{
  "index": "1938-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle \\( \\theta \\) along a horizontal line through the axis of the cylinder. If \\( \\theta \\) is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal.",
  "solution": "First Solution. Suppose \\( 0 \\leq \\alpha_{1}<\\alpha_{2}<\\alpha_{2}+\\theta<\\pi / 2 \\); then the wedgeshaped solid between the planes at angles \\( \\alpha_{1} \\) and \\( \\alpha_{1}+\\theta \\) is smaller than the wedge between \\( \\alpha_{2} \\) and \\( \\alpha_{2}+\\theta \\), because a simple rotation of the former through an angle \\( \\alpha_{2}-\\alpha_{1} \\) makes it a proper subset of the latter.\n\nConsider now any asymmetrical wedge of angle \\( \\theta \\) with cross-section \\( A O B \\). If \\( A \\) and \\( B \\) are on the same side of the horizontal through \\( O \\), then the above argument shows that the wedge does not have minimal volume.\n\nSuppose then that \\( A \\) is below the horizontal, and \\( B \\) above it. By symmetry we can assume that \\( A O B \\) lies below the symmetrical wedge \\( S O T \\) of angle \\( \\theta \\), as shown. The wedge \\( A O S \\) is congruent by symmetry with the wedge \\( A^{\\prime} O T \\), which is, in turn, larger than wedge \\( B O T \\) (as shown above). Hence\n\\[\n\\begin{aligned}\n\\text { wedge } A O B= & \\text { wedge } A O S+\\text { wedge } S O B \\\\\n& >\\text { wedge } S O B+\\text { wedge } B O T \\\\\n= & \\text { wedge } S O T\n\\end{aligned}\n\\]\n\nThus the symmetrical wedge is a strict minimum.\nSecond Solution. Let \\( a \\) be the radius of the cylindrical tree, and let \\( \\alpha \\) and \\( \\beta \\) be the angles between the planes of the cut and the horizontal;\n\\[\n\\beta=\\alpha+\\theta .\n\\]\n\nThe volume of the wedge is\n\\[\n\\begin{aligned}\nV & =\\int_{0}^{a} 2 x(\\tan \\beta-\\tan \\alpha) \\sqrt{a^{2}-x^{2}} d x \\\\\n& =A(\\tan \\beta-\\tan \\alpha)\n\\end{aligned}\n\\]\n(It is easy, but unnecessary, to evaluate the integral; in fact, \\( A=2 a^{3} / 3 \\).) We seek to minimize \\( V \\) by choice of \\( \\alpha \\). This is equivalent to minimizing\n\\[\nW=\\tan (\\alpha+\\theta)-\\tan \\alpha\n\\]\nfor \\( -\\pi / 2<\\alpha<\\pi / 2-\\theta \\), since \\( -\\pi / 2<\\alpha \\) and \\( \\beta<\\pi / 2 \\). The critical points are found by solving\n\\[\n\\frac{d W}{d \\alpha}=\\sec ^{2}(\\alpha+\\theta)-\\sec ^{2} \\alpha=0\n\\]\n\nSince both \\( \\sec (\\alpha+\\theta) \\) and \\( \\sec \\alpha \\) are positive through the interval in question, \\( \\sec (\\alpha+\\theta)=\\sec \\alpha \\), whence \\( \\alpha+\\theta= \\pm \\alpha \\). Since \\( \\theta \\) is not zero, the only critical point is given by \\( \\alpha=-\\theta / 2 \\). It is easily seen to correspond to a minimum. When \\( \\alpha=-\\theta / 2, \\beta=\\theta / 2 \\) and the horizontal plane bisects the wedge.",
  "vars": [
    "x",
    "V",
    "W",
    "A",
    "O",
    "B",
    "S",
    "T",
    "\\\\alpha_1",
    "\\\\alpha_2",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [
    "a",
    "\\\\theta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "axisdist",
        "V": "wedgevolume",
        "W": "meritfunc",
        "A": "wedgeconst",
        "O": "centerpoint",
        "B": "pointb",
        "S": "points",
        "T": "pointt",
        "\\alpha_1": "angleone",
        "\\alpha_2": "angletwo",
        "\\alpha": "generalangle",
        "\\beta": "secondangle",
        "a": "treeradius",
        "\\theta": "notchangle"
      },
      "question": "4. A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle \\( notchangle \\) along a horizontal line through the axis of the cylinder. If \\( notchangle \\) is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal.",
      "solution": "First Solution. Suppose \\( 0 \\leq angleone < angletwo < angletwo + notchangle < \\pi / 2 \\); then the wedge-shaped solid between the planes at angles \\( angleone \\) and \\( angleone + notchangle \\) is smaller than the wedge between \\( angletwo \\) and \\( angletwo + notchangle \\), because a simple rotation of the former through an angle \\( angletwo - angleone \\) makes it a proper subset of the latter.\n\nConsider now any asymmetrical wedge of angle \\( notchangle \\) with cross-section \\( wedgeconst\\; centerpoint\\; pointb \\). If \\( wedgeconst \\) and \\( pointb \\) are on the same side of the horizontal through \\( centerpoint \\), then the above argument shows that the wedge does not have minimal volume.\n\nSuppose then that \\( wedgeconst \\) is below the horizontal, and \\( pointb \\) above it. By symmetry we can assume that \\( wedgeconst\\; centerpoint\\; pointb \\) lies below the symmetrical wedge \\( points\\; centerpoint\\; pointt \\) of angle \\( notchangle \\), as shown. The wedge \\( wedgeconst\\; centerpoint\\; points \\) is congruent by symmetry with the wedge \\( wedgeconst^{\\prime}\\; centerpoint\\; pointt \\), which is, in turn, larger than wedge \\( pointb\\; centerpoint\\; pointt \\) (as shown above). Hence\n\\[\n\\begin{aligned}\n\\text{wedge } wedgeconst\\; centerpoint\\; pointb &= \\text{wedge } wedgeconst\\; centerpoint\\; points + \\text{wedge } points\\; centerpoint\\; pointb \\\\\n&> \\text{wedge } points\\; centerpoint\\; pointb + \\text{wedge } pointb\\; centerpoint\\; pointt \\\\\n&= \\text{wedge } points\\; centerpoint\\; pointt\n\\end{aligned}\n\\]\n\nThus the symmetrical wedge is a strict minimum.\n\nSecond Solution. Let \\( treeradius \\) be the radius of the cylindrical tree, and let \\( generalangle \\) and \\( secondangle \\) be the angles between the planes of the cut and the horizontal;\n\\[\nsecondangle = generalangle + notchangle .\n\\]\n\nThe volume of the wedge is\n\\[\n\\begin{aligned}\nwedgevolume &= \\int_{0}^{treeradius} 2\\, axisdist (\\tan secondangle - \\tan generalangle) \\sqrt{treeradius^{2} - axisdist^{2}}\\, d axisdist \\\\\n&= wedgeconst (\\tan secondangle - \\tan generalangle)\n\\end{aligned}\n\\]\n(It is easy, but unnecessary, to evaluate the integral; in fact, \\( wedgeconst = 2\\, treeradius^{3} / 3 \\).) We seek to minimize \\( wedgevolume \\) by choice of \\( generalangle \\). This is equivalent to minimizing\n\\[\nmeritfunc = \\tan (generalangle + notchangle) - \\tan generalangle\n\\]\nfor \\( -\\pi / 2 < generalangle < \\pi / 2 - notchangle \\), since \\( -\\pi / 2 < generalangle \\) and \\( secondangle < \\pi / 2 \\). The critical points are found by solving\n\\[\n\\frac{d\\, meritfunc}{d\\, generalangle} = \\sec^{2}(generalangle + notchangle) - \\sec^{2} generalangle = 0 .\n\\]\n\nSince both \\( \\sec (generalangle + notchangle) \\) and \\( \\sec generalangle \\) are positive through the interval in question, \\( \\sec (generalangle + notchangle) = \\sec generalangle \\), whence \\( generalangle + notchangle = \\pm generalangle \\). Since \\( notchangle \\) is not zero, the only critical point is given by \\( generalangle = -\\, notchangle / 2 \\). It is easily seen to correspond to a minimum. When \\( generalangle = -\\, notchangle / 2, \\; secondangle = notchangle / 2 \\) and the horizontal plane bisects the wedge."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanternfish",
        "V": "meadowlark",
        "W": "buttercup",
        "A": "sandstone",
        "O": "pinecones",
        "B": "rainstorm",
        "S": "driftwood",
        "T": "goldcrest",
        "\\alpha_1": "snowdrift",
        "\\alpha_2": "springtide",
        "\\alpha": "riverbank",
        "\\beta": "moonlight",
        "a": "velocity",
        "\\theta": "sunlight"
      },
      "question": "4. A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle \\( sunlight \\) along a horizontal line through the axis of the cylinder. If \\( sunlight \\) is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal.",
      "solution": "First Solution. Suppose \\( 0 \\leq snowdrift<springtide<springtide+sunlight<\\pi / 2 \\); then the wedgeshaped solid between the planes at angles \\( snowdrift \\) and \\( snowdrift+sunlight \\) is smaller than the wedge between \\( springtide \\) and \\( springtide+sunlight \\), because a simple rotation of the former through an angle \\( springtide-snowdrift \\) makes it a proper subset of the latter.\n\nConsider now any asymmetrical wedge of angle \\( sunlight \\) with cross-section \\( sandstone\\, pinecones\\, rainstorm \\). If \\( sandstone \\) and \\( rainstorm \\) are on the same side of the horizontal through \\( pinecones \\), then the above argument shows that the wedge does not have minimal volume.\n\nSuppose then that \\( sandstone \\) is below the horizontal, and \\( rainstorm \\) above it. By symmetry we can assume that \\( sandstone\\, pinecones\\, rainstorm \\) lies below the symmetrical wedge \\( driftwood\\, pinecones\\, goldcrest \\) of angle \\( sunlight \\), as shown. The wedge \\( sandstone\\, pinecones\\, driftwood \\) is congruent by symmetry with the wedge \\( sandstone^{\\prime}\\, pinecones\\, goldcrest \\), which is, in turn, larger than wedge \\( rainstorm\\, pinecones\\, goldcrest \\) (as shown above). Hence\n\\[\n\\begin{aligned}\n\\text { wedge } sandstone\\, pinecones\\, rainstorm= & \\text { wedge } sandstone\\, pinecones\\, driftwood+\\text { wedge } driftwood\\, pinecones\\, rainstorm \\\\ & >\\text { wedge } driftwood\\, pinecones\\, rainstorm+\\text { wedge } rainstorm\\, pinecones\\, goldcrest \\\\ = & \\text { wedge } driftwood\\, pinecones\\, goldcrest\n\\end{aligned}\n\\]\n\nThus the symmetrical wedge is a strict minimum.\n\nSecond Solution. Let \\( velocity \\) be the radius of the cylindrical tree, and let \\( riverbank \\) and \\( moonlight \\) be the angles between the planes of the cut and the horizontal;\n\\[\nmoonlight=riverbank+sunlight .\n\\]\n\nThe volume of the wedge is\n\\[\n\\begin{aligned}\nmeadowlark & =\\int_{0}^{velocity} 2\\, lanternfish(\\tan moonlight-\\tan riverbank) \\sqrt{velocity^{2}-lanternfish^{2}} \\, d\\, lanternfish \\\\ & =sandstone(\\tan moonlight-\\tan riverbank)\n\\end{aligned}\n\\]\n(It is easy, but unnecessary, to evaluate the integral; in fact, \\( sandstone=2\\, velocity^{3} / 3 \\).) We seek to minimize \\( meadowlark \\) by choice of \\( riverbank \\). This is equivalent to minimizing\n\\[\nbuttercup=\\tan (riverbank+sunlight)-\\tan riverbank\n\\]\nfor \\( -\\pi / 2<riverbank<\\pi / 2-sunlight \\), since \\( -\\pi / 2<riverbank \\) and \\( moonlight<\\pi / 2 \\). The critical points are found by solving\n\\[\n\\frac{d buttercup}{d riverbank}=\\sec ^{2}(riverbank+sunlight)-\\sec ^{2} riverbank=0\n\\]\n\nSince both \\( \\sec (riverbank+sunlight) \\) and \\( \\sec riverbank \\) are positive through the interval in question, \\( \\sec (riverbank+sunlight)=\\sec riverbank \\), whence \\( riverbank+sunlight= \\pm riverbank \\). Since \\( sunlight \\) is not zero, the only critical point is given by \\( riverbank=-sunlight / 2 \\). It is easily seen to correspond to a minimum. When \\( riverbank=-sunlight / 2, moonlight=sunlight / 2 \\) and the horizontal plane bisects the wedge."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "V": "voidspace",
        "W": "stillness",
        "A": "perimeter",
        "O": "edgepoint",
        "B": "corepoint",
        "S": "skewpoint",
        "T": "bentpoint",
        "\\alpha_1": "terminalangleone",
        "\\alpha_2": "terminalangletwo",
        "\\alpha": "terminalangle",
        "\\beta": "origindirection",
        "a": "centerheight",
        "\\theta": "flatness"
      },
      "question": "4. A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle \\( flatness \\) along a horizontal line through the axis of the cylinder. If \\( flatness \\) is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal.",
      "solution": "First Solution. Suppose \\( 0 \\leq terminalangleone<terminalangletwo<terminalangletwo+flatness<\\pi / 2 \\); then the wedgeshaped solid between the planes at angles \\( terminalangleone \\) and \\( terminalangleone+flatness \\) is smaller than the wedge between \\( terminalangletwo \\) and \\( terminalangletwo+flatness \\), because a simple rotation of the former through an angle \\( terminalangletwo-terminalangleone \\) makes it a proper subset of the latter.\n\nConsider now any asymmetrical wedge of angle \\( flatness \\) with cross-section \\( perimeter edgepoint corepoint \\). If \\( perimeter \\) and \\( corepoint \\) are on the same side of the horizontal through \\( edgepoint \\), then the above argument shows that the wedge does not have minimal volume.\n\nSuppose then that \\( perimeter \\) is below the horizontal, and \\( corepoint \\) above it. By symmetry we can assume that \\( perimeter edgepoint corepoint \\) lies below the symmetrical wedge \\( skewpoint edgepoint bentpoint \\) of angle \\( flatness \\), as shown. The wedge \\( perimeter edgepoint skewpoint \\) is congruent by symmetry with the wedge \\( perimeter^{\\prime} edgepoint bentpoint \\), which is, in turn, larger than wedge \\( corepoint edgepoint bentpoint \\) (as shown above). Hence\n\\[\n\\begin{aligned}\n\\text { wedge } perimeter edgepoint corepoint= & \\text { wedge } perimeter edgepoint skewpoint+\\text { wedge } skewpoint edgepoint corepoint \\\\\n& >\\text { wedge } skewpoint edgepoint corepoint+\\text { wedge } corepoint edgepoint bentpoint \\\\\n= & \\text { wedge } skewpoint edgepoint bentpoint\n\\end{aligned}\n\\]\n\nThus the symmetrical wedge is a strict minimum.\nSecond Solution. Let \\( centerheight \\) be the radius of the cylindrical tree, and let \\( terminalangle \\) and \\( origindirection \\) be the angles between the planes of the cut and the horizontal;\n\\[\norigindirection=terminalangle+flatness .\n\\]\n\nThe volume of the wedge is\n\\[\n\\begin{aligned}\nvoidspace & =\\int_{0}^{centerheight} 2 constantval(\\tan origindirection-\\tan terminalangle) \\sqrt{centerheight^{2}-constantval^{2}} d constantval \\\\\n& =perimeter(\\tan origindirection-\\tan terminalangle)\n\\end{aligned}\n\\]\n(It is easy, but unnecessary, to evaluate the integral; in fact, \\( perimeter=2 centerheight^{3} / 3 \\).) We seek to minimize \\( voidspace \\) by choice of \\( terminalangle \\). This is equivalent to minimizing\n\\[\nstillness=\\tan (terminalangle+flatness)-\\tan terminalangle\n\\]\nfor \\( -\\pi / 2<terminalangle<\\pi / 2-flatness \\), since \\( -\\pi / 2<terminalangle \\) and \\( origindirection<\\pi / 2 \\). The critical points are found by solving\n\\[\n\\frac{d stillness}{d terminalangle}=\\sec ^{2}(terminalangle+flatness)-\\sec ^{2} terminalangle=0\n\\]\n\nSince both \\( \\sec (terminalangle+flatness) \\) and \\( \\sec terminalangle \\) are positive through the interval in question, \\( \\sec (terminalangle+flatness)=\\sec terminalangle \\), whence \\( terminalangle+flatness= \\pm terminalangle \\). Since \\( flatness \\) is not zero, the only critical point is given by \\( terminalangle=-flatness / 2 \\). It is easily seen to correspond to a minimum. When \\( terminalangle=-flatness / 2, origindirection=flatness / 2 \\) and the horizontal plane bisects the wedge."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "V": "hjgrksla",
        "W": "nkpduolc",
        "A": "sfrdlnke",
        "O": "yxvmbtac",
        "B": "wzcyrhqe",
        "S": "lgopnmsr",
        "T": "bravxkji",
        "\\alpha_1": "cuvthspe",
        "\\alpha_2": "hxpqznor",
        "\\alpha": "jdfgrmks",
        "\\beta": "zekuanlp",
        "a": "lmrstqoi",
        "\\theta": "pxncvhru"
      },
      "question": "A lumberman wishes to cut down a tree whose trunk is cylindrical and whose material is uniform. He will cut a notch, the two sides of which will be planes intersecting at a dihedral angle \\( pxncvhru \\) along a horizontal line through the axis of the cylinder. If \\( pxncvhru \\) is given, show that the least volume of material is cut out when the plane bisecting the dihedral angle is horizontal.",
      "solution": "First Solution. Suppose \\( 0 \\leq cuvthspe<hxpqznor<hxpqznor+pxncvhru<\\pi / 2 \\); then the wedgeshaped solid between the planes at angles \\( cuvthspe \\) and \\( cuvthspe+pxncvhru \\) is smaller than the wedge between \\( hxpqznor \\) and \\( hxpqznor+pxncvhru \\), because a simple rotation of the former through an angle \\( hxpqznor-cuvthspe \\) makes it a proper subset of the latter.\n\nConsider now any asymmetrical wedge of angle \\( pxncvhru \\) with cross-section \\( sfrdlnke\\, yxvmbtac\\, wzcyrhqe \\). If \\( sfrdlnke \\) and \\( wzcyrhqe \\) are on the same side of the horizontal through \\( yxvmbtac \\), then the above argument shows that the wedge does not have minimal volume.\n\nSuppose then that \\( sfrdlnke \\) is below the horizontal, and \\( wzcyrhqe \\) above it. By symmetry we can assume that \\( sfrdlnke\\, yxvmbtac\\, wzcyrhqe \\) lies below the symmetrical wedge \\( lgopnmsr\\, yxvmbtac\\, bravxkji \\) of angle \\( pxncvhru \\), as shown. The wedge \\( sfrdlnke\\, yxvmbtac\\, lgopnmsr \\) is congruent by symmetry with the wedge \\( sfrdlnke^{\\prime}\\, yxvmbtac\\, bravxkji \\), which is, in turn, larger than wedge \\( wzcyrhqe\\, yxvmbtac\\, bravxkji \\) (as shown above). Hence\n\\[\n\\begin{aligned}\n\\text { wedge } sfrdlnke\\, yxvmbtac\\, wzcyrhqe &= \\text { wedge } sfrdlnke\\, yxvmbtac\\, lgopnmsr+\\text { wedge } lgopnmsr\\, yxvmbtac\\, wzcyrhqe \\\\[-2pt]\n& >\\text { wedge } lgopnmsr\\, yxvmbtac\\, wzcyrhqe+\\text { wedge } wzcyrhqe\\, yxvmbtac\\, bravxkji \\\\[-2pt]\n& =\\text { wedge } lgopnmsr\\, yxvmbtac\\, bravxkji\n\\end{aligned}\n\\]\n\nThus the symmetrical wedge is a strict minimum.\n\nSecond Solution. Let \\( lmrstqoi \\) be the radius of the cylindrical tree, and let \\( jdfgrmks \\) and \\( zekuanlp \\) be the angles between the planes of the cut and the horizontal;\n\\[\nzekuanlp=jdfgrmks+pxncvhru .\n\\]\n\nThe volume of the wedge is\n\\[\n\\begin{aligned}\nhjgrksla &=\\int_{0}^{lmrstqoi} 2 qzxwvtnp(\\tan zekuanlp-\\tan jdfgrmks) \\sqrt{lmrstqoi^{2}-qzxwvtnp^{2}} d qzxwvtnp \\\\[-2pt]\n& =sfrdlnke(\\tan zekuanlp-\\tan jdfgrmks)\n\\end{aligned}\n\\]\n(It is easy, but unnecessary, to evaluate the integral; in fact, \\( sfrdlnke=2 lmrstqoi^{3} / 3 \\).) We seek to minimize \\( hjgrksla \\) by choice of \\( jdfgrmks \\). This is equivalent to minimizing\n\\[\nnkpduolc=\\tan (jdfgrmks+pxncvhru)-\\tan jdfgrmks\n\\]\nfor \\( -\\pi / 2<jdfgrmks<\\pi / 2-pxncvhru \\), since \\( -\\pi / 2<jdfgrmks \\) and \\( zekuanlp<\\pi / 2 \\). The critical points are found by solving\n\\[\n\\frac{d nkpduolc}{d jdfgrmks}=\\sec ^{2}(jdfgrmks+pxncvhru)-\\sec ^{2} jdfgrmks=0\n\\]\n\nSince both \\( \\sec (jdfgrmks+pxncvhru) \\) and \\( \\sec jdfgrmks \\) are positive through the interval in question, \\( \\sec (jdfgrmks+pxncvhru)=\\sec jdfgrmks \\), whence \\( jdfgrmks+pxncvhru= \\pm jdfgrmks \\). Since \\( pxncvhru \\) is not zero, the only critical point is given by \\( jdfgrmks=-pxncvhru / 2 \\). It is easily seen to correspond to a minimum. When \\( jdfgrmks=-pxncvhru / 2, zekuanlp=pxncvhru / 2 \\) and the horizontal plane bisects the wedge."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 3$ and fix positive constants  \n\n\\[\nR,\\qquad \\rho_{0},\\qquad\\text{and}\\qquad \\varphi\\quad\\text{with}\\quad 0<\\varphi<\\pi .\n\\]\n\nWrite points of $\\mathbb R^{n}$ in the form $x=(x_{1},\\dots ,x_{n-1},z)$ with $z:=x_{n}$ the vertical\ncoordinate; the $z$–axis is  \n\n\\[\n\\Lambda:=\\{(0,\\dots ,0,z):z\\in\\mathbb R\\}.\n\\]\n\nThe (right–circular) cylinder of radius $R$ about $\\Lambda$ is  \n\n\\[\nC:=\\bigl\\{x\\in\\mathbb R^{n}:x_{1}^{2}+\\dots +x_{\\,n-1}^{2}\\le R^{2}\\bigr\\}. \\tag{1}\n\\]\n\nThe material density depends only on the distance  \n\n\\[\nr:=\\sqrt{x_{1}^{2}+\\dots +x_{\\,n-1}^{2}}\n\\]\n\nfrom the axis and is given by  \n\n\\[\n\\rho(r)=\\rho_{0}\\bigl(1+(r/R)^{p}\\bigr),\\qquad \\rho_{0}>0,\\qquad p>-(n-1). \\tag{2}\n\\]\n\n(Finiteness of all integrals below is guaranteed by $p>-(n-1)$.)\n\nFix a horizontal unit vector $e$; without loss of generality take $e=e_{1}$.\nFor every real parameter  \n\n\\[\n\\alpha\\in I_{\\varphi}:=\\bigl(-\\pi/2+\\varphi/2,\\; \\pi/2-\\varphi/2\\bigr) \\tag{3}\n\\]\n\nintroduce the two non-vertical affine hyperplanes  \n\n\\[\nP_{+}(\\alpha):\\; z=x_{1}\\tan\\bigl(\\alpha+\\varphi/2\\bigr),\\qquad\nP_{-}(\\alpha):\\; z=x_{1}\\tan\\bigl(\\alpha-\\varphi/2\\bigr). \\tag{4}\n\\]\n\nThey intersect in the horizontal $(n-2)$–plane  \n\n\\[\nL:=\\{x_{1}=0,\\; z=0\\}. \\tag{5}\n\\]\n\nThe dihedral angle between $P_{+}(\\alpha)$ and $P_{-}(\\alpha)$ equals $\\varphi$, and for every admissible $\\alpha$ the “wedge’’ inside $C$ cut out by the two planes is  \n\n\\[\nW(\\alpha):=C\\cap\\bigl\\{\\min \\bigl(P_{+}(\\alpha),P_{-}(\\alpha)\\bigr)\n              \\le z\\le\n              \\max \\bigl(P_{+}(\\alpha),P_{-}(\\alpha)\\bigr)\\bigr\\}. \\tag{6}\n\\]\n\n(a)  Show that the total mass of $W(\\alpha)$ is  \n\n\\[\nM(\\alpha)=K_{n,p}(R)\\bigl[\\tan(\\alpha+\\varphi/2)-\\tan(\\alpha-\\varphi/2)\\bigr], \\tag{7}\n\\]\n\nwhere  \n\n\\[\nK_{n,p}(R):=\\rho_{0}\\,\\omega_{n-2}\\,R^{\\,n}\\,\\frac{2n+p}{n(n+p)}, \\tag{8}\n\\]\n\nand $\\displaystyle\\omega_{k}:=\\pi^{k/2}\\bigl/\\Gamma(k/2+1)$ denotes the $k$-dimensional volume of the unit ball.\n\n(b)  Prove that $M(\\alpha)$ attains its unique global minimum precisely at $\\alpha=0$; equivalently, the least mass is removed when the bisecting hyper-plane of the angle $\\varphi$ is horizontal.\n\n(c)  Compute this minimal mass explicitly:\n\n\\[\nM_{\\min}=2\\,\\rho_{0}\\,\\omega_{n-2}\\,R^{\\,n}\\,\n          \\frac{2n+p}{n(n+p)}\\;\\tan\\bigl(\\varphi/2\\bigr). \\tag{9}\n\\]",
      "solution": "Step 1.  Geometry in the vertical $2$-plane  \nLet $\\Sigma:=\\operatorname{span}\\{e_{1},e_{n}\\}$ with orthonormal coordinates $(x,z)$.  \nThe traces of $P_{-}(\\alpha)$ and $P_{+}(\\alpha)$ in $\\Sigma$ are the lines  \n\n\\[\n\\ell_{-}: z=x\\tan\\bigl(\\alpha-\\varphi/2\\bigr),\\qquad\n\\ell_{+}: z=x\\tan\\bigl(\\alpha+\\varphi/2\\bigr). \\tag{10}\n\\]\n\nBecause $\\alpha\\in I_{\\varphi}$ we have\n$\\tan(\\alpha-\\varphi/2)<\\tan(\\alpha+\\varphi/2)$.\nHence, for any abscissa $x$, the vertical length of $W(\\alpha)$ above that $x$ equals  \n\n\\[\n\\Delta z(x)=\\lvert x\\rvert\\,T(\\alpha),\\qquad\nT(\\alpha):=\\tan(\\alpha+\\varphi/2)-\\tan(\\alpha-\\varphi/2)>0. \\tag{11}\n\\]\n\nSince $\\lvert x\\rvert\\le R$, $W(\\alpha)$ is bounded.\n\nStep 2.  Mass element  \nFix $x\\in[0,R]$ and write the remaining horizontal coordinates as $s=(x_{2},\\dots ,x_{n-1})$.  \nSet  \n\n\\[\nB_{n-2}(x):=\\bigl\\{s\\in\\mathbb R^{\\,n-2}:\\lVert s\\rVert^{2}\\le R^{2}-x^{2}\\bigr\\}. \\tag{12}\n\\]\n\nIts $(n-2)$-dimensional volume is  \n\n\\[\n\\operatorname{vol}B_{n-2}(x)=\\omega_{n-2}\\,\\bigl(R^{2}-x^{2}\\bigr)^{(n-2)/2}. \\tag{13}\n\\]\n\nOver $B_{n-2}(x)$ erect a vertical prism of height $\\Delta z(x)$.\nBecause $r^{2}=x^{2}+\\lVert s\\rVert^{2}$, the mass of this prism is  \n\n\\[\n\\mathrm dM(x)=\\rho_{0}\\,\\Delta z(x)\n              \\int_{B_{n-2}(x)}\n              \\Bigl[1+\\bigl(r/R\\bigr)^{p}\\Bigr]\\mathrm ds\n            =\\rho_{0}T(\\alpha)\\,x\\!\n              \\int_{B_{n-2}(x)}\n              \\Bigl[1+\\bigl(r/R\\bigr)^{p}\\Bigr]\\mathrm ds. \\tag{14}\n\\]\n\nStep 3.  Integration over the cross-section  \nBy symmetry in $x\\mapsto -x$,  \n\n\\[\nM(\\alpha)=2\\rho_{0}T(\\alpha)\n          \\int_{0}^{R}x\n          \\int_{B_{n-2}(x)}\n          \\Bigl[1+\\bigl(r/R\\bigr)^{p}\\Bigr]\\mathrm ds\\,\\mathrm dx. \\tag{15}\n\\]\n\nIntroduce cylindrical coordinates $(r,\\theta)$ in the horizontal $(n-1)$-space:\n$r\\in[0,R]$, $\\theta\\in S^{\\,n-2}$, with Jacobian\n$r^{\\,n-2}\\mathrm dr\\,\\mathrm d\\sigma_{n-2}(\\theta)$.\nWrite $\\theta_{1}:=\\langle\\theta,e_{1}\\rangle=\\cos\\vartheta$.\nThe restriction $x\\ge 0$ corresponds to $\\cos\\vartheta\\ge 0$.\nEquation (15) becomes  \n\n\\[\nM(\\alpha)=2\\rho_{0}T(\\alpha)\n          \\int_{0}^{R}r^{\\,n-1}\\Bigl[1+(r/R)^{p}\\Bigr]\\mathrm dr\\;\n          \\int_{S^{\\,n-2},\\,\\cos\\vartheta\\ge 0}\\cos\\vartheta\\,\n               \\mathrm d\\sigma_{n-2}(\\theta). \\tag{16}\n\\]\n\nAngular integral.  \nBecause $\\cos\\vartheta$ depends only on $\\vartheta$,\n\n\\[\n\\int_{\\cos\\vartheta\\ge 0}\\cos\\vartheta\\,\n      \\mathrm d\\sigma_{n-2}(\\theta)\n      =\\sigma_{n-3}\\int_{0}^{\\pi/2}\\cos\\vartheta\\,\n                    \\sin^{\\,n-3}\\vartheta\\,\\mathrm d\\vartheta\n      =\\frac{\\sigma_{n-3}}{n-2}\n      =\\omega_{n-2}. \\tag{17}\n\\]\n\nRadial integral.  With $u=r/R$,\n\n\\[\n\\int_{0}^{R}r^{\\,n-1}\\Bigl[1+(r/R)^{p}\\Bigr]\\mathrm dr\n=R^{\\,n}\\int_{0}^{1}u^{\\,n-1}\\bigl(1+u^{p}\\bigr)\\mathrm du\n=R^{\\,n}\\Bigl(\\frac{1}{n}+\\frac{1}{n+p}\\Bigr)\n=R^{\\,n}\\frac{2n+p}{n(n+p)}. \\tag{18}\n\\]\n\nCombining (16)–(18) yields\n\n\\[\nM(\\alpha)=\\rho_{0}\\,\\omega_{n-2}\\,R^{\\,n}\\,\n          \\frac{2n+p}{n(n+p)}\\;T(\\alpha), \\tag{19}\n\\]\n\ni.e. formula (7) with\n$K_{n,p}(R)$ given by (8).  \nPart (a) is proved.\n\nStep 4.  Optimisation in $\\alpha$  \nSet  \n\n\\[\nF(\\alpha):=T(\\alpha)=\\tan(\\alpha+\\varphi/2)-\\tan(\\alpha-\\varphi/2),\n\\qquad \\alpha\\in I_{\\varphi}. \\tag{20}\n\\]\n\nDifferentiation gives  \n\n\\[\nF'(\\alpha)=\\sec^{2}(\\alpha+\\varphi/2)-\\sec^{2}(\\alpha-\\varphi/2). \\tag{21}\n\\]\n\nA direct sign analysis avoiding the faulty claim from the draft solution:\n\nMethod 1 (parity argument).  \nThe function $g(t):=\\sec^{2}t$ is even, continuous on $(-\\pi/2,\\pi/2)$ and\nstrictly increasing in $\\lvert t\\rvert$ because  \n\n\\[\n\\frac{\\mathrm d}{\\mathrm dt}g(t)=2\\sec^{2}t\\tan t,\n\\]\nwhich has the same sign as $t$ in this interval.\nHence, for $\\alpha>0$ we have\n$\\lvert \\alpha+\\varphi/2\\rvert>\\lvert \\alpha-\\varphi/2\\rvert$ and therefore  \n$g(\\alpha+\\varphi/2)>g(\\alpha-\\varphi/2)$, whence $F'(\\alpha)>0$.\nFor $\\alpha<0$ the inequalities reverse and $F'(\\alpha)<0$.\nConsequently  \n\n\\[\nF'(\\alpha)\\begin{cases}\n<0 &\\text{if }\\alpha<0,\\\\[2pt]\n=0 &\\text{if }\\alpha=0,\\\\[2pt]\n>0 &\\text{if }\\alpha>0.\n\\end{cases} \\tag{22}\n\\]\n\nMethod 2 (integral representation, optional).  \nUsing the fundamental theorem of calculus,\n\n\\[\nF'(\\alpha)=\\int_{\\alpha-\\varphi/2}^{\\alpha+\\varphi/2}\n           2\\sec^{2}t\\tan t\\,\\mathrm dt .\n\\]\n\nBecause $2\\sec^{2}t\\tan t$ has the same sign as $t$, the integral is positive, zero, or negative according as $\\alpha$ is positive, zero, or negative, giving the same conclusion.\n\nEither way, $F$ is strictly decreasing on $(-\\pi/2+\\varphi/2,0]$\nand strictly increasing on $[0,\\pi/2-\\varphi/2)$, so $\\alpha=0$ is the unique global minimiser of $F$ and hence of $M(\\alpha)=K_{n,p}(R)F(\\alpha)$.\nPart (b) is proved.\n\nStep 5.  Minimal mass  \nInsert $\\alpha=0$ into (19):\n\n\\[\nM_{\\min}=K_{n,p}(R)\\bigl[\\tan(\\varphi/2)-\\tan(-\\varphi/2)\\bigr]\n        =2\\,K_{n,p}(R)\\,\\tan(\\varphi/2). \\tag{23}\n\\]\n\nUsing (8) gives (9). \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.337281",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem is lifted from 3-space to ℝⁿ with n ≥ 3, which forces the solver to work with (n–2)-balls and general Beta/Gamma-function evaluations.\n2. Variable density: Instead of uniform density, the mass depends on an arbitrary radial power r^p.  This introduces an additional parameter p and requires non-trivial averaging of the density over (n–2)-dimensional balls.\n3. Multiple concepts: The solution combines classical geometry (dihedral angles and wedges), high-dimensional integration, special functions (Beta and Gamma), and monotonicity arguments for trigonometric functions.\n4. Deeper computations: Determining the constant K_{n,p}(R) demands carrying out two coupled Beta–integrals and justifying the interchange of averaging and integration.\n5. More subtle minimisation: Because of the extra density factor, one must show that the orientation part Δ(α) decouples from the radial integral, and then analyse Δ(α) carefully on its admissible domain.\n\nAll these layers render the enhanced variant substantially more technical and conceptually demanding than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 3 and fix positive numbers  \n\n  R, ρ₀, and φ with 0 < φ < π.  \n\nWrite points of ℝⁿ in the form x = (x₁,…,x_{n-1},z) with z := xₙ the vertical\ncoordinate; the vertical axis of the infinite right–circular cylinder  \n\n  Λ := {(0,…,0,z) : z∈ℝ}  \n\nis the z–axis.  \nThe cylinder itself is  \n\n  C := { x∈ℝⁿ : x₁²+⋯+x_{n-1}² ≤ R² }.                             (1)\n\nAt every point the material has the radial density  \n\n  ρ(r) = ρ₀ (1 + (r/R)^p),  r := √(x₁²+⋯+x_{n-1}²),  ρ₀>0, p>−(n−1).   (2)\n\n(Condition p>−(n−1) guarantees that all integrals occurring below are finite.)\n\nFix a horizontal unit vector e that we take, without loss of generality,\nto be the first coordinate vector e₁.\nFor any real parameter  \n\n  α ∈ I_φ := (−π/2 + φ/2,  π/2 − φ/2)                               (3)\n\ndefine the two (non-vertical) affine hyperplanes\n\n  P₊(α): z = x₁ tan(α+φ/2),  \n  P₋(α): z = x₁ tan(α−φ/2).                                        (4)\n\nP₊(α) and P₋(α) intersect in the horizontal (n−2)-plane  \n\n  L := {x₁=0, z=0}.                                                 (5)\n\nTheir dihedral angle is φ, and for every admissible α the two planes bound\na bounded “wedge’’ inside C,\n\n  W(α) := C ∩ { min(P₊(α),P₋(α)) ≤ z ≤ max(P₊(α),P₋(α)) }.        (6)\n\n(a)  Show that the total mass of W(α) equals  \n\n  M(α) = K_{n,p}(R) [ tan(α+φ/2) − tan(α−φ/2) ],                  (7)\n\nwhere  \n\n  K_{n,p}(R) := ρ₀ ω_{n-2} R^{\\,n} · (2n+p)/(n(n+p)),               (8)\n\nand ω_k := π^{k/2}/Γ(k/2+1) is the k–dimensional volume of the unit ball.\n\n(b)  Prove that M(α) attains its unique global minimum precisely at α = 0;\nequivalently, the least mass is removed when the bisecting\nhyper-plane of the dihedral angle is horizontal\n(each cutting plane makes the same acute angle φ/2 with the horizontal).\n\n(c)  Compute this minimal mass explicitly:\n\n  M_min = 2 ρ₀ ω_{n-2} R^{\\,n} · (2n+p)/(n(n+p)) · tan(φ/2).       (9)",
      "solution": "Step 1. Geometry inside the vertical 2-plane Σ.\n\nLet Σ be the plane spanned by the vectors e₁ (horizontal) and eₙ\n(vertical).  In the orthonormal coordinates (x,z) on Σ the two lines which\nare the traces of P₋(α) and P₊(α) are\n\n  ℓ₋ : z = x tan(α−φ/2),        ℓ₊ : z = x tan(α+φ/2).         (10)\n\nBecause α∈I_φ the slopes satisfy tan(α−φ/2)<tan(α+φ/2); hence,\nfor every abscissa x the z-length of the section of W(α) lying above that\nx equals\n\n  Δz(x) = |x|·T(α),  with  T(α) := tan(α+φ/2) − tan(α−φ/2)>0.   (11)\n\nConsequently W(α) is bounded: even at |x|=R the height does not exceed\n|R| T(α).\n\nStep 2. Mass element.\n\nFix x with 0≤x≤R.\nFor the remaining n−2 horizontal directions introduce the vector\ns=(x₂,…,x_{n-1}) and denote by\n\n  B_{n-2}(x) := { s∈ℝ^{n-2} : ‖s‖² ≤ R²−x² }                    (12)\n\nthe (n−2)-ball of radius √(R²−x²).\nThe (n−2)-dimensional Lebesgue measure of B_{n-2}(x) is\n\n  vol B_{n-2}(x) = ω_{n-2}(R²−x²)^{(n-2)/2}.                      (13)\n\nFor each point (x,s) the distance from the axis is\nr=√(x²+‖s‖²).  A vertical prism over B_{n-2}(x) of height Δz(x)\ntherefore carries the mass\n\n  dM(x) = ρ₀ Δz(x)∫_{B_{n-2}(x)}[1+(r/R)^p]ds                    (14)\n\nand, using (11) and (13),\n\n  dM(x) = ρ₀ T(α)·x∫_{B_{n-2}(x)}[1+(r/R)^p]ds.                 (15)\n\nStep 3. Integration over the whole cross-section.\n\nBecause the integrand only depends on |x| the total mass is twice the\nintegral over x≥0:\n\n  M(α)=2ρ₀T(α)∫_{0}^{R}x\n             ∫_{B_{n-2}(x)}[1+(r/R)^p]ds dx.                      (16)\n\nIntroduce cylindrical coordinates in ℝ^{n-1}.\nA point u=(x,s) is parameterised by its distance\nr∈[0,R] from Λ and a direction θ∈S^{n-2}; write\nθ₁:=⟨θ,e₁⟩=cosϑ where ϑ is the angle with e₁.\nThe Jacobian is r^{\\,n-2}dr dσ_{n-2}(θ), and x= r cosϑ.\nRestricting to x≥0 amounts to keeping only directions with cosϑ≥0.\nHence (16) becomes\n\n  M(α)=2ρ₀T(α)\n         ∫_{0}^{R}r^{\\,n-1}[1+(r/R)^p]dr\n         ∫_{S^{n-2}, cosϑ≥0}cosϑ dσ_{n-2}(θ).                    (17)\n\nAngular integral.\nBecause cosϑ≥0 on the half-sphere {cosϑ≥0} and depends only on ϑ,\n\n  ∫_{cosϑ≥0}cosϑ dσ_{n-2}(θ)\n         = σ_{n-3}∫_{0}^{π/2}cosϑ sin^{\\,n-3}ϑ dϑ\n         = σ_{n-3}/(n-2) = ω_{n-2},                              (18)\n\nwhere we used σ_{k}=2π^{(k+1)/2}/Γ((k+1)/2) and the identity\nσ_{n-3}=(n-2)ω_{n-2}.\n\nRadial integral.  Substituting u=r/R,\n\n  ∫_{0}^{R}r^{\\,n-1}[1+(r/R)^p]dr\n         = R^{\\,n}∫_{0}^{1}u^{\\,n-1}[1+u^{p}]du\n         = R^{\\,n}\\Bigl(\\frac1n+\\frac1{n+p}\\Bigr)\n         = R^{\\,n}\\frac{2n+p}{n(n+p)}.                           (19)\n\nPutting (18) and (19) into (17) yields\n\n  M(α)=ρ₀ ω_{n-2} R^{\\,n}\\frac{2n+p}{n(n+p)}·T(α).              (20)\n\nBy the definition (11) of T(α) this is exactly (7) with the constant\n(8), completing part (a).\n\nStep 4. Optimising the mass with respect to α.\n\nSet  \n\n  F(α):=T(α)=tan(α+φ/2)−tan(α−φ/2), α∈I_φ.                     (21)\n\nDifferentiate:\n\n  F′(α)=sec²(α+φ/2)−sec²(α−φ/2).                                (22)\n\nBecause t↦sec²t is strictly increasing on (−π/2,π/2),\n\n  F′(α)<0 for α<0, and F′(α)>0 for α>0.                         (23)\n\nHence F has a unique critical point at α=0, which by (23) is the\nglobal minimum on I_φ.  Since M(α)=K_{n,p}(R) F(α) with K_{n,p}(R)>0,\nthe same conclusion holds for M.  This proves part (b).\n\nStep 5. The minimal mass.\n\nInsert α=0 into (20):\n\n  M_min = K_{n,p}(R)·[tan(φ/2)−tan(−φ/2)]\n           = 2 K_{n,p}(R) tan(φ/2).                               (24)\n\nUsing (8) gives the announced value (9), completing part (c). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.297777",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem is lifted from 3-space to ℝⁿ with n ≥ 3, which forces the solver to work with (n–2)-balls and general Beta/Gamma-function evaluations.\n2. Variable density: Instead of uniform density, the mass depends on an arbitrary radial power r^p.  This introduces an additional parameter p and requires non-trivial averaging of the density over (n–2)-dimensional balls.\n3. Multiple concepts: The solution combines classical geometry (dihedral angles and wedges), high-dimensional integration, special functions (Beta and Gamma), and monotonicity arguments for trigonometric functions.\n4. Deeper computations: Determining the constant K_{n,p}(R) demands carrying out two coupled Beta–integrals and justifying the interchange of averaging and integration.\n5. More subtle minimisation: Because of the extra density factor, one must show that the orientation part Δ(α) decouples from the radial integral, and then analyse Δ(α) carefully on its admissible domain.\n\nAll these layers render the enhanced variant substantially more technical and conceptually demanding than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}