{
  "index": "1979-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "COMB"
  ],
  "difficulty": "",
  "question": "Problem A-6\nLet \\( 0<p_{i}<1 \\) for \\( i=1,2, \\ldots, n \\). Show that\n\\[\n\\sum_{i=1}^{n} \\frac{1}{\\left|x-p_{i}\\right|}<8 n\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 n-1}\\right)\n\\]\nfor some \\( x \\) satisfying \\( 0<x<1 \\).",
  "solution": "A-6.\nFor \\( k=0,1, \\ldots, 2 n-1 \\) let \\( I_{k} \\) be the open interval \\( (k / 2 n,[k+1] / 2 n) \\). Among the \\( 2 n \\) intervals \\( I_{k} \\) there exist \\( n \\) not containing any of the \\( p_{i} \\) and we place an \\( x_{j} \\) at the center of each of these \\( n \\) intervals. Let \\( \\left|x,-p_{i}\\right|=d_{i j} \\) and\n\\[\nB=8 n\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 n-1}\\right)\n\\]\n\nFor fixed \\( i \\), the \\( d_{i j} \\) satisfy \\( d_{i j} \\geqslant 1 / 4 n \\), at most two of them do not satisfy \\( d_{i j} \\geqslant 3 / 4 n \\), at most four do not satisfy \\( d_{i j} \\geqslant 5 / 4 n \\), etc. Hence\n\\[\n\\sum_{j=1}^{n} \\frac{1}{d_{1 j}} \\leqslant 2 \\sum_{h=0}^{n-1} \\frac{4 n}{1+2 h}=B\n\\]\n(This inequality can be improved.) Thus we have\n\\[\n\\sum_{j=1}^{n}\\left(\\sum_{i=1}^{n} \\frac{1}{d_{i j}}\\right)=\\sum_{i=1}^{n}\\left(\\sum_{j=1}^{n} \\frac{1}{d_{i j}}\\right) \\leqslant n B\n\\]\n\nSo clearly there is a value of \\( j \\) for which \\( \\sum_{i=1}^{n}\\left(1 / d_{i j}\\right) \\leqslant B \\) and the \\( x \\), for such a \\( j \\) can serve as the desired \\( x \\).",
  "vars": [
    "x",
    "x_j",
    "k",
    "p_i",
    "I_k",
    "d_ij",
    "d_1j",
    "h",
    "i",
    "j"
  ],
  "params": [
    "n",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "genericpoint",
        "x_j": "centerpoint",
        "k": "intervalindex",
        "p_i": "givenpoint",
        "I_k": "subinterval",
        "d_ij": "distanceij",
        "d_1j": "distance1j",
        "h": "radiusindex",
        "i": "pointindex",
        "j": "centerindex",
        "n": "totalcount",
        "B": "boundvalue"
      },
      "question": "Problem A-6\nLet \\( 0<givenpoint_{pointindex}<1 \\) for \\( pointindex=1,2, \\ldots, totalcount \\). Show that\n\\[\n\\sum_{pointindex=1}^{totalcount} \\frac{1}{\\left|genericpoint-givenpoint_{pointindex}\\right|}<8\\, totalcount\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2\\, totalcount-1}\\right)\n\\]\nfor some \\( genericpoint \\) satisfying \\( 0<genericpoint<1 \\).",
      "solution": "A-6.\nFor \\( intervalindex=0,1, \\ldots, 2\\, totalcount-1 \\) let \\( subinterval_{intervalindex} \\) be the open interval \\( (intervalindex / 2\\, totalcount,[intervalindex+1] / 2\\, totalcount) \\). Among the \\( 2\\, totalcount \\) intervals \\( subinterval_{intervalindex} \\) there exist \\( totalcount \\) not containing any of the \\( givenpoint_{pointindex} \\) and we place a \\( centerpoint_{centerindex} \\) at the center of each of these \\( totalcount \\) intervals. Let \\( \\left|centerpoint_{centerindex}-givenpoint_{pointindex}\\right|=distanceij \\) and\n\\[\nboundvalue=8\\, totalcount\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2\\, totalcount-1}\\right)\n\\]\n\nFor fixed \\( pointindex \\), the \\( distanceij \\) satisfy \\( distanceij \\geqslant 1 / 4\\, totalcount \\), at most two of them do not satisfy \\( distanceij \\geqslant 3 / 4\\, totalcount \\), at most four do not satisfy \\( distanceij \\geqslant 5 / 4\\, totalcount \\), etc. Hence\n\\[\n\\sum_{centerindex=1}^{totalcount} \\frac{1}{distance1j} \\leqslant 2 \\sum_{radiusindex=0}^{totalcount-1} \\frac{4\\, totalcount}{1+2\\, radiusindex}=boundvalue\n\\]\n(This inequality can be improved.) Thus we have\n\\[\n\\sum_{centerindex=1}^{totalcount}\\left(\\sum_{pointindex=1}^{totalcount} \\frac{1}{distanceij}\\right)=\\sum_{pointindex=1}^{totalcount}\\left(\\sum_{centerindex=1}^{totalcount} \\frac{1}{distanceij}\\right) \\leqslant totalcount\\, boundvalue\n\\]\n\nSo clearly there is a value of \\( centerindex \\) for which \\( \\sum_{pointindex=1}^{totalcount}\\left( \\frac{1}{distanceij}\\right) \\leqslant boundvalue \\) and the \\( genericpoint \\), for such a \\( centerindex \\) can serve as the desired \\( genericpoint \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanternfish",
        "x_j": "candlestick",
        "k": "breadcrumb",
        "p_i": "raincloud",
        "I_k": "sunflower",
        "d_ij": "horseshoe",
        "d_1j": "windchime",
        "h": "marshmallow",
        "i": "sailcloth",
        "j": "applecart",
        "n": "driftwood",
        "B": "toadstool"
      },
      "question": "Problem A-6\nLet \\( 0<raincloud<1 \\) for \\( sailcloth=1,2, \\ldots, driftwood \\). Show that\n\\[\n\\sum_{sailcloth=1}^{driftwood} \\frac{1}{\\left|lanternfish-raincloud\\right|}<8 driftwood\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 driftwood-1}\\right)\n\\]\nfor some \\( lanternfish \\) satisfying \\( 0<lanternfish<1 \\).",
      "solution": "A-6.\nFor \\( breadcrumb=0,1, \\ldots, 2 driftwood-1 \\) let \\( sunflower \\) be the open interval \\( (breadcrumb / 2 driftwood,[breadcrumb+1] / 2 driftwood) \\). Among the \\( 2 driftwood \\) intervals \\( sunflower \\) there exist \\( driftwood \\) not containing any of the \\( raincloud \\) and we place an \\( candlestick \\) at the center of each of these \\( driftwood \\) intervals. Let \\( \\left|lanternfish,-raincloud\\right|=horseshoe \\) and\n\\[\ntoadstool=8 driftwood\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 driftwood-1}\\right)\n\\]\n\nFor fixed \\( sailcloth \\), the \\( horseshoe \\) satisfy \\( horseshoe \\geqslant 1 / 4 driftwood \\), at most two of them do not satisfy \\( horseshoe \\geqslant 3 / 4 driftwood \\), at most four do not satisfy \\( horseshoe \\geqslant 5 / 4 driftwood \\), etc. Hence\n\\[\n\\sum_{applecart=1}^{driftwood} \\frac{1}{windchime} \\leqslant 2 \\sum_{marshmallow=0}^{driftwood-1} \\frac{4 driftwood}{1+2 marshmallow}=toadstool\n\\]\n(This inequality can be improved.) Thus we have\n\\[\n\\sum_{applecart=1}^{driftwood}\\left(\\sum_{sailcloth=1}^{driftwood} \\frac{1}{horseshoe}\\right)=\\sum_{sailcloth=1}^{driftwood}\\left(\\sum_{applecart=1}^{driftwood} \\frac{1}{horseshoe}\\right) \\leqslant driftwood\\ toadstool\n\\]\n\nSo clearly there is a value of \\( applecart \\) for which \\( \\sum_{sailcloth=1}^{driftwood}\\left(1 / horseshoe\\right) \\leqslant toadstool \\) and the \\( lanternfish \\), for such a \\( applecart \\) can serve as the desired \\( lanternfish \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "infinitepoint",
        "x_j": "edgepoint",
        "k": "wholeindex",
        "p_i": "boundarypoint",
        "I_k": "closedsegment",
        "d_ij": "closeness",
        "d_1j": "closenessrow",
        "h": "stillness",
        "i": "collectiveindex",
        "j": "totalityindex",
        "n": "endlessness",
        "B": "boundless"
      },
      "question": "Problem A-6\nLet \\( 0<boundarypoint<1 \\) for \\( collectiveindex=1,2, \\ldots, endlessness \\). Show that\n\\[\n\\sum_{collectiveindex=1}^{endlessness} \\frac{1}{\\left|infinitepoint-boundarypoint\\right|}<8 endlessness\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 endlessness-1}\\right)\n\\]\nfor some \\( infinitepoint \\) satisfying \\( 0<infinitepoint<1 \\).",
      "solution": "A-6.\nFor \\( wholeindex=0,1, \\ldots, 2 endlessness-1 \\) let \\( closedsegment \\) be the open interval \\( (wholeindex / 2 endlessness,[wholeindex+1] / 2 endlessness) \\). Among the \\( 2 endlessness \\) intervals \\( closedsegment \\) there exist \\( endlessness \\) not containing any of the \\( boundarypoint \\) and we place an \\( edgepoint \\) at the center of each of these \\( endlessness \\) intervals. Let \\( \\left|infinitepoint,-boundarypoint\\right|=closeness \\) and\n\\[\nboundless=8 endlessness\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2 endlessness-1}\\right)\n\\]\n\nFor fixed collectiveindex, the closeness satisfy \\( closeness \\geqslant 1 / 4 endlessness \\), at most two of them do not satisfy \\( closeness \\geqslant 3 / 4 endlessness \\), at most four do not satisfy \\( closeness \\geqslant 5 / 4 endlessness \\), etc. Hence\n\\[\n\\sum_{totalityindex=1}^{endlessness} \\frac{1}{closenessrow} \\leqslant 2 \\sum_{stillness=0}^{endlessness-1} \\frac{4 endlessness}{1+2 stillness}=boundless\n\\]\n(This inequality can be improved.) Thus we have\n\\[\n\\sum_{totalityindex=1}^{endlessness}\\left(\\sum_{collectiveindex=1}^{endlessness} \\frac{1}{closeness}\\right)=\\sum_{collectiveindex=1}^{endlessness}\\left(\\sum_{totalityindex=1}^{endlessness} \\frac{1}{closeness}\\right) \\leqslant endlessness\\, boundless\n\\]\n\nSo clearly there is a value of \\( totalityindex \\) for which \\( \\sum_{collectiveindex=1}^{endlessness}\\left(1 / closeness\\right) \\leqslant boundless \\) and the \\( infinitepoint \\), for such a \\( totalityindex \\) can serve as the desired \\( infinitepoint \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "x_j": "hjgrksla",
        "k": "mbdlcrye",
        "p_i": "snvmpqol",
        "I_k": "rcopljzn",
        "d_ij": "afvyqxms",
        "d_1j": "bpktwreh",
        "h": "tfodmnec",
        "i": "wzgkhrbu",
        "j": "yalpezqs",
        "n": "udvepran",
        "B": "gqzshxjw"
      },
      "question": "Problem A-6\nLet \\( 0<snvmpqol<1 \\) for \\( wzgkhrbu=1,2, \\ldots, udvepran \\). Show that\n\\[\n\\sum_{wzgkhrbu=1}^{udvepran} \\frac{1}{\\left|qzxwvtnp-snvmpqol\\right|}<8\\;udvepran\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2\\;udvepran-1}\\right)\n\\]\nfor some \\( qzxwvtnp \\) satisfying \\( 0<qzxwvtnp<1 \\).",
      "solution": "A-6.\nFor \\( mbdlcrye=0,1, \\ldots, 2 udvepran-1 \\) let \\( rcopljzn \\) be the open interval \\( (mbdlcrye / 2 udvepran,[mbdlcrye+1] / 2 udvepran) \\). Among the \\( 2 udvepran \\) intervals \\( rcopljzn \\) there exist \\( udvepran \\) not containing any of the \\( snvmpqol \\) and we place an \\( hjgrksla \\) at the center of each of these \\( udvepran \\) intervals. Let \\( \\left|qzxwvtnp,-snvmpqol\\right|=afvyqxms \\) and\n\\[\n gqzshxjw=8\\;udvepran\\left(1+\\frac{1}{3}+\\frac{1}{5}+\\cdots+\\frac{1}{2\\;udvepran-1}\\right)\n\\]\n\nFor fixed \\( wzgkhrbu \\), the \\( afvyqxms \\) satisfy \\( afvyqxms \\geqslant 1 / 4 udvepran \\), at most two of them do not satisfy \\( afvyqxms \\geqslant 3 / 4 udvepran \\), at most four do not satisfy \\( afvyqxms \\geqslant 5 / 4 udvepran \\), etc. Hence\n\\[\n\\sum_{yalpezqs=1}^{udvepran} \\frac{1}{bpktwreh} \\leqslant 2 \\sum_{tfodmnec=0}^{udvepran-1} \\frac{4 udvepran}{1+2 tfodmnec}=gqzshxjw\n\\]\n(This inequality can be improved.) Thus we have\n\\[\n\\sum_{yalpezqs=1}^{udvepran}\\left(\\sum_{wzgkhrbu=1}^{udvepran} \\frac{1}{afvyqxms}\\right)=\\sum_{wzgkhrbu=1}^{udvepran}\\left(\\sum_{yalpezqs=1}^{udvepran} \\frac{1}{afvyqxms}\\right) \\leqslant udvepran gqzshxjw\n\\]\n\nSo clearly there is a value of \\( yalpezqs \\) for which \\( \\sum_{wzgkhrbu=1}^{udvepran}\\left(1 / afvyqxms\\right) \\leqslant gqzshxjw \\) and the \\( qzxwvtnp \\), for such a \\( yalpezqs \\) can serve as the desired \\( qzxwvtnp \\)."
    },
    "kernel_variant": {
      "question": "Let n be a positive integer and let\n0 < p_{1} < p_{2} < \\dots < p_{n} < 1.\nProve that there exists a real number x with 0 < x < 1 such that\n\n\\[\n\\sum_{i=1}^{n}\\frac{1}{\\lvert x-p_{i}\\rvert}\n\\;<\\;18\\,n\\Bigl(1+\\tfrac14+\\tfrac17+\\cdots+\\tfrac1{3n-2}\\Bigr).\n\\]",
      "solution": "1. Splitting the unit interval.\n   Divide (0,1) into 3n equal open sub-intervals\n   I_k=(k/(3n),(k+1)/(3n)),  k=0,1,\\dots ,3n-1.\n   Because only n of them can contain the numbers p_i, at least 2n of the I_k are empty.\n   Choose any n empty intervals and denote them by J_1,\\dots ,J_n.\n\n2. Choosing the candidate points.\n   If J_j=(a,a+1/(3n)), set\n         x_j=a+1/(9n),\n   i.e. one–third of the way from the left–hand endpoint.\n   Then 0<x_j<1, no x_j coincides with a p_i, and each x_j is at distance at least 1/(9n) from the ends of its J_j.\n\n3. A distance estimate for a fixed p_i.\n   Fix i and let k(i) be the index such that p_i\\in I_{k(i)}.\n   For every positive integer s (s=1,2,\\dots) define the s-th distance band (relative to p_i) to consist of the two intervals\n         I_{k(i)-s}\\quad\\text{and}\\quad I_{k(i)+s}\n   whenever they lie in (0,1).  Each band therefore contains at most two of the J_j.\n\n   Put d_{ij}=|x_j-p_i|.  Assume x_j lies in I_{k(i)+s} with s\\ge 1 (the case on the left is analogous and gives an even larger distance).  Then\n   \\[\n       d_{ij}\\;\\ge\\;\\frac{k(i)+s}{3n}+\\frac{1}{9n}-\\frac{k(i)+1}{3n}\n              =\\frac{s-1}{3n}+\\frac{1}{9n}\n              =\\frac{3s-2}{9n}.\n   \\]\n   If x_j lies to the left, i.e. in I_{k(i)-s}, an almost identical calculation shows d_{ij}\\ge (3s-1)/(9n) \\ge (3s-2)/(9n).  Hence for every s\\ge 1\n   \\[\n        d_{ij}\\;\\ge\\;\\frac{3s-2}{9n}\\quad\\Longrightarrow\\quad\n        \\frac{1}{d_{ij}}\\;\\le\\;\\frac{9n}{3s-2}.  \\tag{1}\n   \\]\n\n4. Bounding the sum for a fixed i.\n   Order the distances d_{ij} for this fixed i in non-decreasing order:\n         d_{i(1)}\\le d_{i(2)}\\le \\dots \\le d_{i(n)}.\n   Because each band accommodates at most two of the points x_j, the t-th smallest distance must arise from a point situated in a band whose index is at most \\lceil t/2 \\rceil.  Consequently,\n   \\[\n       d_{i(t)} \\ge \\frac{3\\lceil t/2 \\rceil -2}{9n}\n       \\quad\\Longrightarrow\\quad\n       \\frac{1}{d_{i(t)}} \\le \\frac{9n}{3\\lceil t/2 \\rceil -2}.\n   \\]\n   Therefore\n   \\[\n       \\sum_{j=1}^{n}\\frac{1}{d_{ij}}\n       = \\sum_{t=1}^{n}\\frac{1}{d_{i(t)}}\n       \\le 2\\sum_{s=1}^{n}\\frac{9n}{3s-2}\n       =18n\\sum_{s=1}^{n}\\frac{1}{3s-2}\n       =18n\\Bigl(1+\\tfrac14+\\tfrac17+\\cdots+\\tfrac1{3n-2}\\Bigr).  \\tag{2}\n   \\]\n   (The factor 2 appears because each band can contribute for at most two indices t.)  Observe that we did not assert that all n points actually lie in the first n bands; rather, to obtain an upper bound we may imagine that they do, since moving any point farther away from p_i can only decrease its contribution 1/d_{ij}.\n\n5. Averaging over all i and choosing x.\n   Summing (2) over i=1,2,\\dots ,n gives\n   \\[\n       \\sum_{i=1}^{n}\\sum_{j=1}^{n}\\frac{1}{|x_j-p_i|}\n       \\;\\le\\;n\\,18n\\Bigl(1+\\tfrac14+\\tfrac17+\\cdots+\\tfrac1{3n-2}\\Bigr).\n   \\]\n   Hence the average of the inner sums is at most the right–hand side, so there exists an index j_0 with\n   \\[\n       \\sum_{i=1}^{n}\\frac{1}{|x_{j_0}-p_i|}\n       \\le 18n\\Bigl(1+\\tfrac14+\\tfrac17+\\cdots+\\tfrac1{3n-2}\\Bigr).\n   \\]\n   Taking x = x_{j_0} (which satisfies 0<x<1) completes the proof.",
      "_meta": {
        "core_steps": [
          "Partition (0,1) into 2n equal subintervals and invoke the pigeonhole principle to obtain n empty subintervals.",
          "Select one test-point x_j inside each empty subinterval (e.g. its midpoint).",
          "For a fixed p_i, count how many x_j lie in successive distance–bands of width 1∕(2n); use those counts to bound Σ_j 1/|x_j−p_i| by a constant multiple of the odd harmonic sum.",
          "Double–count the total Σ_i Σ_j 1/|x_j−p_i|, divide by n, and deduce the existence of at least one x_j whose individual sum meets the required bound."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Number of equal subintervals into which (0,1) is split (must be > n so that at least n are empty).",
            "original": "2n"
          },
          "slot2": {
            "description": "Exact location of the test-point chosen inside each empty subinterval.",
            "original": "center (midpoint)"
          },
          "slot3": {
            "description": "Lower-bound constant used for the smallest possible distance between a p_i and any chosen x_j.",
            "original": "1/(4n)"
          },
          "slot4": {
            "description": "Leading constant in the final bound B that comes from the coarse distance estimate.",
            "original": "8"
          },
          "slot5": {
            "description": "Use of odd-denominator harmonic sum arising from grouping distances in steps of width 1/(2n). A different but comparable grouping would change this series.",
            "original": "1 + 1/3 + ⋯ + 1/(2n−1)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}