{
  "index": "1975-A-2",
  "type": "ALG",
  "tag": [
    "ALG",
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "A-2. For which ordered pairs of real numbers \\( b, c \\) do both roots of the quadratic equation\n\\[\nz^{2}+b z+c=0\n\\]\nlie inside the unit disk \\( \\{|z|<1\\} \\) in the complex plane?\nDraw a reasonably accurate picture (i.e., 'graph') of the region in the real bc-plane for which the above condition holds. Identify precisely the boundary curves of this region.",
  "solution": "A-2.\nThe desired region is the inside of the triangle with vertices \\( (0,-1),(2,1),(-2,1) \\). The boundary segments lie on the lines\n\\[\n\\mathrm{L}_{1}: \\mathrm{c}=1, \\mathrm{~L}_{2}: \\mathrm{c}-\\mathrm{b}+1=0, \\mathrm{~L}_{3}: \\mathrm{c}+\\mathrm{b}+1=0 .\n\\]\n\nTo see this, we let \\( f(z)=z^{2}+b z+c \\) and denote its zeros by \\( r \\) and \\( s \\). Then \\( -b=r+s \\) and \\( c=r s \\). Also\n\\[\n\\begin{array}{l}\n(r+1)(s+1)=r s+r+s+1=c-b+1=f(-1), \\\\\n(r-1)(s-1)=r s-r-s+1=c+b+1=f(1) .\n\\end{array}\n\\]\n\nOn or.below \\( L_{2} \\), at least one zero is real and not greater than -1 ; this follows either from \\( (r+1) \\). \\( (s+1) \\leqq 0 \\) or from \\( f(-1) \\leqq 0 \\) and the fact that the graph of \\( y=f(x) \\), for \\( x \\) real, is an upward opening parabola. Similarly, on or below \\( L_{3} \\) one zero is real and at least 1 . On or above \\( L_{1} \\), at least one zero has absolute value greater than or equal to 1 . Hence the desired points ( \\( b, c \\) ) must be inside the described triangle.\n\nConversely, if \\( (b, c) \\) is inside the triangle, \\( |c|<1 \\) and so \\( |r|<1 \\) or \\( |s|<1 \\) or both. If the zeros are complex, they are conjugates and \\( |r|=|s| \\); then \\( |r|=|s|<1 \\) follows from \\( |c|<1 \\). If the zeros are real, \\( |c|<1 \\) implies that at least one zero is in \\( (-1,1) \\). Then \\( (r+1)(s+1)=f(-1)>0 \\) and \\( (r-1) \\). \\( (s-1)=f(1)>0 \\) imply that the other zero is also in \\( (-1,1) \\).\n\nFor full credit, the region had to be depicted.",
  "vars": [
    "z",
    "r",
    "s",
    "x",
    "y"
  ],
  "params": [
    "b",
    "c",
    "f",
    "L_1",
    "L_2",
    "L_3"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "complexz",
        "r": "firstroot",
        "s": "secondr",
        "x": "realaxis",
        "y": "ordinate",
        "b": "coeffb",
        "c": "coeffc",
        "f": "polyfunc",
        "L_1": "lineone",
        "L_2": "linetwo",
        "L_3": "linethree"
      },
      "question": "A-2. For which ordered pairs of real numbers \\( coeffb, coeffc \\) do both roots of the quadratic equation\n\\[\ncomplexz^{2}+coeffb\\, complexz+coeffc=0\n\\]\nlie inside the unit disk \\( \\{|complexz|<1\\} \\) in the complex plane?\nDraw a reasonably accurate picture (i.e., 'graph') of the region in the real coeffbcoeffc-plane for which the above condition holds. Identify precisely the boundary curves of this region.",
      "solution": "A-2.\nThe desired region is the inside of the triangle with vertices \\( (0,-1),(2,1),(-2,1) \\). The boundary segments lie on the lines\n\\[\n\\mathrm{lineone}: \\mathrm{coeffc}=1, \\mathrm{~linetwo}: \\mathrm{coeffc}-\\mathrm{coeffb}+1=0, \\mathrm{~linethree}: \\mathrm{coeffc}+\\mathrm{coeffb}+1=0 .\n\\]\n\nTo see this, we let \\( polyfunc(complexz)=complexz^{2}+coeffb\\, complexz+coeffc \\) and denote its zeros by \\( firstroot \\) and \\( secondr \\). Then \\( -coeffb=firstroot+secondr \\) and \\( coeffc=firstroot\\, secondr \\). Also\n\\[\n\\begin{array}{l}\n(firstroot+1)(secondr+1)=firstroot\\, secondr+firstroot+secondr+1=coeffc-coeffb+1=polyfunc(-1), \\\\\n(firstroot-1)(secondr-1)=firstroot\\, secondr-firstroot-secondr+1=coeffc+coeffb+1=polyfunc(1) .\n\\end{array}\n\\]\n\nOn or below \\( linetwo \\), at least one zero is real and not greater than \\(-1\\); this follows either from \\( (firstroot+1)(secondr+1) \\leqq 0 \\) or from \\( polyfunc(-1) \\leqq 0 \\) and the fact that the graph of \\( ordinate=polyfunc(realaxis) \\), for \\( realaxis \\) real, is an upward opening parabola. Similarly, on or below \\( linethree \\) one zero is real and at least 1. On or above \\( lineone \\), at least one zero has absolute value greater than or equal to 1. Hence the desired points ( \\( coeffb, coeffc \\) ) must be inside the described triangle.\n\nConversely, if \\( (coeffb, coeffc) \\) is inside the triangle, \\( |coeffc|<1 \\) and so \\( |firstroot|<1 \\) or \\( |secondr|<1 \\) or both. If the zeros are complex, they are conjugates and \\( |firstroot|=|secondr| \\); then \\( |firstroot|=|secondr|<1 \\) follows from \\( |coeffc|<1 \\). If the zeros are real, \\( |coeffc|<1 \\) implies that at least one zero is in \\( (-1,1) \\). Then \\( (firstroot+1)(secondr+1)=polyfunc(-1)>0 \\) and \\( (firstroot-1)(secondr-1)=polyfunc(1)>0 \\) imply that the other zero is also in \\( (-1,1) \\).\n\nFor full credit, the region had to be depicted."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "breadcrumb",
        "r": "lanternfly",
        "s": "hummingtop",
        "x": "windhamper",
        "y": "shoelumber",
        "b": "compassrose",
        "c": "riverstone",
        "f": "paperbloom",
        "L_1": "meadowlark",
        "L_2": "thunderegg",
        "L_3": "coffeegrain"
      },
      "question": "A-2. For which ordered pairs of real numbers \\( compassrose, riverstone \\) do both roots of the quadratic equation\n\\[\nbreadcrumb^{2}+compassrose\\ breadcrumb+riverstone=0\n\\]\nlie inside the unit disk \\( \\{|breadcrumb|<1\\} \\) in the complex plane?\nDraw a reasonably accurate picture (i.e., 'graph') of the region in the real compassroseriverstone-plane for which the above condition holds. Identify precisely the boundary curves of this region.",
      "solution": "A-2.\nThe desired region is the inside of the triangle with vertices \\( (0,-1),(2,1),(-2,1) \\). The boundary segments lie on the lines\n\\[\n\\mathrm{meadowlark}: \\mathrm{riverstone}=1, \\mathrm{~thunderegg}: \\mathrm{riverstone}-\\mathrm{compassrose}+1=0, \\mathrm{~coffeegrain}: \\mathrm{riverstone}+\\mathrm{compassrose}+1=0 .\n\\]\n\nTo see this, we let \\( paperbloom(breadcrumb)=breadcrumb^{2}+compassrose\\ breadcrumb+riverstone \\) and denote its zeros by \\( lanternfly \\) and \\( hummingtop \\). Then \\( -compassrose=lanternfly+hummingtop \\) and \\( riverstone=lanternfly\\ hummingtop \\). Also\n\\[\n\\begin{array}{l}\n(lanternfly+1)(hummingtop+1)=lanternfly\\ hummingtop+lanternfly+hummingtop+1=riverstone-compassrose+1=paperbloom(-1), \\\\\n(lanternfly-1)(hummingtop-1)=lanternfly\\ hummingtop-lanternfly-hummingtop+1=riverstone+compassrose+1=paperbloom(1) .\n\\end{array}\n\\]\n\nOn or.below \\( thunderegg \\), at least one zero is real and not greater than -1 ; this follows either from \\( (lanternfly+1) \\). \\( (hummingtop+1) \\leqq 0 \\) or from \\( paperbloom(-1) \\leqq 0 \\) and the fact that the graph of \\( shoelumber=paperbloom(windhamper) \\), for \\( windhamper \\) real, is an upward opening parabola. Similarly, on or below \\( coffeegrain \\) one zero is real and at least 1 . On or above \\( meadowlark \\), at least one zero has absolute value greater than or equal to 1 . Hence the desired points ( \\( compassrose, riverstone \\) ) must be inside the described triangle.\n\nConversely, if \\( (compassrose, riverstone) \\) is inside the triangle, \\( |riverstone|<1 \\) and so \\( |lanternfly|<1 \\) or \\( |hummingtop|<1 \\) or both. If the zeros are complex, they are conjugates and \\( |lanternfly|=|hummingtop| \\); then \\( |lanternfly|=|hummingtop|<1 \\) follows from \\( |riverstone|<1 \\). If the zeros are real, \\( |riverstone|<1 \\) implies that at least one zero is in \\( (-1,1) \\). Then \\( (lanternfly+1)(hummingtop+1)=paperbloom(-1)>0 \\) and \\( (lanternfly-1) \\). \\( (hummingtop-1)=paperbloom(1)>0 \\) imply that the other zero is also in \\( (-1,1) \\).\n\nFor full credit, the region had to be depicted."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "positivereal",
        "r": "nonrooted",
        "s": "surfaceval",
        "x": "verticalvar",
        "y": "horizontal",
        "b": "imaginarycoef",
        "c": "imaginaryterm",
        "f": "unquadratic",
        "L_1": "curvedpatha",
        "L_2": "curvedpathb",
        "L_3": "curvedpathc"
      },
      "question": "A-2. For which ordered pairs of real numbers \\( imaginarycoef, imaginaryterm \\) do both roots of the quadratic equation\n\\[\npositivereal^{2}+imaginarycoef\\, positivereal+imaginaryterm=0\n\\]\nlie inside the unit disk \\( \\{|positivereal|<1\\} \\) in the complex plane?\nDraw a reasonably accurate picture (i.e., 'graph') of the region in the real imaginarycoefimaginaryterm-plane for which the above condition holds. Identify precisely the boundary curves of this region.",
      "solution": "A-2.\nThe desired region is the inside of the triangle with vertices \\( (0,-1),(2,1),(-2,1) \\). The boundary segments lie on the lines\n\\[\n\\mathrm{curvedpatha}: \\mathrm{imaginaryterm}=1, \\mathrm{~curvedpathb}: \\mathrm{imaginaryterm}-\\mathrm{imaginarycoef}+1=0, \\mathrm{~curvedpathc}: \\mathrm{imaginaryterm}+\\mathrm{imaginarycoef}+1=0 .\n\\]\n\nTo see this, we let \\( unquadratic(positivereal)=positivereal^{2}+imaginarycoef\\, positivereal+imaginaryterm \\) and denote its zeros by \\( nonrooted \\) and \\( surfaceval \\). Then \\( -imaginarycoef=nonrooted+surfaceval \\) and \\( imaginaryterm=nonrooted\\, surfaceval \\). Also\n\\[\n\\begin{array}{l}\n(nonrooted+1)(surfaceval+1)=nonrooted\\, surfaceval+nonrooted+surfaceval+1=imaginaryterm-imaginarycoef+1=unquadratic(-1), \\\\\n(nonrooted-1)(surfaceval-1)=nonrooted\\, surfaceval-nonrooted-surfaceval+1=imaginaryterm+imaginarycoef+1=unquadratic(1) .\n\\end{array}\n\\]\n\nOn or below \\( curvedpathb \\), at least one zero is real and not greater than \\(-1\\); this follows either from \\( (nonrooted+1)(surfaceval+1) \\leqq 0 \\) or from \\( unquadratic(-1) \\leqq 0 \\) and the fact that the graph of \\( horizontal = unquadratic(verticalvar) \\), for \\( verticalvar \\) real, is an upward opening parabola. Similarly, on or below \\( curvedpathc \\) one zero is real and at least 1. On or above \\( curvedpatha \\), at least one zero has absolute value greater than or equal to 1. Hence the desired points ( \\( imaginarycoef, imaginaryterm \\) ) must be inside the described triangle.\n\nConversely, if \\( (imaginarycoef, imaginaryterm) \\) is inside the triangle, \\( |imaginaryterm|<1 \\) and so \\( |nonrooted|<1 \\) or \\( |surfaceval|<1 \\) or both. If the zeros are complex, they are conjugates and \\( |nonrooted|=|surfaceval| \\); then \\( |nonrooted|=|surfaceval|<1 \\) follows from \\( |imaginaryterm|<1 \\). If the zeros are real, \\( |imaginaryterm|<1 \\) implies that at least one zero is in \\((-1,1)\\). Then \\( (nonrooted+1)(surfaceval+1)=unquadratic(-1)>0 \\) and \\( (nonrooted-1)(surfaceval-1)=unquadratic(1)>0 \\) imply that the other zero is also in \\((-1,1)\\).\n\nFor full credit, the region had to be depicted."
    },
    "garbled_string": {
      "map": {
        "z": "qzxwvtnp",
        "r": "hjgrksla",
        "s": "pldmvqre",
        "x": "mntkgfzb",
        "y": "skvoldra",
        "b": "vqsrnloi",
        "c": "tjwhxkpe",
        "f": "zrshcmyg",
        "L_1": "qfwmnzto",
        "L_2": "hrqgxsvm",
        "L_3": "kbvtnfyl"
      },
      "question": "A-2. For which ordered pairs of real numbers \\( vqsrnloi, tjwhxkpe \\) do both roots of the quadratic equation\n\\[\nqzxwvtnp^{2}+vqsrnloi qzxwvtnp+tjwhxkpe=0\n\\]\nlie inside the unit disk \\( \\{|qzxwvtnp|<1\\} \\) in the complex plane?\nDraw a reasonably accurate picture (i.e., 'graph') of the region in the real vqsrnloitjwhxkpe-plane for which the above condition holds. Identify precisely the boundary curves of this region.",
      "solution": "A-2.\nThe desired region is the inside of the triangle with vertices \\( (0,-1),(2,1),(-2,1) \\). The boundary segments lie on the lines\n\\[\n\\mathrm{qfwmnzto}: \\mathrm{tjwhxkpe}=1, \\mathrm{~hrqgxsvm}: \\mathrm{tjwhxkpe}-\\mathrm{vqsrnloi}+1=0, \\mathrm{~kbvtnfyl}: \\mathrm{tjwhxkpe}+\\mathrm{vqsrnloi}+1=0 .\n\\]\n\nTo see this, we let \\( zrshcmyg(qzxwvtnp)=qzxwvtnp^{2}+vqsrnloi qzxwvtnp+tjwhxkpe \\) and denote its zeros by \\( hjgrksla \\) and \\( pldmvqre \\). Then \\( -vqsrnloi=hjgrksla+pldmvqre \\) and \\( tjwhxkpe=hjgrksla pldmvqre \\). Also\n\\[\n\\begin{array}{l}\n(hjgrksla+1)(pldmvqre+1)=hjgrksla pldmvqre+hjgrksla+pldmvqre+1=tjwhxkpe-vqsrnloi+1=zrshcmyg(-1), \\\\\n(hjgrksla-1)(pldmvqre-1)=hjgrksla pldmvqre-hjgrksla-pldmvqre+1=tjwhxkpe+vqsrnloi+1=zrshcmyg(1) .\n\\end{array}\n\\]\n\nOn or.below \\( hrqgxsvm \\), at least one zero is real and not greater than -1; this follows either from \\( (hjgrksla+1)(pldmvqre+1) \\leqq 0 \\) or from \\( zrshcmyg(-1) \\leqq 0 \\) and the fact that the graph of \\( skvoldra=zrshcmyg(mntkgfzb) \\), for \\( mntkgfzb \\) real, is an upward opening parabola. Similarly, on or below \\( kbvtnfyl \\) one zero is real and at least 1. On or above \\( qfwmnzto \\), at least one zero has absolute value greater than or equal to 1. Hence the desired points ( \\( vqsrnloi, tjwhxkpe \\) ) must be inside the described triangle.\n\nConversely, if \\( (vqsrnloi, tjwhxkpe) \\) is inside the triangle, \\( |tjwhxkpe|<1 \\) and so \\( |hjgrksla|<1 \\) or \\( |pldmvqre|<1 \\) or both. If the zeros are complex, they are conjugates and \\( |hjgrksla|=|pldmvqre| \\); then \\( |hjgrksla|=|pldmvqre|<1 \\) follows from \\( |tjwhxkpe|<1 \\). If the zeros are real, \\( |tjwhxkpe|<1 \\) implies that at least one zero is in \\( (-1,1) \\). Then \\( (hjgrksla+1)(pldmvqre+1)=zrshcmyg(-1)>0 \\) and \\( (hjgrksla-1)(pldmvqre-1)=zrshcmyg(1)>0 \\) imply that the other zero is also in \\( (-1,1) \\).\n\nFor full credit, the region had to be depicted."
    },
    "kernel_variant": {
      "question": "Determine all ordered triples of real numbers                     \n\\[\n(b,c,d)\\in\\mathbb R^{3}\n\\]\nfor which every zero of the cubic polynomial                       \n\\[\nP(z)=4z^{3}+bz^{2}+cz+d\n\\]\nlies in the closed disk  \n\\[\n\\mathbb D_{2}:=\\{\\,z\\in\\mathbb C:|z|\\le 2\\}.\n\\]\n\n(a)  Derive a system of explicit algebraic inequalities in the variables \\((b,c,d)\\) that is necessary and sufficient for this property.\n\n(b)  Describe all real–analytic boundary hypersurfaces on which at least one zero satisfies \\(|z|=2\\), and prove that their union is the whole topological boundary of the feasible set.\n\n(c)  Decide whether the feasible set \\(\\mathcal R\\subset\\mathbb R^{3}\\) is convex.  Prove your assertion and, if it is not convex, exhibit two admissible triples whose midpoint is inadmissible.\n\n(d)  Fix a real number \\(d_{0}\\) with \\(|d_{0}|\\le 32\\) and consider the horizontal slice  \n\\[\n\\mathcal R(d_{0})=\\{(b,c,d_{0})\\in\\mathcal R\\}.\n\\]\nShow that \\(\\mathcal R(d_{0})\\) is the intersection of three closed half–planes, hence a convex polygonal region (which degenerates to a half–line when \\(|d_{0}|=32\\)).  Determine the three bounding lines explicitly and give closed‐form coordinates for the vertices when \\(|d_{0}|<32\\).",
      "solution": "Throughout we write \\(\\mathcal R\\subset\\Bbb R^{3}\\) for the desired feasible set.\n\n--------------------------------------------------------------------\n(a)  Reduction to a monic cubic on the unit disk  \n--------------------------------------------------------------------\nPut \\(z=2w\\;(w\\in\\Bbb C)\\).  Then \\(|z|\\le 2\\iff |w|\\le 1\\) and  \n\\[\nP(z)=32\\bigl[w^{3}+{\\textstyle\\frac b8}w^{2}+{\\textstyle\\frac c{16}}w\n            +{\\textstyle\\frac d{32}}\\bigr].\n\\]\nThus every zero of \\(P\\) lies in \\(\\Bbb D_{2}\\) iff the monic cubic  \n\\[\nQ(w)=w^{3}+a_{1}w^{2}+a_{2}w+a_{3},\n\\qquad\na_{1}=\\frac b8,\\;a_{2}=\\frac c{16},\\;a_{3}=\\frac d{32},\n\\]\nhas all its zeros in the closed unit disk \\(\\Bbb D\\).\n\n--------------------------------------------------------------------\nJury–Schur–Cohn criterion for a real monic cubic  \n--------------------------------------------------------------------\nFor a real monic cubic \\(w^{3}+a_{1}w^{2}+a_{2}w+a_{3}\\) the zeros are in\n\\(\\Bbb D\\) iff\n\n(J1) \\(1+a_{1}+a_{2}+a_{3}\\ge 0\\),\n\n(J2) \\(1-a_{1}+a_{2}-a_{3}\\ge 0\\),\n\n(J3) \\(1-a_{2}+a_{1}a_{3}-a_{3}^{2}\\ge 0\\),\n\n(J4) \\(1-a_{3}^{2}\\ge 0\\).\n\nEquality in at least one of (J1)–(J4) is equivalent to the occurrence of a root on the unit circle.\n\n--------------------------------------------------------------------\nTranslation to \\((b,c,d)\\)–coordinates  \n--------------------------------------------------------------------\nInsert \\(a_{1}=b/8,\\;a_{2}=c/16,\\;a_{3}=d/32\\):\n\n(I1)  \\(32+4b+2c+d\\;\\;\\ge 0,\\)\n\n(I2)  \\(32-4b+2c-d\\;\\;\\ge 0,\\)\n\n(I3)  \\(1024-64c+4bd-d^{2}\\ge 0,\\)\n\n(I4)  \\(d^{2}\\le 1024\\;( \\,|d|\\le 32).\\)\n\nHence  \n\\[\n\\boxed{\\;\n\\mathcal R=\\{(b,c,d)\\in\\Bbb R^{3}:\\text{(I1)}\\wedge\\text{(I2)}\\wedge\\text{(I3)}\n        \\wedge\\text{(I4)}\\}\\;}\n\\]\nis the required description.\n\n--------------------------------------------------------------------\n(b)  Boundary hypersurfaces  \n--------------------------------------------------------------------\nReplace “\\(\\ge\\)” by “\\(=\\)” in exactly one of (I1)–(I4) and keep the remaining three inequalities.  This yields the five real–analytic hypersurfaces\n\nS₁ : \\(32+4b+2c+d=0\\)  (plane),\n\nS₂ : \\(32-4b+2c-d=0\\)  (plane),\n\nS₃ : \\(1024-64c+4bd-d^{2}=0\\)  (ruled quadratic surface),\n\nS₄⁺ : \\(d= 32\\) (plane),\n\nS₄⁻ : \\(d=-32\\) (plane).\n\nNecessity.  The coefficients of a polynomial depend continuously on its roots.\nIf a point of \\(\\mathcal R\\) is perturbed so that a root leaves \\(|z|\\le 2\\),\nat least one of the four closed inequalities must be violated, hence the path\nhas crossed one of the hypersurfaces \\(S_{1},S_{2},S_{3},S_{4}^{\\pm}\\).\n\nSufficiency.  Because (I1)–(I4) are jointly necessary **and sufficient**\n(cf. part (a)), crossing any of the hypersurfaces forces a violation of the\nJury inequalities in the monic model, so a zero must indeed leave the disk.\nConsequently  \n\\[\n\\partial\\mathcal R=S_{1}\\cup S_{2}\\cup S_{3}\\cup S_{4}^{+}\\cup S_{4}^{-}.\n\\]\n\n--------------------------------------------------------------------\n(c)  Global non–convexity of \\(\\mathcal R\\)  \n--------------------------------------------------------------------\nThe inequality (I3) contains the bilinear term \\(4bd\\), so convexity is\nunlikely.  We supply an explicit counter–example.\n\nPick  \n\\[\n\\begin{aligned}\nP&=(b_{1},c_{1},d_{1})=(\\;\\;10,\\;20,\\;32),\\\\\nQ&=(b_{2},c_{2},d_{2})=(-10,\\;20,-32).\n\\end{aligned}\n\\]\nVerification for \\(P\\) (the same computations work for \\(Q\\); the signs in\n(I1)–(I3) simply interchange):\n\n• (I4) holds with equality, \\(|d_{1}|=32\\).\n\n• (I1) \\(32+4b_{1}+2c_{1}+d_{1}=32+40+40+32=144>0\\).\n\n• (I2) \\(32-4b_{1}+2c_{1}-d_{1}=32-40+40-32=0\\).\n\n• (I3) \\(1024-64\\!\\cdot\\!20+4\\!\\cdot\\!10\\!\\cdot\\!32-32^{2}\n        =1024-1280+1280-1024=0\\).\n\nHence \\(P,Q\\in\\mathcal R\\).\n\nTheir midpoint is  \n\\[\nM=\\tfrac12(P+Q)=(0,\\,20,\\,0).\n\\]\n\nCheck (I3) at \\(M\\):\n\\[\n1024-64\\!\\cdot\\!20+4\\!\\cdot\\!0\\!\\cdot\\!0-0\n      =1024-1280=-256<0,\n\\]\nso \\(M\\notin\\mathcal R\\) although \\(P,Q\\in\\mathcal R\\).\nTherefore \\(\\mathcal R\\) is **not convex.**\n\n--------------------------------------------------------------------\n(d)  Geometry of a horizontal slice \\(\\mathcal R(d_{0})\\)  \n--------------------------------------------------------------------\nFix \\(d_{0}\\) with \\(|d_{0}|\\le 32\\).  In the plane \\(d=d_{0}\\) the four\ninequalities read\n\n\\[\n\\begin{aligned}\n\\text{(H1)}\\;&  c\\ge -2b-\\frac{d_{0}}2-16,\\\\\n\\text{(H2)}\\;&  c\\ge  \\;2b+\\frac{d_{0}}2-16,\\\\\n\\text{(H3)}\\;&  c\\le \\frac{1024-d_{0}^{2}}{64}+\\frac{d_{0}}{16}\\,b,\\\\\n\\text{(H4)}\\;&  \\text{(automatic, already }|d_{0}|\\le 32\\text{).}\n\\end{aligned}\n\\]\n\nThus \\(\\mathcal R(d_{0})\\) is the intersection of the two lower half–planes\ndefined by (H1) and (H2) and the upper half–plane defined by (H3).  Each is\nclosed, so the slice is a closed convex **polygonal region**.  More precisely\n\n• If \\(|d_{0}|<32\\) the three bounding lines are pairwise non–parallel and\n\\(\\mathcal R(d_{0})\\) is a triangle.\n\n• If \\(d_{0}= 32\\) the lines (H2) and (H3) coincide (\\(c=2b\\)); the region\ndegenerates to the half–line \\(\\{(b,2b):b\\ge -8\\}\\).\n\n• If \\(d_{0}=-32\\) the lines (H1) and (H3) coincide (\\(c=-2b\\)); the slice is\nthe symmetric half–line \\(\\{(b,-2b):b\\ge -8\\}\\).\n\nAssume henceforth \\(|d_{0}|<32\\).  Denote the three lines by  \n\n\\[\n\\begin{aligned}\nL_{1}&:\\;c=-2b-\\frac{d_{0}}2-16,\\\\\nL_{2}&:\\;c= \\;2b+\\frac{d_{0}}2-16,\\\\\nL_{3}&:\\;c=\\frac{1024-d_{0}^{2}}{64}+\\frac{d_{0}}{16}\\,b.\n\\end{aligned}\n\\]\n\nVertices (obtained by pairwise intersection):\n\n\\[\n\\begin{aligned}\nV_{1}&=L_{1}\\cap L_{2}: &\n      &\\;b=-\\dfrac{d_{0}}4,\\quad c=-16,\\\\[4pt]\nV_{2}&=L_{1}\\cap L_{3}: &\n      &\\;b=\\dfrac{d_{0}^{2}-32d_{0}-2048}{4(d_{0}+32)},\n        \\quad\n        c=-2b-\\dfrac{d_{0}}2-16,\\\\[6pt]\nV_{3}&=L_{2}\\cap L_{3}: &\n      &\\;b=\\dfrac{d_{0}^{2}+32d_{0}-2048}{4(d_{0}-32)},\n        \\quad\n        c= 2b+\\dfrac{d_{0}}2-16.\n\\end{aligned}\n\\]\n\nFor \\(|d_{0}|<32\\) the three points are distinct and the feasible section is\nthe triangle \\(\\triangle V_{1}V_{2}V_{3}\\).\nAll three coordinates depend rationally on \\(d_{0}\\) and continuously extend\nto the limiting half–line cases when \\(|d_{0}|\\to 32\\).\n\n--------------------------------------------------------------------\nAll parts (a)–(d) are now rigorously settled.  \n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.618532",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimension jump The problem now lives in \\(\\mathbb R^{3}\\) instead of the \\((b,c)\\)-plane, so the answer is a 3-D solid bounded by several nonlinear algebraic surfaces, not just by three straight line segments.\n\n2.  Higher-degree polynomial A cubic replaces the quadratic, so the root-to-coefficient relations are no longer merely Vieta’s two equations but involve three roots and considerably more complicated stability criteria.\n\n3.  Advanced machinery While the quadratic case can be handled by elementary inequalities or simple geometric observations, the cubic requires the full Schur–Cohn (or Jury) stability test, including one iteration of the Schur transform and an embedded quadratic stability check.  This introduces nested fractions, absolute-value inequalities, and a quartic determinant.\n\n4.  Boundary description The boundary of the admissible set is now the union of planes, quadrics, and a quartic surface; finding and interpreting them demand skill with algebraic manipulation and some multivariable geometry.\n\n5.  Multi-stage reasoning Necessity and sufficiency split into four separate inequalities, each arising from a different stage in the Schur algorithm.  The solver must track how each stage transforms the coefficients and why the resulting inequalities are sharp.\n\nAll these features make the enhanced variant substantially more technical and conceptually deeper than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Determine all ordered triples of real numbers                     \n\\[\n(b,c,d)\\in\\mathbb R^{3}\n\\]\nfor which every zero of the cubic polynomial                       \n\\[\nP(z)=4z^{3}+bz^{2}+cz+d\n\\]\nlies in the closed disk  \n\\[\n\\mathbb D_{2}:=\\{\\,z\\in\\mathbb C:|z|\\le 2\\}.\n\\]\n\n(a)  Derive a system of explicit algebraic inequalities in the variables \\((b,c,d)\\) that is necessary and sufficient for this property.\n\n(b)  Describe all real–analytic boundary hypersurfaces on which at least one zero satisfies \\(|z|=2\\), and prove that their union is the whole topological boundary of the feasible set.\n\n(c)  Decide whether the feasible set \\(\\mathcal R\\subset\\mathbb R^{3}\\) is convex.  Prove your assertion and, if it is not convex, exhibit two admissible triples whose midpoint is inadmissible.\n\n(d)  Fix a real number \\(d_{0}\\) with \\(|d_{0}|\\le 32\\) and consider the horizontal slice  \n\\[\n\\mathcal R(d_{0})=\\{(b,c,d_{0})\\in\\mathcal R\\}.\n\\]\nShow that \\(\\mathcal R(d_{0})\\) is the intersection of three closed half–planes, hence a convex polygonal region (which degenerates to a half–line when \\(|d_{0}|=32\\)).  Determine the three bounding lines explicitly and give closed‐form coordinates for the vertices when \\(|d_{0}|<32\\).",
      "solution": "Throughout we write \\(\\mathcal R\\subset\\Bbb R^{3}\\) for the desired feasible set.\n\n--------------------------------------------------------------------\n(a)  Reduction to a monic cubic on the unit disk  \n--------------------------------------------------------------------\nPut \\(z=2w\\;(w\\in\\Bbb C)\\).  Then \\(|z|\\le 2\\iff |w|\\le 1\\) and  \n\\[\nP(z)=32\\bigl[w^{3}+{\\textstyle\\frac b8}w^{2}+{\\textstyle\\frac c{16}}w\n            +{\\textstyle\\frac d{32}}\\bigr].\n\\]\nThus every zero of \\(P\\) lies in \\(\\Bbb D_{2}\\) iff the monic cubic  \n\\[\nQ(w)=w^{3}+a_{1}w^{2}+a_{2}w+a_{3},\n\\qquad\na_{1}=\\frac b8,\\;a_{2}=\\frac c{16},\\;a_{3}=\\frac d{32},\n\\]\nhas all its zeros in the closed unit disk \\(\\Bbb D\\).\n\n--------------------------------------------------------------------\nJury–Schur–Cohn criterion for a real monic cubic  \n--------------------------------------------------------------------\nFor a real monic cubic \\(w^{3}+a_{1}w^{2}+a_{2}w+a_{3}\\) the zeros are in\n\\(\\Bbb D\\) iff\n\n(J1) \\(1+a_{1}+a_{2}+a_{3}\\ge 0\\),\n\n(J2) \\(1-a_{1}+a_{2}-a_{3}\\ge 0\\),\n\n(J3) \\(1-a_{2}+a_{1}a_{3}-a_{3}^{2}\\ge 0\\),\n\n(J4) \\(1-a_{3}^{2}\\ge 0\\).\n\nEquality in at least one of (J1)–(J4) is equivalent to the occurrence of a root on the unit circle.\n\n--------------------------------------------------------------------\nTranslation to \\((b,c,d)\\)–coordinates  \n--------------------------------------------------------------------\nInsert \\(a_{1}=b/8,\\;a_{2}=c/16,\\;a_{3}=d/32\\):\n\n(I1)  \\(32+4b+2c+d\\;\\;\\ge 0,\\)\n\n(I2)  \\(32-4b+2c-d\\;\\;\\ge 0,\\)\n\n(I3)  \\(1024-64c+4bd-d^{2}\\ge 0,\\)\n\n(I4)  \\(d^{2}\\le 1024\\;( \\,|d|\\le 32).\\)\n\nHence  \n\\[\n\\boxed{\\;\n\\mathcal R=\\{(b,c,d)\\in\\Bbb R^{3}:\\text{(I1)}\\wedge\\text{(I2)}\\wedge\\text{(I3)}\n        \\wedge\\text{(I4)}\\}\\;}\n\\]\nis the required description.\n\n--------------------------------------------------------------------\n(b)  Boundary hypersurfaces  \n--------------------------------------------------------------------\nReplace “\\(\\ge\\)” by “\\(=\\)” in exactly one of (I1)–(I4) and keep the remaining three inequalities.  This yields the five real–analytic hypersurfaces\n\nS₁ : \\(32+4b+2c+d=0\\)  (plane),\n\nS₂ : \\(32-4b+2c-d=0\\)  (plane),\n\nS₃ : \\(1024-64c+4bd-d^{2}=0\\)  (ruled quadratic surface),\n\nS₄⁺ : \\(d= 32\\) (plane),\n\nS₄⁻ : \\(d=-32\\) (plane).\n\nNecessity.  The coefficients of a polynomial depend continuously on its roots.\nIf a point of \\(\\mathcal R\\) is perturbed so that a root leaves \\(|z|\\le 2\\),\nat least one of the four closed inequalities must be violated, hence the path\nhas crossed one of the hypersurfaces \\(S_{1},S_{2},S_{3},S_{4}^{\\pm}\\).\n\nSufficiency.  Because (I1)–(I4) are jointly necessary **and sufficient**\n(cf. part (a)), crossing any of the hypersurfaces forces a violation of the\nJury inequalities in the monic model, so a zero must indeed leave the disk.\nConsequently  \n\\[\n\\partial\\mathcal R=S_{1}\\cup S_{2}\\cup S_{3}\\cup S_{4}^{+}\\cup S_{4}^{-}.\n\\]\n\n--------------------------------------------------------------------\n(c)  Global non–convexity of \\(\\mathcal R\\)  \n--------------------------------------------------------------------\nThe inequality (I3) contains the bilinear term \\(4bd\\), so convexity is\nunlikely.  We supply an explicit counter–example.\n\nPick  \n\\[\n\\begin{aligned}\nP&=(b_{1},c_{1},d_{1})=(\\;\\;10,\\;20,\\;32),\\\\\nQ&=(b_{2},c_{2},d_{2})=(-10,\\;20,-32).\n\\end{aligned}\n\\]\nVerification for \\(P\\) (the same computations work for \\(Q\\); the signs in\n(I1)–(I3) simply interchange):\n\n• (I4) holds with equality, \\(|d_{1}|=32\\).\n\n• (I1) \\(32+4b_{1}+2c_{1}+d_{1}=32+40+40+32=144>0\\).\n\n• (I2) \\(32-4b_{1}+2c_{1}-d_{1}=32-40+40-32=0\\).\n\n• (I3) \\(1024-64\\!\\cdot\\!20+4\\!\\cdot\\!10\\!\\cdot\\!32-32^{2}\n        =1024-1280+1280-1024=0\\).\n\nHence \\(P,Q\\in\\mathcal R\\).\n\nTheir midpoint is  \n\\[\nM=\\tfrac12(P+Q)=(0,\\,20,\\,0).\n\\]\n\nCheck (I3) at \\(M\\):\n\\[\n1024-64\\!\\cdot\\!20+4\\!\\cdot\\!0\\!\\cdot\\!0-0\n      =1024-1280=-256<0,\n\\]\nso \\(M\\notin\\mathcal R\\) although \\(P,Q\\in\\mathcal R\\).\nTherefore \\(\\mathcal R\\) is **not convex.**\n\n--------------------------------------------------------------------\n(d)  Geometry of a horizontal slice \\(\\mathcal R(d_{0})\\)  \n--------------------------------------------------------------------\nFix \\(d_{0}\\) with \\(|d_{0}|\\le 32\\).  In the plane \\(d=d_{0}\\) the four\ninequalities read\n\n\\[\n\\begin{aligned}\n\\text{(H1)}\\;&  c\\ge -2b-\\frac{d_{0}}2-16,\\\\\n\\text{(H2)}\\;&  c\\ge  \\;2b+\\frac{d_{0}}2-16,\\\\\n\\text{(H3)}\\;&  c\\le \\frac{1024-d_{0}^{2}}{64}+\\frac{d_{0}}{16}\\,b,\\\\\n\\text{(H4)}\\;&  \\text{(automatic, already }|d_{0}|\\le 32\\text{).}\n\\end{aligned}\n\\]\n\nThus \\(\\mathcal R(d_{0})\\) is the intersection of the two lower half–planes\ndefined by (H1) and (H2) and the upper half–plane defined by (H3).  Each is\nclosed, so the slice is a closed convex **polygonal region**.  More precisely\n\n• If \\(|d_{0}|<32\\) the three bounding lines are pairwise non–parallel and\n\\(\\mathcal R(d_{0})\\) is a triangle.\n\n• If \\(d_{0}= 32\\) the lines (H2) and (H3) coincide (\\(c=2b\\)); the region\ndegenerates to the half–line \\(\\{(b,2b):b\\ge -8\\}\\).\n\n• If \\(d_{0}=-32\\) the lines (H1) and (H3) coincide (\\(c=-2b\\)); the slice is\nthe symmetric half–line \\(\\{(b,-2b):b\\ge -8\\}\\).\n\nAssume henceforth \\(|d_{0}|<32\\).  Denote the three lines by  \n\n\\[\n\\begin{aligned}\nL_{1}&:\\;c=-2b-\\frac{d_{0}}2-16,\\\\\nL_{2}&:\\;c= \\;2b+\\frac{d_{0}}2-16,\\\\\nL_{3}&:\\;c=\\frac{1024-d_{0}^{2}}{64}+\\frac{d_{0}}{16}\\,b.\n\\end{aligned}\n\\]\n\nVertices (obtained by pairwise intersection):\n\n\\[\n\\begin{aligned}\nV_{1}&=L_{1}\\cap L_{2}: &\n      &\\;b=-\\dfrac{d_{0}}4,\\quad c=-16,\\\\[4pt]\nV_{2}&=L_{1}\\cap L_{3}: &\n      &\\;b=\\dfrac{d_{0}^{2}-32d_{0}-2048}{4(d_{0}+32)},\n        \\quad\n        c=-2b-\\dfrac{d_{0}}2-16,\\\\[6pt]\nV_{3}&=L_{2}\\cap L_{3}: &\n      &\\;b=\\dfrac{d_{0}^{2}+32d_{0}-2048}{4(d_{0}-32)},\n        \\quad\n        c= 2b+\\dfrac{d_{0}}2-16.\n\\end{aligned}\n\\]\n\nFor \\(|d_{0}|<32\\) the three points are distinct and the feasible section is\nthe triangle \\(\\triangle V_{1}V_{2}V_{3}\\).\nAll three coordinates depend rationally on \\(d_{0}\\) and continuously extend\nto the limiting half–line cases when \\(|d_{0}|\\to 32\\).\n\n--------------------------------------------------------------------\nAll parts (a)–(d) are now rigorously settled.  \n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.494399",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimension jump The problem now lives in \\(\\mathbb R^{3}\\) instead of the \\((b,c)\\)-plane, so the answer is a 3-D solid bounded by several nonlinear algebraic surfaces, not just by three straight line segments.\n\n2.  Higher-degree polynomial A cubic replaces the quadratic, so the root-to-coefficient relations are no longer merely Vieta’s two equations but involve three roots and considerably more complicated stability criteria.\n\n3.  Advanced machinery While the quadratic case can be handled by elementary inequalities or simple geometric observations, the cubic requires the full Schur–Cohn (or Jury) stability test, including one iteration of the Schur transform and an embedded quadratic stability check.  This introduces nested fractions, absolute-value inequalities, and a quartic determinant.\n\n4.  Boundary description The boundary of the admissible set is now the union of planes, quadrics, and a quartic surface; finding and interpreting them demand skill with algebraic manipulation and some multivariable geometry.\n\n5.  Multi-stage reasoning Necessity and sufficiency split into four separate inequalities, each arising from a different stage in the Schur algorithm.  The solver must track how each stage transforms the coefficients and why the resulting inequalities are sharp.\n\nAll these features make the enhanced variant substantially more technical and conceptually deeper than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}