        "P_n+1": "followingpoly",
        "k": "snipindex",
        "n": "stepindex",
        "u": "firstedge",
        "v": "secondedge",
        "a": "coeffa",
        "b": "coeffb"
      },
      "question": "Problem B-6\nA sequence of convex polygons \\( \\left\\{nthpoly\\right\\}, stepindex \\geqslant 0 \\), is defined inductively as follows. \\( initpoly \\) is an equilateral triangle with sides of length 1 . Once \\( nthpoly \\) has been determined, its sides are trisected; the vertices of \\( followingpoly \\) are the interior trisection points of the sides of \\( nthpoly \\). Thus, \\( followingpoly \\) is obtained by cutting corners off \\( nthpoly \\), and \\( nthpoly \\) has \\( 3 \\cdot 2^{stepindex} \\) sides. ( \\( firstpoly \\) is a regular hexagon with sides of length \\( 1 / 3 \\).)\n\nExpress \\( \\lim _{stepindex \\rightarrow \\infty} \\operatorname{Area}\\left(nthpoly\\right) \\) in the form \\( \\sqrt{coeffa} / coeffb \\), where \\( coeffa \\) and \\( coeffb \\) are positive integers.",
      "solution": "B-6.\nSuppose that \\( \\vec{firstedge} \\) and \\( \\vec{secondedge} \\) are consecutive edges in \\( nthpoly \\). Then \\( \\vec{firstedge} / 3,(\\vec{firstedge}+\\vec{secondedge}) / 3 \\), and \\( \\vec{secondedge} / 3 \\) are consecutive edges in \\( followingpoly \\). Further,\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{firstedge}}{3} \\times \\frac{\\vec{secondedge}}{3}\\right\\|=\\frac{1}{18}\\|\\vec{firstedge} \\times \\vec{secondedge}\\|\n\\]\nis removed at this corner in making \\( followingpoly \\). But at the next step, the amount from these three consecutive edges is\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{firstedge}}{9} \\times \\frac{\\vec{firstedge}+\\vec{secondedge}}{9}\\right\\|+\\frac{1}{2}\\left\\|\\frac{\\vec{firstedge}+\\vec{secondedge}}{9} \\times \\frac{\\vec{secondedge}}{9}\\right\\|=\\frac{1}{81}\\|\\vec{firstedge} \\times \\vec{secondedge}\\| .\n\\]\n\nThus, the amount removed in the \\( (snipindex+1) \\) st snip is \\( 2 / 9 \\) times the amount removed in the \\( snipindex \\) th.\nNote that one-third of the original area is removed at the first step. Thus, the amount removed altogether is\n\\[\n\\frac{1}{3}\\left[1+(2 / 9)+(2 / 9)^{2}+\\cdots\\right]=\\frac{1}{3} \\cdot \\frac{9}{7}=\\frac{3}{7}\n\\]\nof the original area. Since the original area is \\( \\sqrt{3} / 4 \\), we have\n\\[\n\\lim _{stepindex \\rightarrow \\infty} \\text { Area } nthpoly=\\frac{4}{7} \\cdot \\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{7} .\n\\]\n\nThe curve in this problem has been studied extensively by Georges de Rham. (See \"Un peu de mathématiques à propos d'une courbe plane,\" Elem. Math., 2 (1947), 73-76, 89-97; \"Sur une courbe plane,\" J. Math. Pures Appl., 35 (1956), 25-42; and \"Sur les courbes limites de polygones obtenus par trisection,\" Enseign. Math., 5 (1959), 29-43.) Among de Rham's results are the following. The limiting curve is \\( C^{1} \\) with zero curvature almost everywhere, but every subarc contains points where the curvature is infinite. Consequently, the curve is nowhere analytic. De Rham parametrizes pieces of the curve so that the tangent vector is intimately related to the Minkowski ?-function. If the construction is repeated, but with each edge divided in the ratio \\( (1 / 4,1 / 2,1 / 4) \\) rather than \\( (1 / 3,1 / 3,1 / 3) \\), then the resulting limit curve is analytic, consisting of piecewise parabolic arcs."
    },
    "descriptive_long_confusing": {
      "map": {
        "P_0": "pendulum",
        "P_1": "altitude",
        "P_n": "astronomy",
        "P_n+1": "hurricane",
        "k": "backpack",
        "n": "lawnchair",
        "u": "magnetism",
        "v": "notebook",
        "a": "pineapple",
        "b": "strawberry"
      },
      "question": "Problem B-6\nA sequence of convex polygons \\( \\left\\{astronomy_{lawnchair}\\right\\}, lawnchair \\geqslant 0 \\), is defined inductively as follows. \\( pendulum_{0} \\) is an equilateral triangle with sides of length 1 . Once \\( astronomy_{lawnchair} \\) has been determined, its sides are trisected; the vertices of \\( hurricane_{lawnchair+1} \\) are the interior trisection points of the sides of \\( astronomy_{lawnchair} \\). Thus, \\( hurricane_{lawnchair+1} \\) is obtained by cutting corners off \\( astronomy_{lawnchair} \\), and \\( astronomy_{lawnchair} \\) has \\( 3 \\cdot 2^{lawnchair} \\) sides. ( \\( altitude_{1} \\) is a regular hexagon with sides of length \\( 1 / 3 \\).)\n\nExpress \\( \\lim _{lawnchair \\rightarrow \\infty} \\operatorname{Area}\\left(astronomy_{lawnchair}\\right) \\) in the form \\( \\sqrt{pineapple} / strawberry \\), where \\( pineapple \\) and \\( strawberry \\) are positive integers.",
      "solution": "Suppose that \\( \\vec{magnetism} \\) and \\( \\vec{notebook} \\) are consecutive edges in \\( astronomy_{lawnchair} \\). Then \\( \\vec{magnetism} / 3,(\\vec{magnetism}+\\vec{notebook}) / 3 \\), and \\( \\vec{notebook} / 3 \\) are consecutive edges in \\( hurricane_{lawnchair+1} \\). Further,\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{magnetism}}{3} \\times \\frac{\\vec{notebook}}{3}\\right\\|=\\frac{1}{18}\\|\\vec{magnetism} \\times \\vec{notebook}\\|\n\\]\nis removed at this corner in making \\( hurricane_{lawnchair+1} \\). But at the next step, the amount from these three consecutive edges is\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{magnetism}}{9} \\times \\frac{\\vec{magnetism}+\\vec{notebook}}{9}\\right\\|+\\frac{1}{2}\\left\\|\\frac{\\vec{magnetism}+\\vec{notebook}}{9} \\times \\frac{\\vec{notebook}}{9}\\right\\|=\\frac{1}{81}\\|\\vec{magnetism} \\times \\vec{notebook}\\| .\n\\]\n\nThus, the amount removed in the \\( (backpack+1) \\) st snip is \\( 2 / 9 \\) times the amount removed in the \\( backpack \\) th.\nNote that one-third of the original area is removed at the first step. Thus, the amount removed altogether is\n\\[\n\\frac{1}{3}\\left[1+(2 / 9)+(2 / 9)^{2}+\\cdots\\right]=\\frac{1}{3} \\cdot \\frac{9}{7}=\\frac{3}{7}\n\\]\nof the original area. Since the original area is \\( \\sqrt{3} / 4 \\), we have\n\\[\n\\lim _{lawnchair \\rightarrow \\infty} \\text { Area } astronomy_{lawnchair}=\\frac{4}{7} \\cdot \\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{7} .\n\\]\n\nThe curve in this problem has been studied extensively by Georges de Rham. (See \"Un peu de mathématiques à propos d'une courbe plane,\" Elem. Math., 2 (1947), 73-76, 89-97; \"Sur une courbe plane,\" J. Math. Pures Appl., 35 (1956), 25-42; and \"Sur les courbes limites de polygones obtenus par trisection,\" Enseign. Math., 5 (1959), 29-43.) Among de Rham's results are the following. The limiting curve is \\( C^{1} \\) with zero curvature almost everywhere, but every subarc contains points where the curvature is infinite. Consequently, the curve is nowhere analytic. De Rham parametrizes pieces of the curve so that the tangent vector is intimately related to the Minkowski ?-function. If the construction is repeated, but with each edge divided in the ratio \\( (1 / 4,1 / 2,1 / 4) \\) rather than \\( (1 / 3,1 / 3,1 / 3) \\), then the resulting limit curve is analytic, consisting of piecewise parabolic arcs."
    },
    "descriptive_long_misleading": {
      "map": {
        "P_0": "finalshape",
        "P_1": "emptypoly",
        "P_n": "constantshape",
        "P_n+1": "previousshape",
        "k": "steadystate",
        "n": "infinitevalue",
        "u": "scalarlength",
        "v": "scalarbreadth",
        "a": "knownsum",
        "b": "multiprod"
      },
      "question": "Problem B-6\nA sequence of convex polygons \\( \\left\\{constantshape\\right\\}, infinitevalue \\geqslant 0 \\), is defined inductively as follows. \\( finalshape \\) is an equilateral triangle with sides of length 1 . Once \\( constantshape \\) has been determined, its sides are trisected; the vertices of \\( previousshape \\) are the interior trisection points of the sides of \\( constantshape \\). Thus, \\( previousshape \\) is obtained by cutting corners off \\( constantshape \\), and \\( constantshape \\) has \\( 3 \\cdot 2^{infinitevalue} \\) sides. ( \\( emptypoly \\) is a regular hexagon with sides of length \\( 1 / 3 \\).)\n\nExpress \\( \\lim _{infinitevalue \\rightarrow \\infty} \\operatorname{Area}\\left(constantshape\\right) \\) in the form \\( \\sqrt{knownsum} / multiprod \\), where \\( knownsum \\) and \\( multiprod \\) are positive integers.",
      "solution": "B-6.\nSuppose that \\( \\vec{scalarlength} \\) and \\( \\vec{scalarbreadth} \\) are consecutive edges in \\( constantshape \\). Then \\( \\vec{scalarlength} / 3,(\\vec{scalarlength}+\\vec{scalarbreadth}) / 3 \\), and \\( \\vec{scalarbreadth} / 3 \\) are consecutive edges in \\( previousshape \\). Further,\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{scalarlength}}{3} \\times \\frac{\\vec{scalarbreadth}}{3}\\right\\|=\\frac{1}{18}\\|\\vec{scalarlength} \\times \\vec{scalarbreadth}\\|\n\\]\nis removed at this corner in making \\( previousshape \\). But at the next step, the amount from these three consecutive edges is\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{scalarlength}}{9} \\times \\frac{\\vec{scalarlength}+\\vec{scalarbreadth}}{9}\\right\\|+\\frac{1}{2}\\left\\|\\frac{\\vec{scalarlength}+\\vec{scalarbreadth}}{9} \\times \\frac{\\vec{scalarbreadth}}{9}\\right\\|=\\frac{1}{81}\\|\\vec{scalarlength} \\times \\vec{scalarbreadth}\\| .\n\\]\n\nThus, the amount removed in the \\( (steadystate+1) \\) st snip is \\( 2 / 9 \\) times the amount removed in the \\( steadystate \\) th.\nNote that one-third of the original area is removed at the first step. Thus, the amount removed altogether is\n\\[\n\\frac{1}{3}\\left[1+(2 / 9)+(2 / 9)^{2}+\\cdots\\right]=\\frac{1}{3} \\cdot \\frac{9}{7}=\\frac{3}{7}\n\\]\nof the original area. Since the original area is \\( \\sqrt{3} / 4 \\), we have\n\\[\n\\lim _{infinitevalue \\rightarrow \\infty} \\text { Area } constantshape=\\frac{4}{7} \\cdot \\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{7} .\n\\]\n\nThe curve in this problem has been studied extensively by Georges de Rham. (See \"Un peu de mathématiques à propos d'une courbe plane,\" Elem. Math., 2 (1947), 73-76, 89-97; \"Sur une courbe plane,\" J. Math. Pures Appl., 35 (1956), 25-42; and \"Sur les courbes limites de polygones obtenus par trisection,\" Enseign. Math., 5 (1959), 29-43.) Among de Rham's results are the following. The limiting curve is \\( C^{1} \\) with zero curvature almost everywhere, but every subarc contains points where the curvature is infinite. Consequently, the curve is nowhere analytic. De Rham parametrizes pieces of the curve so that the tangent vector is intimately related to the Minkowski ?-function. If the construction is repeated, but with each edge divided in the ratio \\( (1 / 4,1 / 2,1 / 4) \\) rather than \\( (1 / 3,1 / 3,1 / 3) \\), then the resulting limit curve is analytic, consisting of piecewise parabolic arcs."
    },
    "garbled_string": {
      "map": {
        "P_0": "qzxwvtnp",
        "P_1": "hjgrksla",
        "P_n": "vrbdkqwe",
        "P_n+1": "tmskzlqa",
        "k": "fglhrnwe",
        "n": "sdpqjvka",
        "u": "wzcnpyar",
        "v": "kjlmqtsa",
        "a": "frsctgwd",
        "b": "nxmqzvtu"
      },
      "question": "Problem B-6\nA sequence of convex polygons \\( \\left\\{vrbdkqwe\\right\\}, sdpqjvka \\geqslant 0 \\), is defined inductively as follows. \\( qzxwvtnp \\) is an equilateral triangle with sides of length 1 . Once \\( vrbdkqwe \\) has been determined, its sides are trisected; the vertices of \\( tmskzlqa \\) are the interior trisection points of the sides of \\( vrbdkqwe \\). Thus, \\( tmskzlqa \\) is obtained by cutting corners off \\( vrbdkqwe \\), and \\( vrbdkqwe \\) has \\( 3 \\cdot 2^{sdpqjvka} \\) sides. ( \\( hjgrksla \\) is a regular hexagon with sides of length \\( 1 / 3 \\).)\n\nExpress \\( \\lim _{sdpqjvka \\rightarrow \\infty} \\operatorname{Area}\\left(vrbdkqwe\\right) \\) in the form \\( \\sqrt{frsctgwd} / nxmqzvtu \\), where \\( frsctgwd \\) and \\( nxmqzvtu \\) are positive integers.",
      "solution": "B-6.\nSuppose that \\( \\vec{wzcnpyar} \\) and \\( \\vec{kjlmqtsa} \\) are consecutive edges in \\( vrbdkqwe \\). Then \\( \\vec{wzcnpyar} / 3,(\\vec{wzcnpyar}+\\vec{kjlmqtsa}) / 3 \\), and \\( \\vec{kjlmqtsa} / 3 \\) are consecutive edges in \\( tmskzlqa \\). Further,\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{wzcnpyar}}{3} \\times \\frac{\\vec{kjlmqtsa}}{3}\\right\\|=\\frac{1}{18}\\|\\vec{wzcnpyar} \\times \\vec{kjlmqtsa}\\|\n\\]\nis removed at this corner in making \\( tmskzlqa \\). But at the next step, the amount from these three consecutive edges is\n\\[\n\\frac{1}{2}\\left\\|\\frac{\\vec{wzcnpyar}}{9} \\times \\frac{\\vec{wzcnpyar}+\\vec{kjlmqtsa}}{9}\\right\\|+\\frac{1}{2}\\left\\|\\frac{\\vec{wzcnpyar}+\\vec{kjlmqtsa}}{9} \\times \\frac{\\vec{kjlmqtsa}}{9}\\right\\|=\\frac{1}{81}\\|\\vec{wzcnpyar} \\times \\vec{kjlmqtsa}\\| .\n\\]\n\nThus, the amount removed in the \\( (fglhrnwe+1) \\) st snip is \\( 2 / 9 \\) times the amount removed in the \\( fglhrnwe \\) th.\nNote that one-third of the original area is removed at the first step. Thus, the amount removed altogether is\n\\[\n\\frac{1}{3}\\left[1+(2 / 9)+(2 / 9)^{2}+\\cdots\\right]=\\frac{1}{3} \\cdot \\frac{9}{7}=\\frac{3}{7}\n\\]\nof the original area. Since the original area is \\( \\sqrt{3} / 4 \\), we have\n\\[\n\\lim _{sdpqjvka \\rightarrow \\infty} \\text { Area } vrbdkqwe=\\frac{4}{7} \\cdot \\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{7} .\n\\]\n\nThe curve in this problem has been studied extensively by Georges de Rham. (See \"Un peu de mathématiques à propos d'une courbe plane,\" Elem. Math., 2 (1947), 73-76, 89-97; \"Sur une courbe plane,\" J. Math. Pures Appl., 35 (1956), 25-42; and \"Sur les courbes limites de polygones obtenus par trisection,\" Enseign. Math., 5 (1959), 29-43.) Among de Rham's results are the following. The limiting curve is \\( C^{1} \\) with zero curvature almost everywhere, but every subarc contains points where the curvature is infinite. Consequently, the curve is nowhere analytic. De Rham parametrizes pieces of the curve so that the tangent vector is intimately related to the Minkowski ?-function. If the construction is repeated, but with each edge divided in the ratio \\( (1 / 4,1 / 2,1 / 4) \\) rather than \\( (1 / 3,1 / 3,1 / 3) \\), then the resulting limit curve is analytic, consisting of piecewise parabolic arcs."
    },
    "kernel_variant": {
      "question": "────────────────────────\nLet $(Q_{n})_{n\\ge 0}$ be a sequence of convex polyhedra constructed as follows.  \n\n• $Q_{0}$ is the unit cube, so ${\\rm Vol}(Q_{0})=1$.  \n• Assume $Q_{n}$ is known.  On every edge of $Q_{n}$ mark the points situated one–quarter and three–quarters of the way from one end; equivalently each edge is internally divided in the ratio $1:2:1$.  Define $Q_{n+1}$ to be the convex hull of all these interior division points.  (Geometrically $Q_{n+1}$ is obtained from $Q_{n}$ by slicing off every trihedral corner.)\n\nEvaluate\n\\[\n\\lim_{n\\to\\infty}{\\rm Vol}(Q_{n})\n\\]\nand present your answer in the form $\\displaystyle\\frac{\\sqrt{a}}{b}$ with positive integers $a$ and $b$.\n\n────────────────────────",
      "solution": "────────────────────────\nWe keep track of the volume eliminated at one fixed corner through the successive snips.\n\nNotation.  \nAt some stage let the outward‐pointing, pairwise independent edge–vectors of the corner be $\\vec a,\\vec b,\\vec c$.  Then\n\\[\n\\frac16\\,|\\vec a\\cdot(\\vec b\\times\\vec c)|\n\\]\nis the volume of the orthogonal tetrahedron they span; throughout we write $\\Delta:=|\\vec a\\cdot(\\vec b\\times\\vec c)|$.\n\nStep 1 – the very first cut  \nThe points $\\tfrac14\\vec a,\\tfrac14\\vec b,\\tfrac14\\vec c$ determine the slicing plane, so the excised tetrahedron has edge–vectors $\\vec a/4,\\vec b/4,\\vec c/4$.  Hence\n\\[\nV_{1}\\;=\\;\\frac16\\Bigl|\\frac{\\vec a}{4}\\!\\cdot\\!\\Bigl(\\frac{\\vec b}{4}\\times\\frac{\\vec c}{4}\\Bigr)\\Bigr|\n         =\\frac{\\Delta}{384}.\n\\]\n\nStep 2 – the three grandchildren of that corner  \nAfter the first snip the original vertex is replaced by the three points\n$\\vec a/4,\\vec b/4,\\vec c/4$.  Concentrate on $A:=\\vec a/4$; the incident edge–vectors there are\n\\[\n\\frac{\\vec a}{2},\\qquad\\frac{\\vec b-\\vec a}{4},\\qquad\\frac{\\vec c-\\vec a}{4}.\n\\]\nDividing each of these new edges again in the ratio $1:2:1$ and removing the tiny tetrahedron they span erases the volume\n\\[\n\\frac16\\Bigl|\\frac{\\vec a}{8}\\!\\cdot\\!\\Bigl(\\frac{\\vec b-\\vec a}{16}\\times\\frac{\\vec c-\\vec a}{16}\\Bigr)\\Bigr|\n           =\\frac{\\Delta}{12\\,288}.\n\\]\n(The identity $\\vec a\\cdot((\\vec b-\\vec a)\\times(\\vec c-\\vec a))=\\Delta$ is immediate because determinants with repeated columns vanish.)  \nBy symmetry the same amount disappears at the two sibling vertices $\\vec b/4$ and $\\vec c/4$, so altogether\n\\[\nV_{2}\\;=\\;3\\cdot\\frac{\\Delta}{12\\,288}=\\frac{3}{32}\\,V_{1}.\n\\]\n\nInductive step – the constant ratio $3/32$  \nSuppose an ancestral corner burns off the volume $V$.  Its three children sit on the quarter–points of the parental edges, and the computation carried out above shows that, regardless of the specific vectors, the three grandchildren together delete exactly a factor\n\\[\n\\frac{3}{32}\n\\]\nof $V$.  Consequently the infinite geometric series of losses from the initial cube corner is\n\\[\nV_{1}\\Bigl(1+\\tfrac{3}{32}+\\bigl(\\tfrac{3}{32}\\bigr)^{2}+\\dots\\Bigr)\n      =\\frac{V_{1}}{1-\\tfrac{3}{32}}\n      =\\frac{\\Delta}{384}\\cdot\\frac{32}{29}\n      =\\frac{\\Delta}{348}.\n\\]\n\nApplication to the unit cube  \nFor a cube corner choose $\\vec a=\\langle1,0,0\\rangle$, $\\vec b=\\langle0,1,0\\rangle$, $\\vec c=\\langle0,0,1\\rangle$; then $\\Delta=1$.  Hence one corner ultimately loses $1/348$ of volume, and all eight corners together lose\n\\[\n8\\cdot\\frac1{348}=\\frac2{87}.\n\\]\nIt follows that\n\\[\n\\lim_{n\\to\\infty}{\\rm Vol}(Q_{n})\n        =1-\\frac2{87}\n        =\\frac{85}{87}\n        =\\frac{\\sqrt{7225}}{87}.\n\\]\n\nAnswer.  \n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}{\\rm Vol}(Q_{n})=\\frac{\\sqrt{7225}}{87}}.\n\\]\n\n────────────────────────",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.081563",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}