{
  "index": "1963-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "5. (i) Prove that if a function \\( f \\) is continuous on the closed interval \\( [0, \\pi] \\) and if\n\\[\n\\int_{0}^{\\pi} f(\\theta) \\cos \\theta d \\theta=\\int_{0}^{\\pi} f(\\theta) \\sin \\theta d \\theta=0\n\\]\nthen there exist points \\( \\alpha \\) and \\( \\beta \\) such that\n\\[\n0<\\alpha<\\beta<\\pi \\quad \\text { and } f(\\alpha)=f(\\beta)=0\n\\]\n(ii) Let \\( R \\) be any bounded convex open region in the Euclidean plane (that is, \\( R \\) is a connected open set contained in some circular disk, and the line segment joining any two points of \\( R \\) lies entirely in \\( R \\) ). Prove with the help of part (i) that the centroid (center of gravity) of \\( R \\) bisects at least three distinct chords of the boundary of \\( R \\).",
  "solution": "Solution. (i) We may assume that \\( f \\not \\equiv 0 \\). Then since \\( \\int_{0}^{\\pi} f(\\theta) \\sin \\theta d \\theta=0 \\) and \\( \\sin \\theta>0 \\) for \\( 0<\\theta<\\pi, f \\) must change sign somewhere on \\( (0, \\pi) \\), say at \\( \\alpha \\). If \\( \\alpha \\) is the only zero of \\( f \\) on \\( (0, \\pi) \\), then \\( f \\) has one sign on \\( (0, \\alpha) \\) and the opposite sign on \\( (\\alpha, \\pi) \\). In the latter event,\n\\[\n\\int_{0}^{\\pi} f(\\theta) \\sin (\\theta-\\alpha) d \\theta \\neq 0 .\n\\]\n\nBut\n\\[\n\\int_{0}^{\\pi} f(\\theta) \\sin (\\theta-\\alpha) d \\theta=\\cos \\alpha \\int_{0}^{\\pi} f(\\theta) \\sin \\theta d \\theta-\\sin \\alpha \\int_{0}^{\\pi} f(\\theta) \\cos \\theta d \\theta=0 .\n\\]\n\nThis contradiction implies the existence of a second point \\( \\beta \\) with \\( f(\\beta)=0 \\).\n(ii) Take polar coordinates with pole at the centroid \\( P \\) of the bounded convex open region \\( R \\), and write the equation of the bounding curve \\( \\Gamma \\) of \\( R \\) as \\( \\rho=g(\\theta) \\). [Since \\( R \\) is convex, each ray from \\( P \\) meets \\( \\Gamma \\) just once. Thus \\( \\Gamma \\) has such an equation.]\n\\( \\Gamma \\) is compact and the mapping \\( (\\rho, \\theta) \\rightarrow(1, \\theta) \\) (polar coordinates) of \\( \\Gamma \\) into the unit circle is continuous and bijective; therefore, the inverse mapping \\( (1, \\theta) \\rightarrow(g(\\theta), \\theta) \\) is continuous. So \\( g \\) is continuous.\n\nThe moments of the region about the lines \\( \\theta=0 \\) and \\( \\theta=\\pi / 2 \\) are given by\n\\[\n\\iint_{R} \\rho \\cos \\theta \\rho d \\rho d \\theta \\text { and } \\iint_{R} \\rho \\sin \\theta \\rho d \\rho d \\theta .\n\\]\n\nThese are both zero since we selected the origin at the centroid. Integration with respect to \\( \\rho \\) gives\n\\[\n\\frac{1}{3} \\int_{0}^{2 \\pi}[g(\\theta)]^{3} \\cos \\theta d \\theta=0=\\frac{1}{3} \\int_{0}^{2 \\pi}[g(\\theta)]^{3} \\sin \\theta d \\theta .\n\\]\n\nSince \\( \\cos (\\theta+\\pi)=-\\cos \\theta, \\sin (\\theta+\\pi)=-\\sin \\theta \\) this gives\n\\[\n\\begin{aligned}\n0 & =\\int_{0}^{\\pi}[g(\\theta)]^{3} \\cos \\theta d \\theta-\\int_{0}^{\\pi}[g(\\theta+\\pi)]^{3} \\cos \\theta d \\theta \\\\\n& =\\int_{0}^{\\pi}\\left\\{[g(\\theta)]^{3}-[g(\\theta+\\pi)]^{3}\\right\\} \\cos \\theta d \\theta=0,\n\\end{aligned}\n\\]\nand similarly\n\\[\n0=\\int_{0}^{\\pi}\\left\\{[g(\\theta)]^{3}-[g(\\theta+\\pi)]^{3}\\right\\} \\sin \\theta d \\theta=0 .\n\\]\n\nNow according to part (i),\n\\[\n[g(\\theta)]^{3}-[g(\\theta+\\pi)]^{3}=0\n\\]\nholds for at least two values of \\( \\theta \\) in \\( (0, \\pi) \\). For these values\n\\[\ng(\\theta)=g(\\theta+\\pi),\n\\]\nwhich means that the centroid bisects the chords having these two directions.\nNow since we know that \\( P \\) must bisect at least one chord, we may as well assume that the polar axis was chosen in such a direction. Then the argument above shows that \\( P \\) also bisects at least two other chords.",
  "vars": [
    "f",
    "R",
    "P",
    "g",
    "\\\\Gamma",
    "\\\\theta",
    "\\\\alpha",
    "\\\\beta",
    "\\\\rho"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "funcval",
        "R": "convexrg",
        "P": "centroid",
        "g": "boundfun",
        "\\Gamma": "boundary",
        "\\theta": "anglevar",
        "\\alpha": "firstzero",
        "\\beta": "secondzero",
        "\\rho": "radialvr"
      },
      "question": "5. (i) Prove that if a function \\( funcval \\) is continuous on the closed interval \\( [0, \\pi] \\) and if\n\\[\n\\int_{0}^{\\pi} funcval(anglevar) \\cos anglevar d anglevar=\\int_{0}^{\\pi} funcval(anglevar) \\sin anglevar d anglevar=0\n\\]\nthen there exist points \\( firstzero \\) and \\( secondzero \\) such that\n\\[\n0<firstzero<secondzero<\\pi \\quad \\text { and } funcval(firstzero)=funcval(secondzero)=0\n\\]\n(ii) Let \\( convexrg \\) be any bounded convex open region in the Euclidean plane (that is, \\( convexrg \\) is a connected open set contained in some circular disk, and the line segment joining any two points of \\( convexrg \\) lies entirely in \\( convexrg \\) ). Prove with the help of part (i) that the centroid (center of gravity) of \\( convexrg \\) bisects at least three distinct chords of the boundary of \\( convexrg \\).",
      "solution": "Solution. (i) We may assume that \\( funcval \\not \\equiv 0 \\). Then since \\( \\int_{0}^{\\pi} funcval(anglevar) \\sin anglevar \\, d anglevar=0 \\) and \\( \\sin anglevar>0 \\) for \\( 0<anglevar<\\pi, funcval \\) must change sign somewhere on \\( (0, \\pi) \\), say at \\( firstzero \\). If \\( firstzero \\) is the only zero of \\( funcval \\) on \\( (0, \\pi) \\), then \\( funcval \\) has one sign on \\( (0, firstzero) \\) and the opposite sign on \\( (firstzero, \\pi) \\). In the latter event,\n\\[\n\\int_{0}^{\\pi} funcval(anglevar) \\sin (anglevar-firstzero) d anglevar \\neq 0 .\n\\]\n\nBut\n\\[\n\\int_{0}^{\\pi} funcval(anglevar) \\sin (anglevar-firstzero) d anglevar=\\cos firstzero \\int_{0}^{\\pi} funcval(anglevar) \\sin anglevar d anglevar-\\sin firstzero \\int_{0}^{\\pi} funcval(anglevar) \\cos anglevar d anglevar=0 .\n\\]\n\nThis contradiction implies the existence of a second point \\( secondzero \\) with \\( funcval(secondzero)=0 \\).\n(ii) Take polar coordinates with pole at the centroid \\( centroid \\) of the bounded convex open region \\( convexrg \\), and write the equation of the bounding curve \\( boundary \\) of \\( convexrg \\) as \\( radialvr=boundfun(anglevar) \\). [Since \\( convexrg \\) is convex, each ray from \\( centroid \\) meets \\( boundary \\) just once. Thus \\( boundary \\) has such an equation.]\n\\( boundary \\) is compact and the mapping \\( (radialvr, anglevar) \\rightarrow(1, anglevar) \\) (polar coordinates) of \\( boundary \\) into the unit circle is continuous and bijective; therefore, the inverse mapping \\( (1, anglevar) \\rightarrow(boundfun(anglevar), anglevar) \\) is continuous. So \\( boundfun \\) is continuous.\n\nThe moments of the region about the lines \\( anglevar=0 \\) and \\( anglevar=\\pi / 2 \\) are given by\n\\[\n\\iint_{convexrg} radialvr \\cos anglevar \\, radialvr \\, d radialvr \\, d anglevar \\text { and } \\iint_{convexrg} radialvr \\sin anglevar \\, radialvr \\, d radialvr \\, d anglevar .\n\\]\n\nThese are both zero since we selected the origin at the centroid. Integration with respect to \\( radialvr \\) gives\n\\[\n\\frac{1}{3} \\int_{0}^{2 \\pi}[boundfun(anglevar)]^{3} \\cos anglevar d anglevar=0=\\frac{1}{3} \\int_{0}^{2 \\pi}[boundfun(anglevar)]^{3} \\sin anglevar d anglevar .\n\\]\n\nSince \\( \\cos (anglevar+\\pi)=-\\cos anglevar, \\sin (anglevar+\\pi)=-\\sin anglevar \\) this gives\n\\[\n\\begin{aligned}\n0 & =\\int_{0}^{\\pi}[boundfun(anglevar)]^{3} \\cos anglevar d anglevar-\\int_{0}^{\\pi}[boundfun(anglevar+\\pi)]^{3} \\cos anglevar d anglevar \\\\\n& =\\int_{0}^{\\pi}\\left\\{[boundfun(anglevar)]^{3}-[boundfun(anglevar+\\pi)]^{3}\\right\\} \\cos anglevar d anglevar=0,\n\\end{aligned}\n\\]\nand similarly\n\\[\n0=\\int_{0}^{\\pi}\\left\\{[boundfun(anglevar)]^{3}-[boundfun(anglevar+\\pi)]^{3}\\right\\} \\sin anglevar d anglevar=0 .\n\\]\n\nNow according to part (i),\n\\[\n[boundfun(anglevar)]^{3}-[boundfun(anglevar+\\pi)]^{3}=0\n\\]\nholds for at least two values of \\( anglevar \\) in \\( (0, \\pi) \\). For these values\n\\[\nboundfun(anglevar)=boundfun(anglevar+\\pi),\n\\]\nwhich means that the centroid bisects the chords having these two directions.\nNow since we know that \\( centroid \\) must bisect at least one chord, we may as well assume that the polar axis was chosen in such a direction. Then the argument above shows that \\( centroid \\) also bisects at least two other chords."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "copperpipe",
        "R": "gardenplot",
        "P": "silkquartz",
        "g": "elmseedling",
        "\\Gamma": "mangrovese",
        "\\theta": "cobblestone",
        "\\alpha": "butterleaf",
        "\\beta": "sandcastle",
        "\\rho": "brookfield"
      },
      "question": "5. (i) Prove that if a function \\( copperpipe \\) is continuous on the closed interval \\( [0, \\pi] \\) and if\n\\[\n\\int_{0}^{\\pi} copperpipe(cobblestone) \\cos cobblestone d cobblestone=\\int_{0}^{\\pi} copperpipe(cobblestone) \\sin cobblestone d cobblestone=0\n\\]\nthen there exist points \\( butterleaf \\) and \\( sandcastle \\) such that\n\\[\n0<butterleaf<sandcastle<\\pi \\quad \\text { and } copperpipe(butterleaf)=copperpipe(sandcastle)=0\n\\]\n(ii) Let \\( gardenplot \\) be any bounded convex open region in the Euclidean plane (that is, \\( gardenplot \\) is a connected open set contained in some circular disk, and the line segment joining any two points of \\( gardenplot \\) lies entirely in \\( gardenplot \\) ). Prove with the help of part (i) that the centroid (center of gravity) of \\( gardenplot \\) bisects at least three distinct chords of the boundary of \\( gardenplot \\).",
      "solution": "Solution. (i) We may assume that \\( copperpipe \\not \\equiv 0 \\). Then since \\( \\int_{0}^{\\pi} copperpipe(cobblestone) \\sin cobblestone d cobblestone=0 \\) and \\( \\sin cobblestone>0 \\) for \\( 0<cobblestone<\\pi, copperpipe \\) must change sign somewhere on \\( (0, \\pi) \\), say at \\( butterleaf \\). If \\( butterleaf \\) is the only zero of \\( copperpipe \\) on \\( (0, \\pi) \\), then \\( copperpipe \\) has one sign on \\( (0, butterleaf) \\) and the opposite sign on \\( (butterleaf, \\pi) \\). In the latter event,\n\\[\n\\int_{0}^{\\pi} copperpipe(cobblestone) \\sin (cobblestone-butterleaf) d cobblestone \\neq 0 .\n\\]\n\nBut\n\\[\n\\int_{0}^{\\pi} copperpipe(cobblestone) \\sin (cobblestone-butterleaf) d cobblestone=\\cos butterleaf \\int_{0}^{\\pi} copperpipe(cobblestone) \\sin cobblestone d cobblestone-\\sin butterleaf \\int_{0}^{\\pi} copperpipe(cobblestone) \\cos cobblestone d cobblestone=0 .\n\\]\n\nThis contradiction implies the existence of a second point \\( sandcastle \\) with \\( copperpipe(sandcastle)=0 \\).\n(ii) Take polar coordinates with pole at the centroid \\( silkquartz \\) of the bounded convex open region \\( gardenplot \\), and write the equation of the bounding curve \\( mangrovese \\) of \\( gardenplot \\) as \\( brookfield=elmseedling(cobblestone) \\). [Since \\( gardenplot \\) is convex, each ray from \\( silkquartz \\) meets \\( mangrovese \\) just once. Thus \\( mangrovese \\) has such an equation.]\n\\( mangrovese \\) is compact and the mapping \\( (brookfield, cobblestone) \\rightarrow(1, cobblestone) \\) (polar coordinates) of \\( mangrovese \\) into the unit circle is continuous and bijective; therefore, the inverse mapping \\( (1, cobblestone) \\rightarrow(elmseedling(cobblestone), cobblestone) \\) is continuous. So \\( elmseedling \\) is continuous.\n\nThe moments of the region about the lines \\( cobblestone=0 \\) and \\( cobblestone=\\pi / 2 \\) are given by\n\\[\n\\iint_{gardenplot} brookfield \\cos cobblestone \\, brookfield d brookfield d cobblestone \\text { and } \\iint_{gardenplot} brookfield \\sin cobblestone \\, brookfield d brookfield d cobblestone .\n\\]\n\nThese are both zero since we selected the origin at the centroid. Integration with respect to \\( brookfield \\) gives\n\\[\n\\frac{1}{3} \\int_{0}^{2 \\pi}[elmseedling(cobblestone)]^{3} \\cos cobblestone d cobblestone=0=\\frac{1}{3} \\int_{0}^{2 \\pi}[elmseedling(cobblestone)]^{3} \\sin cobblestone d cobblestone .\n\\]\n\nSince \\( \\cos (cobblestone+\\pi)=-\\cos cobblestone, \\sin (cobblestone+\\pi)=-\\sin cobblestone \\) this gives\n\\[\n\\begin{aligned}\n0 & =\\int_{0}^{\\pi}[elmseedling(cobblestone)]^{3} \\cos cobblestone d cobblestone-\\int_{0}^{\\pi}[elmseedling(cobblestone+\\pi)]^{3} \\cos cobblestone d cobblestone \\\\\n& =\\int_{0}^{\\pi}\\left\\{[elmseedling(cobblestone)]^{3}-[elmseedling(cobblestone+\\pi)]^{3}\\right\\} \\cos cobblestone d cobblestone=0,\n\\end{aligned}\n\\]\nand similarly\n\\[\n0=\\int_{0}^{\\pi}\\left\\{[elmseedling(cobblestone)]^{3}-[elmseedling(cobblestone+\\pi)]^{3}\\right\\} \\sin cobblestone d cobblestone=0 .\n\\]\n\nNow according to part (i),\n\\[\n[elmseedling(cobblestone)]^{3}-[elmseedling(cobblestone+\\pi)]^{3}=0\n\\]\nholds for at least two values of \\( cobblestone \\) in \\( (0, \\pi) \\). For these values\n\\[\nelmseedling(cobblestone)=elmseedling(cobblestone+\\pi),\n\\]\nwhich means that the centroid bisects the chords having these two directions.\nNow since we know that \\( silkquartz \\) must bisect at least one chord, we may as well assume that the polar axis was chosen in such a direction. Then the argument above shows that \\( silkquartz \\) also bisects at least two other chords."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "constantval",
        "R": "emptiness",
        "P": "offcenter",
        "g": "flatfunc",
        "\\Gamma": "linepiece",
        "\\theta": "fixedangle",
        "\\alpha": "endingval",
        "\\beta": "latevalue",
        "\\rho": "diameter"
      },
      "question": "5. (i) Prove that if a function \\( constantval \\) is continuous on the closed interval \\( [0, \\pi] \\) and if\n\\[\n\\int_{0}^{\\pi} constantval(fixedangle) \\cos fixedangle d fixedangle=\\int_{0}^{\\pi} constantval(fixedangle) \\sin fixedangle d fixedangle=0\n\\]\nthen there exist points \\( endingval \\) and \\( latevalue \\) such that\n\\[\n0<endingval<latevalue<\\pi \\quad \\text { and } constantval(endingval)=constantval(latevalue)=0\n\\]\n(ii) Let \\( emptiness \\) be any bounded convex open region in the Euclidean plane (that is, \\( emptiness \\) is a connected open set contained in some circular disk, and the line segment joining any two points of \\( emptiness \\) lies entirely in \\( emptiness \\) ). Prove with the help of part (i) that the centroid (center of gravity) of \\( emptiness \\) bisects at least three distinct chords of the boundary of \\( emptiness \\).",
      "solution": "Solution. (i) We may assume that \\( constantval \\not \\equiv 0 \\). Then since \\( \\int_{0}^{\\pi} constantval(fixedangle) \\sin fixedangle d fixedangle=0 \\) and \\( \\sin fixedangle>0 \\) for \\( 0<fixedangle<\\pi, constantval \\) must change sign somewhere on \\( (0, \\pi) \\), say at \\( endingval \\). If \\( endingval \\) is the only zero of \\( constantval \\) on \\( (0, \\pi) \\), then \\( constantval \\) has one sign on \\( (0, endingval) \\) and the opposite sign on \\( (endingval, \\pi) \\). In the latter event,\n\\[\n\\int_{0}^{\\pi} constantval(fixedangle) \\sin (fixedangle-endingval) d fixedangle \\neq 0 .\n\\]\n\nBut\n\\[\n\\int_{0}^{\\pi} constantval(fixedangle) \\sin (fixedangle-endingval) d fixedangle=\\cos endingval \\int_{0}^{\\pi} constantval(fixedangle) \\sin fixedangle d fixedangle-\\sin endingval \\int_{0}^{\\pi} constantval(fixedangle) \\cos fixedangle d fixedangle=0 .\n\\]\n\nThis contradiction implies the existence of a second point \\( latevalue \\) with \\( constantval(latevalue)=0 \\).\n(ii) Take polar coordinates with pole at the centroid \\( offcenter \\) of the bounded convex open region \\( emptiness \\), and write the equation of the bounding curve \\( linepiece \\) of \\( emptiness \\) as \\( diameter=flatfunc(fixedangle) \\). [Since \\( emptiness \\) is convex, each ray from \\( offcenter \\) meets \\( linepiece \\) just once. Thus \\( linepiece \\) has such an equation.]\n\\( linepiece \\) is compact and the mapping \\( (diameter, fixedangle) \\rightarrow(1, fixedangle) \\) (polar coordinates) of \\( linepiece \\) into the unit circle is continuous and bijective; therefore, the inverse mapping \\( (1, fixedangle) \\rightarrow(flatfunc(fixedangle), fixedangle) \\) is continuous. So \\( flatfunc \\) is continuous.\n\nThe moments of the region about the lines \\( fixedangle=0 \\) and \\( fixedangle=\\pi / 2 \\) are given by\n\\[\n\\iint_{emptiness} diameter \\cos fixedangle diameter d diameter d fixedangle \\text { and } \\iint_{emptiness} diameter \\sin fixedangle diameter d diameter d fixedangle .\n\\]\n\nThese are both zero since we selected the origin at the centroid. Integration with respect to \\( diameter \\) gives\n\\[\n\\frac{1}{3} \\int_{0}^{2 \\pi}[flatfunc(fixedangle)]^{3} \\cos fixedangle d fixedangle=0=\\frac{1}{3} \\int_{0}^{2 \\pi}[flatfunc(fixedangle)]^{3} \\sin fixedangle d fixedangle .\n\\]\n\nSince \\( \\cos (fixedangle+\\pi)=-\\cos fixedangle, \\sin (fixedangle+\\pi)=-\\sin fixedangle \\) this gives\n\\[\n\\begin{aligned}\n0 & =\\int_{0}^{\\pi}[flatfunc(fixedangle)]^{3} \\cos fixedangle d fixedangle-\\int_{0}^{\\pi}[flatfunc(fixedangle+\\pi)]^{3} \\cos fixedangle d fixedangle \\\\\n& =\\int_{0}^{\\pi}\\left\\{[flatfunc(fixedangle)]^{3}-[flatfunc(fixedangle+\\pi)]^{3}\\right\\} \\cos fixedangle d fixedangle=0,\n\\end{aligned}\n\\]\nand similarly\n\\[\n0=\\int_{0}^{\\pi}\\left\\{[flatfunc(fixedangle)]^{3}-[flatfunc(fixedangle+\\pi)]^{3}\\right\\} \\sin fixedangle d fixedangle=0 .\n\\]\n\nNow according to part (i),\n\\[\n[flatfunc(fixedangle)]^{3}-[flatfunc(fixedangle+\\pi)]^{3}=0\n\\]\n holds for at least two values of \\( fixedangle \\) in \\( (0, \\pi) \\). For these values\n\\[\nflatfunc(fixedangle)=flatfunc(fixedangle+\\pi),\n\\]\nwhich means that the centroid bisects the chords having these two directions.\nNow since we know that \\( offcenter \\) must bisect at least one chord, we may as well assume that the polar axis was chosen in such a direction. Then the argument above shows that \\( offcenter \\) also bisects at least two other chords."
    },
    "garbled_string": {
      "map": {
        "f": "qlmzxtrb",
        "R": "kdpsrjgh",
        "P": "ztqfmxla",
        "g": "bxrmqozv",
        "\\Gamma": "lvchqent",
        "\\theta": "qzxwvtnp",
        "\\alpha": "hjgrksla",
        "\\beta": "tnxwzrgh",
        "\\rho": "vlsrhzpd"
      },
      "question": "5. (i) Prove that if a function \\( qlmzxtrb \\) is continuous on the closed interval \\( [0, \\pi] \\) and if\n\\[\n\\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\cos qzxwvtnp d qzxwvtnp=\\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\sin qzxwvtnp d qzxwvtnp=0\n\\]\nthen there exist points \\( hjgrksla \\) and \\( tnxwzrgh \\) such that\n\\[\n0<hjgrksla<tnxwzrgh<\\pi \\quad \\text { and } qlmzxtrb(hjgrksla)=qlmzxtrb(tnxwzrgh)=0\n\\]\n(ii) Let \\( kdpsrjgh \\) be any bounded convex open region in the Euclidean plane (that is, \\( kdpsrjgh \\) is a connected open set contained in some circular disk, and the line segment joining any two points of \\( kdpsrjgh \\) lies entirely in \\( kdpsrjgh \\) ). Prove with the help of part (i) that the centroid (center of gravity) of \\( kdpsrjgh \\) bisects at least three distinct chords of the boundary of \\( kdpsrjgh \\).",
      "solution": "Solution. (i) We may assume that \\( qlmzxtrb \\not \\equiv 0 \\). Then since \\( \\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\sin qzxwvtnp d qzxwvtnp=0 \\) and \\( \\sin qzxwvtnp>0 \\) for \\( 0<qzxwvtnp<\\pi, qlmzxtrb \\) must change sign somewhere on \\( (0, \\pi) \\), say at \\( hjgrksla \\). If \\( hjgrksla \\) is the only zero of \\( qlmzxtrb \\) on \\( (0, \\pi) \\), then \\( qlmzxtrb \\) has one sign on \\( (0, hjgrksla) \\) and the opposite sign on \\( (hjgrksla, \\pi) \\). In the latter event,\n\\[\n\\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\sin (qzxwvtnp-hjgrksla) d qzxwvtnp \\neq 0 .\n\\]\n\nBut\n\\[\n\\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\sin (qzxwvtnp-hjgrksla) d qzxwvtnp=\\cos hjgrksla \\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\sin qzxwvtnp d qzxwvtnp-\\sin hjgrksla \\int_{0}^{\\pi} qlmzxtrb(qzxwvtnp) \\cos qzxwvtnp d qzxwvtnp=0 .\n\\]\n\nThis contradiction implies the existence of a second point \\( tnxwzrgh \\) with \\( qlmzxtrb(tnxwzrgh)=0 \\).\n(ii) Take polar coordinates with pole at the centroid \\( ztqfmxla \\) of the bounded convex open region \\( kdpsrjgh \\), and write the equation of the bounding curve \\( lvchqent \\) of \\( kdpsrjgh \\) as \\( vlsrhzpd=bxrmqozv(qzxwvtnp) \\). [Since \\( kdpsrjgh \\) is convex, each ray from \\( ztqfmxla \\) meets \\( lvchqent \\) just once. Thus \\( lvchqent \\) has such an equation.]\n\\( lvchqent \\) is compact and the mapping \\( (vlsrhzpd, qzxwvtnp) \\rightarrow(1, qzxwvtnp) \\) (polar coordinates) of \\( lvchqent \\) into the unit circle is continuous and bijective; therefore, the inverse mapping \\( (1, qzxwvtnp) \\rightarrow(bxrmqozv(qzxwvtnp), qzxwvtnp) \\) is continuous. So \\( bxrmqozv \\) is continuous.\n\nThe moments of the region about the lines \\( qzxwvtnp=0 \\) and \\( qzxwvtnp=\\pi / 2 \\) are given by\n\\[\n\\iint_{kdpsrjgh} vlsrhzpd \\cos qzxwvtnp vlsrhzpd d vlsrhzpd d qzxwvtnp \\text { and } \\iint_{kdpsrjgh} vlsrhzpd \\sin qzxwvtnp vlsrhzpd d vlsrhzpd d qzxwvtnp .\n\\]\n\nThese are both zero since we selected the origin at the centroid. Integration with respect to \\( vlsrhzpd \\) gives\n\\[\n\\frac{1}{3} \\int_{0}^{2 \\pi}[bxrmqozv(qzxwvtnp)]^{3} \\cos qzxwvtnp d qzxwvtnp=0=\\frac{1}{3} \\int_{0}^{2 \\pi}[bxrmqozv(qzxwvtnp)]^{3} \\sin qzxwvtnp d qzxwvtnp .\n\\]\n\nSince \\( \\cos (qzxwvtnp+\\pi)=-\\cos qzxwvtnp, \\sin (qzxwvtnp+\\pi)=-\\sin qzxwvtnp \\) this gives\n\\[\n\\begin{aligned}\n0 & =\\int_{0}^{\\pi}[bxrmqozv(qzxwvtnp)]^{3} \\cos qzxwvtnp d qzxwvtnp-\\int_{0}^{\\pi}[bxrmqozv(qzxwvtnp+\\pi)]^{3} \\cos qzxwvtnp d qzxwvtnp \\\\\n& =\\int_{0}^{\\pi}\\left\\{[bxrmqozv(qzxwvtnp)]^{3}-[bxrmqozv(qzxwvtnp+\\pi)]^{3}\\right\\} \\cos qzxwvtnp d qzxwvtnp=0,\n\\end{aligned}\n\\]\nand similarly\n\\[\n0=\\int_{0}^{\\pi}\\left\\{[bxrmqozv(qzxwvtnp)]^{3}-[bxrmqozv(qzxwvtnp+\\pi)]^{3}\\right\\} \\sin qzxwvtnp d qzxwvtnp=0 .\n\\]\n\nNow according to part (i),\n\\[\n[bxrmqozv(qzxwvtnp)]^{3}-[bxrmqozv(qzxwvtnp+\\pi)]^{3}=0\n\\]\nholds for at least two values of \\( qzxwvtnp \\) in \\( (0, \\pi) \\). For these values\n\\[\nbxrmqozv(qzxwvtnp)=bxrmqozv(qzxwvtnp+\\pi),\n\\]\nwhich means that the centroid bisects the chords having these two directions.\nNow since we know that \\( ztqfmxla \\) must bisect at least one chord, we may as well assume that the polar axis was chosen in such a direction. Then the argument above shows that \\( ztqfmxla \\) also bisects at least two other chords."
    },
    "kernel_variant": {
      "question": "Let I:=[-\\pi/6,5\\pi/6] and define \n  u(\\theta)=\\sin(\\theta+\\pi/6),\\qquad  v(\\theta)=\\cos(\\theta+\\pi/6).\nA direct calculation shows that the two functions are orthogonal on I:\n      \\[\\int_{I}u(\\theta)\\,v(\\theta)\\,d\\theta =0.\\]\n\ni)  (Two–zero lemma with shifted weights)\nLet \\(f\\) be a continuous function on I which satisfies\n\\[\\int_{I}f(\\theta)u(\\theta)\\,d\\theta=0,\\qquad\\int_{I}f(\\theta)v(\\theta)\\,d\\theta=0 .\\]\nProve that there exist two distinct points \\(\\alpha,\\beta\\in I\\) such that \\(f(\\alpha)=f(\\beta)=0\\).\n\n(ii)  (Bisected boundary chords for a uniform lamina)\nLet R be a bounded open region in the plane that is strictly star–shaped with respect to a point P: every ray issuing from P meets \\(\\partial R\\) in exactly one point and the distance \\(\\rho =|PQ|\\) of this point from P depends continuously on the polar angle \\(\\theta\\).  Equip the lamina occupying R with constant surface–density  \\(\\delta(Q)\\equiv 1\\).  Denote by \\(P_{m}\\) the centre of mass of the lamina.  Show that, if the reference point is chosen so that \\(P=P_{m}\\), then there are at least three different straight lines through P whose two boundary points are equidistant from P (equivalently: there are at least three distinct directions \\(\\theta\\) for which \\(\\rho(\\theta)=\\rho(\\theta+\\pi)\\)).  In other words, the centroid bisects at least three different boundary chords of the region.",
      "solution": "PART  (i)\nPut\n    \\[\\varphi:=\\theta+\\pi/6\\;(0\\le\\varphi\\le\\pi),\\qquad h(\\varphi):=f(\\varphi-\\pi/6).\\]\nWith this change of variable the two hypotheses become\n\\[\\int_{0}^{\\pi}h(\\varphi)\\sin \\varphi\\,d\\varphi=\\int_{0}^{\\pi}h(\\varphi)\\cos \\varphi\\,d\\varphi=0.\\]\nBecause \\(\\sin\\varphi>0\\;(0<\\varphi<\\pi)\\), the familiar two–zero lemma (Putnam 1982 B-5) yields two numbers\n          \\[0<\\varphi_{1}<\\varphi_{2}<\\pi\\quad\\text{with }\\;h(\\varphi_{1})=h(\\varphi_{2})=0.\\]\nReturning to the original variable gives the required zeros\n          \\[\\alpha=\\varphi_{1}-\\tfrac{\\pi}{6},\\qquad \\beta =\\varphi_{2}-\\tfrac{\\pi}{6}\\in I.\\]\n\n--------------------------------------------------------------------\nPART  (ii)  (uniform density)\n\nThroughout we place the pole of the polar coordinate system at the centre of mass \\(P\\).  Let\n          \\[\\rho=g(\\theta)\\qquad(0\\le\\theta<2\\pi)\\]\nbe the polar equation of the boundary.  Since R is strictly star–shaped, \\(g\\) is positive and continuous.\n\n0.  Preliminaries.\nFor the uniform density the elementary area element is\n    \\[dA = \\rho\\,d\\rho\\,d\\theta.\\]\nIntegrating first with respect to \\(\\rho\\) we obtain the linear mass element\n    \\[dM = \\tfrac12 g(\\theta)^{2}\\,d\\theta\\]\nand hence the total mass is \\(M=\\tfrac12\\int_{0}^{2\\pi}g(\\theta)^{2}d\\theta\\).\n\nBecause the pole is the centre of mass, the first moments about the coordinate axes are zero:\n    \\[\\int_{0}^{2\\pi}\\frac{g(\\theta)^{3}}{3}\\cos\\theta\\,d\\theta=0,\\qquad\n      \\int_{0}^{2\\pi}\\frac{g(\\theta)^{3}}{3}\\sin\\theta\\,d\\theta=0.\\tag{1}\\]\n\n1.  A first bisected chord.\nFor a direction \\(\\theta\\) put\n    \\[A(\\theta):=g(\\theta)-g(\\theta+\\pi).\\]\nBecause \\(A(\\theta+\\pi)=-A(\\theta)\\) and \\(A\\) is continuous, there is a number\n\\[\\theta_{0}\\in(0,\\pi)\\quad\\text{with }A(\\theta_{0})=0,\\]\nthat is, the chord through P in direction \\(\\theta_{0}\\) is bisected by P.\n\nRotate the angular coordinate so that this direction becomes the new zero axis.  With the same symbol \\(g\\) we now have\n\\[g(0)=g(\\pi).\\tag{2}\\]\n\n2.  A function to which Part (i) applies.\nDefine\n\\[F(\\theta):=g(\\theta)^{3}-g(\\theta+\\pi)^{3}\\qquad(0\\le\\theta\\le\\pi).\\tag{3}\\]\nThen\n\\[F(0)=F(\\pi)=0,\\qquad F(\\theta+\\pi)=-F(\\theta).\\tag{4}\\]\n\n3.  Orthogonality relations for \\(F\\).\nUsing (3) and the change of variable \\(\\phi=\\theta+\\pi\\) for the second integrals we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi} F(\\theta)\\cos\\theta\\,d\\theta \n&= \\int_{0}^{\\pi} g(\\theta)^{3}\\cos\\theta\\,d\\theta-\\int_{0}^{\\pi} g(\\theta+\\pi)^{3}\\cos\\theta\\,d\\theta\\\\[2mm]\n&= \\int_{0}^{\\pi} g(\\theta)^{3}\\cos\\theta\\,d\\theta+\\int_{\\pi}^{2\\pi} g(\\phi)^{3}\\cos\\phi\\,d\\phi\\\\[2mm]\n&= \\int_{0}^{2\\pi} g(u)^{3}\\cos u\\,du = 0\\quad\\text{by (1).}\n\\end{aligned}\n\\]\nExactly the same computation with \\(\\sin\\theta\\) in place of \\(\\cos\\theta\\) gives\n\\[\\int_{0}^{\\pi} F(\\theta)\\sin\\theta\\,d\\theta=0.\\tag{5}\\]\nThus \\(F\\) satisfies on the interval \\([0,\\pi]\\) the hypotheses of Part (i).\n\n4.  Two further bisected chords.\nApplying Part (i) to the function \\(F\\) on \\([0,\\pi]\\) we obtain two points\n\\[0<\\alpha<\\beta<\\pi\\quad\\text{with }F(\\alpha)=F(\\beta)=0.\\]\nFor these angles (3) gives \\(g(\\alpha)=g(\\alpha+\\pi)\\) and \\(g(\\beta)=g(\\beta+\\pi)\\); hence the chords through P in the directions \\n\\alpha and \\beta are also bisected by P.\n\n5.  Counting the chords.\nBecause of (2) we already had the bisected chord in direction 0.  Together with the directions \\(\\alpha\\) and \\(\\beta\\) we have found at least three distinct directions through P along which the boundary points are equidistant from P.  Consequently the centroid bisects at least three different boundary chords of the region.\n\nThe theorem is proved.",
      "_meta": {
        "core_steps": [
          "Orthogonality ∫f sinθ = ∫f cosθ = 0 forces f to change sign, giving at least one interior zero.",
          "Assuming a single zero α, use ∫f sin(θ−α)=0 to contradict the sign pattern; hence a second zero β exists.",
          "Place the origin at the centroid of R, write the boundary as ρ = g(θ); zero first moments ⇒ ∫[g³(θ)−g³(θ+π)] sinθ = ∫[g³(θ)−g³(θ+π)] cosθ = 0.",
          "Apply the two-zero lemma to h(θ)=g³(θ)−g³(θ+π) to obtain two directions where g(θ)=g(θ+π).",
          "Together with one obvious bisected chord, these two give a total of three centroid-bisected chords."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Angular interval of length π on which the weight sinθ is positive",
            "original": "[0, π]"
          },
          "slot2": {
            "description": "Specific orthogonal weight functions used in the integral conditions",
            "original": "sin θ and cos θ"
          },
          "slot3": {
            "description": "Odd exponent arising from the ρ-integration in the moment calculation",
            "original": "3 (as in g(θ)³)"
          },
          "slot4": {
            "description": "Geometric assumption ensuring each ray from the centroid meets the boundary exactly once",
            "original": "‘bounded convex open region’"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}