{
  "index": "1992-B-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "the set of numbers that occur as averages of two distinct elements of\n$S$. For a given $n \\geq 2$, what is the smallest possible number of\nelements in $A_S$?",
  "solution": "Solution. Let \\( x_{1}<x_{2}<\\cdots<x_{n} \\) represent the elements of \\( S \\). Then\n\\[\n\\frac{x_{1}+x_{2}}{2}<\\frac{x_{1}+x_{3}}{2}<\\cdots<\\frac{x_{1}+x_{n}}{2}<\\frac{x_{2}+x_{n}}{2}<\\frac{x_{3}+x_{n}}{2}<\\cdots<\\frac{x_{n-1}+x_{n}}{2}\n\\]\nrepresent \\( 2 n-3 \\) distinct elements of \\( A_{S} \\), so \\( A_{S} \\) has at least \\( 2 n-3 \\) distinct elements.\nOn the other hand, if we take \\( S=\\{1,2, \\ldots, n\\} \\), the elements of \\( A_{S} \\) are \\( \\frac{3}{2}, \\frac{4}{2}, \\frac{5}{2} \\), \\( \\ldots, \\frac{2 n-1}{2} \\). There are only \\( 2 n-3 \\) such numbers; thus there is a set \\( A_{S} \\) with at most \\( 2 n-3 \\) distinct elements.",
  "vars": [
    "x_1",
    "x_2",
    "x_3",
    "x_n-1",
    "x_n",
    "S",
    "A_S"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x_1": "elemone",
        "x_2": "elemtwo",
        "x_3": "elemthree",
        "x_n-1": "elempenult",
        "x_n": "elemfinal",
        "S": "baselist",
        "A_S": "averagelist",
        "n": "elemscount"
      },
      "question": "the set of numbers that occur as averages of two distinct elements of $baselist$. For a given $elemscount \\geq 2$, what is the smallest possible number of elements in $averagelist$?",
      "solution": "Solution. Let \\( elemone<elemtwo<\\cdots<elemfinal \\) represent the elements of \\( baselist \\). Then\n\\[\n\\frac{elemone+elemtwo}{2}<\\frac{elemone+elemthree}{2}<\\cdots<\\frac{elemone+elemfinal}{2}<\\frac{elemtwo+elemfinal}{2}<\\frac{elemthree+elemfinal}{2}<\\cdots<\\frac{elempenult+elemfinal}{2}\n\\]\nrepresent \\( 2\\,elemscount-3 \\) distinct elements of \\( averagelist \\), so \\( averagelist \\) has at least \\( 2\\,elemscount-3 \\) distinct elements.\nOn the other hand, if we take \\( baselist=\\{1,2, \\ldots, elemscount\\} \\), the elements of \\( averagelist \\) are \\( \\frac{3}{2}, \\frac{4}{2}, \\frac{5}{2} \\), \\( \\ldots, \\frac{2\\,elemscount-1}{2} \\). There are only \\( 2\\,elemscount-3 \\) such numbers; thus there is a set \\( averagelist \\) with at most \\( 2\\,elemscount-3 \\) distinct elements."
    },
    "descriptive_long_confusing": {
      "map": {
        "x_1": "pebblestone",
        "x_2": "marshmallow",
        "x_3": "gingerbread",
        "x_{n-1}": "dandelion",
        "x_n": "butterscotch",
        "S": "tumbleweed",
        "A_S": "chandelier",
        "n": "peppercorn"
      },
      "question": "the set of numbers that occur as averages of two distinct elements of\n$tumbleweed$. For a given $peppercorn \\geq 2$, what is the smallest possible number of\nelements in $chandelier$?",
      "solution": "Solution. Let \\( pebblestone<marshmallow<\\cdots<butterscotch \\) represent the elements of \\( tumbleweed \\). Then\n\\[\n\\frac{pebblestone+marshmallow}{2}<\\frac{pebblestone+gingerbread}{2}<\\cdots<\\frac{pebblestone+butterscotch}{2}<\\frac{marshmallow+butterscotch}{2}<\\frac{gingerbread+butterscotch}{2}<\\cdots<\\frac{dandelion+butterscotch}{2}\n\\]\nrepresent \\( 2 peppercorn-3 \\) distinct elements of \\( chandelier \\), so \\( chandelier \\) has at least \\( 2 peppercorn-3 \\) distinct elements.\nOn the other hand, if we take \\( tumbleweed=\\{1,2, \\ldots, peppercorn\\} \\), the elements of \\( chandelier \\) are \\( \\frac{3}{2}, \\frac{4}{2}, \\frac{5}{2} \\), \\( \\ldots, \\frac{2 peppercorn-1}{2} \\). There are only \\( 2 peppercorn-3 \\) such numbers; thus there is a set \\( chandelier \\) with at most \\( 2 peppercorn-3 \\) distinct elements."
    },
    "descriptive_long_misleading": {
      "map": {
        "x_1": "finalvalue",
        "x_2": "latervalue",
        "x_3": "subsequent",
        "x_n-1": "earlyvalue",
        "x_n": "initialvalue",
        "S": "voidcollection",
        "A_S": "sumcollection",
        "n": "singularity"
      },
      "question": "the set of numbers that occur as averages of two distinct elements of\n$voidcollection$. For a given $singularity \\geq 2$, what is the smallest possible number of\nelements in $sumcollection$?",
      "solution": "Solution. Let \\( finalvalue<latervalue<\\cdots<initialvalue \\) represent the elements of \\( voidcollection \\). Then\n\\[\n\\frac{finalvalue+latervalue}{2}<\\frac{finalvalue+subsequent}{2}<\\cdots<\\frac{finalvalue+initialvalue}{2}<\\frac{latervalue+initialvalue}{2}<\\frac{subsequent+initialvalue}{2}<\\cdots<\\frac{earlyvalue+initialvalue}{2}\n\\]\nrepresent \\( 2 singularity-3 \\) distinct elements of \\( sumcollection \\), so \\( sumcollection \\) has at least \\( 2 singularity-3 \\) distinct elements.\nOn the other hand, if we take \\( voidcollection=\\{1,2, \\ldots, singularity\\} \\), the elements of \\( sumcollection \\) are \\( \\frac{3}{2}, \\frac{4}{2}, \\frac{5}{2} \\), \\( \\ldots, \\frac{2 singularity-1}{2} \\). There are only \\( 2 singularity-3 \\) such numbers; thus there is a set \\( sumcollection \\) with at most \\( 2 singularity-3 \\) distinct elements."
    },
    "garbled_string": {
      "map": {
        "x_1": "qzxwvtnp",
        "x_2": "hjgrksla",
        "x_3": "mlfdngye",
        "x_n-1": "bfztsqwe",
        "x_n": "vrhjkplm",
        "S": "ctwlmnop",
        "A_S": "dkrbvxyz",
        "n": "pldrfqae"
      },
      "question": "the set of numbers that occur as averages of two distinct elements of\n$ctwlmnop$. For a given $pldrfqae \\geq 2$, what is the smallest possible number of\nelements in $dkrbvxyz$?",
      "solution": "Solution. Let \\( qzxwvtnp_{1}<hjgrksla_{2}<\\cdots<vrhjkplm_{pldrfqae} \\) represent the elements of \\( ctwlmnop \\). Then\n\\[\n\\frac{qzxwvtnp_{1}+hjgrksla_{2}}{2}<\\frac{qzxwvtnp_{1}+mlfdngye_{3}}{2}<\\cdots<\\frac{qzxwvtnp_{1}+vrhjkplm_{pldrfqae}}{2}<\\frac{hjgrksla_{2}+vrhjkplm_{pldrfqae}}{2}<\\frac{mlfdngye_{3}+vrhjkplm_{pldrfqae}}{2}<\\cdots<\\frac{bfztsqwe_{pldrfqae-1}+vrhjkplm_{pldrfqae}}{2}\n\\]\nrepresent \\( 2 pldrfqae-3 \\) distinct elements of \\( dkrbvxyz \\), so \\( dkrbvxyz \\) has at least \\( 2 pldrfqae-3 \\) distinct elements.\nOn the other hand, if we take \\( ctwlmnop=\\{1,2, \\ldots, pldrfqae\\} \\), the elements of \\( dkrbvxyz \\) are \\( \\frac{3}{2}, \\frac{4}{2}, \\frac{5}{2} \\), \\( \\ldots, \\frac{2 pldrfqae-1}{2} \\). There are only \\( 2 pldrfqae-3 \\) such numbers; thus there is a set \\( dkrbvxyz \\) with at most \\( 2 pldrfqae-3 \\) distinct elements."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer.  For a finite set $S$ of $n$ distinct real numbers define\\[M_S=\\Bigl\\{\\frac{s_i+s_j}{2}\\;:\\;s_i,s_j\\in S,\\;i\\ne j\\Bigr\\}\\]to be the collection of all mid-points of unordered pairs of elements of $S$.  Determine, as an explicit function of $n$, the smallest possible value of $|M_S|$.",
      "solution": "Write the elements of S in increasing order x₁<x₂<⋯<xₙ. We exhibit 2n−3 distinct midpoints in M_S and then give an example attaining that size.\n\n1. (Lower bound) Consider the “first block” of n−1 midpoints  \n   A_i = (x₁ + x_{n−i+1})/2,  i=1,2,…,n−1;  \nand the “second block” of n−2 midpoints  \n   B_j = (x_{j+1} + xₙ)/2,  j=1,2,…,n−2.\n\n  • Within the first block, as i increases from 1 to n−1, x_{n−i+1} decreases from xₙ down to x₂, so A₁ > A₂ > ⋯ > A_{n−1}. Thus the A_i are n−1 distinct numbers.\n\n  • Within the second block, as j increases from 1 to n−2, x_{j+1} increases from x₂ up to x_{n−1}, so B₁ < B₂ < ⋯ < B_{n−2}. Thus the B_j are n−2 distinct numbers.\n\n  • To show no A_i can equal any B_j, compare the largest A (namely A₁=(x₁+xₙ)/2) with the smallest B (namely B₁=(x₂+xₙ)/2). Since x₁<x₂, we have\n     (x₁ + xₙ)/2 < (x₂ + xₙ)/2,\n  hence every A_i ≤ A₁ < B₁ ≤ every B_j. Therefore the two blocks are disjoint, yielding in all\n     |M_S| ≥ (n−1)+(n−2) = 2n−3.\n\n2. (Sharpness) Take for instance the arithmetic progression\n   S₀ = {0,3,6,…,3(n−1)}.\nAny midpoint of two distinct terms 3a<3b is (3a+3b)/2 = (3/2)(a+b). Since a<b run over pairs in {0,…,n−1}, the sum a+b takes exactly the 2n−3 integer values 1,2,…,2n−3. Thus M_{S₀} = {(3/2)⋅k : k=1,2,…,2n−3} has size 2n−3.\n\nCombining the lower bound |M_S|≥2n−3 with this example shows that the minimum possible size of M_S is\n   2n−3.\n\nAnswer: 2n−3.",
      "_meta": {
        "core_steps": [
          "Sort S as x1 < x2 < … < xn.",
          "Apply monotonicity of averages to extract 2n−3 distinct midpoints: (x1+xi)/2 (i=2..n) and (xi+xn)/2 (i=2..n−1), yielding |AS| ≥ 2n−3.",
          "Give an explicit example (consecutive integers) whose average–set has exactly 2n−3 elements, so the lower bound is sharp."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Concrete set used to hit the upper bound; any arithmetic progression of length n works equally well.",
            "original": "{1,2,…,n}"
          },
          "slot2": {
            "description": "Exact list/order of the 2n−3 averages employed in the lower-bound chain; any injective selection of 2n−3 averages based on extreme elements suffices.",
            "original": "(x1+x2)/2, (x1+x3)/2, …, (x1+xn)/2, (x2+xn)/2, …, (x_{n−1}+xn)/2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}