{
  "index": "1958-2-B-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "3. If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than \\( \\sqrt{5} / 2 \\). Show also that no larger number will do.",
  "solution": "Solution. Suppose the square is \\( A B C D \\) (with unit side) and the midpoints of the sides \\( A B \\) and \\( B C \\) are \\( E \\) and \\( F \\), respectively. Then \\( |A F|= \\) \\( |D F|=|D E|=|C E|=\\sqrt{5} / 2 \\).\n\nSuppose that the square is partitioned into two sets \\( S \\) and \\( T \\) of diameter less than \\( \\sqrt{5} / 2 \\), and choose the notation so that \\( A \\in S \\). Then \\( F \\in T, D \\in S \\), \\( E \\in T, C \\in S \\), since \\( A \\) and \\( F \\), for example, are too far apart to be both\nmembers of \\( S \\). Thus \\( A \\) and \\( C \\) are in same subset \\( S \\), but \\( |A C|=\\sqrt{2}>\\sqrt{5} / 2 \\), contradicting the fact that the diameter of \\( S \\) is \\( <\\sqrt{5} / 2 \\).\n\nOn the other hand, one can clearly partition the square into two rectangular sets of diameter \\( \\sqrt{5} / 2 \\), as indicated.\n\nThe points of the dividing segment \\( E G \\) can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.)",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "G",
    "S",
    "T"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "vertexa",
        "B": "vertexb",
        "C": "vertexc",
        "D": "vertexd",
        "E": "midpointab",
        "F": "midpointbc",
        "G": "dividingpt",
        "S": "subsetone",
        "T": "subsettwo"
      },
      "question": "3. If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than \\( \\sqrt{5} / 2 \\). Show also that no larger number will do.",
      "solution": "Solution. Suppose the square is \\( vertexa vertexb vertexc vertexd \\) (with unit side) and the midpoints of the sides \\( vertexa vertexb \\) and \\( vertexb vertexc \\) are \\( midpointab \\) and \\( midpointbc \\), respectively. Then \\( |vertexa midpointbc|= \\) \\( |vertexd midpointbc|=|vertexd midpointab|=|vertexc midpointab|=\\sqrt{5} / 2 \\).\n\nSuppose that the square is partitioned into two sets \\( subsetone \\) and \\( subsettwo \\) of diameter less than \\( \\sqrt{5} / 2 \\), and choose the notation so that \\( vertexa \\in subsetone \\). Then \\( midpointbc \\in subsettwo, vertexd \\in subsetone \\), \\( midpointab \\in subsettwo, vertexc \\in subsetone \\), since \\( vertexa \\) and \\( midpointbc \\), for example, are too far apart to be both\nmembers of \\( subsetone \\). Thus \\( vertexa \\) and \\( vertexc \\) are in same subset \\( subsetone \\), but \\( |vertexa vertexc|=\\sqrt{2}>\\sqrt{5} / 2 \\), contradicting the fact that the diameter of \\( subsetone \\) is \\( <\\sqrt{5} / 2 \\).\n\nOn the other hand, one can clearly partition the square into two rectangular sets of diameter \\( \\sqrt{5} / 2 \\), as indicated.\n\nThe points of the dividing segment \\( midpointab dividingpt \\) can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "beverage",
        "B": "lighthouse",
        "C": "harmonica",
        "D": "junction",
        "E": "parchment",
        "F": "carousel",
        "G": "molecule",
        "S": "tapestry",
        "T": "asteroid"
      },
      "question": "<<<\n3. If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than \\( \\sqrt{5} / 2 \\). Show also that no larger number will do.\n>>>",
      "solution": "<<<\nSolution. Suppose the square is \\( beverage lighthouse harmonica junction \\) (with unit side) and the midpoints of the sides \\( beverage lighthouse \\) and \\( lighthouse harmonica \\) are \\( parchment \\) and \\( carousel \\), respectively. Then \\( |beverage carousel|= \\) \\( |junction carousel|=|junction parchment|=|harmonica parchment|=\\sqrt{5} / 2 \\).\n\nSuppose that the square is partitioned into two sets \\( tapestry \\) and \\( asteroid \\) of diameter less than \\( \\sqrt{5} / 2 \\), and choose the notation so that \\( beverage \\in tapestry \\). Then \\( carousel \\in asteroid, junction \\in tapestry \\), \\( parchment \\in asteroid, harmonica \\in tapestry \\), since \\( beverage \\) and \\( carousel \\), for example, are too far apart to be both\nmembers of \\( tapestry \\). Thus \\( beverage \\) and \\( harmonica \\) are in same subset \\( tapestry \\), but \\( |beverage harmonica|=\\sqrt{2}>\\sqrt{5} / 2 \\), contradicting the fact that the diameter of \\( tapestry \\) is \\( <\\sqrt{5} / 2 \\).\n\nOn the other hand, one can clearly partition the square into two rectangular sets of diameter \\( \\sqrt{5} / 2 \\), as indicated.\n\nThe points of the dividing segment \\( parchment molecule \\) can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.)\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "centerpoint",
        "B": "midsection",
        "C": "midregion",
        "D": "coreplace",
        "E": "extremepoint",
        "F": "terminalspot",
        "G": "undividedspot",
        "S": "entiregroup",
        "T": "totalset"
      },
      "question": "3. If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than \\( \\sqrt{5} / 2 \\). Show also that no larger number will do.",
      "solution": "Solution. Suppose the square is \\( centerpoint midsection midregion coreplace \\) (with unit side) and the midpoints of the sides \\( centerpoint midsection \\) and \\( midsection midregion \\) are \\( extremepoint \\) and \\( terminalspot \\), respectively. Then \\( |centerpoint terminalspot|= \\) \\( |coreplace terminalspot|=|coreplace extremepoint|=|midregion extremepoint|=\\sqrt{5} / 2 \\).\n\nSuppose that the square is partitioned into two sets \\( entiregroup \\) and \\( totalset \\) of diameter less than \\( \\sqrt{5} / 2 \\), and choose the notation so that \\( centerpoint \\in entiregroup \\). Then \\( terminalspot \\in totalset, coreplace \\in entiregroup \\), \\( extremepoint \\in totalset, midregion \\in entiregroup \\), since \\( centerpoint \\) and \\( terminalspot \\), for example, are too far apart to be both\nmembers of \\( entiregroup \\). Thus \\( centerpoint \\) and \\( midregion \\) are in same subset \\( entiregroup \\), but \\( |centerpoint midregion|=\\sqrt{2}>\\sqrt{5} / 2 \\), contradicting the fact that the diameter of \\( entiregroup \\) is \\( <\\sqrt{5} / 2 \\).\n\nOn the other hand, one can clearly partition the square into two rectangular sets of diameter \\( \\sqrt{5} / 2 \\), as indicated.\n\nThe points of the dividing segment \\( extremepoint undividedspot \\) can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.)"
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mkdlsevr",
        "D": "tbaxjroo",
        "E": "vnyshgzc",
        "F": "pdruqkwe",
        "G": "lofmcatr",
        "S": "zehugpni",
        "T": "wdbyvark"
      },
      "question": "3. If a square of unit side be partitioned into two sets, then the diameter (least upper bound of the distances between pairs of points) of one of the sets is not less than \\( \\sqrt{5} / 2 \\). Show also that no larger number will do.",
      "solution": "Solution. Suppose the square is \\( qzxwvtnp hjgrksla mkdlsevr tbaxjroo \\) (with unit side) and the midpoints of the sides \\( qzxwvtnp hjgrksla \\) and \\( hjgrksla mkdlsevr \\) are \\( vnyshgzc \\) and \\( pdruqkwe \\), respectively. Then \\( |qzxwvtnp pdruqkwe|= \\) \\( |tbaxjroo pdruqkwe|=|tbaxjroo vnyshgzc|=|mkdlsevr vnyshgzc|=\\sqrt{5} / 2 \\).\n\nSuppose that the square is partitioned into two sets \\( zehugpni \\) and \\( wdbyvark \\) of diameter less than \\( \\sqrt{5} / 2 \\), and choose the notation so that \\( qzxwvtnp \\in zehugpni \\). Then \\( pdruqkwe \\in wdbyvark, tbaxjroo \\in zehugpni \\), \\( vnyshgzc \\in wdbyvark, mkdlsevr \\in zehugpni \\), since \\( qzxwvtnp \\) and \\( pdruqkwe \\), for example, are too far apart to be both members of \\( zehugpni \\). Thus \\( qzxwvtnp \\) and \\( mkdlsevr \\) are in same subset \\( zehugpni \\), but \\( |qzxwvtnp mkdlsevr|=\\sqrt{2}>\\sqrt{5} / 2 \\), contradicting the fact that the diameter of \\( zehugpni \\) is \\( <\\sqrt{5} / 2 \\).\n\nOn the other hand, one can clearly partition the square into two rectangular sets of diameter \\( \\sqrt{5} / 2 \\), as indicated.\n\nThe points of the dividing segment \\( vnyshgzc lofmcatr \\) can be apportioned to either subset. (It is obvious that the diameter of a rectangle is the length of its diagonal, but an analytic proof can easily be given.)"
    },
    "kernel_variant": {
      "question": "Let $PQRS$ be a square of side $3$ in the plane (vertices listed counter-clockwise).  For any set $X\\subset\\mathbb R^{2}$ define its diameter by \n\\[\\operatorname{diam}(X)=\\sup\\{|xy|:x,y\\in X\\}.\\]\nProve that whenever the square is written as a union of two (possibly non–closed) sets, at least one of the two sets has diameter not smaller than $\\dfrac{3\\sqrt5}{2}$.  Then give an explicit partition of the square into two sets, each of diameter exactly $\\dfrac{3\\sqrt5}{2}$, to show that the constant $\\dfrac{3\\sqrt5}{2}$ is best possible.",
      "solution": "1. Lower bound.  Label the vertices of the square so that P=(0,0), Q=(3,0), R=(3,3), S=(0,3).  Let E be the midpoint of QR and F the midpoint of RS, so\n\n    E = (3,1.5),   F = (1.5,3).\n\nA direct calculation shows\n\n    |QF| = √[(3−1.5)² + (0−3)²] = √(2.25+9) = √11.25 = 3√5/2,\n    |PF| = √[(0−1.5)² + (0−3)²] = √11.25 = 3√5/2,\n    |SE| = √[(0−3)² + (3−1.5)²] = √11.25 = 3√5/2,\n    |PE| = √[(0−3)² + (0−1.5)²] = √11.25 = 3√5/2.\n\nNow suppose for contradiction that the square is partitioned into two sets A and B with diam(A)<3√5/2 and diam(B)<3√5/2.  Place Q in A.  Since |QF|=3√5/2, F cannot lie in A, so F∈B.  Since |PF|=3√5/2, P∉B, so P∈A.  Since |PE|=3√5/2, E∉A, so E∈B.  Since |SE|=3√5/2, S∉B, so S∈A.  Thus Q and S both lie in A, but\n\n    |QS| = √[(3−0)² + (0−3)²] = √18 = 3√2 > 3√5/2,\n\ncontradicting diam(A)<3√5/2.  Therefore in any two‐set partition of the square, at least one part has diameter ≥3√5/2.\n\n2. Sharp example.  Let M=(0,1.5) and N=(3,1.5) be the midpoints of SP and QR, respectively.  The horizontal segment MN splits the 3×3 square into two congruent rectangles PMNQ and SMNR, each of size 3 by 1.5.  The diagonal of a 3×1.5 rectangle is\n\n    √(3² + 1.5²) = √(9 + 2.25) = √11.25 = 3√5/2.\n\nHence each rectangle has diameter exactly 3√5/2.  Assign one rectangle to A and the other to B to achieve diam(A)=diam(B)=3√5/2.  This shows the bound 3√5/2 is best possible.",
      "_meta": {
        "core_steps": [
          "Pick midpoints of two adjacent sides to create four special points with distance √5⁄2 from the adjoining corners.",
          "Assume both subsets have diameter < √5⁄2 and place one corner in the first subset.",
          "Distance constraints then force the two opposite corners into the same subset, giving a corner-to-corner distance √2 > √5⁄2 — contradiction.",
          "Construct an explicit two-rectangle partition whose diagonal length is exactly √5⁄2 to show the bound is sharp."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Scale of the square (side length). All distances, including the claimed bound, scale proportionally.",
            "original": "side = 1"
          },
          "slot2": {
            "description": "Which pair of adjacent sides one takes midpoints from; any adjacent pair works by symmetry.",
            "original": "sides AB and BC"
          },
          "slot3": {
            "description": "The numerical bound itself, which is forced to be the distance from a vertex to the chosen midpoint; if slot1 is changed to side = s, this becomes √5·s⁄2.",
            "original": "√5⁄2"
          },
          "slot4": {
            "description": "Orientation/location of the separating segment for the sharpness construction (e.g. joining the two midpoints vs. a parallel translate); as long as it yields two rectangles with the critical diagonal, the reasoning is unaffected.",
            "original": "segment EG joining the two midpoints"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}