{
  "index": "1964-B-6",
  "type": "GEO",
  "tag": [
    "GEO"
  ],
  "difficulty": "",
  "question": "6. Show that the unit disk in the plane cannot be partitioned into two disjoint congruent subsets.",
  "solution": "Solution. Let \\( D \\) be the unit disk, \\( p \\) its center and \\( \\rho \\) the usual distance function.\nSuppose \\( D \\) is the union of two disjoint congruent sets \\( A, B \\), where the notation is so chosen that \\( p \\in A \\). Let \\( x \\rightarrow x^{*} \\) be the congruence map from \\( \\boldsymbol{A} \\) to \\( B \\). Then \\( \\boldsymbol{p}^{*} \\) is in \\( B \\). Let \\( r \\) and \\( s \\) be the endpoints of the diameter of \\( \\boldsymbol{D} \\) perpendicular to \\( p p^{*} \\).\n\nSince \\( \\rho(p . a) \\leq 1 \\) for every point \\( a \\in A, \\rho\\left(p^{*}, b\\right) \\leq 1 \\) for every \\( b \\in B \\). Clearly \\( \\rho\\left(p^{*}, r\\right)=\\rho\\left(p^{*}, s\\right)>1 \\); so we conclude that \\( r \\) and \\( s \\) belong to \\( A \\). Then \\( \\rho\\left(r^{*}, s^{*}\\right)=\\rho(r, s)=2 \\). So \\( r^{*} \\) and \\( s^{*} \\) are the endpoints of another diameter of \\( D \\). Then \\( p^{*} \\) must be the midpoint of this diameter since \\( \\rho\\left(p^{*}, r^{*}\\right)=\\rho(p, r)=1 \\) and similarly \\( \\rho\\left(p^{*}, s^{*}\\right)=1 \\). Thus \\( p^{*}=p \\), a contradiction, since \\( A \\cap B=\\emptyset \\). Therefore \\( D \\) is not the union of two disjoint congruent sets.",
  "vars": [
    "x",
    "a",
    "b"
  ],
  "params": [
    "D",
    "p",
    "A",
    "B",
    "r",
    "s",
    "\\\\rho"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "a": "variablea",
        "b": "variableb",
        "D": "unitdisk",
        "p": "centerpt",
        "A": "subsetone",
        "B": "subsettwo",
        "r": "endpointone",
        "s": "endpointtwo",
        "\\rho": "distance"
      },
      "question": "6. Show that the unit disk in the plane cannot be partitioned into two disjoint congruent subsets.",
      "solution": "Solution. Let \\( unitdisk \\) be the unit disk, \\( centerpt \\) its center and \\( distance \\) the usual distance function.\nSuppose \\( unitdisk \\) is the union of two disjoint congruent sets \\( subsetone, subsettwo \\), where the notation is so chosen that \\( centerpt \\in subsetone \\). Let \\( variablex \\rightarrow variablex^{*} \\) be the congruence map from \\( \\boldsymbol{subsetone} \\) to \\( subsettwo \\). Then \\( \\boldsymbol{centerpt}^{*} \\) is in \\( subsettwo \\). Let \\( endpointone \\) and \\( endpointtwo \\) be the endpoints of the diameter of \\( \\boldsymbol{unitdisk} \\) perpendicular to \\( centerpt\\,centerpt^{*} \\).\n\nSince \\( distance(centerpt . variablea) \\leq 1 \\) for every point \\( variablea \\in subsetone, distance\\left(centerpt^{*}, variableb\\right) \\leq 1 \\) for every \\( variableb \\in subsettwo \\). Clearly \\( distance\\left(centerpt^{*}, endpointone\\right)=distance\\left(centerpt^{*}, endpointtwo\\right)>1 \\); so we conclude that \\( endpointone \\) and \\( endpointtwo \\) belong to \\( subsetone \\). Then \\( distance\\left(endpointone^{*}, endpointtwo^{*}\\right)=distance(endpointone, endpointtwo)=2 \\). So \\( endpointone^{*} \\) and \\( endpointtwo^{*} \\) are the endpoints of another diameter of \\( unitdisk \\). Then \\( centerpt^{*} \\) must be the midpoint of this diameter since \\( distance\\left(centerpt^{*}, endpointone^{*}\\right)=distance(centerpt, endpointone)=1 \\) and similarly \\( distance\\left(centerpt^{*}, endpointtwo^{*}\\right)=1 \\). Thus \\( centerpt^{*}=centerpt \\), a contradiction, since \\( subsetone \\cap subsettwo=\\emptyset \\). Therefore \\( unitdisk \\) is not the union of two disjoint congruent sets."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "a": "mountain",
        "b": "lanterns",
        "D": "umbrella",
        "p": "seashell",
        "A": "stardust",
        "B": "gemstone",
        "r": "violinist",
        "s": "chandelier",
        "\\rho": "lighthouse"
      },
      "question": "6. Show that the unit disk in the plane cannot be partitioned into two disjoint congruent subsets.",
      "solution": "Solution. Let \\( umbrella \\) be the unit disk, \\( seashell \\) its center and \\( lighthouse \\) the usual distance function.\nSuppose \\( umbrella \\) is the union of two disjoint congruent sets \\( stardust, gemstone \\), where the notation is so chosen that \\( seashell \\in stardust \\). Let \\( sunflower \\rightarrow sunflower^{*} \\) be the congruence map from \\( \\boldsymbol{stardust} \\) to \\( gemstone \\). Then \\( \\boldsymbol{seashell}^{*} \\) is in \\( gemstone \\). Let \\( violinist \\) and \\( chandelier \\) be the endpoints of the diameter of \\( \\boldsymbol{umbrella} \\) perpendicular to \\( seashell seashell^{*} \\).\n\nSince \\( lighthouse(seashell . mountain) \\leq 1 \\) for every point \\( mountain \\in stardust, lighthouse\\left(seashell^{*}, lanterns\\right) \\leq 1 \\) for every \\( lanterns \\in gemstone \\). Clearly \\( lighthouse\\left(seashell^{*}, violinist\\right)=lighthouse\\left(seashell^{*}, chandelier\\right)>1 \\); so we conclude that \\( violinist \\) and \\( chandelier \\) belong to \\( stardust \\). Then \\( lighthouse\\left(violinist^{*}, chandelier^{*}\\right)=lighthouse(violinist, chandelier)=2 \\). So \\( violinist^{*} \\) and \\( chandelier^{*} \\) are the endpoints of another diameter of \\( umbrella \\). Then \\( seashell^{*} \\) must be the midpoint of this diameter since \\( lighthouse\\left(seashell^{*}, violinist^{*}\\right)=lighthouse(seashell, violinist)=1 \\) and similarly \\( lighthouse\\left(seashell^{*}, chandelier^{*}\\right)=1 \\). Thus \\( seashell^{*}=seashell \\), a contradiction, since \\( stardust \\cap gemstone=\\emptyset \\). Therefore \\( umbrella \\) is not the union of two disjoint congruent sets."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stationarypoint",
        "a": "nonmemberpt",
        "b": "outsiderpt",
        "D": "voidregion",
        "p": "edgepoint",
        "A": "wholeset",
        "B": "completeset",
        "r": "interiorpoint",
        "s": "medianpoint",
        "\\rho": "\\proximityfunction"
      },
      "question": "6. Show that the unit disk in the plane cannot be partitioned into two disjoint congruent subsets.",
      "solution": "Solution. Let \\( voidregion \\) be the unit disk, \\( edgepoint \\) its center and \\( \\proximityfunction \\) the usual distance function.\nSuppose \\( voidregion \\) is the union of two disjoint congruent sets \\( wholeset, completeset \\), where the notation is so chosen that \\( edgepoint \\in wholeset \\). Let \\( stationarypoint \\rightarrow stationarypoint^{*} \\) be the congruence map from \\( \\boldsymbol{wholeset} \\) to \\( completeset \\). Then \\( \\boldsymbol{edgepoint}^{*} \\) is in \\( completeset \\). Let \\( interiorpoint \\) and \\( medianpoint \\) be the endpoints of the diameter of \\( \\boldsymbol{voidregion} \\) perpendicular to \\( edgepoint edgepoint^{*} \\).\n\nSince \\( \\proximityfunction(edgepoint . nonmemberpt) \\leq 1 \\) for every point \\( nonmemberpt \\in wholeset, \\proximityfunction\\left(edgepoint^{*}, outsiderpt\\right) \\leq 1 \\) for every \\( outsiderpt \\in completeset \\). Clearly \\( \\proximityfunction\\left(edgepoint^{*}, interiorpoint\\right)=\\proximityfunction\\left(edgepoint^{*}, medianpoint\\right)>1 \\); so we conclude that \\( interiorpoint \\) and \\( medianpoint \\) belong to \\( wholeset \\). Then \\( \\proximityfunction\\left(interiorpoint^{*}, medianpoint^{*}\\right)=\\proximityfunction(interiorpoint, medianpoint)=2 \\). So \\( interiorpoint^{*} \\) and \\( medianpoint^{*} \\) are the endpoints of another diameter of \\( voidregion \\). Then \\( edgepoint^{*} \\) must be the midpoint of this diameter since \\( \\proximityfunction\\left(edgepoint^{*}, interiorpoint^{*}\\right)=\\proximityfunction(edgepoint, interiorpoint)=1 \\) and similarly \\( \\proximityfunction\\left(edgepoint^{*}, medianpoint^{*}\\right)=1 \\). Thus \\( edgepoint^{*}=edgepoint \\), a contradiction, since \\( wholeset \\cap completeset=\\emptyset \\). Therefore \\( voidregion \\) is not the union of two disjoint congruent sets."
    },
    "garbled_string": {
      "map": {
        "x": "uxqhtbvl",
        "a": "pkqvmrse",
        "b": "yznlgwca",
        "D": "fnvrqzom",
        "p": "jqlhxstn",
        "A": "mzdgfrcu",
        "B": "hxlnsvke",
        "r": "weskdvao",
        "s": "tiqnbram",
        "\\rho": "makdpwcz"
      },
      "question": "6. Show that the unit disk in the plane cannot be partitioned into two disjoint congruent subsets.",
      "solution": "Solution. Let \\( fnvrqzom \\) be the unit disk, \\( jqlhxstn \\) its center and \\( makdpwcz \\) the usual distance function.\nSuppose \\( fnvrqzom \\) is the union of two disjoint congruent sets \\( mzdgfrcu, hxlnsvke \\), where the notation is so chosen that \\( jqlhxstn \\in mzdgfrcu \\). Let \\( uxqhtbvl \\rightarrow uxqhtbvl^{*} \\) be the congruence map from \\( \\boldsymbol{mzdgfrcu} \\) to \\( hxlnsvke \\). Then \\( \\boldsymbol{jqlhxstn}^{*} \\) is in \\( hxlnsvke \\). Let \\( weskdvao \\) and \\( tiqnbram \\) be the endpoints of the diameter of \\( \\boldsymbol{fnvrqzom} \\) perpendicular to \\( jqlhxstn jqlhxstn^{*} \\).\n\nSince \\( makdpwcz(jqlhxstn . pkqvmrse) \\leq 1 \\) for every point \\( pkqvmrse \\in mzdgfrcu, makdpwcz\\left(jqlhxstn^{*}, yznlgwca\\right) \\leq 1 \\) for every \\( yznlgwca \\in hxlnsvke \\). Clearly \\( makdpwcz\\left(jqlhxstn^{*}, weskdvao\\right)=makdpwcz\\left(jqlhxstn^{*}, tiqnbram\\right)>1 \\); so we conclude that \\( weskdvao \\) and \\( tiqnbram \\) belong to \\( mzdgfrcu \\). Then \\( makdpwcz\\left(weskdvao^{*}, tiqnbram^{*}\\right)=makdpwcz(weskdvao, tiqnbram)=2 \\). So \\( weskdvao^{*} \\) and \\( tiqnbram^{*} \\) are the endpoints of another diameter of \\( fnvrqzom \\). Then \\( jqlhxstn^{*} \\) must be the midpoint of this diameter since \\( makdpwcz\\left(jqlhxstn^{*}, weskdvao^{*}\\right)=makdpwcz(jqlhxstn, weskdvao)=1 \\) and similarly \\( makdpwcz\\left(jqlhxstn^{*}, tiqnbram^{*}\\right)=1 \\). Thus \\( jqlhxstn^{*}=jqlhxstn \\), a contradiction, since \\( mzdgfrcu \\cap hxlnsvke=\\emptyset \\). Therefore \\( fnvrqzom \\) is not the union of two disjoint congruent sets."
    },
    "kernel_variant": {
      "question": "Let\n\\[\nB\\;=\\;\\bigl\\{(x,y,z)\\in\\mathbb R^{3} : x^{2}+y^{2}+z^{2}\\le 9\\bigr\\}\n\\]\nbe the solid ball of radius $3$ in three–dimensional Euclidean space.  Prove that $B$ cannot be written as the union of two disjoint congruent subsets.",
      "solution": "Suppose, toward a contradiction, that the closed ball\n B = { (x,y,z)∈ℝ³ : x²+y²+z² ≤ 9 }\ncan be partitioned into two disjoint congruent subsets A and C.  Then there is a distance‐preserving bijection f: A→C.  Write p=(0,0,0) for the center of B; by symmetry assume p∈A, and set p*=f(p)∈C.  Since p*≠p we have d=|p−p*|>0.\n\n1.  Choose any unit vector v perpendicular to p*–p, and let r=p+3v and s=p–3v.  Then r,s lie on the boundary sphere and the line L through r,s is a diameter perpendicular to pp*.\n\n2.  For any a∈A the image c=f(a)∈C satisfies |c−p*|=|a−p|≤3.  Hence C⊆closed ball of radius 3 about p*.  But\n     |r−p*|² = |r−p|² + |p−p*|² = 9 + d² > 9,\nso |r−p*|>3, forcing r∉C and thus r∈A.  Similarly s∈A.\n\n3.  Set r*=f(r) and s*=f(s).  Then r*,s*∈C⊆B and\n     |r*−s*| = |r−s| = 6.\nThus r*,s* attain the maximum possible distance 6 inside B and so are antipodal boundary points.  Moreover\n     |r*−p*| = |r−p| = 3,\n     |p*−s*| = |p−s| = 3.\nBy the triangle equality,\n     |r*−p*| + |p*−s*| = 3 + 3 = 6 = |r*−s*|,\nso p* lies on the segment from r* to s*.  Equal endpoint–distances force p* to be the unique midpoint of r*s*.\n\n4.  But the midpoint of any pair of antipodal points in B is the center p.  Hence p*=p, contradicting p∈A, p*∈C, A∩C=∅.\n\nTherefore no such decomposition into two disjoint congruent parts exists.",
      "_meta": {
        "core_steps": [
          "Assume a disjoint partition D = A ∪ B with an isometry f : A → B (contradiction setup).",
          "Take the center p ∈ A and its image p* = f(p) ∈ B (distance preservation anchor).",
          "Pick diameter endpoints r, s orthogonal to pp*, so dist(p*, r) and dist(p*, s) exceed the radius ⇒ r, s ∈ A.",
          "Images r* = f(r), s* = f(s) satisfy |r*s*| = |rs| = 2·radius ⇒ r*, s* form a diameter with midpoint p* (geometry of a ball).",
          "Thus p* = center = p ∈ A ∩ B, contradicting disjointness; therefore no such partition exists."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Actual numerical value of the radius; scaling the entire argument changes all occurrences proportionally (|pp*| threshold, diameter length, etc.) but leaves the reasoning intact.",
            "original": "1"
          },
          "slot2": {
            "description": "Ambient Euclidean dimension; the proof works in any n ≥ 2 where a ball has perpendicular diameters.",
            "original": "plane (2-dimensional)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}