{
  "index": "1948-B-6",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. Answer either (i) or (ii):\n(i) Let \\( V_{1}, V_{2}, V_{3} \\), and \\( V \\) denote four vertices of a cube. \\( V_{1}, V_{2} \\), and \\( V_{3} \\) are next neighbors of \\( V \\), that is, the lines \\( V V_{1}, V V_{2} \\), and \\( V V_{3} \\) are edges of the cube. Project the cube orthogonally onto a plane (the \\( z \\)-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of \\( V \\) fall in the origin and the projections of \\( V_{1}, V_{2} \\), and \\( V_{3} \\) in points marked with the complex numbers \\( z_{1}, z_{2} \\), and \\( z_{3} \\), respectively. Show that \\( z_{1}{ }^{2}+z_{2^{2}}+z_{3^{2}}=0 \\).\n(page 261)\n(ii) Let \\( a_{i j} \\) be a determinant in which each diagonal element exceeds in absolute value the sum of the absolute values of the other elements of its row, that is\n\\[\n\\left|a_{i i}\\right|>\\left|a_{i 1}\\right|+\\left|a_{i 2}\\right|+\\cdots+\\left|a_{i, i-1}\\right|+\\left|a_{i, i+1}\\right|+\\cdots+\\left|a_{i n}\\right|\n\\]\n\nShow that the determinant is not equal to zero. (Consider the corresponding system of linear homogeneous equations.)",
  "solution": "Solution. Take Cartesian coordinates \\( w, x, y \\) in space so that the \\( x \\)-axis is the real axis and the \\( y \\)-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point ( \\( w, x, y \\) ) onto \\( (0, x, y) \\).\n\nSuppose ( \\( w_{1}, x_{1}, y_{1} \\) ), \\( \\left(w_{2}, x_{2}, y_{2}\\right) \\) and ( \\( \\left.w_{3}, x_{3}, y_{3}\\right) \\) are mutually orthogonal unit vectors in space. Then the matrix\n\\[\n\\left(\\begin{array}{lll}\nw_{1} & x_{1} & y_{1} \\\\\nw_{2} & x_{2} & y_{2} \\\\\nw_{3} & x_{3} & y_{3}\n\\end{array}\\right)\n\\]\nis an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular\n\\[\n\\begin{array}{c}\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1 \\\\\ny_{1}^{2}+y_{2}^{2}+y_{3}^{2}=1 \\\\\nx_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 .\n\\end{array}\n\\]\n\nThe orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers \\( x_{1}+i y_{1}, x_{2}+i y_{2} \\), and \\( x_{3}+i y_{3} \\), and\n\\[\n\\begin{array}{c}\n\\left(x_{1}+i y_{1}\\right)^{2}+\\left(x_{2}+i y_{2}\\right)^{2}+\\left(x_{3}+i y_{3}\\right)^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-y_{1}^{2} \\\\\n-y_{2}^{2}-y_{3}^{2}+2 i\\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}\\right)=0 .\n\\end{array}\n\\]\n\nSuppose the sides of the given cube are of length \\( a \\). Since it is given that the vertex \\( V \\) projects onto the origin, it must be of the form \\( V=(b, 0,0) \\). There are mutually orthogonal unit vectors ( \\( w_{j}, x_{j}, y_{j} \\) ) such that\n\\[\nV_{j}=(b, 0,0)+a\\left(w_{j}, x_{j}, y_{j}\\right) \\quad \\text { for } j=1,2,3\n\\]\n\nThen the projection of \\( V_{j} \\) into the Gaussian plane is\n\\[\nz_{j}=a\\left(x_{j}+i y_{j}\\right)\n\\]\nand\n\\[\nz_{1}^{2}+z_{2}^{2}+z_{3}^{2}=a^{2}\\left[\\left(x_{1}+i y_{1}\\right)^{2}+\\left(x_{2}+i y_{2}\\right)^{2}+\\left(x_{3}+i y_{3}\\right)^{2}\\right]=0 .\n\\]\n\nSolution. The corresponding system of linear equations is\n\\[\n\\sum_{j=1}^{n} a_{i j} x_{j}=0, \\quad i=1,2, \\ldots, n\n\\]\n\nIf the determinant of the matrix of the coefficients is zero, there exists a non-trivial solution, say ( \\( \\bar{x}_{1}, \\bar{x}_{2}, \\ldots, \\bar{x}_{n} \\) ), of the system. Let \\( m \\) be an index for which \\( \\left|\\bar{x}_{m}\\right| \\) is largest, that is \\( \\left|\\bar{x}_{m}\\right| \\geq\\left|\\bar{x}_{j}\\right| \\) for \\( j=1,2, \\ldots, n \\). Clearly, \\( \\left|\\bar{x}_{m}\\right| \\neq 0 \\), since the solution is non-trivial. Consider the \\( m \\) th equation in the above system written in the form\n\\[\n-a_{m m} \\bar{x}_{m}=\\sum_{j \\neq m} a_{m j} \\bar{x}_{j}\n\\]\n\nWe have\n\\[\n\\left|a_{m m}\\right|\\left|\\bar{x}_{m}\\right| \\leq \\sum_{j \\neq m}\\left|a_{m j}\\right|\\left|\\bar{x}_{j}\\right| \\leq\\left(\\sum_{j \\neq m}\\left|a_{m j}\\right|\\right)\\left|\\bar{x}_{m}\\right|\n\\]\nand therefore\n\\[\n\\left|a_{m m}\\right| \\leq \\sum_{j \\neq m}\\left|a_{m j}\\right|\n\\]\ncontrary to hypothesis. Hence the determinant cannot be zero.\nRemark. This is the same type of argument that was used by Gersgorin to obtain bounds on the eigenvalues of a matrix. See Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, 1964, page 146.",
  "vars": [
    "a_ii",
    "a_ij",
    "a_i1",
    "a_i2",
    "a_in",
    "a_mj",
    "a_mm",
    "j",
    "m",
    "V",
    "V_1",
    "V_2",
    "V_3",
    "w",
    "w_1",
    "w_2",
    "w_3",
    "w_j",
    "x",
    "x_1",
    "x_2",
    "x_3",
    "x_j",
    "x_m",
    "y",
    "y_1",
    "y_2",
    "y_3",
    "y_j",
    "z",
    "z_1",
    "z_2",
    "z_3"
  ],
  "params": [
    "a",
    "b",
    "n"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_ii": "diagentry",
        "a_ij": "genentry",
        "a_i1": "entryfirst",
        "a_i2": "entrysecond",
        "a_in": "entrylast",
        "a_mj": "rowmentry",
        "a_mm": "rowmdiag",
        "j": "indexvar",
        "m": "maxindex",
        "V": "cubevert",
        "V_1": "cubevone",
        "V_2": "cubevtwo",
        "V_3": "cubevthree",
        "w": "coordw",
        "w_1": "vectorwone",
        "w_2": "vectorwtwo",
        "w_3": "vectorwthree",
        "w_j": "vectorwgen",
        "x": "coordx",
        "x_1": "vectorxone",
        "x_2": "vectorxtwo",
        "x_3": "vectorxthree",
        "x_j": "vectorxgen",
        "x_m": "vectorxmax",
        "y": "coordy",
        "y_1": "vectoryone",
        "y_2": "vectorytwo",
        "y_3": "vectorythree",
        "y_j": "vectorygen",
        "z": "complexz",
        "z_1": "complexzone",
        "z_2": "complexztwo",
        "z_3": "complexzthree",
        "a": "sidelength",
        "b": "offsetval",
        "n": "dimsize"
      },
      "question": "6. Answer either (i) or (ii):\n(i) Let \\( cubevone, cubevtwo, cubevthree \\), and \\( cubevert \\) denote four vertices of a cube. \\( cubevone, cubevtwo \\), and \\( cubevthree \\) are next neighbors of \\( cubevert \\), that is, the lines \\( cubevert cubevone, cubevert cubevtwo \\), and \\( cubevert cubevthree \\) are edges of the cube. Project the cube orthogonally onto a plane (the \\( complexz \\)-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of \\( cubevert \\) fall in the origin and the projections of \\( cubevone, cubevtwo \\), and \\( cubevthree \\) in points marked with the complex numbers \\( complexzone, complexztwo \\), and \\( complexzthree \\), respectively. Show that \\( complexzone^{2}+complexztwo^{2}+complexzthree^{2}=0 \\).\n(page 261)\n(ii) Let \\( genentry \\) be a determinant in which each diagonal element exceeds in absolute value the sum of the absolute values of the other elements of its row, that is\n\\[\n\\left|diagentry\\right|>\\left|entryfirst\\right|+\\left|entrysecond\\right|+\\cdots+\\left|genentry_{i, i-1}\\right|+\\left|genentry_{i, i+1}\\right|+\\cdots+\\left|entrylast\\right|\n\\]\n\nShow that the determinant is not equal to zero. (Consider the corresponding system of linear homogeneous equations.)",
      "solution": "Solution. Take Cartesian coordinates \\( coordw, coordx, coordy \\) in space so that the \\( coordx \\)-axis is the real axis and the \\( coordy \\)-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point \\( ( coordw, coordx, coordy ) \\) onto \\( (0, coordx, coordy) \\).\n\nSuppose \\( ( vectorwone, vectorxone, vectoryone ) , ( vectorwtwo, vectorxtwo, vectorytwo ) \\) and \\( ( vectorwthree, vectorxthree, vectorythree ) \\) are mutually orthogonal unit vectors in space. Then the matrix\n\\[\n\\left(\\begin{array}{lll}\nvectorwone & vectorxone & vectoryone \\\\\nvectorwtwo & vectorxtwo & vectorytwo \\\\\nvectorwthree & vectorxthree & vectorythree\n\\end{array}\\right)\n\\]\nis an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular\n\\[\n\\begin{array}{c}\nvectorxone^{2}+vectorxtwo^{2}+vectorxthree^{2}=1 \\\\\nvectoryone^{2}+vectorytwo^{2}+vectorythree^{2}=1 \\\\\nvectorxone\\,vectoryone+vectorxtwo\\,vectorytwo+vectorxthree\\,vectorythree=0 .\n\\end{array}\n\\]\n\nThe orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers \\( vectorxone+i\\,vectoryone, vectorxtwo+i\\,vectorytwo \\), and \\( vectorxthree+i\\,vectorythree \\), and\n\\[\n\\begin{array}{c}\n\\left(vectorxone+i\\,vectoryone\\right)^{2}+\\left(vectorxtwo+i\\,vectorytwo\\right)^{2}+\\left(vectorxthree+i\\,vectorythree\\right)^{2}=vectorxone^{2}+vectorxtwo^{2}+vectorxthree^{2}-vectoryone^{2} \\\\\n-vectorytwo^{2}-vectorythree^{2}+2 i\\left(vectorxone\\,vectoryone+vectorxtwo\\,vectorytwo+vectorxthree\\,vectorythree\\right)=0 .\n\\end{array}\n\\]\n\nSuppose the sides of the given cube are of length \\( sidelength \\). Since it is given that the vertex \\( cubevert \\) projects onto the origin, it must be of the form \\( cubevert=(offsetval, 0,0) \\). There are mutually orthogonal unit vectors \\( ( vectorwgen, vectorxgen, vectorygen ) \\) such that\n\\[\ncubevone=(offsetval, 0,0)+sidelength( vectorwone, vectorxone, vectoryone ),\\quad\ncubevtwo=(offsetval, 0,0)+sidelength( vectorwtwo, vectorxtwo, vectorytwo ),\\quad\ncubevthree=(offsetval, 0,0)+sidelength( vectorwthree, vectorxthree, vectorythree )\n\\]\n\nThen the projection of \\( cubevone, cubevtwo, cubevthree \\) into the Gaussian plane is\n\\[\ncomplexzone=sidelength\\left(vectorxone+i\\,vectoryone\\right),\\quad\ncomplexztwo=sidelength\\left(vectorxtwo+i\\,vectorytwo\\right),\\quad\ncomplexzthree=sidelength\\left(vectorxthree+i\\,vectorythree\\right)\n\\]\nand\n\\[\ncomplexzone^{2}+complexztwo^{2}+complexzthree^{2}=sidelength^{2}\\left[\\left(vectorxone+i\\,vectoryone\\right)^{2}+\\left(vectorxtwo+i\\,vectorytwo\\right)^{2}+\\left(vectorxthree+i\\,vectorythree\\right)^{2}\\right]=0 .\n\\]\n\nSolution. The corresponding system of linear equations is\n\\[\n\\sum_{indexvar=1}^{dimsize} genentry\\,vectorxgen =0, \\quad i=1,2, \\ldots, dimsize\n\\]\n\nIf the determinant of the matrix of the coefficients is zero, there exists a non-trivial solution, say \\( ( \\bar{vectorxone}, \\bar{vectorxtwo}, \\ldots, \\bar{vectorxgen} ) \\), of the system. Let \\( maxindex \\) be an index for which \\( |\\bar{vectorxmax}| \\) is largest, that is \\( |\\bar{vectorxmax}| \\ge |\\bar{vectorxgen}| \\) for \\( indexvar=1,2, \\ldots, dimsize \\). Clearly, \\( |\\bar{vectorxmax}| \\neq 0 \\), since the solution is non-trivial. Consider the \\( maxindex \\) th equation in the above system written in the form\n\\[\n-rowmdiag \\, \\bar{vectorxmax}=\\sum_{indexvar \\neq maxindex} rowmentry \\, \\bar{vectorxgen}\n\\]\n\nWe have\n\\[\n|rowmdiag|\\,|\\bar{vectorxmax}| \\le \\sum_{indexvar \\neq maxindex}|rowmentry|\\,|\\bar{vectorxgen}| \\le \\left(\\sum_{indexvar \\neq maxindex}|rowmentry|\\right)|\\bar{vectorxmax}|\n\\]\nand therefore\n\\[\n|rowmdiag| \\le \\sum_{indexvar \\neq maxindex}|rowmentry|\n\\]\ncontrary to hypothesis. Hence the determinant cannot be zero.\nRemark. This is the same type of argument that was used by Gersgorin to obtain bounds on the eigenvalues of a matrix. See Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, 1964, page 146."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_ii": "ravenquartz",
        "a_ij": "willowbranch",
        "a_i1": "porcelainjar",
        "a_i2": "meadowlark",
        "a_in": "twilightdew",
        "a_mj": "hazelnutseed",
        "a_mm": "fjordlantern",
        "j": "woodpecker",
        "m": "riverstone",
        "V": "cobblestone",
        "V_1": "thistledrift",
        "V_2": "silverspoon",
        "V_3": "marigoldpetal",
        "w": "glaciermist",
        "w_1": "cranberryfog",
        "w_2": "sunsetember",
        "w_3": "ivorycandle",
        "w_j": "orchidshadow",
        "x": "elmrootvine",
        "x_1": "sagehorizon",
        "x_2": "basilmeadow",
        "x_3": "cedarwhisper",
        "x_j": "pinemurmur",
        "x_m": "birchlantern",
        "y": "rosequartz",
        "y_1": "lilacbreeze",
        "y_2": "peachripple",
        "y_3": "cobaltstream",
        "y_j": "amberglint",
        "z": "primrosegale",
        "z_1": "opalhollow",
        "z_2": "jasperdawn",
        "z_3": "onyxrefuge",
        "a": "lindenbranch",
        "b": "aldercanyon",
        "n": "galaxyorbit"
      },
      "question": "6. Answer either (i) or (ii):\n(i) Let \\( thistledrift, silverspoon, marigoldpetal \\), and \\( cobblestone \\) denote four vertices of a cube. \\( thistledrift, silverspoon \\), and \\( marigoldpetal \\) are next neighbors of \\( cobblestone \\), that is, the lines \\( cobblestone thistledrift, cobblestone silverspoon \\), and \\( cobblestone marigoldpetal \\) are edges of the cube. Project the cube orthogonally onto a plane (the \\( primrosegale \\)-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of \\( cobblestone \\) fall in the origin and the projections of \\( thistledrift, silverspoon \\), and \\( marigoldpetal \\) in points marked with the complex numbers \\( opalhollow, jasperdawn \\), and \\( onyxrefuge \\), respectively. Show that \\( opalhollow^{2}+jasperdawn^{2}+onyxrefuge^{2}=0 \\).\n(page 261)\n(ii) Let \\( willowbranch \\) be a determinant in which each diagonal element exceeds in absolute value the sum of the absolute values of the other elements of its row, that is\n\\[\n\\left|ravenquartz\\right|>\\left|porcelainjar\\right|+\\left|meadowlark\\right|+\\cdots+\\left|twilightdew\\right|\n\\]\nShow that the determinant is not equal to zero. (Consider the corresponding system of linear homogeneous equations.)",
      "solution": "Solution. Take Cartesian coordinates \\( glaciermist, elmrootvine, rosequartz \\) in space so that the \\( elmrootvine \\)-axis is the real axis and the \\( rosequartz \\)-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point ( \\( glaciermist, elmrootvine, rosequartz \\) ) onto \\( (0, elmrootvine, rosequartz) \\).\n\nSuppose ( \\( cranberryfog, sagehorizon, lilacbreeze \\) ), \\( (sunsetember, basilmeadow, peachripple) \\) and ( \\( ivorycandle, cedarwhisper, cobaltstream \\) ) are mutually orthogonal unit vectors in space. Then the matrix\n\\[\n\\left(\\begin{array}{lll}\ncranberryfog & sagehorizon & lilacbreeze \\\\\nsunsetember & basilmeadow & peachripple \\\\\nivorycandle & cedarwhisper & cobaltstream\n\\end{array}\\right)\n\\]\nis an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular\n\\[\n\\begin{array}{c}\nsagehorizon^{2}+basilmeadow^{2}+cedarwhisper^{2}=1 \\\\\nlilacbreeze^{2}+peachripple^{2}+cobaltstream^{2}=1 \\\\\nsagehorizon\\,lilacbreeze+basilmeadow\\,peachripple+cedarwhisper\\,cobaltstream=0 .\n\\end{array}\n\\]\n\nThe orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers \\( sagehorizon+i\\,lilacbreeze, basilmeadow+i\\,peachripple \\), and \\( cedarwhisper+i\\,cobaltstream \\), and\n\\[\n\\begin{array}{c}\n\\left(sagehorizon+i\\,lilacbreeze\\right)^{2}+\\left(basilmeadow+i\\,peachripple\\right)^{2}+\\left(cedarwhisper+i\\,cobaltstream\\right)^{2}=sagehorizon^{2}+basilmeadow^{2}+cedarwhisper^{2}-lilacbreeze^{2}\\\\\n-peachripple^{2}-cobaltstream^{2}+2 i\\left(sagehorizon\\,lilacbreeze+basilmeadow\\,peachripple+cedarwhisper\\,cobaltstream\\right)=0 .\n\\end{array}\n\\]\n\nSuppose the sides of the given cube are of length \\( lindenbranch \\). Since it is given that the vertex \\( cobblestone \\) projects onto the origin, it must be of the form \\( cobblestone=(aldercanyon,0,0) \\). There are mutually orthogonal unit vectors ( \\( glaciermist_{\\woodpecker}, elmrootvine_{\\woodpecker}, rosequartz_{\\woodpecker} \\) ) such that\n\\[\nthistledrift=(aldercanyon,0,0)+lindenbranch\\,(cranberryfog, sagehorizon, lilacbreeze),\\\\\nsilverspoon=(aldercanyon,0,0)+lindenbranch\\,(sunsetember, basilmeadow, peachripple),\\\\\nmarigoldpetal=(aldercanyon,0,0)+lindenbranch\\,(ivorycandle, cedarwhisper, cobaltstream)\n\\]\n\nThen the projection of \\( thistledrift, silverspoon, marigoldpetal \\) into the Gaussian plane is\n\\[\nopalhollow=lindenbranch\\,(sagehorizon+i\\,lilacbreeze),\\quad\njasperdawn=lindenbranch\\,(basilmeadow+i\\,peachripple),\\quad\nonyxrefuge=lindenbranch\\,(cedarwhisper+i\\,cobaltstream)\n\\]\nand\n\\[\nopalhollow^{2}+jasperdawn^{2}+onyxrefuge^{2}=lindenbranch^{2}\\bigl[\\left(sagehorizon+i\\,lilacbreeze\\right)^{2}+\\left(basilmeadow+i\\,peachripple\\right)^{2}+\\left(cedarwhisper+i\\,cobaltstream\\right)^{2}\\bigr]=0 .\n\\]\n\nSolution. The corresponding system of linear equations is\n\\[\n\\sum_{woodpecker=1}^{galaxyorbit} willowbranch\\,x_{woodpecker}=0,\\quad i=1,2,\\ldots,galaxyorbit\n\\]\nIf the determinant of the matrix of the coefficients is zero, there exists a non-trivial solution, say ( \\( \\bar{x}_{1}, \\bar{x}_{2}, \\ldots, \\bar{x}_{galaxyorbit} \\) ), of the system. Let \\( riverstone \\) be an index for which \\( |\\bar{x}_{riverstone}| \\) is largest, that is \\( |\\bar{x}_{riverstone}| \\ge |\\bar{x}_{j}| \\) for \\( j=1,2,\\ldots,galaxyorbit \\). Clearly, \\( |\\bar{x}_{riverstone}| \\neq 0 \\), since the solution is non-trivial. Consider the \\( riverstone \\)th equation in the above system written in the form\n\\[\n-fjordlantern\\,\\bar{x}_{riverstone}=\\sum_{woodpecker \\neq riverstone} hazelnutseed\\,\\bar{x}_{woodpecker}\n\\]\nWe have\n\\[\n|fjordlantern|\\,|\\bar{x}_{riverstone}| \\le \\sum_{woodpecker \\neq riverstone}|hazelnutseed|\\,|\\bar{x}_{woodpecker}| \\le \\left(\\sum_{woodpecker \\neq riverstone}|hazelnutseed|\\right)|\\bar{x}_{riverstone}|\n\\]\nand therefore\n\\[\n|fjordlantern| \\le \\sum_{woodpecker \\neq riverstone}|hazelnutseed|\n\\]\ncontrary to hypothesis. Hence the determinant cannot be zero.\nRemark. This is the same type of argument that was used by Gersgorin to obtain bounds on the eigenvalues of a matrix. See Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, 1964, page 146."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_ii": "offdiagonal",
        "a_ij": "zeroelement",
        "a_i1": "lastcolumn",
        "a_i2": "randomcolumn",
        "a_in": "firstcolumn",
        "a_mj": "fixedvalue",
        "a_mm": "outercorner",
        "j": "constantindex",
        "m": "minimumindex",
        "V": "centerpoint",
        "V_1": "farvertexone",
        "V_2": "farvertextwo",
        "V_3": "farvertexthr",
        "w": "flatcoord",
        "w_1": "flatcoordone",
        "w_2": "flatcoordtwo",
        "w_3": "flatcoordthr",
        "w_j": "flatcoordvar",
        "x": "imagaxis",
        "x_1": "imagaxisone",
        "x_2": "imagaxistwo",
        "x_3": "imagaxisthr",
        "x_j": "imagaxisvar",
        "x_m": "imagaxismin",
        "y": "realaxis",
        "y_1": "realaxisone",
        "y_2": "realaxistwo",
        "y_3": "realaxisthr",
        "y_j": "realaxisvar",
        "z": "realnumber",
        "z_1": "realnumberone",
        "z_2": "realnumbertwo",
        "z_3": "realnumberthr",
        "a": "zeroedge",
        "b": "zeroshift",
        "n": "unitysize"
      },
      "question": "6. Answer either (i) or (ii):\n(i) Let \\( farvertexone, farvertextwo, farvertexthr \\), and \\( centerpoint \\) denote four vertices of a cube. \\( farvertexone, farvertextwo \\), and \\( farvertexthr \\) are next neighbors of \\( centerpoint \\), that is, the lines \\( centerpoint farvertexone, centerpoint farvertextwo \\), and \\( centerpoint farvertexthr \\) are edges of the cube. Project the cube orthogonally onto a plane (the \\( realnumber\\)-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of \\( centerpoint \\) fall in the origin and the projections of \\( farvertexone, farvertextwo \\), and \\( farvertexthr \\) in points marked with the complex numbers \\( realnumberone, realnumbertwo \\), and \\( realnumberthr \\), respectively. Show that \\( realnumberone { }^{2}+realnumbertwo^{2}+realnumberthr^{2}=0 \\).\n(page 261)\n(ii) Let \\( zeroelement \\) be a determinant in which each diagonal element exceeds in absolute value the sum of the absolute values of the other elements of its row, that is\n\\[\n\\left|offdiagonal\\right|>\\left|lastcolumn\\right|+\\left|randomcolumn\\right|+\\cdots+\\left|a_{i, i-1}\\right|+\\left|a_{i, i+1}\\right|+\\cdots+\\left|firstcolumn\\right|\n\\]\n\nShow that the determinant is not equal to zero. (Consider the corresponding system of linear homogeneous equations.)",
      "solution": "Solution. Take Cartesian coordinates \\( flatcoord, imagaxis, realaxis \\) in space so that the \\( imagaxis \\)-axis is the real axis and the \\( realaxis \\)-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point ( \\( flatcoord, imagaxis, realaxis \\) ) onto \\( (0, imagaxis, realaxis) \\).\n\nSuppose ( \\( flatcoordone, imagaxisone, realaxisone \\) ), \\( \\left(flatcoordtwo, imagaxistwo, realaxistwo\\right) \\) and ( \\( \\left.flatcoordthr, imagaxisthr, realaxisthr\\right) \\) are mutually orthogonal unit vectors in space. Then the matrix\n\\[\n\\left(\\begin{array}{lll}\nflatcoordone & imagaxisone & realaxisone \\\\\nflatcoordtwo & imagaxistwo & realaxistwo \\\\\nflatcoordthr & imagaxisthr & realaxisthr\n\\end{array}\\right)\n\\]\nis an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular\n\\[\n\\begin{array}{c}\nimagaxisone^{2}+imagaxistwo^{2}+imagaxisthr^{2}=1 \\\\\nrealaxisone^{2}+realaxistwo^{2}+realaxisthr^{2}=1 \\\\\nimagaxisone\\,realaxisone+imagaxistwo\\,realaxistwo+imagaxisthr\\,realaxisthr=0 .\n\\end{array}\n\\]\n\nThe orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers \\( imagaxisone+i\\,realaxisone, imagaxistwo+i\\,realaxistwo \\), and \\( imagaxisthr+i\\,realaxisthr \\), and\n\\[\n\\begin{array}{c}\n\\left(imagaxisone+i\\,realaxisone\\right)^{2}+\\left(imagaxistwo+i\\,realaxistwo\\right)^{2}+\\left(imagaxisthr+i\\,realaxisthr\\right)^{2}=imagaxisone^{2}+imagaxistwo^{2}+imagaxisthr^{2}-realaxisone^{2} \\\\\n-realaxistwo^{2}-realaxisthr^{2}+2 i\\left(imagaxisone\\,realaxisone+imagaxistwo\\,realaxistwo+imagaxisthr\\,realaxisthr\\right)=0 .\n\\end{array}\n\\]\n\nSuppose the sides of the given cube are of length \\( zeroedge \\). Since it is given that the vertex \\( centerpoint \\) projects onto the origin, it must be of the form \\( centerpoint=(zeroshift, 0,0) \\). There are mutually orthogonal unit vectors ( \\( flatcoordvar, imagaxisvar, realaxisvar \\) ) such that\n\\[\nV_{j}=(zeroshift, 0,0)+zeroedge\\left(flatcoordvar, imagaxisvar, realaxisvar\\right) \\quad \\text { for } j=1,2,3\n\\]\n\nThen the projection of \\( V_{j} \\) into the Gaussian plane is\n\\[\nrealnumber_{j}=zeroedge\\left(imagaxisvar+i\\,realaxisvar\\right)\n\\]\nand\n\\[\nrealnumberone^{2}+realnumbertwo^{2}+realnumberthr^{2}=zeroedge^{2}\\left[\\left(imagaxisone+i\\,realaxisone\\right)^{2}+\\left(imagaxistwo+i\\,realaxistwo\\right)^{2}+\\left(imagaxisthr+i\\,realaxisthr\\right)^{2}\\right]=0 .\n\\]\n\nSolution. The corresponding system of linear equations is\n\\[\n\\sum_{constantindex=1}^{unitysize} zeroelement\\,imagaxisvar=0, \\quad constantindex=1,2, \\ldots, unitysize\n\\]\n\nIf the determinant of the matrix of the coefficients is zero, there exists a non-trivial solution, say ( \\( \\bar{imagaxisone}, \\bar{imagaxistwo}, \\ldots, \\bar{imagaxismin} \\) ), of the system. Let \\( minimumindex \\) be an index for which \\( \\left|\\bar{imagaxismin}\\right| \\) is largest, that is \\( \\left|\\bar{imagaxismin}\\right| \\geq\\left|\\bar{imagaxisvar}\\right| \\) for \\( constantindex=1,2, \\ldots, unitysize \\). Clearly, \\( \\left|\\bar{imagaxismin}\\right| \\neq 0 \\), since the solution is non-trivial. Consider the \\( minimumindex \\) th equation in the above system written in the form\n\\[\n-outercorner \\bar{imagaxismin}=\\sum_{constantindex \\neq minimumindex} fixedvalue \\bar{imagaxisvar}\n\\]\n\nWe have\n\\[\n\\left|outercorner\\right|\\left|\\bar{imagaxismin}\\right| \\leq \\sum_{constantindex \\neq minimumindex}\\left|fixedvalue\\right|\\left|\\bar{imagaxisvar}\\right| \\leq\\left(\\sum_{constantindex \\neq minimumindex}\\left|fixedvalue\\right|\\right)\\left|\\bar{imagaxismin}\\right|\n\\]\nand therefore\n\\[\n\\left|outercorner\\right| \\leq \\sum_{constantindex \\neq minimumindex}\\left|fixedvalue\\right|\n\\]\ncontrary to hypothesis. Hence the determinant cannot be zero.\nRemark. This is the same type of argument that was used by Gersgorin to obtain bounds on the eigenvalues of a matrix. See Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, 1964, page 146."
    },
    "garbled_string": {
      "map": {
        "a_ii": "qzxwvtnp",
        "a_ij": "hjgrksla",
        "a_i1": "fkdlsmnb",
        "a_i2": "prtuvyqe",
        "a_in": "ceghbalu",
        "a_mj": "guorfkzn",
        "a_mm": "yvclirps",
        "j": "omlasfne",
        "m": "qlirepso",
        "V": "wibnexla",
        "V_1": "jrukspad",
        "V_2": "vexlopar",
        "V_3": "cmandoti",
        "w": "iqoskftr",
        "w_1": "exlartqb",
        "w_2": "fqunsdmi",
        "w_3": "gvmopzle",
        "w_j": "ksyqdrav",
        "x": "pfarnedu",
        "x_1": "bzqustla",
        "x_2": "gwelmokn",
        "x_3": "nycravop",
        "x_j": "hmalrefo",
        "x_m": "dzikpuna",
        "y": "lsoravqe",
        "y_1": "orpltune",
        "y_2": "qsdafwom",
        "y_3": "jkarnote",
        "y_j": "vlonskme",
        "z": "tzcrmepl",
        "z_1": "akpldros",
        "z_2": "qrwstnfa",
        "z_3": "zxmorbih",
        "a": "mcfaluzo",
        "b": "nygabrot",
        "n": "plidertas"
      },
      "question": "6. Answer either (i) or (ii):\n(i) Let \\( jrukspad, vexlopar, cmandoti \\), and \\( wibnexla \\) denote four vertices of a cube. \\( jrukspad, vexlopar \\), and \\( cmandoti \\) are next neighbors of \\( wibnexla \\), that is, the lines \\( wibnexla\\, jrukspad, wibnexla\\, vexlopar \\), and \\( wibnexla\\, cmandoti \\) are edges of the cube. Project the cube orthogonally onto a plane (the \\( tzcrmepl \\)-plane, the Gaussian plane) of which the points are marked with complex numbers. Let the projection of \\( wibnexla \\) fall in the origin and the projections of \\( jrukspad, vexlopar \\), and \\( cmandoti \\) in points marked with the complex numbers \\( akpldros, qrwstnfa \\), and \\( zxmorbih \\), respectively. Show that \\( akpldros^{2}+qrwstnfa^{2}+zxmorbih^{2}=0 \\).\n(page 261)\n(ii) Let \\( hjgrksla \\) be a determinant in which each diagonal element exceeds in absolute value the sum of the absolute values of the other elements of its row, that is\n\\[\n\\left|qzxwvtnp\\right|>\\left|fkdlsmnb\\right|+\\left|prtuvyqe\\right|+\\cdots+\\left|a_{i, i-1}\\right|+\\left|a_{i, i+1}\\right|+\\cdots+\\left|ceghbalu\\right|\n\\]\n\nShow that the determinant is not equal to zero. (Consider the corresponding system of linear homogeneous equations.)",
      "solution": "Solution. Take Cartesian coordinates \\( iqoskftr, pfarnedu, lsoravqe \\) in space so that the \\( pfarnedu \\)-axis is the real axis and the \\( lsoravqe \\)-axis is the imaginary axis in the given Gaussian plane. Then the projection maps each point ( \\( iqoskftr, pfarnedu, lsoravqe \\) ) onto \\( (0, pfarnedu, lsoravqe) \\).\n\nSuppose ( \\( exlartqb, bzqustla, orpltune \\) ), \\( \\left(fqunsdmi, gwelmokn, qsdafwom\\right) \\) and ( \\( \\left.gvmopzle, nycravop, jkarnote\\right) \\) are mutually orthogonal unit vectors in space. Then the matrix\n\\[\n\\left(\\begin{array}{lll}\nexlartqb & bzqustla & orpltune \\\\\nfqunsdmi & gwelmokn & qsdafwom \\\\\ngvmopzle & nycravop & jkarnote\n\\end{array}\\right)\n\\]\nis an orthogonal matrix, so its columns are also mutually orthogonal unit vectors, in particular\n\\[\n\\begin{array}{c}\nbzqustla^{2}+gwelmokn^{2}+nycravop^{2}=1 \\\\\norpltune^{2}+qsdafwom^{2}+jkarnote^{2}=1 \\\\\nbzqustla\\, orpltune+gwelmokn\\, qsdafwom+nycravop\\, jkarnote=0 .\n\\end{array}\n\\]\n\nThe orthogonal projections of the unit vectors into the Gaussian plane are the complex numbers \\( bzqustla+i\\, orpltune, gwelmokn+i\\, qsdafwom \\), and \\( nycravop+i\\, jkarnote \\), and\n\\[\n\\begin{array}{c}\n\\left(bzqustla+i\\, orpltune\\right)^{2}+\\left(gwelmokn+i\\, qsdafwom\\right)^{2}+\\left(nycravop+i\\, jkarnote\\right)^{2}=bzqustla^{2}+gwelmokn^{2}+nycravop^{2}-orpltune^{2} \\\\\n-qsdafwom^{2}-jkarnote^{2}+2 i\\left(bzqustla\\, orpltune+gwelmokn\\, qsdafwom+nycravop\\, jkarnote\\right)=0 .\n\\end{array}\n\\]\n\nSuppose the sides of the given cube are of length \\( mcfaluzo \\). Since it is given that the vertex \\( wibnexla \\) projects onto the origin, it must be of the form \\( wibnexla=(nygabrot, 0,0) \\). There are mutually orthogonal unit vectors ( \\( ksyqdrav, hmalrefo, vlonskme \\) ) such that\n\\[\nV_{j}=(nygabrot, 0,0)+mcfaluzo\\left(ksyqdrav, hmalrefo, vlonskme\\right) \\quad \\text { for } j=1,2,3\n\\]\n\nThen the projection of \\( V_{j} \\) into the Gaussian plane is\n\\[\nz_{j}=mcfaluzo\\left(hmalrefo+i\\, vlonskme\\right)\n\\]\nand\n\\[\nakpldros^{2}+qrwstnfa^{2}+zxmorbih^{2}=mcfaluzo^{2}\\left[\\left(hmalrefo+i\\, vlonskme\\right)^{2}+\\left(hmalrefo+i\\, vlonskme\\right)^{2}+\\left(hmalrefo+i\\, vlonskme\\right)^{2}\\right]=0 .\n\\]\n\nSolution. The corresponding system of linear equations is\n\\[\n\\sum_{omlasfne=1}^{plidertas} hjgrksla\\, hmalrefo=0, \\quad i=1,2, \\ldots, plidertas\n\\]\n\nIf the determinant of the matrix of the coefficients is zero, there exists a non-trivial solution, say ( \\( \\bar{bzqustla}, \\bar{gwelmokn}, \\ldots, \\bar{dzikpuna} \\) ), of the system. Let \\( qlirepso \\) be an index for which \\( \\left|\\bar{dzikpuna}\\right| \\) is largest, that is \\( \\left|\\bar{dzikpuna}\\right| \\geq\\left|\\bar{x}_{j}\\right| \\) for \\( j=1,2, \\ldots, plidertas \\). Clearly, \\( \\left|\\bar{dzikpuna}\\right| \\neq 0 \\), since the solution is non-trivial. Consider the \\( qlirepso \\) th equation in the above system written in the form\n\\[\n-yvclirps\\, \\bar{dzikpuna}=\\sum_{j \\neq qlirepso} guorfkzn\\, \\bar{x}_{j}\n\\]\n\nWe have\n\\[\n\\left|yvclirps\\right|\\left|\\bar{dzikpuna}\\right| \\leq \\sum_{j \\neq qlirepso}\\left|guorfkzn\\right|\\left|\\bar{x}_{j}\\right| \\leq\\left(\\sum_{j \\neq qlirepso}\\left|guorfkzn\\right|\\right)\\left|\\bar{dzikpuna}\\right|\n\\]\nand therefore\n\\[\n\\left|yvclirps\\right| \\leq \\sum_{j \\neq qlirepso}\\left|guorfkzn\\right|\n\\]\ncontrary to hypothesis. Hence the determinant cannot be zero.\nRemark. This is the same type of argument that was used by Gersgorin to obtain bounds on the eigenvalues of a matrix. See Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Allyn and Bacon, Boston, 1964, page 146."
    },
    "kernel_variant": {
      "question": "(The statement of the enhanced problem is unchanged, because the error pointed out by the reviewers concerns only the derivation given in the solution of Part (𝓐)(ii).  For convenience we reproduce Part (𝓐) verbatim.)\n\nSolve EITHER Part (𝓐) or Part (𝓑).\n\n(𝓐)  (Geometry in 3–space viewed through complex coordinates)  \nLet 𝐖 be a vertex of a cube in ℝ³; write 𝐖𝐖₁, 𝐖𝐖₂, 𝐖𝐖₃ for the three mutually\nperpendicular edges that emanate from 𝐖 and assume they are oriented so that\n(𝐖𝐖₁,𝐖𝐖₂,𝐖𝐖₃) is a positively oriented orthonormal frame.  \nFix an arbitrary two–dimensional plane Π through 𝐖 and denote by π:ℝ³→Π the\northogonal projection.  \nChoose an orthonormal positively oriented basis (e₁,e₂) of Π, declare e₃:=e₁×e₂, and identify Π with the complex plane ℂ via\n   ξe₁+ηe₂  ↦  ξ+iη .      \nPut\n   ζₖ   =   π(𝐖ₖ) − π(𝐖)   ∈ ℂ,      k=1,2,3,      (1)\nand write\n   aₖ:=|ζₖ| ,  cₖ:=e₃·𝐖𝐖ₖ .               (2)\n(Observe that the (common) edge length of the cube equals a:=‖𝐖𝐖ₖ‖, so that  \naₖ = a√(1−cₖ²).)\n\n(i)  (Quadratic identity) Prove that\n   ζ₁² + ζ₂² + ζ₃² = 0.                 (3)\n\n(ii)  (A sharp upper bound for the cubic product)  \n\n  Establish the exact modulus identity  \n   |ζ₁ ζ₂ ζ₃| = a³ √{ (1−c₁²)(1−c₂²)(1−c₃²) }.    (4)\n\n  Using the constraint  \n   c₁² + c₂² + c₃² = 1                 (5)\n  that follows from the orthonormality of the edges, deduce the sharp\n  inequality\n   |ζ₁ ζ₂ ζ₃| ≤ a³ (2/3)^{3/2}.              (6)\n\n  Show that equality in (6) holds exactly when  \n   |ζ₁| = |ζ₂| = |ζ₃|  ⇔  |c₁| = |c₂| = |c₃| = 1/√3.  (7)\n\n(iii)  (Reconstruction problem) [statement identical to the current enhanced version]\n\n(𝓑)  (Block–operator Gershgorin type theorem – unchanged)  \n[statement identical to the original enhanced version]\n\n————————————————————————————————————————",
      "solution": "Only Part (𝓐)(ii) is altered.  Parts (i) and (iii) were already correct and\nare repeated for completeness.\n\nSolution to Part (𝓐)\n\nThroughout put a:=‖𝐖𝐖ₖ‖.  For k=1,2,3 write\n  uₖ := 𝐖𝐖ₖ /a, so that (u₁,u₂,u₃) is a positively oriented\n  orthonormal basis of ℝ³.                (9)\n\nDenote\n  xₖ = uₖ·e₁ ,  yₖ = uₖ·e₂ ,  cₖ = uₖ·e₃ .      (10)\n\nBecause the uₖ form an orthonormal triple, the 3 × 3 matrix with k-th column\nuₖ is in SO(3); hence its rows\n  X := (x₁,x₂,x₃) ,  Y := (y₁,y₂,y₃) ,  C := (c₁,c₂,c₃) (11)\nconstitute an orthonormal positively oriented frame, i.e.\n  ‖X‖ = ‖Y‖ = ‖C‖ = 1 ,  X·Y = X·C = Y·C = 0.    (12)\n\nFurther, by (1) and (10)\n  ζₖ = a (xₖ + i yₖ) ,  aₖ = a√(1−cₖ²).       (13)\n\n(i)  Quadratic identity (3).  \n ζ₁²+ζ₂²+ζ₃²\n  = a² Σ (xₖ+iyₖ)²\n  = a²[ Σ(xₖ²−yₖ²) + 2i Σ xₖyₖ ]\n  = a²[(‖X‖²−‖Y‖²) + 2i (X·Y)] = 0 by (12).\n\n(ii)  Modulus formula (4) and sharp bound (6).\n\nFrom (13)\n |ζ₁ ζ₂ ζ₃|\n  = a³ |x₁+iy₁| |x₂+iy₂| |x₃+iy₃|\n  = a³ √{(x₁²+y₁²)(x₂²+y₂²)(x₃²+y₃²)}\n  = a³ √{(1−c₁²)(1−c₂²)(1−c₃²)},  which is (4).\n\nIntroduce sₖ := cₖ² (so 0≤sₖ≤1 and s₁+s₂+s₃=1 by (12)).  Then (4) becomes\n |ζ₁ ζ₂ ζ₃| = a³ P(s₁,s₂,s₃)^{1/2},\nwhere\n P(s₁,s₂,s₃):=(1−s₁)(1−s₂)(1−s₃).         (16)\n\nWe must maximise P under the linear constraint s₁+s₂+s₃=1.\n\nMethod 1: Lagrange multipliers  \nLet F(s₁,s₂,s₃,λ)=ln P+λ(s₁+s₂+s₃−1).  Computing\n ∂F/∂s₁ = −1/(1−s₁)+λ, etc.,\nwe obtain critical points only when 1−s₁=1−s₂=1−s₃, i.e. s₁=s₂=s₃=1/3.  \nBecause P(⋅) is symmetric and positive on the open simplex 0<sₖ<1, the\nunique critical point is the (global) maximiser.\n\nMethod 2: sharp AM–GM (shorter)  \nBy AM–GM,\n P = (1−s₁)(1−s₂)(1−s₃)\n  ≤ [(1−s₁)+(1−s₂)+(1−s₃)]³/27\n  = (2/3)³,                      (17′)\nwith equality iff 1−s₁ = 1−s₂ = 1−s₃, i.e. s₁=s₂=s₃=1/3.\n\nEither way,\n P_max = (2/3)³, hence |ζ₁ ζ₂ ζ₃|\n  ≤ a³ (2/3)^{3/2}.                 (6)\n\nEquality holds precisely when s₁=s₂=s₃=1/3, i.e. |c₁|=|c₂|=|c₃|=1/√3,\nwhich by (13) is equivalent to |ζ₁|=|ζ₂|=|ζ₃|.  This is (7).\n\n(iii)  Reconstruction of the cube from prescribed complex projections.  \n[Identical to the previous solution and reproduced here for completeness.]\n\nLet z₁,z₂,z₃ satisfy (8) and define\n X := (Re z₁,Re z₂,Re z₃), Y := (Im z₁,Im z₂,Im z₃),\n L := ‖X‖ = ‖Y‖ (>0).               (18)\n\nFrom (8) we have X·Y = 0.  Put\n Z := (1/L) X×Y.                  (19)\n\nSince X⊥Y and |X×Y| = L², the 3 × 3 matrix\n R := 1/L ·\n   ⎛ Re z₁ Re z₂ Re z₃ ⎞\n   ⎜ Im z₁ Im z₂ Im z₃ ⎟\n   ⎝  Z₁  Z₂  Z₃ ⎠                (20)\nhas orthonormal rows; det R>0 because the third row is the right-hand\ncross‐product of the first two.  Hence R∈SO(3).\n\nLet its columns be u₁,u₂,u₃ and place 𝐖 at the origin.  Setting a:=L and\n𝐖ₖ := a uₖ (k=1,2,3) yields a cube whose edges are 𝐖𝐖₁,𝐖𝐖₂,𝐖𝐖₃ and whose\nprojections are precisely z₁,z₂,z₃.\n\nUniqueness up to rigid motion.  \nIf another oriented cube projects to the same unordered set {z₁,z₂,z₃},\nthe first two rows X,Y of the associated matrix (20) must coincide with or\nbe obtained from the above by a common permutation; consequently Z and the\nentire rotation matrix R agree, so the cubes differ only by translating 𝐖.\n\n————————————————————————————————————————\n\nSolution to Part (𝓑)\n\n[Unchanged.]\n\n————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.422215",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    },
    "original_kernel_variant": {
      "question": "(The statement of the enhanced problem is unchanged, because the error pointed out by the reviewers concerns only the derivation given in the solution of Part (𝓐)(ii).  For convenience we reproduce Part (𝓐) verbatim.)\n\nSolve EITHER Part (𝓐) or Part (𝓑).\n\n(𝓐)  (Geometry in 3–space viewed through complex coordinates)  \nLet 𝐖 be a vertex of a cube in ℝ³; write 𝐖𝐖₁, 𝐖𝐖₂, 𝐖𝐖₃ for the three mutually\nperpendicular edges that emanate from 𝐖 and assume they are oriented so that\n(𝐖𝐖₁,𝐖𝐖₂,𝐖𝐖₃) is a positively oriented orthonormal frame.  \nFix an arbitrary two–dimensional plane Π through 𝐖 and denote by π:ℝ³→Π the\northogonal projection.  \nChoose an orthonormal positively oriented basis (e₁,e₂) of Π, declare e₃:=e₁×e₂, and identify Π with the complex plane ℂ via\n   ξe₁+ηe₂  ↦  ξ+iη .      \nPut\n   ζₖ   =   π(𝐖ₖ) − π(𝐖)   ∈ ℂ,      k=1,2,3,      (1)\nand write\n   aₖ:=|ζₖ| ,  cₖ:=e₃·𝐖𝐖ₖ .               (2)\n(Observe that the (common) edge length of the cube equals a:=‖𝐖𝐖ₖ‖, so that  \naₖ = a√(1−cₖ²).)\n\n(i)  (Quadratic identity) Prove that\n   ζ₁² + ζ₂² + ζ₃² = 0.                 (3)\n\n(ii)  (A sharp upper bound for the cubic product)  \n\n  Establish the exact modulus identity  \n   |ζ₁ ζ₂ ζ₃| = a³ √{ (1−c₁²)(1−c₂²)(1−c₃²) }.    (4)\n\n  Using the constraint  \n   c₁² + c₂² + c₃² = 1                 (5)\n  that follows from the orthonormality of the edges, deduce the sharp\n  inequality\n   |ζ₁ ζ₂ ζ₃| ≤ a³ (2/3)^{3/2}.              (6)\n\n  Show that equality in (6) holds exactly when  \n   |ζ₁| = |ζ₂| = |ζ₃|  ⇔  |c₁| = |c₂| = |c₃| = 1/√3.  (7)\n\n(iii)  (Reconstruction problem) [statement identical to the current enhanced version]\n\n(𝓑)  (Block–operator Gershgorin type theorem – unchanged)  \n[statement identical to the original enhanced version]\n\n————————————————————————————————————————",
      "solution": "Only Part (𝓐)(ii) is altered.  Parts (i) and (iii) were already correct and\nare repeated for completeness.\n\nSolution to Part (𝓐)\n\nThroughout put a:=‖𝐖𝐖ₖ‖.  For k=1,2,3 write\n  uₖ := 𝐖𝐖ₖ /a, so that (u₁,u₂,u₃) is a positively oriented\n  orthonormal basis of ℝ³.                (9)\n\nDenote\n  xₖ = uₖ·e₁ ,  yₖ = uₖ·e₂ ,  cₖ = uₖ·e₃ .      (10)\n\nBecause the uₖ form an orthonormal triple, the 3 × 3 matrix with k-th column\nuₖ is in SO(3); hence its rows\n  X := (x₁,x₂,x₃) ,  Y := (y₁,y₂,y₃) ,  C := (c₁,c₂,c₃) (11)\nconstitute an orthonormal positively oriented frame, i.e.\n  ‖X‖ = ‖Y‖ = ‖C‖ = 1 ,  X·Y = X·C = Y·C = 0.    (12)\n\nFurther, by (1) and (10)\n  ζₖ = a (xₖ + i yₖ) ,  aₖ = a√(1−cₖ²).       (13)\n\n(i)  Quadratic identity (3).  \n ζ₁²+ζ₂²+ζ₃²\n  = a² Σ (xₖ+iyₖ)²\n  = a²[ Σ(xₖ²−yₖ²) + 2i Σ xₖyₖ ]\n  = a²[(‖X‖²−‖Y‖²) + 2i (X·Y)] = 0 by (12).\n\n(ii)  Modulus formula (4) and sharp bound (6).\n\nFrom (13)\n |ζ₁ ζ₂ ζ₃|\n  = a³ |x₁+iy₁| |x₂+iy₂| |x₃+iy₃|\n  = a³ √{(x₁²+y₁²)(x₂²+y₂²)(x₃²+y₃²)}\n  = a³ √{(1−c₁²)(1−c₂²)(1−c₃²)},  which is (4).\n\nIntroduce sₖ := cₖ² (so 0≤sₖ≤1 and s₁+s₂+s₃=1 by (12)).  Then (4) becomes\n |ζ₁ ζ₂ ζ₃| = a³ P(s₁,s₂,s₃)^{1/2},\nwhere\n P(s₁,s₂,s₃):=(1−s₁)(1−s₂)(1−s₃).         (16)\n\nWe must maximise P under the linear constraint s₁+s₂+s₃=1.\n\nMethod 1: Lagrange multipliers  \nLet F(s₁,s₂,s₃,λ)=ln P+λ(s₁+s₂+s₃−1).  Computing\n ∂F/∂s₁ = −1/(1−s₁)+λ, etc.,\nwe obtain critical points only when 1−s₁=1−s₂=1−s₃, i.e. s₁=s₂=s₃=1/3.  \nBecause P(⋅) is symmetric and positive on the open simplex 0<sₖ<1, the\nunique critical point is the (global) maximiser.\n\nMethod 2: sharp AM–GM (shorter)  \nBy AM–GM,\n P = (1−s₁)(1−s₂)(1−s₃)\n  ≤ [(1−s₁)+(1−s₂)+(1−s₃)]³/27\n  = (2/3)³,                      (17′)\nwith equality iff 1−s₁ = 1−s₂ = 1−s₃, i.e. s₁=s₂=s₃=1/3.\n\nEither way,\n P_max = (2/3)³, hence |ζ₁ ζ₂ ζ₃|\n  ≤ a³ (2/3)^{3/2}.                 (6)\n\nEquality holds precisely when s₁=s₂=s₃=1/3, i.e. |c₁|=|c₂|=|c₃|=1/√3,\nwhich by (13) is equivalent to |ζ₁|=|ζ₂|=|ζ₃|.  This is (7).\n\n(iii)  Reconstruction of the cube from prescribed complex projections.  \n[Identical to the previous solution and reproduced here for completeness.]\n\nLet z₁,z₂,z₃ satisfy (8) and define\n X := (Re z₁,Re z₂,Re z₃), Y := (Im z₁,Im z₂,Im z₃),\n L := ‖X‖ = ‖Y‖ (>0).               (18)\n\nFrom (8) we have X·Y = 0.  Put\n Z := (1/L) X×Y.                  (19)\n\nSince X⊥Y and |X×Y| = L², the 3 × 3 matrix\n R := 1/L ·\n   ⎛ Re z₁ Re z₂ Re z₃ ⎞\n   ⎜ Im z₁ Im z₂ Im z₃ ⎟\n   ⎝  Z₁  Z₂  Z₃ ⎠                (20)\nhas orthonormal rows; det R>0 because the third row is the right-hand\ncross‐product of the first two.  Hence R∈SO(3).\n\nLet its columns be u₁,u₂,u₃ and place 𝐖 at the origin.  Setting a:=L and\n𝐖ₖ := a uₖ (k=1,2,3) yields a cube whose edges are 𝐖𝐖₁,𝐖𝐖₂,𝐖𝐖₃ and whose\nprojections are precisely z₁,z₂,z₃.\n\nUniqueness up to rigid motion.  \nIf another oriented cube projects to the same unordered set {z₁,z₂,z₃},\nthe first two rows X,Y of the associated matrix (20) must coincide with or\nbe obtained from the above by a common permutation; consequently Z and the\nentire rotation matrix R agree, so the cubes differ only by translating 𝐖.\n\n————————————————————————————————————————\n\nSolution to Part (𝓑)\n\n[Unchanged.]\n\n————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.366779",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}