{
  "index": "1957-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "5. Given \\( n \\) points in the plane, show that the largest distance determined by these points cannot occur more than \\( n \\) times.",
  "solution": "Solution. Suppose \\( S \\) is a set in the plane. By a diameter of \\( S \\) we mean a segment connecting two points of \\( \\mathcal{S} \\) that are as far apart as any two points of \\( S \\). We are asked to prove (1) a set of \\( n \\) points in a plane can have at most \\( n \\) diameters. We shall use induction on \\( n \\). Clearly, (1) is true for \\( n=2 \\) or 3 .\n\nAssume that (1) is true for \\( n=k \\). Let \\( \\mathcal{E} \\) be a set in a plane with \\( k+1 \\) points.\n\nSuppose some point of \\( \\mathcal{E} \\), say \\( X \\), is the endpoint of at most one diameter of \\( \\mathcal{E} \\). Then \\( \\mathcal{E}-\\{X\\} \\) is a set of \\( k \\) points in the plane and has at most \\( k \\) diameters by the inductive hypothesis. Then \\( \\mathcal{E} \\) has at most \\( k+1 \\) diameters.\n\nSuppose some point of \\( \\mathcal{E} \\), say \\( P \\), is the endpoint of at least three diameters, \\( P Q P R \\), and \\( P S \\). Let \\( r=|P Q| \\). Then \\( Q R \\), and \\( S \\) lie on a circle of radius \\( r \\) about \\( P \\) and, in fact, on a minor arc of that circle. We choose the notation so that \\( R \\) is between \\( Q \\) and \\( S \\) on that arc. Since every two points of \\( \\mathcal{E} \\) are at most \\( r \\) apart, \\( \\mathcal{E} \\) lies in the intersection \\( \\mathcal{J} \\) of three closed circular disks of radius \\( r \\) and centers \\( P, Q \\) and \\( S \\). Now, except for the point \\( P, \\mathcal{J} \\) is inside the circle of radius \\( r \\) about \\( R \\), and therefore \\( R \\) is the end point of just one diameter of \\( \\mathcal{E} \\), namely, \\( R P \\). Therefore, the previous paragraph applies, and \\( \\mathcal{E} \\) has at most \\( k+1 \\) diameters.\n\nFinally, suppose each point of \\( \\mathcal{E} \\) is the endpoint of exactly two diameters. Then the diameters have altogether \\( 2(k+1) \\) endpoints, so there are exactly \\( k+1 \\) of them.\nThus, in any case, \\( \\mathcal{E} \\) has at most \\( k+1 \\) diameters. Hence (1) is true for \\( n=k+1 \\). It follows by induction that (1) is true for all \\( n \\).\n\nRemark. The geometric part of the argument can be completely formalized as follows:\n\nAn endpoint of a diameter of \\( \\mathcal{E} \\) is an extreme point of the convex hull of \\( \\mathcal{E} \\). Therefore, \\( P, Q, R, S \\) are vertices of a convex quadrilateral, and we can choose notation so that \\( \\overparen{P R} \\) separates \\( Q \\) and \\( S \\). Suppose \\( R T \\) is a diameter of \\( \\mathcal{E} \\) with \\( T \\neq P \\). Then \\( T \\) does not lie on \\( \\overrightarrow{P R} \\), so we may assume \\( T \\) and \\( S \\) are on opposite sides of \\( \\overleftrightarrow{P R} \\). Then \\( P, R, S, T \\) are vertices of a convex quadrilateral and \\( P R \\) must be a diagonal. Let \\( P R \\) and \\( S T \\) intersect at \\( U \\). Then\n\\[\n|P R|+|S T|=|P U|+|U S|+|R U|+|U T|>|P S|+|R T| .\n\\]\n\nSince \\( |P R|=|P S|=|R T|=r \\), this shows that \\( |S T|>r \\), which is impossible. Hence there is no such diameter \\( R T \\), and \\( R \\) is the endpoint of just one diameter \\( R P \\), as claimed.\n\nFor other results of a similar nature see P. Erdos, \"On Sets of Distances of \\( n \\) Points,\" American Mathematical Monthly, vol. 53 (1956), pages 248-250.",
  "vars": [
    "n",
    "k",
    "S",
    "X",
    "E",
    "Q",
    "R",
    "P",
    "T",
    "U",
    "r",
    "J"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "pointcount",
        "k": "inductstep",
        "S": "pointset",
        "X": "singlept",
        "E": "planepoints",
        "Q": "pointq",
        "R": "pointr",
        "P": "pointp",
        "T": "pointt",
        "U": "pointu",
        "r": "maxdist",
        "J": "intersec"
      },
      "question": "5. Given \\( pointcount \\) points in the plane, show that the largest distance determined by these points cannot occur more than \\( pointcount \\) times.",
      "solution": "Solution. Suppose \\( pointset \\) is a set in the plane. By a diameter of \\( pointset \\) we mean a segment connecting two points of \\( \\mathcal{pointset} \\) that are as far apart as any two points of \\( pointset \\). We are asked to prove (1) a set of \\( pointcount \\) points in a plane can have at most \\( pointcount \\) diameters. We shall use induction on \\( pointcount \\). Clearly, (1) is true for \\( pointcount=2 \\) or 3.\n\nAssume that (1) is true for \\( pointcount=inductstep \\). Let \\( \\mathcal{planepoints} \\) be a set in a plane with \\( inductstep+1 \\) points.\n\nSuppose some point of \\( \\mathcal{planepoints} \\), say \\( singlept \\), is the endpoint of at most one diameter of \\( \\mathcal{planepoints} \\). Then \\( \\mathcal{planepoints}-\\{singlept\\} \\) is a set of \\( inductstep \\) points in the plane and has at most \\( inductstep \\) diameters by the inductive hypothesis. Then \\( \\mathcal{planepoints} \\) has at most \\( inductstep+1 \\) diameters.\n\nSuppose some point of \\( \\mathcal{planepoints} \\), say \\( pointp \\), is the endpoint of at least three diameters, \\( pointp pointq, pointp pointr \\), and \\( pointp pointset \\). Let \\( maxdist=|pointp pointq| \\). Then \\( pointq, pointr, \\) and \\( pointset \\) lie on a circle of radius \\( maxdist \\) about \\( pointp \\) and, in fact, on a minor arc of that circle. We choose the notation so that \\( pointr \\) is between \\( pointq \\) and \\( pointset \\) on that arc. Since every two points of \\( \\mathcal{planepoints} \\) are at most \\( maxdist \\) apart, \\( \\mathcal{planepoints} \\) lies in the intersection \\( \\mathcal{intersec} \\) of three closed circular disks of radius \\( maxdist \\) and centers \\( pointp, pointq \\) and \\( pointset \\). Now, except for the point \\( pointp, \\mathcal{intersec} \\) is inside the circle of radius \\( maxdist \\) about \\( pointr \\), and therefore \\( pointr \\) is the endpoint of just one diameter of \\( \\mathcal{planepoints} \\), namely, \\( pointr pointp \\). Therefore, the previous paragraph applies, and \\( \\mathcal{planepoints} \\) has at most \\( inductstep+1 \\) diameters.\n\nFinally, suppose each point of \\( \\mathcal{planepoints} \\) is the endpoint of exactly two diameters. Then the diameters have altogether \\( 2(inductstep+1) \\) endpoints, so there are exactly \\( inductstep+1 \\) of them. Thus, in any case, \\( \\mathcal{planepoints} \\) has at most \\( inductstep+1 \\) diameters. Hence (1) is true for \\( pointcount=inductstep+1 \\). It follows by induction that (1) is true for all \\( pointcount \\).\n\nRemark. The geometric part of the argument can be completely formalized as follows:\n\nAn endpoint of a diameter of \\( \\mathcal{planepoints} \\) is an extreme point of the convex hull of \\( \\mathcal{planepoints} \\). Therefore, \\( pointp, pointq, pointr, pointset \\) are vertices of a convex quadrilateral, and we can choose notation so that \\( \\overparen{pointp pointr} \\) separates \\( pointq \\) and \\( pointset \\). Suppose \\( pointr pointt \\) is a diameter of \\( \\mathcal{planepoints} \\) with \\( pointt \\neq pointp \\). Then \\( pointt \\) does not lie on \\( \\overrightarrow{pointp pointr} \\), so we may assume \\( pointt \\) and \\( pointset \\) are on opposite sides of \\( \\overleftrightarrow{pointp pointr} \\). Then \\( pointp, pointr, pointset, pointt \\) are vertices of a convex quadrilateral and \\( pointp pointr \\) must be a diagonal. Let \\( pointp pointr \\) and \\( pointset pointt \\) intersect at \\( pointu \\). Then\n\\[\n|pointp pointr|+|pointset pointt|=|pointp pointu|+|pointu pointset|+|pointr pointu|+|pointu pointt|>|pointp pointset|+|pointr pointt| .\n\\]\n\nSince \\( |pointp pointr|=|pointp pointset|=|pointr pointt|=maxdist \\), this shows that \\( |pointset pointt|>maxdist \\), which is impossible. Hence there is no such diameter \\( pointr pointt \\), and \\( pointr \\) is the endpoint of just one diameter \\( pointr pointp \\), as claimed.\n\nFor other results of a similar nature see P. Erdos, \"On Sets of Distances of \\( pointcount \\) Points,\" American Mathematical Monthly, vol. 53 (1956), pages 248-250."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "sunflower",
        "k": "sandstone",
        "S": "lanterns",
        "X": "quagmire",
        "E": "paintball",
        "Q": "everglade",
        "R": "blackbird",
        "P": "driftwood",
        "T": "moonlight",
        "U": "starshine",
        "r": "buttercup",
        "J": "riverbank"
      },
      "question": "5. Given \\( sunflower \\) points in the plane, show that the largest distance determined by these points cannot occur more than \\( sunflower \\) times.",
      "solution": "Solution. Suppose \\( lanterns \\) is a set in the plane. By a diameter of \\( lanterns \\) we mean a segment connecting two points of \\( \\mathcal{lanterns} \\) that are as far apart as any two points of \\( lanterns \\). We are asked to prove (1) a set of \\( sunflower \\) points in a plane can have at most \\( sunflower \\) diameters. We shall use induction on \\( sunflower \\). Clearly, (1) is true for \\( sunflower=2 \\) or 3 .\n\nAssume that (1) is true for \\( sunflower=sandstone \\). Let \\( \\mathcal{paintball} \\) be a set in a plane with \\( sandstone+1 \\) points.\n\nSuppose some point of \\( \\mathcal{paintball} \\), say \\( quagmire \\), is the endpoint of at most one diameter of \\( \\mathcal{paintball} \\). Then \\( \\mathcal{paintball}-\\{quagmire\\} \\) is a set of \\( sandstone \\) points in the plane and has at most \\( sandstone \\) diameters by the inductive hypothesis. Then \\( \\mathcal{paintball} \\) has at most \\( sandstone+1 \\) diameters.\n\nSuppose some point of \\( \\mathcal{paintball} \\), say \\( driftwood \\), is the endpoint of at least three diameters, \\( driftwood\\, everglade \\), \\( driftwood\\, blackbird \\), and \\( driftwood\\, lanterns \\). Let \\( buttercup=|driftwood\\, everglade| \\). Then \\( everglade, blackbird \\), and \\( lanterns \\) lie on a circle of radius \\( buttercup \\) about \\( driftwood \\) and, in fact, on a minor arc of that circle. We choose the notation so that \\( blackbird \\) is between \\( everglade \\) and \\( lanterns \\) on that arc. Since every two points of \\( \\mathcal{paintball} \\) are at most \\( buttercup \\) apart, \\( \\mathcal{paintball} \\) lies in the intersection \\( \\mathcal{riverbank} \\) of three closed circular disks of radius \\( buttercup \\) and centers \\( driftwood, everglade \\) and \\( lanterns \\). Now, except for the point \\( driftwood \\), \\( \\mathcal{riverbank} \\) is inside the circle of radius \\( buttercup \\) about \\( blackbird \\), and therefore \\( blackbird \\) is the end point of just one diameter of \\( \\mathcal{paintball} \\), namely, \\( blackbird\\, driftwood \\). Therefore, the previous paragraph applies, and \\( \\mathcal{paintball} \\) has at most \\( sandstone+1 \\) diameters.\n\nFinally, suppose each point of \\( \\mathcal{paintball} \\) is the endpoint of exactly two diameters. Then the diameters have altogether \\( 2(sandstone+1) \\) endpoints, so there are exactly \\( sandstone+1 \\) of them.\nThus, in any case, \\( \\mathcal{paintball} \\) has at most \\( sandstone+1 \\) diameters. Hence (1) is true for \\( sunflower=sandstone+1 \\). It follows by induction that (1) is true for all \\( sunflower \\).\n\nRemark. The geometric part of the argument can be completely formalized as follows:\n\nAn endpoint of a diameter of \\( \\mathcal{paintball} \\) is an extreme point of the convex hull of \\( \\mathcal{paintball} \\). Therefore, \\( driftwood, everglade, blackbird, lanterns \\) are vertices of a convex quadrilateral, and we can choose notation so that \\( \\overparen{driftwood\\, blackbird} \\) separates \\( everglade \\) and \\( lanterns \\). Suppose \\( blackbird\\, moonlight \\) is a diameter of \\( \\mathcal{paintball} \\) with \\( moonlight \\neq driftwood \\). Then \\( moonlight \\) does not lie on \\( \\overrightarrow{driftwood\\, blackbird} \\), so we may assume \\( moonlight \\) and \\( lanterns \\) are on opposite sides of \\( \\overleftrightarrow{driftwood\\, blackbird} \\). Then \\( driftwood, blackbird, lanterns, moonlight \\) are vertices of a convex quadrilateral and \\( driftwood\\, blackbird \\) must be a diagonal. Let \\( driftwood\\, blackbird \\) and \\( lanterns\\, moonlight \\) intersect at \\( starshine \\). Then\n\\[\n|driftwood\\, blackbird|+|lanterns\\, moonlight|=|driftwood\\, starshine|+|starshine\\, lanterns|+|blackbird\\, starshine|+|starshine\\, moonlight|>|driftwood\\, lanterns|+|blackbird\\, moonlight| .\n\\]\n\nSince \\( |driftwood\\, blackbird|=|driftwood\\, lanterns|=|blackbird\\, moonlight|= buttercup \\), this shows that \\( |lanterns\\, moonlight|> buttercup \\), which is impossible. Hence there is no such diameter \\( blackbird\\, moonlight \\), and \\( blackbird \\) is the endpoint of just one diameter \\( blackbird\\, driftwood \\), as claimed.\n\nFor other results of a similar nature see P. Erdos, \"On Sets of Distances of \\( sunflower \\) Points,\" American Mathematical Monthly, vol. 53 (1956), pages 248-250."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "boundlessqty",
        "k": "limitlessidx",
        "S": "voidcollection",
        "X": "nothingness",
        "E": "desolation",
        "Q": "anticenter",
        "R": "vanishpoint",
        "P": "nonvertex",
        "T": "nullspot",
        "U": "divergence",
        "r": "diameter",
        "J": "unionzone"
      },
      "question": "5. Given \\( boundlessqty \\) points in the plane, show that the largest distance determined by these points cannot occur more than \\( boundlessqty \\) times.",
      "solution": "Solution. Suppose \\( voidcollection \\) is a set in the plane. By a diameter of \\( voidcollection \\) we mean a segment connecting two points of \\( \\mathcal{voidcollection} \\) that are as far apart as any two points of \\( voidcollection \\). We are asked to prove (1) a set of \\( boundlessqty \\) points in a plane can have at most \\( boundlessqty \\) diameters. We shall use induction on \\( boundlessqty \\). Clearly, (1) is true for \\( boundlessqty =2 \\) or 3.\n\nAssume that (1) is true for \\( boundlessqty = limitlessidx \\). Let \\( \\mathcal{desolation} \\) be a set in a plane with \\( limitlessidx +1 \\) points.\n\nSuppose some point of \\( \\mathcal{desolation} \\), say \\( nothingness \\), is the endpoint of at most one diameter of \\( \\mathcal{desolation} \\). Then \\( \\mathcal{desolation}-\\{nothingness\\} \\) is a set of \\( limitlessidx \\) points in the plane and has at most \\( limitlessidx \\) diameters by the inductive hypothesis. Then \\( \\mathcal{desolation} \\) has at most \\( limitlessidx +1 \\) diameters.\n\nSuppose some point of \\( \\mathcal{desolation} \\), say \\( nonvertex \\), is the endpoint of at least three diameters, \\( nonvertex anticenter \\; nonvertex vanishpoint \\), and \\( nonvertex voidcollection \\). Let \\( diameter = |nonvertex anticenter| \\). Then \\( anticenter vanishpoint \\), and \\( voidcollection \\) lie on a circle of radius \\( diameter \\) about \\( nonvertex \\) and, in fact, on a minor arc of that circle. We choose the notation so that \\( vanishpoint \\) is between \\( anticenter \\) and \\( voidcollection \\) on that arc. Since every two points of \\( \\mathcal{desolation} \\) are at most \\( diameter \\) apart, \\( \\mathcal{desolation} \\) lies in the intersection \\( \\mathcal{unionzone} \\) of three closed circular disks of radius \\( diameter \\) and centers \\( nonvertex, anticenter \\) and \\( voidcollection \\). Now, except for the point \\( nonvertex, \\mathcal{unionzone} \\) is inside the circle of radius \\( diameter \\) about \\( vanishpoint \\), and therefore \\( vanishpoint \\) is the end point of just one diameter of \\( \\mathcal{desolation} \\), namely, \\( vanishpoint nonvertex \\). Therefore, the previous paragraph applies, and \\( \\mathcal{desolation} \\) has at most \\( limitlessidx +1 \\) diameters.\n\nFinally, suppose each point of \\( \\mathcal{desolation} \\) is the endpoint of exactly two diameters. Then the diameters have altogether \\( 2(limitlessidx +1) \\) endpoints, so there are exactly \\( limitlessidx +1 \\) of them.\nThus, in any case, \\( \\mathcal{desolation} \\) has at most \\( limitlessidx +1 \\) diameters. Hence (1) is true for \\( boundlessqty = limitlessidx +1 \\). It follows by induction that (1) is true for all \\( boundlessqty \\).\n\nRemark. The geometric part of the argument can be completely formalized as follows:\n\nAn endpoint of a diameter of \\( \\mathcal{desolation} \\) is an extreme point of the convex hull of \\( \\mathcal{desolation} \\). Therefore, \\( nonvertex, anticenter, vanishpoint, voidcollection \\) are vertices of a convex quadrilateral, and we can choose notation so that \\( \\overparen{nonvertex vanishpoint} \\) separates \\( anticenter \\) and \\( voidcollection \\). Suppose \\( vanishpoint nullspot \\) is a diameter of \\( \\mathcal{desolation} \\) with \\( nullspot \\neq nonvertex \\). Then \\( nullspot \\) does not lie on \\( \\overrightarrow{nonvertex vanishpoint} \\), so we may assume \\( nullspot \\) and \\( voidcollection \\) are on opposite sides of \\( \\overleftrightarrow{nonvertex vanishpoint} \\). Then \\( nonvertex, vanishpoint, voidcollection, nullspot \\) are vertices of a convex quadrilateral and \\( nonvertex vanishpoint \\) must be a diagonal. Let \\( nonvertex vanishpoint \\) and \\( voidcollection nullspot \\) intersect at \\( divergence \\). Then\n\\[\n|nonvertex vanishpoint|+|voidcollection nullspot|=|nonvertex divergence|+|divergence voidcollection|+|vanishpoint divergence|+|divergence nullspot|>|nonvertex voidcollection|+|vanishpoint nullspot| .\n\\]\n\nSince \\( |nonvertex vanishpoint|=|nonvertex voidcollection|=|vanishpoint nullspot|=diameter \\), this shows that \\( |voidcollection nullspot|>diameter \\), which is impossible. Hence there is no such diameter \\( vanishpoint nullspot \\), and \\( vanishpoint \\) is the endpoint of just one diameter \\( vanishpoint nonvertex \\), as claimed.\n\nFor other results of a similar nature see P. Erdos, \"On Sets of Distances of \\( boundlessqty \\) Points,\" American Mathematical Monthly, vol. 53 (1956), pages 248-250."
    },
    "garbled_string": {
      "map": {
        "n": "vczmxpqk",
        "k": "ftrnslab",
        "S": "qzxwvtnp",
        "X": "hjrkdpsl",
        "E": "klmjfsro",
        "Q": "bwkepsmz",
        "R": "dljgvhqc",
        "P": "rczsftwe",
        "T": "pnqslrkc",
        "U": "mzchktvl",
        "r": "gxvplkra",
        "J": "nbswqtrh"
      },
      "question": "5. Given \\( vczmxpqk \\) points in the plane, show that the largest distance determined by these points cannot occur more than \\( vczmxpqk \\) times.",
      "solution": "Solution. Suppose \\( qzxwvtnp \\) is a set in the plane. By a diameter of \\( qzxwvtnp \\) we mean a segment connecting two points of \\( \\mathcal{qzxwvtnp} \\) that are as far apart as any two points of \\( qzxwvtnp \\). We are asked to prove (1) a set of \\( vczmxpqk \\) points in a plane can have at most \\( vczmxpqk \\) diameters. We shall use induction on \\( vczmxpqk \\). Clearly, (1) is true for \\( vczmxpqk=2 \\) or 3.\n\nAssume that (1) is true for \\( vczmxpqk=ftrnslab \\). Let \\( \\mathcal{klmjfsro} \\) be a set in a plane with \\( ftrnslab+1 \\) points.\n\nSuppose some point of \\( \\mathcal{klmjfsro} \\), say \\( hjrkdpsl \\), is the endpoint of at most one diameter of \\( \\mathcal{klmjfsro} \\). Then \\( \\mathcal{klmjfsro}-\\{hjrkdpsl\\} \\) is a set of \\( ftrnslab \\) points in the plane and has at most \\( ftrnslab \\) diameters by the inductive hypothesis. Then \\( \\mathcal{klmjfsro} \\) has at most \\( ftrnslab+1 \\) diameters.\n\nSuppose some point of \\( \\mathcal{klmjfsro} \\), say \\( rczsftwe \\), is the endpoint of at least three diameters, \\( rczsftwe bwkepsmz, rczsftwe dljgvhqc \\), and \\( rczsftwe qzxwvtnp \\). Let \\( gxvplkra=|rczsftwe bwkepsmz| \\). Then \\( bwkepsmz, dljgvhqc \\), and \\( qzxwvtnp \\) lie on a circle of radius \\( gxvplkra \\) about \\( rczsftwe \\) and, in fact, on a minor arc of that circle. We choose the notation so that \\( dljgvhqc \\) is between \\( bwkepsmz \\) and \\( qzxwvtnp \\) on that arc. Since every two points of \\( \\mathcal{klmjfsro} \\) are at most \\( gxvplkra \\) apart, \\( \\mathcal{klmjfsro} \\) lies in the intersection \\( \\mathcal{nbswqtrh} \\) of three closed circular disks of radius \\( gxvplkra \\) and centers \\( rczsftwe, bwkepsmz \\) and \\( qzxwvtnp \\). Now, except for the point \\( rczsftwe, \\mathcal{nbswqtrh} \\) is inside the circle of radius \\( gxvplkra \\) about \\( dljgvhqc \\), and therefore \\( dljgvhqc \\) is the end point of just one diameter of \\( \\mathcal{klmjfsro} \\), namely, \\( dljgvhqc rczsftwe \\). Therefore, the previous paragraph applies, and \\( \\mathcal{klmjfsro} \\) has at most \\( ftrnslab+1 \\) diameters.\n\nFinally, suppose each point of \\( \\mathcal{klmjfsro} \\) is the endpoint of exactly two diameters. Then the diameters have altogether \\( 2(ftrnslab+1) \\) endpoints, so there are exactly \\( ftrnslab+1 \\) of them.\nThus, in any case, \\( \\mathcal{klmjfsro} \\) has at most \\( ftrnslab+1 \\) diameters. Hence (1) is true for \\( vczmxpqk=ftrnslab+1 \\). It follows by induction that (1) is true for all \\( vczmxpqk \\).\n\nRemark. The geometric part of the argument can be completely formalized as follows:\n\nAn endpoint of a diameter of \\( \\mathcal{klmjfsro} \\) is an extreme point of the convex hull of \\( \\mathcal{klmjfsro} \\). Therefore, \\( rczsftwe, bwkepsmz, dljgvhqc, qzxwvtnp \\) are vertices of a convex quadrilateral, and we can choose notation so that \\( \\overparen{rczsftwe dljgvhqc} \\) separates \\( bwkepsmz \\) and \\( qzxwvtnp \\). Suppose \\( dljgvhqc pnqslrkc \\) is a diameter of \\( \\mathcal{klmjfsro} \\) with \\( pnqslrkc \\neq rczsftwe \\). Then \\( pnqslrkc \\) does not lie on \\( \\overrightarrow{rczsftwe dljgvhqc} \\), so we may assume \\( pnqslrkc \\) and \\( qzxwvtnp \\) are on opposite sides of \\( \\overleftrightarrow{rczsftwe dljgvhqc} \\). Then \\( rczsftwe, dljgvhqc, qzxwvtnp, pnqslrkc \\) are vertices of a convex quadrilateral and \\( rczsftwe dljgvhqc \\) must be a diagonal. Let \\( rczsftwe dljgvhqc \\) and \\( qzxwvtnp pnqslrkc \\) intersect at \\( mzchktvl \\). Then\n\\[\n|rczsftwe dljgvhqc|+|qzxwvtnp pnqslrkc|=|rczsftwe mzchktvl|+|mzchktvl qzxwvtnp|+|dljgvhqc mzchktvl|+|mzchktvl pnqslrkc|>|rczsftwe qzxwvtnp|+|dljgvhqc pnqslrkc| .\n\\]\n\nSince \\( |rczsftwe dljgvhqc|=|rczsftwe qzxwvtnp|=|dljgvhqc pnqslrkc|=gxvplkra \\), this shows that \\( |qzxwvtnp pnqslrkc|>gxvplkra \\), which is impossible. Hence there is no such diameter \\( dljgvhqc pnqslrkc \\), and \\( dljgvhqc \\) is the endpoint of just one diameter \\( dljgvhqc rczsftwe \\), as claimed.\n\nFor other results of a similar nature see P. Erdos, \"On Sets of Distances of \\( vczmxpqk \\) Points,\" American Mathematical Monthly, vol. 53 (1956), pages 248-250."
    },
    "kernel_variant": {
      "question": "Let \\(\\mathcal{G}\\) be a set of \\(n\\ge 1\\) distinct points in the Euclidean plane.  Call the (unordered) segment joining two points of \\(\\mathcal{G}\\) a super-chord if its length equals the largest distance \\(\\ell\\) determined by the points of \\(\\mathcal{G}\\).  Prove that there are at most \\(n\\) super-chords.",
      "solution": "We prove by induction on n that any set G of n ≥ 1 points in the plane determines at most n “super‐chords,” i.e. the segments of maximum length ℓ among pairs of points in G. We call a super‐chord a diameter of G.\n\nBase cases. For n = 1 there are 0 diameters. For n = 2 there is 1. For n = 3 there are at most 3 (in fact exactly 3 only in the equilateral case), so in each case the number of diameters ≤ n.\n\nInductive step. Fix m ≥ 3, and assume the statement is true for all sets of size ≤ m. Let G be any set of size m + 1, and let deg(P) be the number of diameters incident with point P. We consider three cases:\n\nCase 1: Some point A of G has deg(A) ≤ 1. Remove A to get G' = G \\ {A}, which has m points. By the inductive hypothesis G' has at most m diameters, and adding back A can contribute at most one new diameter, so G has ≤ m + 1 diameters.\n\nCase 2: Some point B has deg(B) ≥ 3. Let BC, BD, BE be three diameters, all of length ℓ. Then C, D, E lie on the circle of radius ℓ about B, and because no two of C, D, E are more than ℓ apart they must lie on some open semicircle of that circle. Label them in circular order as C, D, E. Since every point X ∈ G satisfies |BX| ≤ ℓ, |CX| ≤ ℓ, |EX| ≤ ℓ, it follows that all of G lies in the intersection of the three closed disks of radius ℓ centered at B, C, E. A standard convexity‐circle (Reuleaux‐triangle) argument shows that (apart from B itself) this intersection lies strictly inside the open disk of radius ℓ about D. Hence no point other than B can lie at distance ℓ from D, so deg(D) = 1. We are back in Case 1 applied to D, and conclude again that G has at most m + 1 diameters.\n\nCase 3: Every point of G has deg = 2. Let k be the total number of diameters. Each diameter contributes 2 to the sum of the degrees, so total degrees = 2k. On the other hand, there are m + 1 points each of degree 2, so total degrees = 2(m + 1). Hence 2k = 2(m + 1) and k = m + 1. In particular k ≤ m + 1.\n\nIn all cases G has at most m + 1 diameters. By induction the result holds for all n. Hence any n‐point set in the plane determines at most n super‐chords (diameters).",
      "_meta": {
        "core_steps": [
          "Induct on the number of points n.",
          "Case A: a point is incident to ≤1 diameter → delete it and invoke induction.",
          "Case B: a point is incident to ≥3 diameters → geometric (circle/convex-hull) argument forces another point with only 1 diameter, so revert to Case A.",
          "Case C: every point is incident to exactly 2 diameters → 2n endpoints ⇒ n diameters.",
          "Thus the maximal number of diameters is n for any finite set of n points."
        ],
        "mutable_slots": {
          "slot_point_labels": {
            "description": "The letters used to denote individual points.",
            "original": "X, P, Q, R, S, T"
          },
          "slot_set_label": {
            "description": "Symbol chosen for the set of points.",
            "original": "𝔈 (\\mathcal{E})"
          },
          "slot_induction_symbols": {
            "description": "Letters for the induction index and its successor.",
            "original": "k and k+1"
          },
          "slot_distance_symbol": {
            "description": "Letter representing the maximal distance (radius of the circle).",
            "original": "r"
          },
          "slot_base_case_range": {
            "description": "Exact small values checked before induction starts (any finite set where the claim is obvious).",
            "original": "n = 2 or 3"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}