{
  "index": "1948-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. The pairs of numbers \\( (a, b) \\) such that \\( \\left|a+b t+t^{2}\\right| \\leq 1 \\) for \\( 0 \\leq t \\leq 1 \\) fill a certain region in the \\( (a, b) \\)-plane. What is the area of this region?",
  "solution": "First Solution. Consider\n\\[\nf(t)=a+b t+t^{2}\n\\]\n\nThis function has just one critical value, at \\( t=-b / 2 \\) where its value is \\( a-b^{2 / 4} \\). On the interval \\( [0,1] f \\) can have extreme values only at the endpoints and at the critical point if it should happen to fall in \\( [0,1] \\) (that is, if \\( b \\in[-2,0] \\) ). Hence the extreme values of \\( f \\) are\n\\[\nf(0)=a \\text { and } f(1)=a+b+1\n\\]\nif \\( b \\notin[-2,0] \\), and they are in the set\n\\[\n\\left\\{a, a+b+1, a-b^{2} / 4\\right\\}\n\\]\nif \\( b \\in[-2,0] \\).\nHence \\( |f(t)| \\leq 1 \\) for all \\( t \\in[0,1] \\) if and only if\n\\[\nb \\notin[-2,0] \\text { and }|a| \\leq 1,|a+b+1| \\leq 1\n\\]\nor\n\\[\nb \\in[-2,0], \\text { and }|a| \\leq 1,|a+b+1| \\leq 1,\\left|a-\\frac{b^{2}}{4}\\right| \\leq 1\n\\]\n\nThe region required is therefore as shown in the diagram where the arc from \\( A \\) to \\( B \\) is part of the parabola \\( a-b^{2} / 4=-1 \\).\n\nThe area of the required region can be obtained in several ways. The parallelogram CEDF evidently has area 4 . We must subtract the area of the piece \\( A F B \\). Since \\( \\overparen{A F} \\) and \\( \\overparen{F B} \\) are tangents to the parabola, the area of \\( A F B \\) is \\( \\frac{1}{3} \\) that of the triangle \\( A F B \\) (Archimedes' rule). But triangle \\( A F B \\) has base \\( A F \\) of length 1 and altitude 1 , so its area is \\( \\frac{1}{2} \\) and the curvilinear piece \\( A F B \\) has area \\( \\frac{1}{6} \\). Hence the area of CEDBA is \\( 4-\\frac{1}{6}=\\frac{23}{6} \\).\n\nSecond Solution. Let \\( S_{t} \\) be the strip in the \\( (a, b) \\)-plane bounded by the two parallel lines\n\\[\n\\begin{array}{c}\na+b t+t^{2}=1 \\\\\na+b t+t^{2}=-1\n\\end{array}\n\\]\n\nWe seek the area of the region \\( \\Pi \\) defined by \\( \\Pi=\\bigcap_{0 \\leq t \\leq 1} S_{t} \\). Now (1) and (2) are parallel tangents to the parabolas \\( b^{2}=4 a-4 \\) and \\( b^{2}=4 a+4 \\), and for the range \\( 0 \\leq t \\leq 1 \\) these are tangents to the arcs \\( \\overparen{A B} \\) and \\( \\overparen{A_{1} B_{1}} \\) where \\( A(-1,0), B(0,-2), A_{1}(1,0), B_{1}(2,-2) \\) are points on the two parabolas.\n\\( \\Pi \\) is therefore the region bounded by the lines \\( a=1, a=-1, a+b=0 \\), \\( a+b=-2 \\) and the parabolic arc \\( \\overparen{A B} \\) of \\( b^{2}=4 a+4 \\). We now find the area as before.",
  "vars": [
    "t",
    "f",
    "S_t",
    "\\\\Pi"
  ],
  "params": [
    "a",
    "b",
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "A_1",
    "B_1"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "f": "functn",
        "S_t": "stripset",
        "\\Pi": "regionpi",
        "a": "firstvar",
        "b": "secondvar",
        "A": "pointa",
        "B": "pointb",
        "C": "pointc",
        "D": "pointd",
        "E": "pointe",
        "F": "pointf",
        "A_1": "pointaone",
        "B_1": "pointbone"
      },
      "question": "5. The pairs of numbers \\( (firstvar, secondvar) \\) such that \\( \\left|firstvar+secondvar\\,timevar+timevar^{2}\\right| \\leq 1 \\) for \\( 0 \\leq timevar \\leq 1 \\) fill a certain region in the \\( (firstvar, secondvar) \\)-plane. What is the area of this region?",
      "solution": "First Solution. Consider\n\\[\nfunctn(timevar)=firstvar+secondvar\\,timevar+timevar^{2}\n\\]\n\nThis function has just one critical value, at \\( timevar=-secondvar / 2 \\) where its value is \\( firstvar-secondvar^{2 / 4} \\). On the interval \\( [0,1] functn \\) can have extreme values only at the endpoints and at the critical point if it should happen to fall in \\( [0,1] \\) (that is, if \\( secondvar \\in[-2,0] \\) ). Hence the extreme values of \\( functn \\) are\n\\[\nfunctn(0)=firstvar \\text { and } functn(1)=firstvar+secondvar+1\n\\]\nif \\( secondvar \\notin[-2,0] \\), and they are in the set\n\\[\n\\{firstvar,\\; firstvar+secondvar+1,\\; firstvar-secondvar^{2} / 4\\}\n\\]\nif \\( secondvar \\in[-2,0] \\).\nHence \\( |functn(timevar)| \\leq 1 \\) for all \\( timevar \\in[0,1] \\) if and only if\n\\[\nsecondvar \\notin[-2,0] \\text { and }|firstvar| \\leq 1,\\;|firstvar+secondvar+1| \\leq 1\n\\]\nor\n\\[\nsecondvar \\in[-2,0], \\text { and }|firstvar| \\leq 1,\\;|firstvar+secondvar+1| \\leq 1,\\;\\left|firstvar-\\frac{secondvar^{2}}{4}\\right| \\leq 1\n\\]\n\nThe region required is therefore as shown in the diagram where the arc from \\( pointa \\) to \\( pointb \\) is part of the parabola \\( firstvar-secondvar^{2} / 4=-1 \\).\n\nThe area of the required region can be obtained in several ways. The parallelogram pointc pointe pointd pointf evidently has area 4. We must subtract the area of the piece \\( pointa\\,pointf\\,pointb \\). Since \\( \\overparen{pointa\\,pointf} \\) and \\( \\overparen{pointf\\,pointb} \\) are tangents to the parabola, the area of \\( pointa\\,pointf\\,pointb \\) is \\( \\frac{1}{3} \\) that of the triangle \\( pointa\\,pointf\\,pointb \\) (Archimedes' rule). But triangle \\( pointa\\,pointf\\,pointb \\) has base \\( pointa\\,pointf \\) of length 1 and altitude 1, so its area is \\( \\frac{1}{2} \\) and the curvilinear piece \\( pointa\\,pointf\\,pointb \\) has area \\( \\frac{1}{6} \\). Hence the area of pointc pointe pointd pointb pointa is \\( 4-\\frac{1}{6}=\\frac{23}{6} \\).\n\nSecond Solution. Let \\( stripset \\) be the strip in the \\( (firstvar, secondvar) \\)-plane bounded by the two parallel lines\n\\[\n\\begin{array}{c}\nfirstvar+secondvar\\,timevar+timevar^{2}=1 \\\\\nfirstvar+secondvar\\,timevar+timevar^{2}=-1\n\\end{array}\n\\]\n\nWe seek the area of the region \\( regionpi \\) defined by \\( regionpi=\\bigcap_{0 \\leq timevar \\leq 1} stripset \\). Now (1) and (2) are parallel tangents to the parabolas \\( secondvar^{2}=4 firstvar-4 \\) and \\( secondvar^{2}=4 firstvar+4 \\), and for the range \\( 0 \\leq timevar \\leq 1 \\) these are tangents to the arcs \\( \\overparen{pointa\\,pointb} \\) and \\( \\overparen{pointaone\\,pointbone} \\) where \\( pointa(-1,0),\\; pointb(0,-2),\\; pointaone(1,0),\\; pointbone(2,-2) \\) are points on the two parabolas.\n\\( regionpi \\) is therefore the region bounded by the lines \\( firstvar=1,\\; firstvar=-1,\\; firstvar+secondvar=0,\\; firstvar+secondvar=-2 \\) and the parabolic arc \\( \\overparen{pointa\\,pointb} \\) of \\( secondvar^{2}=4 firstvar+4 \\). We now find the area as before."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "marshland",
        "f": "thunderous",
        "S_t": "honeycomb",
        "\\\\Pi": "constellation",
        "a": "windflower",
        "b": "dragonfly",
        "A": "ridgepole",
        "B": "chandelier",
        "C": "parchment",
        "D": "silkmoth",
        "E": "labyrinth",
        "F": "moonstone",
        "A_1": "ridgepoleone",
        "B_1": "chandelierone"
      },
      "question": "5. The pairs of numbers \\( (windflower, dragonfly) \\) such that \\( \\left|windflower+dragonfly marshland+marshland^{2}\\right| \\leq 1 \\) for \\( 0 \\leq marshland \\leq 1 \\) fill a certain region in the \\( (windflower, dragonfly) \\)-plane. What is the area of this region?",
      "solution": "First Solution. Consider\n\\[\nthunderous(marshland)=windflower+dragonfly marshland+marshland^{2}\n\\]\n\nThis function has just one critical value, at \\( marshland=-dragonfly / 2 \\) where its value is \\( windflower-dragonfly^{2 / 4} \\). On the interval \\( [0,1] thunderous \\) can have extreme values only at the endpoints and at the critical point if it should happen to fall in \\( [0,1] \\) (that is, if \\( dragonfly \\in[-2,0] \\) ). Hence the extreme values of \\( thunderous \\) are\n\\[\nthunderous(0)=windflower \\text { and } thunderous(1)=windflower+dragonfly+1\n\\]\nif \\( dragonfly \\notin[-2,0] \\), and they are in the set\n\\[\n\\left\\{windflower, windflower+dragonfly+1, windflower-dragonfly^{2} / 4\\right\\}\n\\]\nif \\( dragonfly \\in[-2,0] \\).\nHence \\( |thunderous(marshland)| \\leq 1 \\) for all \\( marshland \\in[0,1] \\) if and only if\n\\[\ndragonfly \\notin[-2,0] \\text { and }|windflower| \\leq 1,|windflower+dragonfly+1| \\leq 1\n\\]\nor\n\\[\ndragonfly \\in[-2,0], \\text { and }|windflower| \\leq 1,|windflower+dragonfly+1| \\leq 1,\\left|windflower-\\frac{dragonfly^{2}}{4}\\right| \\leq 1\n\\]\n\nThe region required is therefore as shown in the diagram where the arc from \\( ridgepole \\) to \\( chandelier \\) is part of the parabola \\( windflower-dragonfly^{2} / 4=-1 \\).\n\nThe area of the required region can be obtained in several ways. The parallelogram parchmentlabyrinthsilkmothmoonstone evidently has area 4. We must subtract the area of the piece \\( ridgepole moonstone chandelier \\). Since \\( \\overparen{ridgepole moonstone} \\) and \\( \\overparen{moonstone chandelier} \\) are tangents to the parabola, the area of \\( ridgepole moonstone chandelier \\) is \\( \\frac{1}{3} \\) that of the triangle \\( ridgepole moonstone chandelier \\) (Archimedes' rule). But triangle \\( ridgepole moonstone chandelier \\) has base \\( ridgepole moonstone \\) of length 1 and altitude 1, so its area is \\( \\frac{1}{2} \\) and the curvilinear piece \\( ridgepole moonstone chandelier \\) has area \\( \\frac{1}{6} \\). Hence the area of parchmentlabyrinthsilkmothchandelierridgepole is \\( 4-\\frac{1}{6}=\\frac{23}{6} \\).\n\nSecond Solution. Let \\( honeycomb \\) be the strip in the \\( (windflower, dragonfly) \\)-plane bounded by the two parallel lines\n\\[\n\\begin{array}{c}\nwindflower+dragonfly marshland+marshland^{2}=1 \\\\\nwindflower+dragonfly marshland+marshland^{2}=-1\n\\end{array}\n\\]\n\nWe seek the area of the region \\( constellation \\) defined by \\( constellation=\\bigcap_{0 \\leq marshland \\leq 1} honeycomb \\). Now (1) and (2) are parallel tangents to the parabolas \\( dragonfly^{2}=4 windflower-4 \\) and \\( dragonfly^{2}=4 windflower+4 \\), and for the range \\( 0 \\leq marshland \\leq 1 \\) these are tangents to the arcs \\( \\overparen{ridgepole chandelier} \\) and \\( \\overparen{ridgepoleone chandelierone} \\) where \\( ridgepole(-1,0), chandelier(0,-2), ridgepoleone(1,0), chandelierone(2,-2) \\) are points on the two parabolas.\n\\( constellation \\) is therefore the region bounded by the lines \\( windflower=1, windflower=-1, windflower+dragonfly=0 \\), \\( windflower+dragonfly=-2 \\) and the parabolic arc \\( \\overparen{ridgepole chandelier} \\) of \\( dragonfly^{2}=4 windflower+4 \\). We now find the area as before."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "eternity",
        "f": "constant",
        "S_t": "pinpoint",
        "\\\\Pi": "linearity",
        "a": "volatile",
        "b": "steadfast",
        "A": "nowhereloc",
        "B": "anywhere",
        "C": "outsider",
        "D": "midpoints",
        "E": "interior",
        "F": "originless",
        "A_1": "nowheretwo",
        "B_1": "anywheretwo"
      },
      "question": "5. The pairs of numbers \\( (volatile, steadfast) \\) such that \\( \\left|volatile+steadfast eternity+eternity^{2}\\right| \\leq 1 \\) for \\( 0 \\leq eternity \\leq 1 \\) fill a certain region in the \\( (volatile, steadfast) \\)-plane. What is the area of this region?",
      "solution": "First Solution. Consider\n\\[\nconstant( eternity)=volatile+steadfast eternity+eternity^{2}\n\\]\n\nThis function has just one critical value, at \\( eternity=-steadfast / 2 \\) where its value is \\( volatile-steadfast^{2 / 4} \\). On the interval \\( [0,1] constant \\) can have extreme values only at the endpoints and at the critical point if it should happen to fall in \\( [0,1] \\) (that is, if \\( steadfast \\in[-2,0] \\) ). Hence the extreme values of \\( constant \\) are\n\\[\nconstant(0)=volatile \\text { and } constant(1)=volatile+steadfast+1\n\\]\nif \\( steadfast \\notin[-2,0] \\), and they are in the set\n\\[\n\\{volatile, volatile+steadfast+1, volatile-steadfast^{2} / 4\\}\n\\]\nif \\( steadfast \\in[-2,0] \\).\nHence \\( |constant(eternity)| \\leq 1 \\) for all \\( eternity \\in[0,1] \\) if and only if\n\\[\nsteadfast \\notin[-2,0] \\text { and }|volatile| \\leq 1,|volatile+steadfast+1| \\leq 1\n\\]\nor\n\\[\nsteadfast \\in[-2,0], \\text { and }|volatile| \\leq 1,|volatile+steadfast+1| \\leq 1,\\left|volatile-\\frac{steadfast^{2}}{4}\\right| \\leq 1\n\\]\n\nThe region required is therefore as shown in the diagram where the arc from \\( nowhereloc \\) to \\( anywhere \\) is part of the parabola \\( volatile-steadfast^{2} / 4=-1 \\).\n\nThe area of the required region can be obtained in several ways. The parallelogram outsiderinteriormidpointsoriginless evidently has area 4 . We must subtract the area of the piece nowhereloc originless anywhere. Since \\( \\overparen{ nowhereloc originless } \\) and \\( \\overparen{ originless anywhere } \\) are tangents to the parabola, the area of nowhereloc originless anywhere is \\( \\frac{1}{3} \\) that of the triangle nowhereloc originless anywhere (Archimedes' rule). But triangle nowhereloc originless anywhere has base nowhereloc originless of length 1 and altitude 1 , so its area is \\( \\frac{1}{2} \\) and the curvilinear piece nowhereloc originless anywhere has area \\( \\frac{1}{6} \\). Hence the area of outsider interiormidpoints anywhere nowhereloc is \\( 4-\\frac{1}{6}=\\frac{23}{6} \\).\n\nSecond Solution. Let \\( pinpoint \\) be the strip in the \\( (volatile, steadfast) \\)-plane bounded by the two parallel lines\n\\[\n\\begin{array}{c}\nvolatile+steadfast eternity+eternity^{2}=1 \\\\\nvolatile+steadfast eternity+eternity^{2}=-1\n\\end{array}\n\\]\n\nWe seek the area of the region \\( linearity \\) defined by \\( linearity=\\bigcap_{0 \\leq eternity \\leq 1} pinpoint \\). Now (1) and (2) are parallel tangents to the parabolas \\( steadfast^{2}=4 volatile-4 \\) and \\( steadfast^{2}=4 volatile+4 \\), and for the range \\( 0 \\leq eternity \\leq 1 \\) these are tangents to the arcs \\( \\overparen{ nowhereloc anywhere } \\) and \\( \\overparen{ nowheretwo anywheretwo } \\) where \\( nowhereloc(-1,0), anywhere(0,-2), nowheretwo(1,0), anywheretwo(2,-2) \\) are points on the two parabolas.\n\\( linearity \\) is therefore the region bounded by the lines \\( volatile=1, volatile=-1, volatile+steadfast=0 \\), \\( volatile+steadfast=-2 \\) and the parabolic arc \\( \\overparen{ nowhereloc anywhere } \\) of \\( steadfast^{2}=4 volatile+4 \\). We now find the area as before."
    },
    "garbled_string": {
      "map": {
        "t": "qzxwvtnp",
        "f": "hjgrksla",
        "S_t": "lugpstan",
        "\\Pi": "mnqsdwer",
        "a": "kfjhnvms",
        "b": "gyswkrqp",
        "A": "vceodmhi",
        "B": "nstkuzla",
        "C": "wpmxgqre",
        "D": "lfrcazto",
        "E": "xaeqnzub",
        "F": "odlmvqpi",
        "A_1": "cbdrajmx",
        "B_1": "svtkunwe"
      },
      "question": "5. The pairs of numbers \\( (kfjhnvms, gyswkrqp) \\) such that \\( \\left|kfjhnvms+gyswkrqp qzxwvtnp+qzxwvtnp^{2}\\right| \\leq 1 \\) for \\( 0 \\leq qzxwvtnp \\leq 1 \\) fill a certain region in the \\( (kfjhnvms, gyswkrqp) \\)-plane. What is the area of this region?",
      "solution": "First Solution. Consider\n\\[\nhjgrksla(qzxwvtnp)=kfjhnvms+gyswkrqp qzxwvtnp+qzxwvtnp^{2}\n\\]\n\nThis function has just one critical value, at \\( qzxwvtnp=-gyswkrqp / 2 \\) where its value is \\( kfjhnvms-gyswkrqp^{2 / 4} \\). On the interval \\([0,1] hjgrksla\\) can have extreme values only at the endpoints and at the critical point if it should happen to fall in \\([0,1]\\) (that is, if \\( gyswkrqp \\in[-2,0] \\)). Hence the extreme values of \\( hjgrksla \\) are\n\\[\nhjgrksla(0)=kfjhnvms \\text { and } hjgrksla(1)=kfjhnvms+gyswkrqp+1\n\\]\nif \\( gyswkrqp \\notin[-2,0] \\), and they are in the set\n\\[\n\\left\\{kfjhnvms, kfjhnvms+gyswkrqp+1, kfjhnvms-gyswkrqp^{2} / 4\\right\\}\n\\]\nif \\( gyswkrqp \\in[-2,0] \\).\nHence \\( |hjgrksla(qzxwvtnp)| \\leq 1 \\) for all \\( qzxwvtnp \\in[0,1] \\) if and only if\n\\[\ngyswkrqp \\notin[-2,0] \\text { and }|kfjhnvms| \\leq 1,|kfjhnvms+gyswkrqp+1| \\leq 1\n\\]\nor\n\\[\ngyswkrqp \\in[-2,0], \\text { and }|kfjhnvms| \\leq 1,|kfjhnvms+gyswkrqp+1| \\leq 1,\\left|kfjhnvms-\\frac{gyswkrqp^{2}}{4}\\right| \\leq 1\n\\]\n\nThe region required is therefore as shown in the diagram where the arc from \\( vceodmhi \\) to \\( nstkuzla \\) is part of the parabola \\( kfjhnvms-gyswkrqp^{2} / 4=-1 \\).\n\nThe area of the required region can be obtained in several ways. The parallelogram wpmxgqre xaeqnzub lfrcazto odlmvqpi evidently has area 4. We must subtract the area of the piece \\( vceodmhi odlmvqpi nstkuzla \\). Since \\( \\overparen{vceodmhi odlmvqpi} \\) and \\( \\overparen{odlmvqpi nstkuzla} \\) are tangents to the parabola, the area of \\( vceodmhi odlmvqpi nstkuzla \\) is \\( \\frac{1}{3} \\) that of the triangle \\( vceodmhi odlmvqpi nstkuzla \\) (Archimedes' rule). But triangle \\( vceodmhi odlmvqpi nstkuzla \\) has base \\( vceodmhi odlmvqpi \\) of length 1 and altitude 1, so its area is \\( \\frac{1}{2} \\) and the curvilinear piece \\( vceodmhi odlmvqpi nstkuzla \\) has area \\( \\frac{1}{6} \\). Hence the area of wpmxgqre lfrcazto xaeqnzub nstkuzla vceodmhi is \\( 4-\\frac{1}{6}=\\frac{23}{6} \\).\n\nSecond Solution. Let \\( lugpstan \\) be the strip in the \\( (kfjhnvms, gyswkrqp) \\)-plane bounded by the two parallel lines\n\\[\n\\begin{array}{c}\nkfjhnvms+gyswkrqp qzxwvtnp+qzxwvtnp^{2}=1 \\\\\nkfjhnvms+gyswkrqp qzxwvtnp+qzxwvtnp^{2}=-1\n\\end{array}\n\\]\n\nWe seek the area of the region \\( mnqsdwer \\) defined by \\( mnqsdwer=\\bigcap_{0 \\leq qzxwvtnp \\leq 1} lugpstan \\). Now (1) and (2) are parallel tangents to the parabolas \\( gyswkrqp^{2}=4 kfjhnvms-4 \\) and \\( gyswkrqp^{2}=4 kfjhnvms+4 \\), and for the range \\( 0 \\leq qzxwvtnp \\leq 1 \\) these are tangents to the arcs \\( \\overparen{vceodmhi nstkuzla} \\) and \\( \\overparen{cbdrajmx svtkunwe} \\) where \\( vceodmhi(-1,0), nstkuzla(0,-2), cbdrajmx(1,0), svtkunwe(2,-2) \\) are points on the two parabolas.\n\\( mnqsdwer \\) is therefore the region bounded by the lines \\( kfjhnvms=1, kfjhnvms=-1, kfjhnvms+gyswkrqp=0 \\), \\( kfjhnvms+gyswkrqp=-2 \\) and the parabolic arc \\( \\overparen{vceodmhi nstkuzla} \\) of \\( gyswkrqp^{2}=4 kfjhnvms+4 \\). We now find the area as before."
    },
    "kernel_variant": {
      "question": "Let  \n  Ω := { (a,b,c) ∈ ℝ³ : |a + bt + ct²| ≤ 1 for every t ∈ [−1,1] }.  \nDetermine the exact Euclidean volume (Lebesgue 3-measure) of Ω in (a,b,c)–space.",
      "solution": "Throughout we write  \n g(t)=bt+ct², x:=|b|≥0, w:=|c|≥0.\n\nStep 1 – Eliminating a.  \nFor fixed (b,c) the condition |a+g(t)|≤1 (∀ t∈[−1,1]) is equivalent to  \n −1−g(t) ≤ a ≤ 1−g(t) (∀ t).  \nHence admissible a’s exist iff  \n δ(b,c):=max_{[−1,1]}g − min_{[−1,1]}g ≤ 2.                         (1)  \nIf (1) holds, the permissible a-segment has length  \n L(b,c)=2−δ(b,c).                                                   (2)  \nThus  \n Vol(Ω)=∬_{δ(b,c)≤2}L(b,c) db dc.                                  (3)\n\nStep 2 – The spread δ.  Put g(t)=bt+ct².\n\nA. Endpoint (monotone) regime (c=0 or |b|>2|c|):  \n δ=|g(1)−g(−1)|=2|b|=2x.                                           (4)\n\nB. Interior-extremum regime (c≠0 and |b|≤2|c|).  \n Without loss of generality take c>0,b≥0 (restore all signs later).  \n The stationary point t₀=−b/(2c) lies in [−1,1].  \n min g=g(t₀)=−x²/(4w), max g=g(1)=x+w,  \n δ=w+x+x²/(4w).                                                    (5)\n\nStep 3 – Admissible region in the first quadrant (x,w≥0).  \nFrom δ≤2 we immediately obtain x≤1.\n\nRegion A (endpoint): 0≤x≤1, 0≤w<x/2.  \nRegion B (interior): 0≤x≤1, x/2≤w≤w_max(x),  \nwhere w_max(x) is the positive root of (5)=2:  \n w²+(x−2)w+x²/4=0 ⇒ w_max(x)=1−x/2+√(1−x).                      (6)\n\nStep 4 – Volume over Region A.  \nHere δ=2x, so L=2−2x.  Exploiting the 4 sign-choices of (b,c),  \n\nVol_A = 4∫₀¹∫₀^{x/2}(2−2x) dw dx  \n       = 4∫₀¹(2−2x)(x/2) dx  \n       = 4∫₀¹(x−x²) dx = 4·(1/6)=2/3.                               (7)\n\nStep 5 – Volume over Region B.  \nThe integrand is L=2−δ=2−(w+x+x²/(4w)).  Again multiplying by 4 for all sign-choices,\n\nVol_B = 4∫_{x=0}^{1}∫_{w=x/2}^{w_max(x)}\n              [2−(w+x+x²/(4w))] dw dx.                               (8)\n\nAn antiderivative with respect to w is  \n\n F(w)=2w−w²/2−xw−(x²/4)ln w,\n\nwhence\n\nI(x)=F(w_max)−F(x/2)  \n    =1−2x+x²+(1−x/2)√(1−x)  \n      −(x²/4)ln[w_max/(x/2)].                                       (9)\n\nSet J:=∫₀¹I(x)dx and split it into three parts.\n\n(i) ∫₀¹(1−2x+x²)dx = 1/3.  \n\n(ii) ∫₀¹(1−x/2)√(1−x)dx.  Put x=1−y² (0≤y≤1): dx=−2y dy,  \n\n  =∫₀¹(y²+y⁴)dy = 1/3+1/5 = 8/15.\n\n(iii) The logarithmic term  \n\n S:=−∫₀¹(x²/4)ln[w_max/(x/2)]dx.                                   (10)\n\nCorrect evaluation of S.  \nInsert x=1−y² ⇒ dx=−2y dy (0≤y≤1) and observe  \n\n w_max(x)=1−x/2+√(1−x)=½(1+y)²,  \n w_max/(x/2)=[(1+y)²]/(1−y²)=(1+y)/(1−y).  \n\nHence\n\n S=−∫₀¹(1−y²)²y ln[(1+y)/(1−y)]/2 dy.                             (11)\n\nFor |y|<1 one has ln[(1+y)/(1−y)]=2 artanh y\n =2∑_{k=0}^{∞}y^{2k+1}/(2k+1).  \nSubstituting this series into (11) and integrating term-wise (justified by absolute\nconvergence) gives  \n\n S=−∑_{k=0}^{∞}8/[(2k+1)(2k+3)(2k+5)(2k+7)].                       (12)\n\nA telescoping device.  Define  \n\n A_k:=\\frac{4/3}{(2k+1)(2k+3)(2k+5)} (k≥0).                        (13)\n\nThen  \n A_k−A_{k+1}=\\frac{4/3}{(2k+3)(2k+5)}\n              \\Bigl(\\frac1{2k+1}-\\frac1{2k+7}\\Bigr)\n           =\\frac{8}{(2k+1)(2k+3)(2k+5)(2k+7)}.                     (14)\n\nThus the series in (12) is −∑_{k=0}^{∞}(A_k−A_{k+1}) = −A_0, i.e.  \n\n S=−A_0=−\\frac{4/3}{1·3·5}=−\\frac{4}{45}.                          (15)\n\nCollecting (i)–(iii):\n\n J=1/3+8/15−4/45=7/9, so Vol_B=4J=28/9.                           (16)\n\nStep 6 – Total volume.  \n\n Vol(Ω)=Vol_A+Vol_B=2/3+28/9=34/9.                                 (17)\n\nAnswer.  \nThe Euclidean volume of Ω is\n\n  Vol(Ω)=34⁄9.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.421159",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the unknown parameter space is ℝ³ instead of ℝ², so the required quantity is a volume rather than an area.  \n• Additional constraints: the inequality must hold for every t in the two-sided interval [−1,1], doubling the analytic work and creating symmetry that must be exploited.  \n• More sophisticated structures: the feasible region is the infinite intersection of half-spaces in ℝ³; analysing its shape demands convex–geometric reasoning, careful case–splitting, and non-trivial optimisation.  \n• Deeper theory and techniques: solving the problem uses calculus (extrema of quadratics), convexity, piece-wise integration in two variables, and delicate logarithmic integrals; the appearance and exact cancellation of log-terms show that ad-hoc pattern matching is insufficient.  \n• Increased step count: compared with the original 2-D area problem, an extra variable, an extra critical-point case, and an additional definite integral are all necessary before the concise rational answer 67⁄30 emerges."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n  Ω := { (a,b,c) ∈ ℝ³ : |a + bt + ct²| ≤ 1 for every t ∈ [−1,1] }.  \nDetermine the exact Euclidean volume (Lebesgue 3-measure) of Ω in (a,b,c)–space.",
      "solution": "Throughout we write  \n g(t)=bt+ct², x:=|b|≥0, w:=|c|≥0.\n\nStep 1 – Eliminating a.  \nFor fixed (b,c) the condition |a+g(t)|≤1 (∀ t∈[−1,1]) is equivalent to  \n −1−g(t) ≤ a ≤ 1−g(t) (∀ t).  \nHence admissible a’s exist iff  \n δ(b,c):=max_{[−1,1]}g − min_{[−1,1]}g ≤ 2.                         (1)  \nIf (1) holds, the permissible a-segment has length  \n L(b,c)=2−δ(b,c).                                                   (2)  \nThus  \n Vol(Ω)=∬_{δ(b,c)≤2}L(b,c) db dc.                                  (3)\n\nStep 2 – The spread δ.  Put g(t)=bt+ct².\n\nA. Endpoint (monotone) regime (c=0 or |b|>2|c|):  \n δ=|g(1)−g(−1)|=2|b|=2x.                                           (4)\n\nB. Interior-extremum regime (c≠0 and |b|≤2|c|).  \n Without loss of generality take c>0,b≥0 (restore all signs later).  \n The stationary point t₀=−b/(2c) lies in [−1,1].  \n min g=g(t₀)=−x²/(4w), max g=g(1)=x+w,  \n δ=w+x+x²/(4w).                                                    (5)\n\nStep 3 – Admissible region in the first quadrant (x,w≥0).  \nFrom δ≤2 we immediately obtain x≤1.\n\nRegion A (endpoint): 0≤x≤1, 0≤w<x/2.  \nRegion B (interior): 0≤x≤1, x/2≤w≤w_max(x),  \nwhere w_max(x) is the positive root of (5)=2:  \n w²+(x−2)w+x²/4=0 ⇒ w_max(x)=1−x/2+√(1−x).                      (6)\n\nStep 4 – Volume over Region A.  \nHere δ=2x, so L=2−2x.  Exploiting the 4 sign-choices of (b,c),  \n\nVol_A = 4∫₀¹∫₀^{x/2}(2−2x) dw dx  \n       = 4∫₀¹(2−2x)(x/2) dx  \n       = 4∫₀¹(x−x²) dx = 4·(1/6)=2/3.                               (7)\n\nStep 5 – Volume over Region B.  \nThe integrand is L=2−δ=2−(w+x+x²/(4w)).  Again multiplying by 4 for all sign-choices,\n\nVol_B = 4∫_{x=0}^{1}∫_{w=x/2}^{w_max(x)}\n              [2−(w+x+x²/(4w))] dw dx.                               (8)\n\nAn antiderivative with respect to w is  \n\n F(w)=2w−w²/2−xw−(x²/4)ln w,\n\nwhence\n\nI(x)=F(w_max)−F(x/2)  \n    =1−2x+x²+(1−x/2)√(1−x)  \n      −(x²/4)ln[w_max/(x/2)].                                       (9)\n\nSet J:=∫₀¹I(x)dx and split it into three parts.\n\n(i) ∫₀¹(1−2x+x²)dx = 1/3.  \n\n(ii) ∫₀¹(1−x/2)√(1−x)dx.  Put x=1−y² (0≤y≤1): dx=−2y dy,  \n\n  =∫₀¹(y²+y⁴)dy = 1/3+1/5 = 8/15.\n\n(iii) The logarithmic term  \n\n S:=−∫₀¹(x²/4)ln[w_max/(x/2)]dx.                                   (10)\n\nCorrect evaluation of S.  \nInsert x=1−y² ⇒ dx=−2y dy (0≤y≤1) and observe  \n\n w_max(x)=1−x/2+√(1−x)=½(1+y)²,  \n w_max/(x/2)=[(1+y)²]/(1−y²)=(1+y)/(1−y).  \n\nHence\n\n S=−∫₀¹(1−y²)²y ln[(1+y)/(1−y)]/2 dy.                             (11)\n\nFor |y|<1 one has ln[(1+y)/(1−y)]=2 artanh y\n =2∑_{k=0}^{∞}y^{2k+1}/(2k+1).  \nSubstituting this series into (11) and integrating term-wise (justified by absolute\nconvergence) gives  \n\n S=−∑_{k=0}^{∞}8/[(2k+1)(2k+3)(2k+5)(2k+7)].                       (12)\n\nA telescoping device.  Define  \n\n A_k:=\\frac{4/3}{(2k+1)(2k+3)(2k+5)} (k≥0).                        (13)\n\nThen  \n A_k−A_{k+1}=\\frac{4/3}{(2k+3)(2k+5)}\n              \\Bigl(\\frac1{2k+1}-\\frac1{2k+7}\\Bigr)\n           =\\frac{8}{(2k+1)(2k+3)(2k+5)(2k+7)}.                     (14)\n\nThus the series in (12) is −∑_{k=0}^{∞}(A_k−A_{k+1}) = −A_0, i.e.  \n\n S=−A_0=−\\frac{4/3}{1·3·5}=−\\frac{4}{45}.                          (15)\n\nCollecting (i)–(iii):\n\n J=1/3+8/15−4/45=7/9, so Vol_B=4J=28/9.                           (16)\n\nStep 6 – Total volume.  \n\n Vol(Ω)=Vol_A+Vol_B=2/3+28/9=34/9.                                 (17)\n\nAnswer.  \nThe Euclidean volume of Ω is\n\n  Vol(Ω)=34⁄9.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.366076",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the unknown parameter space is ℝ³ instead of ℝ², so the required quantity is a volume rather than an area.  \n• Additional constraints: the inequality must hold for every t in the two-sided interval [−1,1], doubling the analytic work and creating symmetry that must be exploited.  \n• More sophisticated structures: the feasible region is the infinite intersection of half-spaces in ℝ³; analysing its shape demands convex–geometric reasoning, careful case–splitting, and non-trivial optimisation.  \n• Deeper theory and techniques: solving the problem uses calculus (extrema of quadratics), convexity, piece-wise integration in two variables, and delicate logarithmic integrals; the appearance and exact cancellation of log-terms show that ad-hoc pattern matching is insufficient.  \n• Increased step count: compared with the original 2-D area problem, an extra variable, an extra critical-point case, and an additional definite integral are all necessary before the concise rational answer 67⁄30 emerges."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}