{
  "index": "1967-A-6",
  "type": "ALG",
  "tag": [
    "ALG",
    "GEO"
  ],
  "difficulty": "",
  "question": "A-6. Given real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\},(i=1,2,3,4) \\), such that \\( a_{1} b_{2}-a_{2} b_{1} \\neq 0 \\). Consider the set of all solutions ( \\( x_{1}, x_{2}, x_{8}, x_{4} \\) ) of the simultaneous equations\n\\[\na_{1} x_{1}+a_{2} x_{2}+a_{2} x_{3}+a_{4} x_{4}=0 \\text { and } b_{1} x_{1}+b_{2} x_{2}+b_{4} x_{3}+b_{4} x_{4}=0\n\\]\nfor which no \\( x_{i}(i=1,2,3,4) \\) is zero. Each such solution generates a 4-tuple of plus and minus signs (signum \\( x_{1} \\), signum \\( x_{2} \\), signum \\( x_{3} \\), signum \\( x_{4} \\) ).\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\} \\) such that the above maximum number of 4 -tuples is attained.",
  "solution": "A-6 Solving the given equations in terms of \\( x_{3} \\) and \\( x_{4} \\), leads to the equivalent system: \\( x_{1}=A_{1} x_{3}+B_{1} x_{4}, \\quad x_{2}=A_{2} x_{3}+B_{2} x_{4}, \\quad x_{3}=x_{3}, \\quad x_{4}=x_{4} \\), where \\( A_{1}=\\left(a_{2} b_{3}-a_{3} b_{2}\\right) /\\left(a_{1} b_{2}-a_{2} b_{1}\\right) \\), etc.\n\nEach point in the \\( x_{3}, x_{4} \\)-plane corresponds uniquely to a solution ( \\( x_{1}, x_{2} \\), \\( x_{3}, x_{4} \\) ). Signum \\( x_{1} \\) is positive for ( \\( x_{3}, x_{4} \\) ) on one side of \\( A_{1} x_{3}+B_{1} x_{4}=0 \\) and negative on the other side. Similarly for signum \\( x_{2} \\), signum \\( x_{3} \\), and signum \\( x_{4} \\), using \\( A_{2} x_{3}+B_{2} x_{3}=0, x_{3}=0 \\) and \\( x_{4}=0 \\), respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4 -tuple of signum values. Hence the maximum number of distinct 4 -tuples is eight.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions \\( A_{1} \\neq 0, A_{2} \\neq 0, B_{1} \\neq 0 \\), \\( B_{2} \\neq 0 \\) and \\( A_{1} B_{2}-A_{2} B_{1} \\neq 0 \\) or, simply, \\( a_{i} b_{j}-a_{j} b_{i} \\neq 0 \\) for \\( i, j=1,2,3,4 \\) and \\( i<j \\).",
  "vars": [
    "x_1",
    "x_2",
    "x_3",
    "x_4"
  ],
  "params": [
    "a_1",
    "a_2",
    "a_3",
    "a_4",
    "b_1",
    "b_2",
    "b_3",
    "b_4",
    "A_1",
    "A_2",
    "B_1",
    "B_2"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x_1": "firstcoord",
        "x_2": "secondcoord",
        "x_3": "thirdcoord",
        "x_4": "fourthcoord",
        "a_1": "acoeffone",
        "a_2": "acoefftwo",
        "a_3": "acoeffthree",
        "a_4": "acoefffour",
        "b_1": "bcoeffone",
        "b_2": "bcoefftwo",
        "b_3": "bcoeffthree",
        "b_4": "bcoefffour",
        "A_1": "matrixaone",
        "A_2": "matrixatwo",
        "B_1": "matrixbone",
        "B_2": "matrixbtwo"
      },
      "question": "A-6. Given real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\},(i=1,2,3,4) \\), such that \\( acoeffone bcoefftwo-acoefftwo bcoeffone \\neq 0 \\). Consider the set of all solutions ( \\( firstcoord, secondcoord, x_{8}, fourthcoord \\) ) of the simultaneous equations\n\\[\nacoeffone firstcoord+acoefftwo secondcoord+acoefftwo thirdcoord+acoefffour fourthcoord=0 \\text { and } bcoeffone firstcoord+bcoefftwo secondcoord+bcoefffour thirdcoord+bcoefffour fourthcoord=0\n\\]\nfor which no \\( x_{i}(i=1,2,3,4) \\) is zero. Each such solution generates a 4-tuple of plus and minus signs (signum \\( firstcoord \\), signum \\( secondcoord \\), signum \\( thirdcoord \\), signum \\( fourthcoord \\) ).\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\} \\) such that the above maximum number of 4 -tuples is attained.",
      "solution": "A-6 Solving the given equations in terms of thirdcoord and fourthcoord, leads to the equivalent system: \\( firstcoord=matrixaone thirdcoord+matrixbone fourthcoord, \\quad secondcoord=matrixatwo thirdcoord+matrixbtwo fourthcoord, \\quad thirdcoord=thirdcoord, \\quad fourthcoord=fourthcoord \\), where \\( matrixaone=(acoefftwo bcoeffthree-acoeffthree bcoefftwo)/(acoeffone bcoefftwo-acoefftwo bcoeffone) \\), etc.\n\nEach point in the thirdcoord, fourthcoord-plane corresponds uniquely to a solution ( firstcoord, secondcoord, thirdcoord, fourthcoord ). Signum firstcoord is positive for ( thirdcoord, fourthcoord ) on one side of \\( matrixaone thirdcoord+matrixbone fourthcoord=0 \\) and negative on the other side. Similarly for signum secondcoord, signum thirdcoord, and signum fourthcoord, using \\( matrixatwo thirdcoord+matrixbtwo thirdcoord=0, thirdcoord=0 \\) and \\( fourthcoord=0 \\), respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4 -tuple of signum values. Hence the maximum number of distinct 4 -tuples is eight.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions \\( matrixaone \\neq 0, matrixatwo \\neq 0, matrixbone \\neq 0, matrixbtwo \\neq 0 \\) and \\( matrixaone matrixbtwo-matrixatwo matrixbone \\neq 0 \\) or, simply, \\( a_{i} b_{j}-a_{j} b_{i} \\neq 0 \\) for \\( i, j=1,2,3,4 \\) and \\( i<j \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x_1": "sunflower",
        "x_2": "raincloud",
        "x_3": "bookshelf",
        "x_4": "paintbrush",
        "a_1": "gemstone",
        "a_2": "sailboat",
        "a_3": "starlight",
        "a_4": "driftwood",
        "b_1": "thunderclap",
        "b_2": "parchment",
        "b_3": "waterfall",
        "b_4": "coppermine",
        "A_1": "moonbeams",
        "A_2": "sandcastle",
        "B_1": "fireflies",
        "B_2": "periscope"
      },
      "question": "A-6. Given real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\},(i=1,2,3,4) \\), such that \\( gemstone\\,parchment-sailboat\\,thunderclap \\neq 0 \\). Consider the set of all solutions ( \\( sunflower, raincloud, x_{8}, paintbrush \\) ) of the simultaneous equations\n\\[\ngemstone\\,sunflower+sailboat\\,raincloud+sailboat\\,bookshelf+driftwood\\,paintbrush=0 \\text { and } thunderclap\\,sunflower+parchment\\,raincloud+coppermine\\,bookshelf+coppermine\\,paintbrush=0\n\\]\nfor which no \\( x_{i}(i=1,2,3,4) \\) is zero. Each such solution generates a 4-tuple of plus and minus signs (signum \\( sunflower \\), signum \\( raincloud \\), signum \\( bookshelf \\), signum \\( paintbrush \\) ).\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\} \\) such that the above maximum number of 4 -tuples is attained.",
      "solution": "A-6 Solving the given equations in terms of \\( bookshelf \\) and \\( paintbrush \\), leads to the equivalent system: \\( sunflower=moonbeams\\,bookshelf+fireflies\\,paintbrush, \\quad raincloud=sandcastle\\,bookshelf+periscope\\,paintbrush, \\quad bookshelf=bookshelf, \\quad paintbrush=paintbrush \\), where \\( moonbeams=\\left(sailboat\\,waterfall-starlight\\,parchment\\right) /\\left(gemstone\\,parchment-sailboat\\,thunderclap\\right) \\), etc.\n\nEach point in the \\( bookshelf, paintbrush \\)-plane corresponds uniquely to a solution ( \\( sunflower, raincloud \\), \\( bookshelf, paintbrush \\) ). Signum \\( sunflower \\) is positive for ( \\( bookshelf, paintbrush \\) ) on one side of \\( moonbeams\\,bookshelf+fireflies\\,paintbrush=0 \\) and negative on the other side. Similarly for signum \\( raincloud \\), signum \\( bookshelf \\), and signum \\( paintbrush \\), using \\( sandcastle\\,bookshelf+periscope\\,bookshelf=0, bookshelf=0 \\) and \\( paintbrush=0 \\), respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4 -tuple of signum values. Hence the maximum number of distinct 4 -tuples is eight.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions \\( moonbeams \\neq 0, sandcastle \\neq 0, fireflies \\neq 0 \\), \\( periscope \\neq 0 \\) and \\( moonbeams\\,periscope-sandcastle\\,fireflies \\neq 0 \\) or, simply, \\( a_{i} b_{j}-a_{j} b_{i} \\neq 0 \\) for \\( i, j=1,2,3,4 \\) and \\( i<j \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x_1": "knownvalue",
        "x_2": "fixedvalue",
        "x_3": "determined",
        "x_4": "specified",
        "a_1": "variableone",
        "a_2": "variabletwo",
        "a_3": "variablethree",
        "a_4": "variablefour",
        "b_1": "mutableone",
        "b_2": "mutabletwo",
        "b_3": "mutablethree",
        "b_4": "mutablefour",
        "A_1": "horizontalone",
        "A_2": "horizontaltwo",
        "B_1": "verticalone",
        "B_2": "verticaltwo"
      },
      "question": "A-6. Given real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\},(i=1,2,3,4) \\), such that \\( variableone mutabletwo-variabletwo mutableone \\neq 0 \\). Consider the set of all solutions ( \\( knownvalue, fixedvalue, x_{8}, specified \\) ) of the simultaneous equations\n\\[\nvariableone knownvalue+variabletwo fixedvalue+variabletwo determined+variablefour specified=0 \\text { and } mutableone knownvalue+mutabletwo fixedvalue+mutablefour determined+mutablefour specified=0\n\\]\nfor which no \\( x_{i}(i=1,2,3,4) \\) is zero. Each such solution generates a 4-tuple of plus and minus signs (signum \\( knownvalue \\), signum \\( fixedvalue \\), signum \\( determined \\), signum \\( specified \\) ).\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\} \\) such that the above maximum number of 4 -tuples is attained.",
      "solution": "A-6 Solving the given equations in terms of \\( determined \\) and \\( specified \\), leads to the equivalent system: \\( knownvalue=horizontalone determined+verticalone specified, \\quad fixedvalue=horizontaltwo determined+verticaltwo specified, \\quad determined=determined, \\quad specified=specified \\), where \\( horizontalone=\\left(variabletwo mutablethree-variablethree mutabletwo\\right) /\\left(variableone mutabletwo-variabletwo mutableone\\right) \\), etc.\n\nEach point in the \\( determined, specified \\)-plane corresponds uniquely to a solution ( \\( knownvalue, fixedvalue \\), \\( determined, specified \\) ). Signum \\( knownvalue \\) is positive for ( \\( determined, specified \\) ) on one side of \\( horizontalone determined+verticalone specified=0 \\) and negative on the other side. Similarly for signum \\( fixedvalue \\), signum \\( determined \\), and signum \\( specified \\), using \\( horizontaltwo determined+verticaltwo determined=0, determined=0 \\) and \\( specified=0 \\), respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4 -tuple of signum values. Hence the maximum number of distinct 4 -tuples is eight.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions \\( horizontalone \\neq 0, horizontaltwo \\neq 0, verticalone \\neq 0, verticaltwo \\neq 0 \\) and \\( horizontalone verticaltwo-horizontaltwo verticalone \\neq 0 \\) or, simply, \\( a_{i} b_{j}-a_{j} b_{i} \\neq 0 \\) for \\( i, j=1,2,3,4 \\) and \\( i<j \\)."
    },
    "garbled_string": {
      "map": {
        "x_1": "flaripton",
        "x_2": "moscabled",
        "x_3": "niewtrag",
        "x_4": "spoldurc",
        "a_1": "grentavos",
        "a_2": "quasperli",
        "a_3": "dukravlon",
        "a_4": "yimsethor",
        "b_1": "clovairn",
        "b_2": "pandruxel",
        "b_3": "thalkorin",
        "b_4": "sibmorant",
        "A_1": "velquasar",
        "A_2": "wembrioth",
        "B_1": "huxamedor",
        "B_2": "zoldriven"
      },
      "question": "A-6. Given real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\},(i=1,2,3,4) \\), such that \\( grentavos pandruxel-quasperli clovairn \\neq 0 \\). Consider the set of all solutions ( \\( flaripton, moscabled, x_{8}, spoldurc \\) ) of the simultaneous equations\n\\[\ngrentavos flaripton+quasperli moscabled+quasperli niewtrag+yimsethor spoldurc=0 \\text { and } clovairn flaripton+pandruxel moscabled+sibmorant niewtrag+sibmorant spoldurc=0\n\\]\nfor which no \\( x_{i}(i=1,2,3,4) \\) is zero. Each such solution generates a 4-tuple of plus and minus signs (signum \\( flaripton \\), signum \\( moscabled \\), signum \\( niewtrag \\), signum \\( spoldurc \\) ).\n(a) Determine, with a proof, the maximum number of distinct 4-tuples possible.\n(b) Investigate necessary and sufficient conditions on the real numbers \\( \\left\\{a_{i}\\right\\} \\) and \\( \\left\\{b_{i}\\right\\} \\) such that the above maximum number of 4 -tuples is attained.",
      "solution": "A-6 Solving the given equations in terms of \\( niewtrag \\) and \\( spoldurc \\), leads to the equivalent system: \\( flaripton=velquasar niewtrag+huxamedor spoldurc, \\quad moscabled=wembrioth niewtrag+zoldriven spoldurc, \\quad niewtrag=niewtrag, \\quad spoldurc=spoldurc \\), where \\( velquasar=\\left(quasperli thalkorin-dukravlon pandruxel\\right) /\\left(grentavos pandruxel-quasperli clovairn\\right) \\), etc.\n\nEach point in the \\( niewtrag, spoldurc \\)-plane corresponds uniquely to a solution ( \\( flaripton, moscabled \\), \\( niewtrag, spoldurc \\) ). Signum \\( flaripton \\) is positive for ( \\( niewtrag, spoldurc \\) ) on one side of \\( velquasar niewtrag+huxamedor spoldurc=0 \\) and negative on the other side. Similarly for signum \\( moscabled \\), signum \\( niewtrag \\), and signum \\( spoldurc \\), using \\( wembrioth niewtrag+zoldriven niewtrag=0, niewtrag=0 \\) and \\( spoldurc=0 \\), respectively. These four lines through the origin, in general, divide the plane into eight regions, each having a different 4 -tuple of signum values. Hence the maximum number of distinct 4 -tuples is eight.\n\nThe maximum number of eight will occur if and only if there are actually four distinct lines. This is equivalent to the conditions \\( velquasar \\neq 0, wembrioth \\neq 0, huxamedor \\neq 0 \\), \\( zoldriven \\neq 0 \\) and \\( velquasar zoldriven-wembrioth huxamedor \\neq 0 \\) or, simply, \\( a_{i} b_{j}-a_{j} b_{i} \\neq 0 \\) for \\( i, j=1,2,3,4 \\) and \\( i<j \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nA=\\bigl(a_{ij}\\bigr)_{1\\le i\\le 3,\\;1\\le j\\le 6}\n\\]\nbe a real $3\\times 6$ matrix of rank $3$.  Denote by  \n\\[\n\\mathcal N(A)=\\bigl\\{x=(x_{1},\\dots ,x_{6})\\in\\mathbb R^{6}\\; \\bigl|\\; Ax=0\\bigr\\}\n\\]\nits $3$–dimensional null–space and, for every non–zero $x\\in\\mathcal N(A)$\nwhose six coordinates are all non–vanishing, set  \n\\[\n\\sigma(x)=\\bigl(\\operatorname{sgn}x_{1},\\dots ,\\operatorname{sgn}x_{6}\\bigr)\\in\\{\\pm1\\}^{6}.\n\\]\n\n(a)  Determine the largest possible number $Q_{\\max}$ of distinct sign–patterns $\\sigma(x)$ that can be realised by vectors\n$x\\in\\mathcal N(A)\\setminus\\{0\\}$ with all coordinates non–zero.\n\n(b)  Prove that this maximum is attained if and only if\n\n\\[\n\\text{(GP)}\\qquad\n\\det\\bigl(a_{i,\\alpha_j}\\bigr)_{1\\le i\\le 3,\\;1\\le j\\le 3}\\neq 0\n\\quad\\text{for every three–element subset }  \n\\{\\alpha_1,\\alpha_2,\\alpha_3\\}\\subset\\{1,\\dots ,6\\},\n\\]\nequivalently, every choice of three columns of $A$ is linearly independent.\n\n(c)  Suppose $\\operatorname{rank}A=3$, every entry of $A$ is non–zero, and\nexactly one among the $20$ determinants occurring in (GP) vanishes while the\nother $19$ remain non–zero.  Determine the maximal possible number of\ndistinct sign–patterns in this situation.",
      "solution": "\\textbf{Preliminaries – reduction to an arrangement of great circles.}  \nThe space $\\mathcal N(A)$ is $3$–dimensional.  For $j=1,\\dots ,6$ define the\ncoordinate hyperplanes\n\\[\nH_{j}:=\\mathcal N(A)\\cap\\{x_{j}=0\\}\\subset\\mathcal N(A).\n\\]\nWhenever the $j$–th column of $A$ is not the zero vector (this will always\nbe the case because $\\operatorname{rank}A=3$), $H_{j}$ is a $2$–plane\nthrough the origin.  The connected components of\n\\[\n\\mathcal N(A)\\setminus\\bigl(\\cup_{j=1}^{6}H_{j}\\bigr)\\tag{$\\star$}\n\\]\nare exactly the sets on which the six coordinates keep fixed signs, hence\nthey are in bijection with the attainable sign–patterns.\n\nBecause each component of ($\\star$) is a cone with apex $0$, radial\nprojection onto the unit sphere $S\\bigl(\\mathcal N(A)\\bigr)\\cong S^{2}$\nsends every $H_{j}$ to a great circle\n\\[\nC_{j}:=H_{j}\\cap S^{2},\n\\]\nand it bijects the components of ($\\star$) with the $2$–cells\n(``chambers'') of the arrangement\n\\[\n\\mathcal C=\\{C_{1},\\dots ,C_{6}\\}\n\\]\nof six great circles on $S^{2}$.\n\n--------------------------------------------------------------------\n1.  \\emph{The general–position count.}  \nFor $k$ great circles on $S^{2}$ in \\emph{general position}\n(no two coincide and no three pass through the same pair of antipodal\npoints) the numbers of vertices ($V$), edges ($E$) and regions ($R$) are\nwell known and can be proved by induction or by Euler’s formula:\n\\[\nV=2\\binom{k}{2},\\qquad E=2k(k-1),\\qquad\nR=E-V+2=k(k-1)+2.\\tag{1}\n\\]\nWith $k=6$ one obtains\n\\[\nR=6\\cdot5+2=32,\\tag{2}\n\\]\nso\n\\[\nQ_{\\max}=32.\\qquad\\text{(answer to (a))}\n\\]\n\n--------------------------------------------------------------------\n2.  \\emph{Sufficiency of (GP).}  \nAssume (GP).  Then\n\n(i)  No two planes $H_{i},H_{j}$ coincide, for otherwise the $i$–th and\n$j$–th columns of $A$ would be proportional, and together with any third\ncolumn they would violate (GP).\n\n(ii)  No three planes $H_{\\alpha},H_{\\beta},H_{\\gamma}$ share a common\nline.  Indeed, if such a line existed, its direction vector\n$v\\in H_{\\alpha}\\cap H_{\\beta}\\cap H_{\\gamma}\\setminus\\{0\\}$ would have the\ncoordinates $v_{\\alpha}=v_{\\beta}=v_{\\gamma}=0$, which forces the\nremaining three columns of $A$ to be linearly dependent, contradicting\n(GP).\n\nHence the six great circles are in general position, and (2) shows that\n$32$ chambers—and therefore $32$ sign–patterns—are realised.\n\n--------------------------------------------------------------------\n3.  \\emph{Necessity of (GP).}  \nSuppose that a $3\\times3$ minor vanishes; relabel so that the columns\n$\\beta=\\{\\beta_{1},\\beta_{2},\\beta_{3}\\}=\\{1,2,3\\}$ are dependent.\nThus there exist scalars $t_{1},t_{2},t_{3}$, not all zero, such that  \n\\[\nt_{1}C_{1}+t_{2}C_{2}+t_{3}C_{3}=0.\n\\]\nLet $v\\in\\mathbb R^{6}$ be the vector with coordinates\n\\[\nv_{1}=t_{1},\\;v_{2}=t_{2},\\;v_{3}=t_{3},\\;v_{4}=v_{5}=v_{6}=0.\n\\]\nThen $A v=0$, \\emph{i.e.} $v\\in\\mathcal N(A)$, and\n\\[\nv\\in H_{4}\\cap H_{5}\\cap H_{6}\\setminus\\{0\\}.\n\\]\nConsequently $H_{4},H_{5},H_{6}$ all contain the line $\\mathbb R\\cdot v$,\nso the corresponding great circles $C_{4},C_{5},C_{6}$ meet in the same\npair of antipodal points; the arrangement $\\mathcal C$ is \\emph{not} in\ngeneral position.  Near such a triple intersection, three of the six\nsectors of a small spherical disc collapse, and the number of regions is\nreduced by at least $2$.  Therefore the maximum value $32$ cannot occur\nunless all $3\\times3$ minors are non–zero, that is, unless (GP) holds.  \nThis completes the proof of (b).\n\n--------------------------------------------------------------------\n4.  \\emph{Exactly one vanishing $3\\times3$ minor.}\n\n\\underline{Upper bound.}  \nAssume columns $1,2,3$ are dependent while every other triple is\nindependent.  Put $\\Gamma=\\{1,2,3\\}$ and $\\Delta=\\{4,5,6\\}$.\nBy the argument in Step~3, the three planes $H_{4},H_{5},H_{6}$ intersect\nin a single line $L$, whereas every other pair of distinct planes meets in\na different line and no four planes share a line.  \nExcept for this single triple concurrence the arrangement is generic.\n\nStart from the general–position values (1) with $k=6$:\n$V_{0}=30,\\;E_{0}=60,\\;R_{0}=32$.  \nLet $p$ and $-p$ be the antipodal points in which $C_{4},C_{5},C_{6}$ meet.\n\n$\\bullet$ \\emph{Vertices.}  \nIn general position the three pairs $(C_{4},C_{5}),(C_{4},C_{6}),\n(C_{5},C_{6})$ yield six distinct vertices near $p$ and six near $-p$.\nAfter the collapse these six vertices merge into one at $p$ and one at\n$-p$; thus $4$ vertices disappear:\n\\[\nV=V_{0}-4=26.\\tag{3}\n\\]\n\n$\\bullet$ \\emph{Edges.}  \nAlong each of $C_{4},C_{5},C_{6}$ two edges vanish, so the edge count\ndrops by $6$:\n\\[\nE=E_{0}-6=54.\\tag{4}\n\\]\n\n$\\bullet$ \\emph{Regions.}  \nEuler’s formula for $S^{2}$ gives\n\\[\nR=E-V+2=54-26+2=30.\\tag{5}\n\\]\nHence at most $30$ sign–patterns are possible.\n\n\\underline{Sharpness.}  \nWe construct a $3\\times6$ matrix realising exactly $30$ patterns.\n\nChoose two linearly independent vectors $c_{1},c_{2}\\in\\mathbb R^{3}$ with\nno zero coordinate and set\n\\[\nc_{3}=c_{1}+c_{2}\\quad\\bigl(\\text{so }c_{1},c_{2},c_{3}\\text{ are dependent}\\bigr).\n\\]\nNow pick three column vectors depending on a parameter $t>0$:\n\\[\nc_{4}=(1,t,t^{2}),\\qquad\nc_{5}=(1,t^{2},t^{3}),\\qquad\nc_{6}=(1,t^{3},t^{4}).\\tag{6}\n\\]\nLet $A(t)=[c_{1}\\;c_{2}\\;c_{3}\\;c_{4}\\;c_{5}\\;c_{6}]$.\nFor every triple $\\tau\\neq\\Gamma$, the determinant\n$\\det A_{\\tau}(t)$ is a non–zero polynomial in $t$; if it were the zero\npolynomial, the corresponding columns would be dependent for every\n$t>0$, which is impossible because their last coordinates grow with\ndifferent degrees in $t$.  Each polynomial therefore has only finitely\nmany real roots, and since there are finitely many such polynomials, one\ncan choose $t$ outside their union.  For such a value of $t$ we have\n\\[\n\\operatorname{rank}A(t)=3,\\quad\n\\text{all entries of }A(t)\\text{ are non–zero},\\quad\n\\det A_{\\Gamma}(t)=0,\\quad\n\\det A_{\\tau}(t)\\neq0\\;\\;(\\tau\\neq\\Gamma).\n\\]\nThe associated circle arrangement is precisely the one analysed above, so\n$30$ chambers—and thus $30$ sign–patterns—are realised.\n\nConsequently, under the condition that exactly one $3\\times3$ minor\nvanishes, the maximal number of different sign–patterns equals\n\\[\n\\boxed{30}.\n\\]\nThis completes part (c).\n\n--------------------------------------------------------------------\n\\[\n\\boxed{Q_{\\max}=32\\quad\\text{and, with exactly one vanishing\nminor, at most and in fact }30\\text{ patterns can occur}.}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.573184",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension jump – The ambient ℝ^6 and a 3-dimensional null-space require familiarity with arrangements of planes in 3-space instead of lines in a plane, forcing combinatorial geometry of higher dimension.  \n• Combinatorial geometry – The solver must know (or derive) formulas for the number of regions cut out by central hyperplane arrangements and cope with degeneracies.  \n• Algebraic characterisation – “General position’’ is translated into the vanishing/non-vanishing of every 3×3 minor of a 3×6 matrix, linking linear-algebraic conditions to geometric ones; proving equivalence is substantially subtler than in the original 4-variable, 2-line case.  \n• Degenerate case (part (c)) – Beyond the maximal case, the problem demands a precise chamber count under a specified singularity, requiring either topological arguments or advanced counting theorems (e.g. Zaslavsky’s theorem with Möbius invariants).  \nThese added layers of geometry, combinatorics and algebra make the variant significantly more intricate than the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nA=\\bigl(a_{ij}\\bigr)_{1\\le i\\le 3,\\;1\\le j\\le 6}\n\\]\nbe a real $3\\times 6$ matrix of rank $3$.  Denote by  \n\\[\n\\mathcal N(A)=\\bigl\\{x=(x_{1},\\dots ,x_{6})\\in\\mathbb R^{6}\\; \\bigl|\\; Ax=0\\bigr\\}\n\\]\nits $3$–dimensional null–space and, for every non–zero $x\\in\\mathcal N(A)$\nwhose six coordinates are all non–vanishing, set  \n\\[\n\\sigma(x)=\\bigl(\\operatorname{sgn}x_{1},\\dots ,\\operatorname{sgn}x_{6}\\bigr)\\in\\{\\pm1\\}^{6}.\n\\]\n\n(a)  Determine the largest possible number $Q_{\\max}$ of distinct sign–patterns $\\sigma(x)$ that can be realised by vectors\n$x\\in\\mathcal N(A)\\setminus\\{0\\}$ with all coordinates non–zero.\n\n(b)  Prove that this maximum is attained if and only if\n\n\\[\n\\text{(GP)}\\qquad\n\\det\\bigl(a_{i,\\alpha_j}\\bigr)_{1\\le i\\le 3,\\;1\\le j\\le 3}\\neq 0\n\\quad\\text{for every three–element subset }  \n\\{\\alpha_1,\\alpha_2,\\alpha_3\\}\\subset\\{1,\\dots ,6\\},\n\\]\nequivalently, every choice of three columns of $A$ is linearly independent.\n\n(c)  Suppose $\\operatorname{rank}A=3$, every entry of $A$ is non–zero, and\nexactly one among the $20$ determinants occurring in (GP) vanishes while the\nother $19$ remain non–zero.  Determine the maximal possible number of\ndistinct sign–patterns in this situation.",
      "solution": "\\textbf{Preliminaries – reduction to an arrangement of great circles.}  \nThe space $\\mathcal N(A)$ is $3$–dimensional.  For $j=1,\\dots ,6$ define the\ncoordinate hyperplanes\n\\[\nH_{j}:=\\mathcal N(A)\\cap\\{x_{j}=0\\}\\subset\\mathcal N(A).\n\\]\nWhenever the $j$–th column of $A$ is not the zero vector (this will always\nbe the case because $\\operatorname{rank}A=3$), $H_{j}$ is a $2$–plane\nthrough the origin.  The connected components of\n\\[\n\\mathcal N(A)\\setminus\\bigl(\\cup_{j=1}^{6}H_{j}\\bigr)\\tag{$\\star$}\n\\]\nare exactly the sets on which the six coordinates keep fixed signs, hence\nthey are in bijection with the attainable sign–patterns.\n\nBecause each component of ($\\star$) is a cone with apex $0$, radial\nprojection onto the unit sphere $S\\bigl(\\mathcal N(A)\\bigr)\\cong S^{2}$\nsends every $H_{j}$ to a great circle\n\\[\nC_{j}:=H_{j}\\cap S^{2},\n\\]\nand it bijects the components of ($\\star$) with the $2$–cells\n(``chambers'') of the arrangement\n\\[\n\\mathcal C=\\{C_{1},\\dots ,C_{6}\\}\n\\]\nof six great circles on $S^{2}$.\n\n--------------------------------------------------------------------\n1.  \\emph{The general–position count.}  \nFor $k$ great circles on $S^{2}$ in \\emph{general position}\n(no two coincide and no three pass through the same pair of antipodal\npoints) the numbers of vertices ($V$), edges ($E$) and regions ($R$) are\nwell known and can be proved by induction or by Euler’s formula:\n\\[\nV=2\\binom{k}{2},\\qquad E=2k(k-1),\\qquad\nR=E-V+2=k(k-1)+2.\\tag{1}\n\\]\nWith $k=6$ one obtains\n\\[\nR=6\\cdot5+2=32,\\tag{2}\n\\]\nso\n\\[\nQ_{\\max}=32.\\qquad\\text{(answer to (a))}\n\\]\n\n--------------------------------------------------------------------\n2.  \\emph{Sufficiency of (GP).}  \nAssume (GP).  Then\n\n(i)  No two planes $H_{i},H_{j}$ coincide, for otherwise the $i$–th and\n$j$–th columns of $A$ would be proportional, and together with any third\ncolumn they would violate (GP).\n\n(ii)  No three planes $H_{\\alpha},H_{\\beta},H_{\\gamma}$ share a common\nline.  Indeed, if such a line existed, its direction vector\n$v\\in H_{\\alpha}\\cap H_{\\beta}\\cap H_{\\gamma}\\setminus\\{0\\}$ would have the\ncoordinates $v_{\\alpha}=v_{\\beta}=v_{\\gamma}=0$, which forces the\nremaining three columns of $A$ to be linearly dependent, contradicting\n(GP).\n\nHence the six great circles are in general position, and (2) shows that\n$32$ chambers—and therefore $32$ sign–patterns—are realised.\n\n--------------------------------------------------------------------\n3.  \\emph{Necessity of (GP).}  \nSuppose that a $3\\times3$ minor vanishes; relabel so that the columns\n$\\beta=\\{\\beta_{1},\\beta_{2},\\beta_{3}\\}=\\{1,2,3\\}$ are dependent.\nThus there exist scalars $t_{1},t_{2},t_{3}$, not all zero, such that  \n\\[\nt_{1}C_{1}+t_{2}C_{2}+t_{3}C_{3}=0.\n\\]\nLet $v\\in\\mathbb R^{6}$ be the vector with coordinates\n\\[\nv_{1}=t_{1},\\;v_{2}=t_{2},\\;v_{3}=t_{3},\\;v_{4}=v_{5}=v_{6}=0.\n\\]\nThen $A v=0$, \\emph{i.e.} $v\\in\\mathcal N(A)$, and\n\\[\nv\\in H_{4}\\cap H_{5}\\cap H_{6}\\setminus\\{0\\}.\n\\]\nConsequently $H_{4},H_{5},H_{6}$ all contain the line $\\mathbb R\\cdot v$,\nso the corresponding great circles $C_{4},C_{5},C_{6}$ meet in the same\npair of antipodal points; the arrangement $\\mathcal C$ is \\emph{not} in\ngeneral position.  Near such a triple intersection, three of the six\nsectors of a small spherical disc collapse, and the number of regions is\nreduced by at least $2$.  Therefore the maximum value $32$ cannot occur\nunless all $3\\times3$ minors are non–zero, that is, unless (GP) holds.  \nThis completes the proof of (b).\n\n--------------------------------------------------------------------\n4.  \\emph{Exactly one vanishing $3\\times3$ minor.}\n\n\\underline{Upper bound.}  \nAssume columns $1,2,3$ are dependent while every other triple is\nindependent.  Put $\\Gamma=\\{1,2,3\\}$ and $\\Delta=\\{4,5,6\\}$.\nBy the argument in Step~3, the three planes $H_{4},H_{5},H_{6}$ intersect\nin a single line $L$, whereas every other pair of distinct planes meets in\na different line and no four planes share a line.  \nExcept for this single triple concurrence the arrangement is generic.\n\nStart from the general–position values (1) with $k=6$:\n$V_{0}=30,\\;E_{0}=60,\\;R_{0}=32$.  \nLet $p$ and $-p$ be the antipodal points in which $C_{4},C_{5},C_{6}$ meet.\n\n$\\bullet$ \\emph{Vertices.}  \nIn general position the three pairs $(C_{4},C_{5}),(C_{4},C_{6}),\n(C_{5},C_{6})$ yield six distinct vertices near $p$ and six near $-p$.\nAfter the collapse these six vertices merge into one at $p$ and one at\n$-p$; thus $4$ vertices disappear:\n\\[\nV=V_{0}-4=26.\\tag{3}\n\\]\n\n$\\bullet$ \\emph{Edges.}  \nAlong each of $C_{4},C_{5},C_{6}$ two edges vanish, so the edge count\ndrops by $6$:\n\\[\nE=E_{0}-6=54.\\tag{4}\n\\]\n\n$\\bullet$ \\emph{Regions.}  \nEuler’s formula for $S^{2}$ gives\n\\[\nR=E-V+2=54-26+2=30.\\tag{5}\n\\]\nHence at most $30$ sign–patterns are possible.\n\n\\underline{Sharpness.}  \nWe construct a $3\\times6$ matrix realising exactly $30$ patterns.\n\nChoose two linearly independent vectors $c_{1},c_{2}\\in\\mathbb R^{3}$ with\nno zero coordinate and set\n\\[\nc_{3}=c_{1}+c_{2}\\quad\\bigl(\\text{so }c_{1},c_{2},c_{3}\\text{ are dependent}\\bigr).\n\\]\nNow pick three column vectors depending on a parameter $t>0$:\n\\[\nc_{4}=(1,t,t^{2}),\\qquad\nc_{5}=(1,t^{2},t^{3}),\\qquad\nc_{6}=(1,t^{3},t^{4}).\\tag{6}\n\\]\nLet $A(t)=[c_{1}\\;c_{2}\\;c_{3}\\;c_{4}\\;c_{5}\\;c_{6}]$.\nFor every triple $\\tau\\neq\\Gamma$, the determinant\n$\\det A_{\\tau}(t)$ is a non–zero polynomial in $t$; if it were the zero\npolynomial, the corresponding columns would be dependent for every\n$t>0$, which is impossible because their last coordinates grow with\ndifferent degrees in $t$.  Each polynomial therefore has only finitely\nmany real roots, and since there are finitely many such polynomials, one\ncan choose $t$ outside their union.  For such a value of $t$ we have\n\\[\n\\operatorname{rank}A(t)=3,\\quad\n\\text{all entries of }A(t)\\text{ are non–zero},\\quad\n\\det A_{\\Gamma}(t)=0,\\quad\n\\det A_{\\tau}(t)\\neq0\\;\\;(\\tau\\neq\\Gamma).\n\\]\nThe associated circle arrangement is precisely the one analysed above, so\n$30$ chambers—and thus $30$ sign–patterns—are realised.\n\nConsequently, under the condition that exactly one $3\\times3$ minor\nvanishes, the maximal number of different sign–patterns equals\n\\[\n\\boxed{30}.\n\\]\nThis completes part (c).\n\n--------------------------------------------------------------------\n\\[\n\\boxed{Q_{\\max}=32\\quad\\text{and, with exactly one vanishing\nminor, at most and in fact }30\\text{ patterns can occur}.}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.465311",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension jump – The ambient ℝ^6 and a 3-dimensional null-space require familiarity with arrangements of planes in 3-space instead of lines in a plane, forcing combinatorial geometry of higher dimension.  \n• Combinatorial geometry – The solver must know (or derive) formulas for the number of regions cut out by central hyperplane arrangements and cope with degeneracies.  \n• Algebraic characterisation – “General position’’ is translated into the vanishing/non-vanishing of every 3×3 minor of a 3×6 matrix, linking linear-algebraic conditions to geometric ones; proving equivalence is substantially subtler than in the original 4-variable, 2-line case.  \n• Degenerate case (part (c)) – Beyond the maximal case, the problem demands a precise chamber count under a specified singularity, requiring either topological arguments or advanced counting theorems (e.g. Zaslavsky’s theorem with Möbius invariants).  \nThese added layers of geometry, combinatorics and algebra make the variant significantly more intricate than the original."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}