{
  "index": "1940-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Prove that the simultaneous equations\n\\[\nx^{4}-x^{2}=y^{4}-y^{2}=z^{4}-z^{2}\n\\]\nare satisfied by the points of four straight lines and six ellipses, and by no other points.",
  "solution": "Solution. Let \\( L \\) denote the locus of the given equations. Then a point is on \\( L \\) if and only if its coordinates ( \\( x, y, z \\) ) satisfy\n\\[\n\\begin{array}{l}\n\\left(x^{2}+y^{2}-1\\right)\\left(x^{2}-y^{2}\\right)=0 \\\\\n\\left(y^{2}+z^{2}-1\\right)\\left(y^{2}-z^{2}\\right)=0 \\\\\n\\left(z^{2}+x^{2}-1\\right)\\left(z^{2}-x^{2}\\right)=0\n\\end{array}\n\\]\n\nConsider the loci \\( A, B, C, D \\) defined as follows:\n\\[\n\\left.\\begin{array}{l}\nA:\\left\\{\\begin{array}{ll}\nx^{2}+y^{2}-1 & =0 \\\\\ny^{2}-z^{2} & =0,\n\\end{array}\\right. \\\\\nB:\\left\\{\\begin{array}{ll}\ny^{2}+z^{2}-1 & =0 \\\\\nz^{2}-x^{2} & =0\n\\end{array}\\right. \\\\\nC:\\left\\{\\begin{array}{l}\nz^{2}+x^{2}-1=0 \\\\\nx^{2}-y^{2}\n\\end{array}=0,\\right.\n\\end{array}\\right] \\begin{array}{l}\n\\text { D: } x^{2}=y^{2}=z^{2}\n\\end{array}\n\\]\n\\( A \\) is the intersection of a right circular cylinder with the union of the planes \\( z=y \\) and \\( z=-y \\). Hence \\( A \\) is the union of two ellipses. Similarly \\( B \\) and \\( C \\) are each the union of two ellipses. \\( D \\), on the other hand, is the union of four straight lines, namely:\n\\[\nx=y=z, \\quad x=y=-z, \\quad x=-y=z \\quad \\text { and } \\quad x=-y=-z\n\\]\n\nAny point common to \\( A \\) and \\( B \\) is in fact also on \\( D \\), so the ellipses of \\( A \\) and \\( B \\) are different. Similarly for \\( B \\) and \\( C \\) and for \\( C \\) and \\( A \\). Hence \\( A \\cup B \\cup C \\cup D \\) consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.\n\nWe now show that \\( L=A \\cup B \\cup C \\cup D \\). If \\( (x, y, z) \\in A \\), then evidently ( \\( x, y, z \\) ) satisfies (1), (2), and (3), the latter because\n\\[\nz^{2}+x^{2}-1=\\left(x^{2}+y^{2}-1\\right)+\\left(z^{2}-y^{2}\\right)=0\n\\]\n\nThus \\( A \\subseteq L \\). Similarly \\( B \\subseteq L \\) and \\( C \\subseteq L \\). It is immediate that \\( D \\subseteq L \\). So \\( A \\cup B \\cup C \\cup D \\subseteq L \\).\n\nNow consider a point \\( p(x, y, z) \\) of \\( L \\) that is not on \\( D \\). Assume, therefore, \\( x^{2} \\neq y^{2} \\) and \\( x^{2} \\neq z^{2} \\). Since \\( p \\) satisfies (1) and (3), we have\n\\[\n\\begin{array}{l}\nx^{2}+y^{2}-1=0 \\\\\nx^{2}+z^{2}-1=0\n\\end{array}\n\\]\nand therefore \\( y^{2}=z^{2} \\), so \\( p \\in A \\). The other cases of inequalities lead to \\( p \\in B \\) or \\( p \\in C \\) by the same argument. Hence \\( L \\subseteq A \\cup B \\cup C \\cup D \\) and indeed \\( L=A \\cup B \\cup C \\cup D \\) is the union of 4 lines and 6 ellipses.",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "A",
    "B",
    "C",
    "D",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "absciss",
        "y": "ordinate",
        "z": "applicate",
        "A": "locusalpha",
        "B": "locusbeta",
        "C": "locusgamma",
        "D": "locusdelta",
        "L": "locusmain"
      },
      "question": "5. Prove that the simultaneous equations\n\\[\nabsciss^{4}-absciss^{2}=ordinate^{4}-ordinate^{2}=applicate^{4}-applicate^{2}\n\\]\nare satisfied by the points of four straight lines and six ellipses, and by no other points.",
      "solution": "Solution. Let \\( locusmain \\) denote the locus of the given equations. Then a point is on \\( locusmain \\) if and only if its coordinates ( \\( absciss, ordinate, applicate \\) ) satisfy\n\\[\n\\begin{array}{l}\n\\left(absciss^{2}+ordinate^{2}-1\\right)\\left(absciss^{2}-ordinate^{2}\\right)=0 \\\\\n\\left(ordinate^{2}+applicate^{2}-1\\right)\\left(ordinate^{2}-applicate^{2}\\right)=0 \\\\\n\\left(applicate^{2}+absciss^{2}-1\\right)\\left(applicate^{2}-absciss^{2}\\right)=0\n\\end{array}\n\\]\n\nConsider the loci \\( locusalpha, locusbeta, locusgamma, locusdelta \\) defined as follows:\n\\[\n\\left.\\begin{array}{l}\n locusalpha:\\left\\{\\begin{array}{ll}\n absciss^{2}+ordinate^{2}-1 & =0 \\\\\n ordinate^{2}-applicate^{2} & =0,\n\\end{array}\\right. \\\\\n locusbeta:\\left\\{\\begin{array}{ll}\n ordinate^{2}+applicate^{2}-1 & =0 \\\\\n applicate^{2}-absciss^{2} & =0\n\\end{array}\\right. \\\\\n locusgamma:\\left\\{\\begin{array}{l}\n applicate^{2}+absciss^{2}-1=0 \\\\\n absciss^{2}-ordinate^{2}\n\\end{array}=0,\\right.\n\\end{array}\\right] \\begin{array}{l}\n \\text { locusdelta: } absciss^{2}=ordinate^{2}=applicate^{2}\n\\end{array}\n\\]\n\\( locusalpha \\) is the intersection of a right circular cylinder with the union of the planes \\( applicate=ordinate \\) and \\( applicate=-ordinate \\). Hence \\( locusalpha \\) is the union of two ellipses. Similarly \\( locusbeta \\) and \\( locusgamma \\) are each the union of two ellipses. \\( locusdelta \\), on the other hand, is the union of four straight lines, namely:\n\\[\nabsciss=ordinate=applicate, \\quad absciss=ordinate=-applicate, \\quad absciss=-ordinate=applicate \\quad \\text { and } \\quad absciss=-ordinate=-applicate\n\\]\n\nAny point common to \\( locusalpha \\) and \\( locusbeta \\) is in fact also on \\( locusdelta \\), so the ellipses of \\( locusalpha \\) and \\( locusbeta \\) are different. Similarly for \\( locusbeta \\) and \\( locusgamma \\) and for \\( locusgamma \\) and \\( locusalpha \\). Hence \\( locusalpha \\cup locusbeta \\cup locusgamma \\cup locusdelta \\) consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.\n\nWe now show that \\( locusmain=locusalpha \\cup locusbeta \\cup locusgamma \\cup locusdelta \\). If \\( (absciss, ordinate, applicate) \\in locusalpha \\), then evidently ( \\( absciss, ordinate, applicate \\) ) satisfies (1), (2), and (3), the latter because\n\\[\napplicate^{2}+absciss^{2}-1=\\left(absciss^{2}+ordinate^{2}-1\\right)+\\left(applicate^{2}-ordinate^{2}\\right)=0\n\\]\n\nThus \\( locusalpha \\subseteq locusmain \\). Similarly \\( locusbeta \\subseteq locusmain \\) and \\( locusgamma \\subseteq locusmain \\). It is immediate that \\( locusdelta \\subseteq locusmain \\). So \\( locusalpha \\cup locusbeta \\cup locusgamma \\cup locusdelta \\subseteq locusmain \\).\n\nNow consider a point \\( p(absciss, ordinate, applicate) \\) of \\( locusmain \\) that is not on \\( locusdelta \\). Assume, therefore, \\( absciss^{2} \\neq ordinate^{2} \\) and \\( absciss^{2} \\neq applicate^{2} \\). Since \\( p \\) satisfies (1) and (3), we have\n\\[\n\\begin{array}{l}\nabsciss^{2}+ordinate^{2}-1=0 \\\\\nabsciss^{2}+applicate^{2}-1=0\n\\end{array}\n\\]\nand therefore \\( ordinate^{2}=applicate^{2} \\), so \\( p \\in locusalpha \\). The other cases of inequalities lead to \\( p \\in locusbeta \\) or \\( p \\in locusgamma \\) by the same argument. Hence \\( locusmain \\subseteq locusalpha \\cup locusbeta \\cup locusgamma \\cup locusdelta \\) and indeed \\( locusmain=locusalpha \\cup locusbeta \\cup locusgamma \\cup locusdelta \\) is the union of 4 lines and 6 ellipses."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "moonlight",
        "y": "sunflower",
        "z": "butterfly",
        "A": "pineapple",
        "B": "strawberry",
        "C": "raspberry",
        "D": "blueberry",
        "L": "watercress"
      },
      "question": "5. Prove that the simultaneous equations\n\\[\nmoonlight^{4}-moonlight^{2}=sunflower^{4}-sunflower^{2}=butterfly^{4}-butterfly^{2}\n\\]\nare satisfied by the points of four straight lines and six ellipses, and by no other points.",
      "solution": "Solution. Let \\( watercress \\) denote the locus of the given equations. Then a point is on \\( watercress \\) if and only if its coordinates ( \\( moonlight, sunflower, butterfly \\) ) satisfy\n\\[\n\\begin{array}{l}\n\\left(moonlight^{2}+sunflower^{2}-1\\right)\\left(moonlight^{2}-sunflower^{2}\\right)=0 \\\\\n\\left(sunflower^{2}+butterfly^{2}-1\\right)\\left(sunflower^{2}-butterfly^{2}\\right)=0 \\\\\n\\left(butterfly^{2}+moonlight^{2}-1\\right)\\left(butterfly^{2}-moonlight^{2}\\right)=0\n\\end{array}\n\\]\n\nConsider the loci \\( pineapple, strawberry, raspberry, blueberry \\) defined as follows:\n\\[\n\\left.\\begin{array}{l}\npineapple:\\left\\{\\begin{array}{ll}\nmoonlight^{2}+sunflower^{2}-1 & =0 \\\\\nsunflower^{2}-butterfly^{2} & =0,\n\\end{array}\\right. \\\\\nstrberry:\\left\\{\\begin{array}{ll}\nsunflower^{2}+butterfly^{2}-1 & =0 \\\\\nbutterfly^{2}-moonlight^{2} & =0\n\\end{array}\\right. \\\\\nraspberry:\\left\\{\\begin{array}{l}\nbutterfly^{2}+moonlight^{2}-1=0 \\\\\nmoonlight^{2}-sunflower^{2}\n\\end{array}=0,\\right.\n\\end{array}\\right] \\begin{array}{l}\n\\text { blueberry: } moonlight^{2}=sunflower^{2}=butterfly^{2}\n\\end{array}\n\\]\n\\( pineapple \\) is the intersection of a right circular cylinder with the union of the planes \\( butterfly=sunflower \\) and \\( butterfly=-sunflower \\). Hence \\( pineapple \\) is the union of two ellipses. Similarly \\( strawberry \\) and \\( raspberry \\) are each the union of two ellipses. \\( blueberry \\), on the other hand, is the union of four straight lines, namely:\n\\[\nmoonlight=sunflower=butterfly, \\quad moonlight=sunflower=-butterfly, \\quad moonlight=-sunflower=butterfly \\quad \\text { and } \\quad moonlight=-sunflower=-butterfly\n\\]\n\nAny point common to \\( pineapple \\) and \\( strawberry \\) is in fact also on \\( blueberry \\), so the ellipses of \\( pineapple \\) and \\( strawberry \\) are different. Similarly for \\( strawberry \\) and \\( raspberry \\) and for \\( raspberry \\) and \\( pineapple \\). Hence \\( pineapple \\cup strawberry \\cup raspberry \\cup blueberry \\) consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.\n\nWe now show that \\( watercress=pineapple \\cup strawberry \\cup raspberry \\cup blueberry \\). If \\( (moonlight, sunflower, butterfly) \\in pineapple \\), then evidently ( \\( moonlight, sunflower, butterfly \\) ) satisfies (1), (2), and (3), the latter because\n\\[\nbutterfly^{2}+moonlight^{2}-1=\\left(moonlight^{2}+sunflower^{2}-1\\right)+\\left(butterfly^{2}-sunflower^{2}\\right)=0\n\\]\n\nThus \\( pineapple \\subseteq watercress \\). Similarly \\( strawberry \\subseteq watercress \\) and \\( raspberry \\subseteq watercress \\). It is immediate that \\( blueberry \\subseteq watercress \\). So \\( pineapple \\cup strawberry \\cup raspberry \\cup blueberry \\subseteq watercress \\).\n\nNow consider a point \\( p(moonlight, sunflower, butterfly) \\) of \\( watercress \\) that is not on \\( blueberry \\). Assume, therefore, \\( moonlight^{2} \\neq sunflower^{2} \\) and \\( moonlight^{2} \\neq butterfly^{2} \\). Since \\( p \\) satisfies (1) and (3), we have\n\\[\n\\begin{array}{l}\nmoonlight^{2}+sunflower^{2}-1=0 \\\\\nmoonlight^{2}+butterfly^{2}-1=0\n\\end{array}\n\\]\nand therefore \\( sunflower^{2}=butterfly^{2} \\), so \\( p \\in pineapple \\). The other cases of inequalities lead to \\( p \\in strawberry \\) or \\( p \\in raspberry \\) by the same argument. Hence \\( watercress \\subseteq pineapple \\cup strawberry \\cup raspberry \\cup blueberry \\) and indeed \\( watercress=pineapple \\cup strawberry \\cup raspberry \\cup blueberry \\) is the union of 4 lines and 6 ellipses."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "constantvalue",
        "z": "planaraxis",
        "A": "widearea",
        "B": "voidzone",
        "C": "solidfield",
        "D": "curvedpath",
        "L": "discretepoint"
      },
      "question": "5. Prove that the simultaneous equations\n\\[\nverticalaxis^{4}-verticalaxis^{2}=constantvalue^{4}-constantvalue^{2}=planaraxis^{4}-planaraxis^{2}\n\\]\nare satisfied by the points of four straight lines and six ellipses, and by no other points.",
      "solution": "Solution. Let \\( discretepoint \\) denote the locus of the given equations. Then a point is on \\( discretepoint \\) if and only if its coordinates ( \\( verticalaxis, constantvalue, planaraxis \\) ) satisfy\n\\[\n\\begin{array}{l}\n\\left(verticalaxis^{2}+constantvalue^{2}-1\\right)\\left(verticalaxis^{2}-constantvalue^{2}\\right)=0 \\\\\n\\left(constantvalue^{2}+planaraxis^{2}-1\\right)\\left(constantvalue^{2}-planaraxis^{2}\\right)=0 \\\\\n\\left(planaraxis^{2}+verticalaxis^{2}-1\\right)\\left(planaraxis^{2}-verticalaxis^{2}\\right)=0\n\\end{array}\n\\]\n\nConsider the loci \\( widearea, voidzone, solidfield, curvedpath \\) defined as follows:\n\\[\n\\left.\\begin{array}{l}\nwidearea:\\left\\{\\begin{array}{ll}\nverticalaxis^{2}+constantvalue^{2}-1 & =0 \\\\\nconstantvalue^{2}-planaraxis^{2} & =0,\n\\end{array}\\right. \\\\\nvoidzone:\\left\\{\\begin{array}{ll}\nconstantvalue^{2}+planaraxis^{2}-1 & =0 \\\\\nplanaraxis^{2}-verticalaxis^{2} & =0\n\\end{array}\\right. \\\\\nsolidfield:\\left\\{\\begin{array}{l}\nplanaraxis^{2}+verticalaxis^{2}-1=0 \\\\\nverticalaxis^{2}-constantvalue^{2}\n\\end{array}=0,\\right.\n\\end{array}\\right] \\begin{array}{l}\n\\text { curvedpath: } verticalaxis^{2}=constantvalue^{2}=planaraxis^{2}\n\\end{array}\n\\]\n\\( widearea \\) is the intersection of a right circular cylinder with the union of the planes \\( planaraxis=constantvalue \\) and \\( planaraxis=-constantvalue \\). Hence \\( widearea \\) is the union of two ellipses. Similarly \\( voidzone \\) and \\( solidfield \\) are each the union of two ellipses. \\( curvedpath \\), on the other hand, is the union of four straight lines, namely:\n\\[\nverticalaxis=constantvalue=planaraxis, \\quad verticalaxis=constantvalue=-planaraxis, \\quad verticalaxis=-constantvalue=planaraxis \\quad \\text { and } \\quad verticalaxis=-constantvalue=-planaraxis\n\\]\n\nAny point common to \\( widearea \\) and \\( voidzone \\) is in fact also on \\( curvedpath \\), so the ellipses of \\( widearea \\) and \\( voidzone \\) are different. Similarly for \\( voidzone \\) and \\( solidfield \\) and for \\( solidfield \\) and \\( widearea \\). Hence \\( widearea \\cup voidzone \\cup solidfield \\cup curvedpath \\) consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.\n\nWe now show that \\( discretepoint=widearea \\cup voidzone \\cup solidfield \\cup curvedpath \\). If \\( (verticalaxis, constantvalue, planaraxis) \\in widearea \\), then evidently ( \\( verticalaxis, constantvalue, planaraxis \\) ) satisfies (1), (2), and (3), the latter because\n\\[\nplanaraxis^{2}+verticalaxis^{2}-1=\\left(verticalaxis^{2}+constantvalue^{2}-1\\right)+\\left(planaraxis^{2}-constantvalue^{2}\\right)=0\n\\]\n\nThus \\( widearea \\subseteq discretepoint \\). Similarly \\( voidzone \\subseteq discretepoint \\) and \\( solidfield \\subseteq discretepoint \\). It is immediate that \\( curvedpath \\subseteq discretepoint \\). So \\( widearea \\cup voidzone \\cup solidfield \\cup curvedpath \\subseteq discretepoint \\).\n\nNow consider a point \\( p(verticalaxis, constantvalue, planaraxis) \\) of \\( discretepoint \\) that is not on \\( curvedpath \\). Assume, therefore, \\( verticalaxis^{2} \\neq constantvalue^{2} \\) and \\( verticalaxis^{2} \\neq planaraxis^{2} \\). Since \\( p \\) satisfies (1) and (3), we have\n\\[\n\\begin{array}{l}\nverticalaxis^{2}+constantvalue^{2}-1=0 \\\\\nverticalaxis^{2}+planaraxis^{2}-1=0\n\\end{array}\n\\]\nand therefore \\( constantvalue^{2}=planaraxis^{2} \\), so \\( p \\in widearea \\). The other cases of inequalities lead to \\( p \\in voidzone \\) or \\( p \\in solidfield \\) by the same argument. Hence \\( discretepoint \\subseteq widearea \\cup voidzone \\cup solidfield \\cup curvedpath \\) and indeed \\( discretepoint=widearea \\cup voidzone \\cup solidfield \\cup curvedpath \\) is the union of 4 lines and 6 ellipses."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mdfplrqe",
        "A": "ksjdhfla",
        "B": "vbncxzpq",
        "C": "fghrtyui",
        "D": "lkjhgfas",
        "L": "asdfqwer"
      },
      "question": "Problem:\n<<<\n5. Prove that the simultaneous equations\n\\[\nqzxwvtnp^{4}-qzxwvtnp^{2}=hjgrksla^{4}-hjgrksla^{2}=mdfplrqe^{4}-mdfplrqe^{2}\n\\]\nare satisfied by the points of four straight lines and six ellipses, and by no other points.\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. Let \\( asdfqwer \\) denote the locus of the given equations. Then a point is on \\( asdfqwer \\) if and only if its coordinates ( \\( qzxwvtnp, hjgrksla, mdfplrqe \\) ) satisfy\n\\[\n\\begin{array}{l}\n\\left(qzxwvtnp^{2}+hjgrksla^{2}-1\\right)\\left(qzxwvtnp^{2}-hjgrksla^{2}\\right)=0 \\\\\n\\left(hjgrksla^{2}+mdfplrqe^{2}-1\\right)\\left(hjgrksla^{2}-mdfplrqe^{2}\\right)=0 \\\\\n\\left(mdfplrqe^{2}+qzxwvtnp^{2}-1\\right)\\left(mdfplrqe^{2}-qzxwvtnp^{2}\\right)=0\n\\end{array}\n\\]\n\nConsider the loci \\( ksjdhfla, vbncxzpq, fghrtyui, lkjhgfas \\) defined as follows:\n\\[\n\\left.\\begin{array}{l}\nksjdhfla:\\left\\{\\begin{array}{ll}\nqzxwvtnp^{2}+hjgrksla^{2}-1 & =0 \\\\\nhjgrksla^{2}-mdfplrqe^{2} & =0,\n\\end{array}\\right. \\\\\nvbncxzpq:\\left\\{\\begin{array}{ll}\nhjgrksla^{2}+mdfplrqe^{2}-1 & =0 \\\\\nmdfplrqe^{2}-qzxwvtnp^{2} & =0\n\\end{array}\\right. \\\\\nfghrtyui:\\left\\{\\begin{array}{l}\nmdfplrqe^{2}+qzxwvtnp^{2}-1=0 \\\\\nqzxwvtnp^{2}-hjgrksla^{2}\n\\end{array}=0,\\right.\n\\end{array}\\right] \\begin{array}{l}\n\\text { lkjhgfas: } qzxwvtnp^{2}=hjgrksla^{2}=mdfplrqe^{2}\n\\end{array}\n\\]\n\\( ksjdhfla \\) is the intersection of a right circular cylinder with the union of the planes \\( mdfplrqe=hjgrksla \\) and \\( mdfplrqe=-hjgrksla \\). Hence \\( ksjdhfla \\) is the union of two ellipses. Similarly \\( vbncxzpq \\) and \\( fghrtyui \\) are each the union of two ellipses. \\( lkjhgfas \\), on the other hand, is the union of four straight lines, namely:\n\\[\nqzxwvtnp=hjgrksla=mdfplrqe, \\quad qzxwvtnp=hjgrksla=-mdfplrqe, \\quad qzxwvtnp=-hjgrksla=mdfplrqe \\quad \\text { and } \\quad qzxwvtnp=-hjgrksla=-mdfplrqe\n\\]\n\nAny point common to \\( ksjdhfla \\) and \\( vbncxzpq \\) is in fact also on \\( lkjhgfas \\), so the ellipses of \\( ksjdhfla \\) and \\( vbncxzpq \\) are different. Similarly for \\( vbncxzpq \\) and \\( fghrtyui \\) and for \\( fghrtyui \\) and \\( ksjdhfla \\). Hence \\( ksjdhfla \\cup vbncxzpq \\cup fghrtyui \\cup lkjhgfas \\) consists of the union of 6 (distinct) ellipses and 4 (distinct) lines.\n\nWe now show that \\( asdfqwer=ksjdhfla \\cup vbncxzpq \\cup fghrtyui \\cup lkjhgfas \\). If \\( (qzxwvtnp, hjgrksla, mdfplrqe) \\in ksjdhfla \\), then evidently ( \\( qzxwvtnp, hjgrksla, mdfplrqe \\) ) satisfies (1), (2), and (3), the latter because\n\\[\nmdfplrqe^{2}+qzxwvtnp^{2}-1=\\left(qzxwvtnp^{2}+hjgrksla^{2}-1\\right)+\\left(mdfplrqe^{2}-hjgrksla^{2}\\right)=0\n\\]\n\nThus \\( ksjdhfla \\subseteq asdfqwer \\). Similarly \\( vbncxzpq \\subseteq asdfqwer \\) and \\( fghrtyui \\subseteq asdfqwer \\). It is immediate that \\( lkjhgfas \\subseteq asdfqwer \\). So \\( ksjdhfla \\cup vbncxzpq \\cup fghrtyui \\cup lkjhgfas \\subseteq asdfqwer \\).\n\nNow consider a point \\( p(qzxwvtnp, hjgrksla, mdfplrqe) \\) of \\( asdfqwer \\) that is not on \\( lkjhgfas \\). Assume, therefore, \\( qzxwvtnp^{2} \\neq hjgrksla^{2} \\) and \\( qzxwvtnp^{2} \\neq mdfplrqe^{2} \\). Since \\( p \\) satisfies (1) and (3), we have\n\\[\n\\begin{array}{l}\nqzxwvtnp^{2}+hjgrksla^{2}-1=0 \\\\\nqzxwvtnp^{2}+mdfplrqe^{2}-1=0\n\\end{array}\n\\]\nand therefore \\( hjgrksla^{2}=mdfplrqe^{2} \\), so \\( p \\in ksjdhfla \\). The other cases of inequalities lead to \\( p \\in vbncxzpq \\) or \\( p \\in fghrtyui \\) by the same argument. Hence \\( asdfqwer \\subseteq ksjdhfla \\cup vbncxzpq \\cup fghrtyui \\cup lkjhgfas \\) and indeed \\( asdfqwer=ksjdhfla \\cup vbncxzpq \\cup fghrtyui \\cup lkjhgfas \\) is the union of 4 lines and 6 ellipses.\n>>>\n"
    },
    "kernel_variant": {
      "question": "Let\n\na(x)=x^{4}-2x^{2}, \\qquad x\\in\\mathbb R.\n\nIn \\(\\mathbb R^{3}\\) consider the locus\n\n\\[L=\\{(x,y,z)\\in\\mathbb R^{3}:a(x)=a(y)=a(z)\\}.\\]\n\nShow that L is the union of exactly four straight lines and six ellipses, and that no other points belong to L.  Give explicit equations for the four lines and the six ellipses.\n\n(You should obtain the four lines\n\\[\\ell_{1}:x=y=z,\\;\\;\\ell_{2}:x=y=-z,\\;\\;\\ell_{3}:x=-y=z,\\;\\;\\ell_{4}:x=-y=-z,\\]\nand the six ellipses\n\\[\nA_{\\pm}:\\begin{cases}y^{2}+z^{2}=2,\\\\ z=\\pm x,\\end{cases}\\qquad\nB_{\\pm}:\\begin{cases}z^{2}+x^{2}=2,\\\\ x=\\pm y,\\end{cases}\\qquad\nC_{\\pm}:\\begin{cases}x^{2}+y^{2}=2,\\\\ y=\\pm z.\\end{cases}\n\\])",
      "solution": "Throughout write a(t)=t^{4}-2t^{2}\\;(t\\in\\mathbb R).  A point (x,y,z) is in L precisely when\n\n\\[a(x)=a(y)=a(z).\\]\n\nFactorisation of the pairwise differences.\nFor any real numbers u,v\n\\[\n a(u)-a(v)=u^{4}-v^{4}-2(u^{2}-v^{2})=(u^{2}-v^{2})(u^{2}+v^{2}-2).\n\\]\nHence the three equalities a(x)=a(y), a(y)=a(z), a(z)=a(x) are equivalent to\n\\[\n\\begin{cases}\n(x^{2}-y^{2})(x^{2}+y^{2}-2)=0,\\\\\n(y^{2}-z^{2})(y^{2}+z^{2}-2)=0,\\\\\n(z^{2}-x^{2})(z^{2}+x^{2}-2)=0.\n\\end{cases}\\tag{1}\n\\]\nFor every product at least one factor vanishes.  We distinguish cases.\n\n1.  All three difference–factors vanish:  \\(x^{2}=y^{2}=z^{2}.\\)\nThen x=\\pm y and x=\\pm z.  The eight sign possibilities collapse into four geometric lines (the common factor −1 yields the same set):\n\\[\n\\ell_{1}:x=y=z,\\;\\;\\ell_{2}:x=y=-z,\\;\\;\\ell_{3}:x=-y=z,\\;\\;\\ell_{4}:x=-y=-z.\\tag{2}\n\\]\nEvery point of the union \\(\\ell_{1}\\cup\\dots\\cup\\ell_{4}\\) satisfies (1).\n\n2.  At least one pair of squares is different.\nAssume first\n\\[x^{2}\\ne y^{2}.\\]\nThen for the first equation of (1) the factor \\(x^{2}-y^{2}\\) is non–zero, so the other factor must vanish:\n\\[x^{2}+y^{2}=2.\\tag{3}\\]\n\n2a.  Suppose in addition \\(y^{2}\\ne z^{2}.\\)\nThe second equation of (1) now yields \\(y^{2}+z^{2}=2.\\)  Subtracting this from (3) gives \\(x^{2}=z^{2}\\), i.e.  \\(z=\\pm x\\).  Combining with (3) we obtain the two ellipses\n\\[\nA_{\\pm}:\\;\\begin{cases}y^{2}+z^{2}=2,\\\\ z=\\pm x.\\end{cases}\n\\]\nThey lie in the planes \\(z=\\pm x\\) and are the sections of the circular cylinder \\(y^{2}+z^{2}=2\\) by those planes.\n\n2b.  Suppose instead \\(y^{2}=z^{2}.\\)  (Because we are not on the lines (2) we still have \\(x^{2}\\ne z^{2}.\\))  The second equation of (1) is automatically satisfied, while the third equation forces\n\\[z^{2}+x^{2}=2.\\]\nTogether with \\(y^{2}=z^{2}\\) this gives\n\\[\n x^{2}+y^{2}=2,\\qquad y=\\pm z.\n\\]\nThus we obtain the two ellipses\n\\[\nC_{\\pm}:\\;\\begin{cases}x^{2}+y^{2}=2,\\\\ y=\\pm z.\\end{cases}\n\\]\n\nUp to this point we have produced four of the six ellipses (A_{\\pm}, C_{\\pm}).\n\n3.  Symmetric repetitions.\nThe defining equations (1) are completely symmetric in the variables x,y,z.  Cyclically permuting the roles of the coordinates in the analysis of Section 2 produces the remaining pair of ellipses\n\\[\nB_{\\pm}:\\;\\begin{cases}z^{2}+x^{2}=2,\\\\ x=\\pm y,\\end{cases}\n\\]\nobtained by starting with the assumption \\(y^{2}\\ne z^{2}\\) (or \\(z^{2}\\ne x^{2}\\)).  Because the arguments are identical after the permutation, no new case–work is necessary.\n\n4.  Exhaustion of the locus.\nWe have exhibited ten sets – the four lines (2) and the six ellipses A_{\\pm},B_{\\pm},C_{\\pm}.  Every point on each of these ten sets satisfies (1), hence belongs to L:\n\\[\n \\ell_{1}\\cup\\ell_{2}\\cup\\ell_{3}\\cup\\ell_{4}\\cup A_{+}\\cup A_{-}\\cup B_{+}\\cup B_{-}\\cup C_{+}\\cup C_{-}\\subseteq L.\n\\]\nConversely, the discussion in Sections 1–3 shows that a point of L either has all three squares equal (giving the four lines) or, after possibly relabelling the coordinates, satisfies the hypotheses of Section 2, which place it on exactly one of the six ellipses.  Therefore\n\\[\n L=(\\ell_{1}\\cup\\ell_{2}\\cup\\ell_{3}\\cup\\ell_{4})\\cup(A_{+}\\cup A_{-}\\cup B_{+}\\cup B_{-}\\cup C_{+}\\cup C_{-}).\n\\]\n\n5.  Intersections.\nThe two ellipses in the same family meet in two points; for instance\n\\(A_{+}\\cap A_{-}=\\{(0,\\pm\\sqrt2,0)\\}.\\)  Intersections between ellipses from different families occur precisely along the four lines (2).  These intersection facts do not alter the classification of L as the union of four distinct lines and six distinct ellipses.\n\nHence the locus of the equation a(x)=a(y)=a(z) consists of exactly the ten sets listed above, completing the proof.",
      "_meta": {
        "core_steps": [
          "Take pairwise differences of the equal expressions so they equal 0.",
          "Factor each difference as (x²−y²)(x²+y²−1)=0 (difference of squares trick).",
          "Split the locus according to which factor vanishes, defining four sets A,B,C,D.",
          "Recognize A,B,C as cylinder-plane intersections (ellipses) and D as coordinate-equality lines.",
          "Show the union A∪B∪C∪D equals the whole locus by double containment."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The constant in the quadratic factors that gives the cylinder radius; x²+y²−1 can be replaced by x²+y²−c with any positive c.",
            "original": 1
          },
          "slot2": {
            "description": "Which coordinate pair is used in each factorisation step (i.e., any permutation of (x,y,z) when naming A,B,C).",
            "original": "(x,y,z) order"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}