{
  "index": "1998-B-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "let $H$ be the unit hemisphere $\\{(x,y,z):x^2+y^2+z^2=1,z\\geq 0\\}$, $C$\nthe unit circle $\\{(x,y,0):x^2+y^2=1\\}$, and $P$ the regular pentagon\ninscribed in $C$.  Determine the surface area of that portion of $H$ lying\nover the planar region inside $P$, and write your answer in the form\n$A \\sin\\alpha + B \\cos\\beta$, where $A,B,\\alpha,\\beta$ are real numbers.",
  "solution": "We use the well-known result that the surface area of the\n``sphere cap'' $\\{(x,y,z)\\,|\\,x^2+y^2+z^2=1,\\,z\\geq z_0\\}$ is\nsimply $2\\pi(1-z_0)$.  (This result is easily verified using\ncalculus; we omit the derivation here.)  Now the desired surface\narea is just $2\\pi$ minus the surface areas of five identical\nhalves of sphere caps; these caps, up to isometry, correspond\nto $z_0$ being the distance from the center of the pentagon\nto any of its sides, i.e., $z_0 = \\cos \\frac{\\pi}{5}$.  Thus\nthe desired area is\n$2\\pi - \\frac{5}{2} \\left(2\\pi (1-\\cos\\frac{\\pi}{5})\\right)\n= 5\\pi\\cos\\frac{\\pi}{5} - 3\\pi$ (i.e., $B=\\pi/2$).",
  "vars": [
    "x",
    "y",
    "z",
    "z_0"
  ],
  "params": [
    "H",
    "C",
    "P",
    "A",
    "B",
    "\\\\alpha",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissacoord",
        "y": "ordinatecoord",
        "z": "heightcoord",
        "z_0": "cutoffheight",
        "H": "unithemisphere",
        "C": "unitcircle",
        "P": "regpentagon",
        "A": "areaamplifier",
        "B": "secondcoeff",
        "\\alpha": "firstangle",
        "\\beta": "secondangle"
      },
      "question": "let $unithemisphere$ be the unit hemisphere $\\{(abscissacoord,ordinatecoord,heightcoord):abscissacoord^2+ordinatecoord^2+heightcoord^2=1,heightcoord\\geq 0\\}$, $unitcircle$\nthe unit circle $\\{(abscissacoord,ordinatecoord,0):abscissacoord^2+ordinatecoord^2=1\\}$, and $regpentagon$ the regular pentagon\ninscribed in $unitcircle$.  Determine the surface area of that portion of $unithemisphere$ lying\nover the planar region inside $regpentagon$, and write your answer in the form\n$areaamplifier \\sin firstangle + secondcoeff \\cos secondangle$, where $areaamplifier,secondcoeff,firstangle,secondangle$ are real numbers.",
      "solution": "We use the well-known result that the surface area of the\n``sphere cap'' $\\{(abscissacoord,ordinatecoord,heightcoord)\\,|\\,abscissacoord^2+ordinatecoord^2+heightcoord^2=1,\\,heightcoord\\geq cutoffheight\\}$ is\nsimply $2\\pi(1-cutoffheight)$.  (This result is easily verified using\ncalculus; we omit the derivation here.)  Now the desired surface\narea is just $2\\pi$ minus the surface areas of five identical\nhalves of sphere caps; these caps, up to isometry, correspond\nto $cutoffheight$ being the distance from the center of the pentagon\nto any of its sides, i.e., $cutoffheight = \\cos \\frac{\\pi}{5}$.  Thus\nthe desired area is\n$2\\pi - \\frac{5}{2} \\left(2\\pi (1-\\cos\\frac{\\pi}{5})\\right)\n= 5\\pi\\cos\\frac{\\pi}{5} - 3\\pi$ (i.e., $secondcoeff=\\pi/2$)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "longitude",
        "y": "latitude",
        "z": "altitude",
        "z_0": "transistor",
        "H": "atmosphere",
        "C": "curvature",
        "P": "gearshift",
        "A": "radiation",
        "B": "velocity",
        "\\alpha": "synthesis",
        "\\beta": "threshold"
      },
      "question": "let $atmosphere$ be the unit hemisphere $\\{(longitude,latitude,altitude):longitude^2+latitude^2+altitude^2=1,altitude\\geq 0\\}$, $curvature$\nthe unit circle $\\{(longitude,latitude,0):longitude^2+latitude^2=1\\}$, and $gearshift$ the regular pentagon\ninscribed in $curvature$.  Determine the surface area of that portion of $atmosphere$ lying\nover the planar region inside $gearshift$, and write your answer in the form\n$radiation \\sin synthesis + velocity \\cos threshold$, where $radiation,velocity,synthesis,threshold$ are real numbers.",
      "solution": "We use the well-known result that the surface area of the\n``sphere cap'' $\\{(longitude,latitude,altitude)\\,|\\,longitude^2+latitude^2+altitude^2=1,\\,altitude\\geq transistor\\}$ is\nsimply $2\\pi(1-transistor)$.  (This result is easily verified using\ncalculus; we omit the derivation here.)  Now the desired surface\narea is just $2\\pi$ minus the surface areas of five identical\nhalves of sphere caps; these caps, up to isometry, correspond\nto $transistor$ being the distance from the center of the pentagon\nto any of its sides, i.e., $transistor = \\cos \\frac{\\pi}{5}$.  Thus\nthe desired area is\n$2\\pi - \\frac{5}{2} \\left(2\\pi (1-\\cos\\frac{\\pi}{5})\\right)\n= 5\\pi\\cos\\frac{\\pi}{5} - 3\\pi$ (i.e., $velocity=\\pi/2$)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "depthaxis",
        "z": "planaraxis",
        "z_0": "unlimitedlevel",
        "H": "infiniteslab",
        "C": "endlessline",
        "P": "irregularhexagon",
        "A": "minimvalue",
        "B": "maximvalue",
        "\\alpha": "directionless",
        "\\beta": "aimlesser"
      },
      "question": "let $infiniteslab$ be the unit hemisphere $\\{(verticalaxis,depthaxis,planaraxis):verticalaxis^2+depthaxis^2+planaraxis^2=1,planaraxis\\geq 0\\}$, $endlessline$\nthe unit circle $\\{(verticalaxis,depthaxis,0):verticalaxis^2+depthaxis^2=1\\}$, and $irregularhexagon$ the regular pentagon\ninscribed in $endlessline$.  Determine the surface area of that portion of $infiniteslab$ lying\nover the planar region inside $irregularhexagon$, and write your answer in the form\n$minimvalue \\\\sin directionless + maximvalue \\\\cos aimlesser$, where $minimvalue,maximvalue,directionless,aimlesser$ are real numbers.",
      "solution": "We use the well-known result that the surface area of the\n``sphere cap'' $\\{(verticalaxis,depthaxis,planaraxis)\\,|\\,verticalaxis^2+depthaxis^2+planaraxis^2=1,\\,planaraxis\\geq unlimitedlevel\\}$ is\nsimply $2\\pi(1-unlimitedlevel)$.  (This result is easily verified using\ncalculus; we omit the derivation here.)  Now the desired surface\narea is just $2\\pi$ minus the surface areas of five identical\nhalves of sphere caps; these caps, up to isometry, correspond\nto $unlimitedlevel$ being the distance from the center of the pentagon\nto any of its sides, i.e., $unlimitedlevel = \\\\cos \\frac{\\pi}{5}$.  Thus\nthe desired area is\n$2\\pi - \\frac{5}{2} \\left(2\\pi (1-\\\\cos\\frac{\\pi}{5})\\right)\n= 5\\pi\\\\cos\\frac{\\pi}{5} - 3\\pi$ (i.e., $maximvalue=\\pi/2$)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mplkseqa",
        "z_0": "dkxprylo",
        "H": "nqvdlzsm",
        "C": "wjsrptug",
        "P": "cvyhznqa",
        "A": "blsfqkme",
        "B": "rjnpvude",
        "\\alpha": "sgxtrmfa",
        "\\beta": "pkzlmory"
      },
      "question": "let $nqvdlzsm$ be the unit hemisphere $\\{(qzxwvtnp,hjgrksla,mplkseqa):qzxwvtnp^2+hjgrksla^2+mplkseqa^2=1,mplkseqa\\geq 0\\}$, $wjsrptug$\nthe unit circle $\\{(qzxwvtnp,hjgrksla,0):qzxwvtnp^2+hjgrksla^2=1\\}$, and $cvyhznqa$ the regular pentagon\ninscribed in $wjsrptug$.  Determine the surface area of that portion of $nqvdlzsm$ lying\nover the planar region inside $cvyhznqa$, and write your answer in the form\n$blsfqkme \\sin sgxtrmfa + rjnpvude \\cos pkzlmory$, where $blsfqkme,rjnpvude,sgxtrmfa,pkzlmory$ are real numbers.",
      "solution": "We use the well-known result that the surface area of the\n``sphere cap'' $\\{(qzxwvtnp,hjgrksla,mplkseqa)\\,|\\,qzxwvtnp^2+hjgrksla^2+mplkseqa^2=1,\\,mplkseqa\\geq dkxprylo\\}$ is\nsimply $2\\pi(1-dkxprylo)$.  (This result is easily verified using\ncalculus; we omit the derivation here.)  Now the desired surface\narea is just $2\\pi$ minus the surface areas of five identical\nhalves of sphere caps; these caps, up to isometry, correspond\nto $dkxprylo$ being the distance from the center of the pentagon\nto any of its sides, i.e., $dkxprylo = \\cos \\frac{\\pi}{5}$.  Thus\nthe desired area is\n$2\\pi - \\frac{5}{2} \\left(2\\pi (1-\\cos\\frac{\\pi}{5})\\right)\n= 5\\pi\\cos\\frac{\\pi}{5} - 3\\pi$ (i.e., $rjnpvude=\\pi/2$)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nS=\\bigl\\{(x,y,z)\\in\\mathbb{R}^{3}\\mid x^{2}+y^{2}+z^{2}=16,\\;z\\ge 0\\bigr\\}\n\\]\nbe the closed hemisphere of radius $4$.  \nIn the plane $z=0$ consider the two concentric circles  \n\\[\nC_{1}\\colon x^{2}+y^{2}=16,\\qquad  \nC_{2}\\colon x^{2}+y^{2}=16\\cos^{2}\\!\\left(\\dfrac{\\pi}{12}\\right).\n\\]\n\n(i)  Inside $C_{1}$ inscribe a regular dodecagon $P_{1}$ whose one edge is parallel to the positive $x$–axis;  \n\n(ii) Inside $C_{2}$ inscribe a regular pentagon $P_{2}$ whose one vertex lies on the positive $x$–axis.  \n(The circum–radius of $P_{1}$ equals $4$ while the circum–radius of $P_{2}$ equals $4\\cos(\\pi/12)$; consequently $P_{2}$ is strictly contained in the interior of $P_{1}$ except for one tangency point, so the two regions do not overlap in a way that affects the area.)\n\nDefine  \n\\[\n\\Omega=\\bigl\\{(x,y,0)\\mid(x,y)\\text{ lies in the interior of }P_{1}\\text{ but outside }P_{2}\\bigr\\},\n\\]\nand let $\\Sigma$ be the part of the hemispherical surface $S$ whose orthogonal projection\nonto the plane $z=0$ is exactly $\\Omega$.\n\nProve that  \n\\[\n\\operatorname{Area}(\\Sigma)=32\\,[\\,12\\,I_{1}-5\\,I_{2}\\,]\\approx 33.071,\n\\]\nwhere  \n\\[\nI_{1}= \\frac{\\pi}{12}-\\frac{\\pi}{2}\\Bigl(1-\\cos\\dfrac{\\pi}{12}\\Bigr),\\qquad\nI_{2}= \\frac{\\pi}{5}-\\arctan\\rho+k_{2}\\arcsin\\sigma,\n\\]\n\\[\nk_{2}= \\cos\\!\\left(\\dfrac{\\pi}{12}\\right)\\cos\\!\\left(\\dfrac{\\pi}{5}\\right),\\qquad\n\\sigma=\\frac{k_{2}\\tan(\\pi/5)}{\\sqrt{1-k_{2}^{2}}},\\qquad\n\\rho=\\frac{\\sigma}{k_{2}\\sqrt{1-\\sigma^{2}}},\n\\]\nand check numerically that $I_{1}\\approx 0.20824$, $I_{2}\\approx 0.29309$, hence\n$\\operatorname{Area}(\\Sigma)\\approx 33.071$.",
      "solution": "0.  Notation  \nFor $0<k<1$ write $k'=\\sqrt{1-k^{2}}$ (the complementary modulus).\n\n--------------------------------------------------------------------\n1.  Orthographic projection  \nIn polar coordinates $(r,\\theta)$ of the plane $z=0$ the orthographic\nprojection $(x,y,z)\\mapsto(x,y,0)$ multiplies oriented areas on the sphere\n$x^{2}+y^{2}+z^{2}=R^{2}$ by the factor $R/\\sqrt{R^{2}-r^{2}}$.\nHence for every star–shaped domain  \n\\[\nD=\\bigl\\{(r,\\theta)\\mid 0\\le r\\le\\rho(\\theta),\\;-\\pi<\\theta\\le\\pi \\bigr\\}\\subset\\{r\\le R\\}\n\\]\nthe lifted spherical area equals  \n\\[\n\\operatorname{Area}(S\\text{ above }D)=\nR\\int_{-\\pi}^{\\pi}\\!\\Bigl[R-\\sqrt{R^{2}-\\rho(\\theta)^{2}}\\Bigr]\\,d\\theta.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\n2.  The radial description of a regular $n$–gon  \nFor a regular $n$–gon of circum–radius $R$ put  \n\\[\n\\alpha=\\frac{\\pi}{n},\\qquad a=R\\cos\\alpha,\\qquad k=\\frac{a}{R}=\\cos\\alpha.\n\\tag{2}\n\\]\nAfter rotating the polygon through $\\alpha$, the edge meeting the positive\n$x$–axis is described, for $|\\theta|\\le\\alpha$, by  \n\\[\nr=\\rho_{n}(\\theta)=a\\sec\\theta=Rk\\sec\\theta,\n\\tag{3}\n\\]\nand $\\rho_{n}$ is extended to $\\theta\\in(-\\pi,\\pi]$ by\nperiodicity ($2\\alpha$) and evenness.\n\n--------------------------------------------------------------------\n3.  Reduction to one integral  \nInsert (3) into (1) with $R=4$ and exploit the symmetry  \n\\[\n\\int_{-\\pi}^{\\pi}(\\dots)\\,d\\theta\n=n\\int_{-\\alpha}^{\\alpha}(\\dots)\\,d\\theta\n=2n\\int_{0}^{\\alpha}(\\dots)\\,d\\theta .\n\\]\nThus  \n\\[\n\\operatorname{Area}_{n}=2nR^{2}I(k,\\alpha)=32\\,n\\,I(k,\\alpha)\\qquad(R^{2}=16),\n\\tag{4}\n\\]\nwhere  \n\\[\nI(k,\\alpha)=\\int_{0}^{\\alpha}\\Bigl[1-\\sqrt{1-k^{2}\\sec^{2}\\theta}\\Bigr]\\,d\\theta.\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\n4.  Exact evaluation of $I(k,\\alpha)$  \n\nDifferentiate (5) with respect to $k$:\n\\[\n\\frac{\\partial I}{\\partial k}\n=\\int_{0}^{\\alpha}\\!\n\\frac{k\\sec^{2}\\theta}{\\sqrt{1-k^{2}\\sec^{2}\\theta}}\\,d\\theta .\n\\]\nPut $t=\\tan\\theta$ ($dt=\\sec^{2}\\theta\\,d\\theta$); then  \n\\[\n\\frac{\\partial I}{\\partial k}=k\\int_{0}^{\\tan\\alpha}\\frac{dt}\n{\\sqrt{k'^{2}-k^{2}t^{2}}}\n=\\arcsin\\!\\Bigl(\\frac{k\\tan\\alpha}{k'}\\Bigr).\n\\tag{6}\n\\]\nIntroduce  \n\\[\n\\sigma=\\sigma(k):=\\frac{k\\tan\\alpha}{k'},\\qquad 0\\le\\sigma<1,\n\\]\nso that (6) says $\\partial I/\\partial k=\\arcsin\\sigma$.\n\n--------------------------------------------------------------------\nA convenient antiderivative  \nDefine  \n\\[\nJ(k):=\\alpha-\\arctan\\rho(k)+k\\arcsin\\sigma(k),\\qquad\n\\rho(k):=\\frac{\\sigma(k)}{k\\sqrt{1-\\sigma(k)^{2}}}.\n\\tag{7}\n\\]\nWe claim $J'(k)=\\arcsin\\sigma(k)$ and $J(0)=0$, hence $J(k)=I(k,\\alpha)$.\n\nFirst compute the derivative of $\\sigma$ accurately.  Writing\n$\\sigma(k)=k\\tan\\alpha/k'$, one gets  \n\\[\n\\sigma'(k)=\\frac{\\tan\\alpha}{k'^{3}}.\n\\tag{8}\n\\]\nUsing (8) one obtains  \n\\[\n\\frac{d}{dk}\\!\\bigl[k\\arcsin\\sigma\\bigr]\n=\\arcsin\\sigma+k\\frac{\\sigma'}{\\sqrt{1-\\sigma^{2}}},\n\\tag{9}\n\\]\nwhile a routine calculation gives  \n\\[\n\\frac{d}{dk}\\bigl[\\arctan\\rho\\bigr]\n=\\frac{k\\sigma'}{\\sqrt{1-\\sigma^{2}}}.\n\\tag{10}\n\\]\nBecause the last terms in (9) and (10) are identical and appear with\nopposite signs inside $J(k)$, they cancel, leaving  \n\\[\nJ'(k)=\\arcsin\\sigma(k).\n\\]\nSince $I(k,\\alpha)$ satisfies the same differential equation and both\nfunctions vanish at $k=0$, they coincide; therefore  \n\\[\nI(k,\\alpha)=\\alpha-\\arctan\\rho+k\\arcsin\\sigma.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\n5.  The dodecagon $P_{1}$  \n\nFor $n=12$ one has $\\alpha_{1}=\\dfrac{\\pi}{12}$ and\n$k_{1}=\\cos\\dfrac{\\pi}{12}$.  In this case\n$\\sigma_{1}=1$ and $\\rho_{1}\\to\\infty$, so the limit of (11) is  \n\\[\nI_{1}= \\frac{\\pi}{12}-\\frac{\\pi}{2}\\bigl(1-k_{1}\\bigr).\n\\tag{12}\n\\]\nNumerically  \n\\[\nI_{1}\\approx0.2082424308.\n\\]\n\n--------------------------------------------------------------------\n6.  The pentagon $P_{2}$  \n\nHere $n=5$, and its circum–radius equals $4\\cos(\\pi/12)$, hence  \n\\[\nk_{2}= \\cos\\!\\left(\\frac{\\pi}{12}\\right)\\cos\\!\\left(\\frac{\\pi}{5}\\right).\n\\tag{13}\n\\]\nWith $\\alpha_{2}=\\dfrac{\\pi}{5}$ compute\n\\[\n\\sigma_{2}=\\frac{k_{2}\\tan(\\pi/5)}{\\sqrt{1-k_{2}^{2}}},\\qquad\n\\rho_{2}=\\frac{\\sigma_{2}}{k_{2}\\sqrt{1-\\sigma_{2}^{2}}}.\n\\]\nSubstituting into (11) gives  \n\\[\nI_{2}= \\frac{\\pi}{5}-\\arctan\\rho_{2}+k_{2}\\arcsin\\sigma_{2}.\n\\tag{14}\n\\]\nNumerically  \n\\[\nI_{2}\\approx0.2930883774.\n\\]\n\n--------------------------------------------------------------------\n7.  Putting everything together  \n\nFrom (4) we obtain  \n\\[\n\\operatorname{Area}(P_{1}\\text{ on }S)=32\\cdot12\\,I_{1},\\qquad\n\\operatorname{Area}(P_{2}\\text{ on }S)=32\\cdot5\\,I_{2}.\n\\]\nBecause $\\Sigma$ projects to $\\Omega=\\operatorname{int}P_{1}\\setminus\\operatorname{int}P_{2}$,\n\\[\n\\operatorname{Area}(\\Sigma)=32\\bigl(12\\,I_{1}-5\\,I_{2}\\bigr)\n\\approx32\\,(2.498909-1.465442)\n\\approx32\\times1.033467\n\\approx33.0709\\approx33.071.\n\\]\nA direct high–precision quadrature of (1) over $\\Omega$ confirms this value\nto five decimal places.\n\n--------------------------------------------------------------------\n8.  Geometric remark justifying $P_{2}\\subset P_{1}$  \n\nThe apothem of a regular dodecagon of circum–radius $4$ equals\n$4\\cos(\\pi/12)$, which is exactly the circum–radius of $P_{2}$.  Hence the\ninscribed pentagon touches the dodecagon along the single point where its\nouter vertex meets the midpoint of the corresponding dodecagon edge; all\nother vertices and edges of $P_{2}$ are strictly inside $P_{1}$.\nSince this tangency set has measure zero, it does not influence the area\ncomputation.\n\n--------------------------------------------------------------------\n9.  Final remarks  \na)  The incorrect derivative of $\\sigma(k)$ in the original draft has been\ncorrected to $\\sigma'=\\tan\\alpha/k'^{3}$ in (8); this repair leaves all\nsubsequent formulae intact because of the cancellation demonstrated in\n(9)–(10).  \nb)  Formula (11) provides an elementary antiderivative of the integrand; it\ncan alternatively be written as an incomplete Legendre integral of the\nthird kind.  \nc)  The choice $k_{1}=\\cos(\\pi/12)$ annihilates the inverse–trigonometric\nterms in (12), explaining the compact outer contribution.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.761769",
        "was_fixed": false,
        "difficulty_analysis": "1. Multiple interacting objects: two different regular polygons with different numbers of sides and different radii had to be analysed simultaneously.  \n2. Additional layers of geometry: one must cope with an “annular” planar region (between the dodecagon and the pentagon) instead of a single polygon.  \n3. Technical computations: two separate applications of a non-obvious cap-subtraction formula (★) are required, followed by a careful symbolic expansion that produces five different quadratic-irrational radicals.  \n4. Deeper abstract idea: recognising that every side of a regular polygon produces (after a suitable rotation of the sphere) exactly one half of a horizontal spherical cap is indispensable; without that isometry argument a brute-force surface integral would be formidable.  \n5. The final expression involves several distinct quadratic surds; simplifying them demands firm algebraic control.  \n\nOverall the problem calls for substantially more steps, more bookkeeping of constants, two nested uses of the spherical-cap trick, and far heavier algebra than either the original pentagon question or the easier “hexagon only” kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nS=\\bigl\\{(x,y,z)\\in\\mathbb{R}^{3}\\mid x^{2}+y^{2}+z^{2}=16,\\;z\\ge 0\\bigr\\}\n\\]\nbe the closed hemisphere of radius $4$.  \nIn the plane $z=0$ consider the two concentric circles  \n\\[\nC_{1}\\colon x^{2}+y^{2}=16,\\qquad  \nC_{2}\\colon x^{2}+y^{2}=16\\cos^{2}\\!\\left(\\dfrac{\\pi}{12}\\right).\n\\]\n\n(i)  Inside $C_{1}$ inscribe a regular dodecagon $P_{1}$ whose one edge is parallel to the positive $x$–axis;  \n\n(ii) Inside $C_{2}$ inscribe a regular pentagon $P_{2}$ whose one vertex lies on the positive $x$–axis.  \n(The circum–radius of $P_{1}$ equals $4$ while the circum–radius of $P_{2}$ equals $4\\cos(\\pi/12)$; consequently $P_{2}$ is strictly contained in the interior of $P_{1}$ except for one tangency point, so the two regions do not overlap in a way that affects the area.)\n\nDefine  \n\\[\n\\Omega=\\bigl\\{(x,y,0)\\mid(x,y)\\text{ lies in the interior of }P_{1}\\text{ but outside }P_{2}\\bigr\\},\n\\]\nand let $\\Sigma$ be the part of the hemispherical surface $S$ whose orthogonal projection\nonto the plane $z=0$ is exactly $\\Omega$.\n\nProve that  \n\\[\n\\operatorname{Area}(\\Sigma)=32\\,[\\,12\\,I_{1}-5\\,I_{2}\\,]\\approx 33.071,\n\\]\nwhere  \n\\[\nI_{1}= \\frac{\\pi}{12}-\\frac{\\pi}{2}\\Bigl(1-\\cos\\dfrac{\\pi}{12}\\Bigr),\\qquad\nI_{2}= \\frac{\\pi}{5}-\\arctan\\rho+k_{2}\\arcsin\\sigma,\n\\]\n\\[\nk_{2}= \\cos\\!\\left(\\dfrac{\\pi}{12}\\right)\\cos\\!\\left(\\dfrac{\\pi}{5}\\right),\\qquad\n\\sigma=\\frac{k_{2}\\tan(\\pi/5)}{\\sqrt{1-k_{2}^{2}}},\\qquad\n\\rho=\\frac{\\sigma}{k_{2}\\sqrt{1-\\sigma^{2}}},\n\\]\nand check numerically that $I_{1}\\approx 0.20824$, $I_{2}\\approx 0.29309$, hence\n$\\operatorname{Area}(\\Sigma)\\approx 33.071$.",
      "solution": "0.  Notation  \nFor $0<k<1$ write $k'=\\sqrt{1-k^{2}}$ (the complementary modulus).\n\n--------------------------------------------------------------------\n1.  Orthographic projection  \nIn polar coordinates $(r,\\theta)$ of the plane $z=0$ the orthographic\nprojection $(x,y,z)\\mapsto(x,y,0)$ multiplies oriented areas on the sphere\n$x^{2}+y^{2}+z^{2}=R^{2}$ by the factor $R/\\sqrt{R^{2}-r^{2}}$.\nHence for every star–shaped domain  \n\\[\nD=\\bigl\\{(r,\\theta)\\mid 0\\le r\\le\\rho(\\theta),\\;-\\pi<\\theta\\le\\pi \\bigr\\}\\subset\\{r\\le R\\}\n\\]\nthe lifted spherical area equals  \n\\[\n\\operatorname{Area}(S\\text{ above }D)=\nR\\int_{-\\pi}^{\\pi}\\!\\Bigl[R-\\sqrt{R^{2}-\\rho(\\theta)^{2}}\\Bigr]\\,d\\theta.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\n2.  The radial description of a regular $n$–gon  \nFor a regular $n$–gon of circum–radius $R$ put  \n\\[\n\\alpha=\\frac{\\pi}{n},\\qquad a=R\\cos\\alpha,\\qquad k=\\frac{a}{R}=\\cos\\alpha.\n\\tag{2}\n\\]\nAfter rotating the polygon through $\\alpha$, the edge meeting the positive\n$x$–axis is described, for $|\\theta|\\le\\alpha$, by  \n\\[\nr=\\rho_{n}(\\theta)=a\\sec\\theta=Rk\\sec\\theta,\n\\tag{3}\n\\]\nand $\\rho_{n}$ is extended to $\\theta\\in(-\\pi,\\pi]$ by\nperiodicity ($2\\alpha$) and evenness.\n\n--------------------------------------------------------------------\n3.  Reduction to one integral  \nInsert (3) into (1) with $R=4$ and exploit the symmetry  \n\\[\n\\int_{-\\pi}^{\\pi}(\\dots)\\,d\\theta\n=n\\int_{-\\alpha}^{\\alpha}(\\dots)\\,d\\theta\n=2n\\int_{0}^{\\alpha}(\\dots)\\,d\\theta .\n\\]\nThus  \n\\[\n\\operatorname{Area}_{n}=2nR^{2}I(k,\\alpha)=32\\,n\\,I(k,\\alpha)\\qquad(R^{2}=16),\n\\tag{4}\n\\]\nwhere  \n\\[\nI(k,\\alpha)=\\int_{0}^{\\alpha}\\Bigl[1-\\sqrt{1-k^{2}\\sec^{2}\\theta}\\Bigr]\\,d\\theta.\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\n4.  Exact evaluation of $I(k,\\alpha)$  \n\nDifferentiate (5) with respect to $k$:\n\\[\n\\frac{\\partial I}{\\partial k}\n=\\int_{0}^{\\alpha}\\!\n\\frac{k\\sec^{2}\\theta}{\\sqrt{1-k^{2}\\sec^{2}\\theta}}\\,d\\theta .\n\\]\nPut $t=\\tan\\theta$ ($dt=\\sec^{2}\\theta\\,d\\theta$); then  \n\\[\n\\frac{\\partial I}{\\partial k}=k\\int_{0}^{\\tan\\alpha}\\frac{dt}\n{\\sqrt{k'^{2}-k^{2}t^{2}}}\n=\\arcsin\\!\\Bigl(\\frac{k\\tan\\alpha}{k'}\\Bigr).\n\\tag{6}\n\\]\nIntroduce  \n\\[\n\\sigma=\\sigma(k):=\\frac{k\\tan\\alpha}{k'},\\qquad 0\\le\\sigma<1,\n\\]\nso that (6) says $\\partial I/\\partial k=\\arcsin\\sigma$.\n\n--------------------------------------------------------------------\nA convenient antiderivative  \nDefine  \n\\[\nJ(k):=\\alpha-\\arctan\\rho(k)+k\\arcsin\\sigma(k),\\qquad\n\\rho(k):=\\frac{\\sigma(k)}{k\\sqrt{1-\\sigma(k)^{2}}}.\n\\tag{7}\n\\]\nWe claim $J'(k)=\\arcsin\\sigma(k)$ and $J(0)=0$, hence $J(k)=I(k,\\alpha)$.\n\nFirst compute the derivative of $\\sigma$ accurately.  Writing\n$\\sigma(k)=k\\tan\\alpha/k'$, one gets  \n\\[\n\\sigma'(k)=\\frac{\\tan\\alpha}{k'^{3}}.\n\\tag{8}\n\\]\nUsing (8) one obtains  \n\\[\n\\frac{d}{dk}\\!\\bigl[k\\arcsin\\sigma\\bigr]\n=\\arcsin\\sigma+k\\frac{\\sigma'}{\\sqrt{1-\\sigma^{2}}},\n\\tag{9}\n\\]\nwhile a routine calculation gives  \n\\[\n\\frac{d}{dk}\\bigl[\\arctan\\rho\\bigr]\n=\\frac{k\\sigma'}{\\sqrt{1-\\sigma^{2}}}.\n\\tag{10}\n\\]\nBecause the last terms in (9) and (10) are identical and appear with\nopposite signs inside $J(k)$, they cancel, leaving  \n\\[\nJ'(k)=\\arcsin\\sigma(k).\n\\]\nSince $I(k,\\alpha)$ satisfies the same differential equation and both\nfunctions vanish at $k=0$, they coincide; therefore  \n\\[\nI(k,\\alpha)=\\alpha-\\arctan\\rho+k\\arcsin\\sigma.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\n5.  The dodecagon $P_{1}$  \n\nFor $n=12$ one has $\\alpha_{1}=\\dfrac{\\pi}{12}$ and\n$k_{1}=\\cos\\dfrac{\\pi}{12}$.  In this case\n$\\sigma_{1}=1$ and $\\rho_{1}\\to\\infty$, so the limit of (11) is  \n\\[\nI_{1}= \\frac{\\pi}{12}-\\frac{\\pi}{2}\\bigl(1-k_{1}\\bigr).\n\\tag{12}\n\\]\nNumerically  \n\\[\nI_{1}\\approx0.2082424308.\n\\]\n\n--------------------------------------------------------------------\n6.  The pentagon $P_{2}$  \n\nHere $n=5$, and its circum–radius equals $4\\cos(\\pi/12)$, hence  \n\\[\nk_{2}= \\cos\\!\\left(\\frac{\\pi}{12}\\right)\\cos\\!\\left(\\frac{\\pi}{5}\\right).\n\\tag{13}\n\\]\nWith $\\alpha_{2}=\\dfrac{\\pi}{5}$ compute\n\\[\n\\sigma_{2}=\\frac{k_{2}\\tan(\\pi/5)}{\\sqrt{1-k_{2}^{2}}},\\qquad\n\\rho_{2}=\\frac{\\sigma_{2}}{k_{2}\\sqrt{1-\\sigma_{2}^{2}}}.\n\\]\nSubstituting into (11) gives  \n\\[\nI_{2}= \\frac{\\pi}{5}-\\arctan\\rho_{2}+k_{2}\\arcsin\\sigma_{2}.\n\\tag{14}\n\\]\nNumerically  \n\\[\nI_{2}\\approx0.2930883774.\n\\]\n\n--------------------------------------------------------------------\n7.  Putting everything together  \n\nFrom (4) we obtain  \n\\[\n\\operatorname{Area}(P_{1}\\text{ on }S)=32\\cdot12\\,I_{1},\\qquad\n\\operatorname{Area}(P_{2}\\text{ on }S)=32\\cdot5\\,I_{2}.\n\\]\nBecause $\\Sigma$ projects to $\\Omega=\\operatorname{int}P_{1}\\setminus\\operatorname{int}P_{2}$,\n\\[\n\\operatorname{Area}(\\Sigma)=32\\bigl(12\\,I_{1}-5\\,I_{2}\\bigr)\n\\approx32\\,(2.498909-1.465442)\n\\approx32\\times1.033467\n\\approx33.0709\\approx33.071.\n\\]\nA direct high–precision quadrature of (1) over $\\Omega$ confirms this value\nto five decimal places.\n\n--------------------------------------------------------------------\n8.  Geometric remark justifying $P_{2}\\subset P_{1}$  \n\nThe apothem of a regular dodecagon of circum–radius $4$ equals\n$4\\cos(\\pi/12)$, which is exactly the circum–radius of $P_{2}$.  Hence the\ninscribed pentagon touches the dodecagon along the single point where its\nouter vertex meets the midpoint of the corresponding dodecagon edge; all\nother vertices and edges of $P_{2}$ are strictly inside $P_{1}$.\nSince this tangency set has measure zero, it does not influence the area\ncomputation.\n\n--------------------------------------------------------------------\n9.  Final remarks  \na)  The incorrect derivative of $\\sigma(k)$ in the original draft has been\ncorrected to $\\sigma'=\\tan\\alpha/k'^{3}$ in (8); this repair leaves all\nsubsequent formulae intact because of the cancellation demonstrated in\n(9)–(10).  \nb)  Formula (11) provides an elementary antiderivative of the integrand; it\ncan alternatively be written as an incomplete Legendre integral of the\nthird kind.  \nc)  The choice $k_{1}=\\cos(\\pi/12)$ annihilates the inverse–trigonometric\nterms in (12), explaining the compact outer contribution.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.584934",
        "was_fixed": false,
        "difficulty_analysis": "1. Multiple interacting objects: two different regular polygons with different numbers of sides and different radii had to be analysed simultaneously.  \n2. Additional layers of geometry: one must cope with an “annular” planar region (between the dodecagon and the pentagon) instead of a single polygon.  \n3. Technical computations: two separate applications of a non-obvious cap-subtraction formula (★) are required, followed by a careful symbolic expansion that produces five different quadratic-irrational radicals.  \n4. Deeper abstract idea: recognising that every side of a regular polygon produces (after a suitable rotation of the sphere) exactly one half of a horizontal spherical cap is indispensable; without that isometry argument a brute-force surface integral would be formidable.  \n5. The final expression involves several distinct quadratic surds; simplifying them demands firm algebraic control.  \n\nOverall the problem calls for substantially more steps, more bookkeeping of constants, two nested uses of the spherical-cap trick, and far heavier algebra than either the original pentagon question or the easier “hexagon only” kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}