{
  "index": "1952-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to \\( 45^{\\circ} \\) South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude \\( 30^{\\circ} \\mathrm{S} \\). In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?",
  "solution": "Solution. Let a point on the earth with co-latitude \\( \\alpha \\) and longitude \\( \\theta \\) be represented by a point on the flag map with polar coordinates \\( k \\alpha \\) and \\( \\theta \\) where \\( k \\) is a constant of proportionality.\nLet \\( R \\) be the radius of the earth. A meridian on the earth has length \\( \\pi R \\) and on the (complete) flag map is represented by a segment of length \\( \\pi k \\). Hence a short distance \\( d \\) in the North-South direction is represented on the flag map by the distance \\( d_{1}=k d / R \\).\nOn the earth the small circle of latitude \\( 30^{\\circ} S(= \\) co-latitude \\( 2 \\pi / 3 \\) ) is represented on the flag map by a circle of radius \\( 2 \\pi k / 3 \\) and of circumference \\( 4 \\pi^{2} k / 3 \\). On the earth that circle has length \\( 2 \\pi R \\sin (2 \\pi / 3)= \\) \\( \\pi R \\sqrt{3} \\).\n\nHence a short distance \\( d \\) in the East-West direction is represented on the flag map by a distance\n\\[\nd \\frac{\\frac{4 \\pi^{2} k}{3}}{\\pi R \\sqrt{3}}=\\frac{4 \\pi k d}{3 \\sqrt{3} R}=\\frac{4 \\pi}{3 \\sqrt{3}} d_{1} .\n\\]\n\nTherefore distances in the East-West direction at latitude \\( 30^{\\circ} \\) S are magnified relative to distances in the North-South direction by the factor \\( 4 \\pi / 3 \\sqrt{3} \\sim 2.42 \\).",
  "vars": [
    "\\\\alpha",
    "\\\\theta",
    "d",
    "d_1"
  ],
  "params": [
    "k",
    "R"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "\\alpha": "anglecolat",
        "\\theta": "anglelong",
        "d": "smallstep",
        "d_1": "mappedstep",
        "k": "scalefactor",
        "R": "earthradius"
      },
      "question": "4. The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to \\( 45^{\\circ} \\) South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude \\( 30^{\\circ} \\mathrm{S} \\). In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?",
      "solution": "Solution. Let a point on the earth with co-latitude \\( anglecolat \\) and longitude \\( anglelong \\) be represented by a point on the flag map with polar coordinates \\( scalefactor \\, anglecolat \\) and \\( anglelong \\) where \\( scalefactor \\) is a constant of proportionality.\nLet \\( earthradius \\) be the radius of the earth. A meridian on the earth has length \\( \\pi \\, earthradius \\) and on the (complete) flag map is represented by a segment of length \\( \\pi \\, scalefactor \\). Hence a short distance \\( smallstep \\) in the North–South direction is represented on the flag map by the distance \\( mappedstep = scalefactor \\, smallstep / earthradius \\).\nOn the earth the small circle of latitude \\( 30^{\\circ} S (= \\) co-latitude \\( 2 \\pi / 3 ) \\) is represented on the flag map by a circle of radius \\( 2 \\pi \\, scalefactor / 3 \\) and of circumference \\( 4 \\pi^{2} \\, scalefactor / 3 \\). On the earth that circle has length \\( 2 \\pi \\, earthradius \\sin(2 \\pi / 3) = \\pi \\, earthradius \\sqrt{3} \\).\n\nHence a short distance \\( smallstep \\) in the East–West direction is represented on the flag map by a distance\n\\[\nsmallstep \\frac{\\frac{4 \\pi^{2} \\, scalefactor}{3}}{\\pi \\, earthradius \\sqrt{3}} = \\frac{4 \\pi \\, scalefactor \\, smallstep}{3 \\sqrt{3} \\, earthradius} = \\frac{4 \\pi}{3 \\sqrt{3}} \\, mappedstep .\n\\]\n\nTherefore distances in the East–West direction at latitude \\( 30^{\\circ} \\) S are magnified relative to distances in the North–South direction by the factor \\( 4 \\pi / (3 \\sqrt{3}) \\approx 2.42 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "\\alpha": "sunflower",
        "\\theta": "backpack",
        "d": "windstorm",
        "d_1": "watermelon",
        "k": "shoelaces",
        "R": "paintbrush"
      },
      "question": "4. The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to \\( 45^{\\circ} \\) South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude \\( 30^{\\circ} \\mathrm{S} \\). In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?",
      "solution": "Solution. Let a point on the earth with co-latitude \\( sunflower \\) and longitude \\( backpack \\) be represented by a point on the flag map with polar coordinates \\( shoelaces \\, sunflower \\) and \\( backpack \\) where \\( shoelaces \\) is a constant of proportionality.\nLet \\( paintbrush \\) be the radius of the earth. A meridian on the earth has length \\( \\pi paintbrush \\) and on the (complete) flag map is represented by a segment of length \\( \\pi shoelaces \\). Hence a short distance \\( windstorm \\) in the North-South direction is represented on the flag map by the distance \\( watermelon=shoelaces \\, windstorm / paintbrush \\).\nOn the earth the small circle of latitude \\( 30^{\\circ} S(= \\) co-latitude \\( 2 \\pi / 3 \\) ) is represented on the flag map by a circle of radius \\( 2 \\pi shoelaces / 3 \\) and of circumference \\( 4 \\pi^{2} shoelaces / 3 \\). On the earth that circle has length \\( 2 \\pi paintbrush \\sin (2 \\pi / 3)= \\) \\( \\pi paintbrush \\sqrt{3} \\).\n\nHence a short distance \\( windstorm \\) in the East-West direction is represented on the flag map by a distance\n\\[\nwindstorm \\frac{\\frac{4 \\pi^{2} shoelaces}{3}}{\\pi paintbrush \\sqrt{3}}=\\frac{4 \\pi shoelaces \\, windstorm}{3 \\sqrt{3} paintbrush}=\\frac{4 \\pi}{3 \\sqrt{3}} watermelon .\n\\]\n\nTherefore distances in the East-West direction at latitude \\( 30^{\\circ} \\) S are magnified relative to distances in the North-South direction by the factor \\( 4 \\pi / 3 \\sqrt{3} \\sim 2.42 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "\\alpha": "lastvalue",
        "\\theta": "straightline",
        "d": "largesegment",
        "d_1": "ultimatepiece",
        "k": "discrepancy",
        "R": "depthpoint"
      },
      "question": "Problem: <<<\\n4. The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to \\( 45^{\\\\circ} \\) South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude \\( 30^{\\\\circ} \\\\mathrm{S} \\). In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?\\n>>>",
      "solution": "Solution:\\n<<<\\nSolution. Let a point on the earth with co-latitude \\( lastvalue \\) and longitude \\( straightline \\) be represented by a point on the flag map with polar coordinates \\( discrepancy lastvalue \\) and \\( straightline \\) where \\( discrepancy \\) is a constant of proportionality.\\nLet \\( depthpoint \\) be the radius of the earth. A meridian on the earth has length \\( \\pi depthpoint \\) and on the (complete) flag map is represented by a segment of length \\( \\pi discrepancy \\). Hence a short distance \\( largesegment \\) in the North-South direction is represented on the flag map by the distance \\( ultimatepiece=discrepancy largesegment / depthpoint \\).\\nOn the earth the small circle of latitude \\( 30^{\\\\circ} S(= \\) co-latitude \\( 2 \\pi / 3 \\) ) is represented on the flag map by a circle of radius \\( 2 \\pi discrepancy / 3 \\) and of circumference \\( 4 \\pi^{2} discrepancy / 3 \\). On the earth that circle has length \\( 2 \\pi depthpoint \\\\sin (2 \\pi / 3)= \\pi depthpoint \\\\sqrt{3} \\).\\n\\nHence a short distance \\( largesegment \\) in the East-West direction is represented on the flag map by a distance\\n\\\\[\\nlargesegment \\\\frac{\\\\frac{4 \\pi^{2} discrepancy}{3}}{\\pi depthpoint \\\\sqrt{3}}=\\\\frac{4 \\pi discrepancy largesegment}{3 \\\\sqrt{3} depthpoint}=\\\\frac{4 \\pi}{3 \\\\sqrt{3}} ultimatepiece .\\n\\\\]\\n\\nTherefore distances in the East-West direction at latitude \\( 30^{\\\\circ} \\) S are magnified relative to distances in the North-South direction by the factor \\( 4 \\pi / 3 \\\\sqrt{3} \\\\sim 2.42 .\\n>>>"
    },
    "garbled_string": {
      "map": {
        "\\alpha": "finrqsam",
        "\\theta": "gxbmdlke",
        "d": "mzptsaro",
        "d_1": "jhfqenud",
        "k": "vqrzpemo",
        "R": "bsrtmgdw"
      },
      "question": "4. The flag of the United Nations consists of a polar map of the world, with the North Pole as center, extending approximately to \\( 45^{\\circ} \\) South Latitude. The parallels of latitude are concentric circles with radii proportional to their co-latitudes. Australia is near the periphery of the map and is intersected by the parallel of latitude \\( 30^{\\circ} \\mathrm{S} \\). In the very close vicinity of this parallel how much are East and West distances exaggerated as compared to North and South distances?",
      "solution": "Solution. Let a point on the earth with co-latitude \\( finrqsam \\) and longitude \\( gxbmdlke \\) be represented by a point on the flag map with polar coordinates \\( vqrzpemo finrqsam \\) and \\( gxbmdlke \\) where \\( vqrzpemo \\) is a constant of proportionality.\nLet \\( bsrtmgdw \\) be the radius of the earth. A meridian on the earth has length \\( \\pi bsrtmgdw \\) and on the (complete) flag map is represented by a segment of length \\( \\pi vqrzpemo \\). Hence a short distance \\( mzptsaro \\) in the North-South direction is represented on the flag map by the distance \\( jhfqenud = vqrzpemo mzptsaro / bsrtmgdw \\).\nOn the earth the small circle of latitude \\( 30^{\\circ} S(= \\) co-latitude \\( 2 \\pi / 3 \\) ) is represented on the flag map by a circle of radius \\( 2 \\pi vqrzpemo / 3 \\) and of circumference \\( 4 \\pi^{2} vqrzpemo / 3 \\). On the earth that circle has length \\( 2 \\pi bsrtmgdw \\sin (2 \\pi / 3)= \\) \\( \\pi bsrtmgdw \\sqrt{3} \\).\n\nHence a short distance \\( mzptsaro \\) in the East-West direction is represented on the flag map by a distance\n\\[\nmzptsaro \\frac{\\frac{4 \\pi^{2} vqrzpemo}{3}}{\\pi bsrtmgdw \\sqrt{3}}=\\frac{4 \\pi vqrzpemo mzptsaro}{3 \\sqrt{3} bsrtmgdw}=\\frac{4 \\pi}{3 \\sqrt{3}} jhfqenud .\n\\]\n\nTherefore distances in the East-West direction at latitude \\( 30^{\\circ} \\) S are magnified relative to distances in the North-South direction by the factor \\( 4 \\pi / 3 \\sqrt{3} \\sim 2.42 \\)."
    },
    "kernel_variant": {
      "question": "A commemorative logo for an international cartographic congress depicts the Earth by means of the oblique azimuthal–equidistant projection whose centre  \n$C$ has geographic latitude $\\varphi_{0}=35^{\\circ}\\text{ N}$ and longitude $\\lambda_{0}=40^{\\circ}\\text{ E}$.  \nFor any point $P(\\varphi,\\lambda)$ on the spherical Earth (radius $R=6371\\text{ km}$) put  \n\\[\nc(\\varphi,\\lambda)=\\arccos\\!\\bigl[\\sin\\varphi_{0}\\sin\\varphi+\n\\cos\\varphi_{0}\\cos\\varphi\\cos\\bigl(\\lambda-\\lambda_{0}\\bigr)\\bigr],\\qquad\n0\\le c\\le\\pi,\n\\]\n\\[\nA(\\varphi,\\lambda)=\\operatorname{atan2}\\!\\Bigl(\n\\cos\\varphi\\,\\sin\\bigl(\\lambda-\\lambda_{0}\\bigr),\\;\n\\cos\\varphi_{0}\\sin\\varphi-\\sin\\varphi_{0}\\cos\\varphi\\cos\\bigl(\\lambda-\\lambda_{0}\\bigr)\n\\Bigr).\n\\]\n\nThe map–plane coordinates are  \n\\[\nx   =R\\,c\\sin A,\\qquad\ny   =R\\,c\\cos A.\\tag{1}\n\\]\n\na)  Obtain fully–simplified expressions for the four partial derivatives  \n\\[\nx_{\\varphi},\\;x_{\\lambda},\\;y_{\\varphi},\\;y_{\\lambda}\\qquad\n(\\text{i.e.\\ for the Jacobian }J=\\partial(x,y)/\\partial(\\varphi,\\lambda))\n\\]\nin terms only of the elementary functions\n$\\varphi,\\lambda,\\varphi_{0},\\lambda_{0},c$ and $A$\n— \\emph{no extra shorthand symbols such as $\\Delta\\lambda$, $s_{0}$, \n$\\bar c$, etc.\\ may appear in the final formulas.}\n\nb)  At the point  \n\\[\nP^{\\star}:\\;\n\\varphi^{\\star}=-5^{\\circ}\\;(5^{\\circ}\\text{ S}),\\qquad\n\\lambda^{\\star}=100^{\\circ}\\text{ E},\n\\]\nevaluate $J$ numerically and compute  \n• the principal scale factors $k_{\\max}\\ge k_{\\min}$ \n(the singular values of $J/R$);  \n• the azimuth, measured clockwise from true North, of the major\naxis of the Tissot indicatrix at $P^{\\star}$.\n\nc)  From $J$ derive the local linear scales  \n\\[\nk_{E}\\quad(\\text{true East–West}),\\qquad\nk_{N}\\quad(\\text{true North–South}),\\qquad\nk_{NE}\\quad(\\text{bearing }45^{\\circ}\\text{ east of North}),\n\\]\nthen give the ratios $k_{E}/k_{N}$ and $k_{E}/k_{NE}$ correct to\nfour significant digits.\n\nd)  The designer removes every map point whose planar distance\n$\\rho=\\sqrt{x^{2}+y^{2}}$ from $C$ exceeds $13\\,000\\text{ km}$.\nWill $P^{\\star}$ still appear on the logo? Give a quantitative argument.\n\ne)  A $100$-km geodesic segment at $P^{\\star}$ is oriented NE–SW.\nTo the nearest kilometre, what length will it have on the map?",
      "solution": "Throughout write $\\delta\\lambda=\\lambda-\\lambda_{0}$ for convenience\n\\emph{inside the calculations}; the final answers of part (a) will be\nre-expressed without this abbreviation.\n\nStep 1 – derivatives of $c$.  \nPut\n\\[\nu(\\varphi,\\lambda)=\\sin\\varphi_{0}\\sin\\varphi+\n\\cos\\varphi_{0}\\cos\\varphi\\cos\\delta\\lambda=\\cos c.\n\\]\nBecause $c=\\arccos u$,  \n\\[\n\\partial c=-\\frac{\\partial u}{\\sin c}.\n\\]\nDifferentiating $u$ gives\n\\begin{align}\nc_{\\varphi}&=\n\\frac{\\cos\\varphi_{0}\\sin\\varphi\\cos\\delta\\lambda-\\sin\\varphi_{0}\\cos\\varphi}\n{\\sin c},\\tag{2a}\\\\\nc_{\\lambda}&=\n\\frac{\\cos\\varphi_{0}\\cos\\varphi\\sin\\delta\\lambda}{\\sin c}.\\tag{2b}\n\\end{align}\n\nStep 2 – derivatives of $A$.  \nDefine\n\\[\nB=\\cos\\varphi\\,\\sin\\delta\\lambda,\\qquad\nD=\\cos\\varphi_{0}\\sin\\varphi-\\sin\\varphi_{0}\\cos\\varphi\\cos\\delta\\lambda,\n\\qquad A=\\operatorname{atan2}(B,D).\n\\]\nBecause $B^{2}+D^{2}=(\\sin c)^{2}$, straightforward differentiation\nyields\n\\begin{align}\nA_{\\varphi}&=\n-\\frac{\\cos\\varphi_{0}\\sin\\delta\\lambda}{\\sin^{2}c},\\tag{3a}\\\\\nA_{\\lambda}&=\n\\frac{\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos\\delta\\lambda-\n\\sin\\varphi_{0}\\cos\\varphi\\bigr)\\cos\\varphi}{\\sin^{2}c}.\\tag{3b}\n\\end{align}\n\nStep 3 – the Jacobian $J$.  \nFrom (1)\n\\[\n\\begin{aligned}\nx_{\\varphi}&=R\\bigl(c_{\\varphi}\\sin A+c\\cos A\\,A_{\\varphi}\\bigr),\\\\\nx_{\\lambda}&=R\\bigl(c_{\\lambda}\\sin A+c\\cos A\\,A_{\\lambda}\\bigr),\\\\\ny_{\\varphi}&=R\\bigl(c_{\\varphi}\\cos A-c\\sin A\\,A_{\\varphi}\\bigr),\\\\\ny_{\\lambda}&=R\\bigl(c_{\\lambda}\\cos A-c\\sin A\\,A_{\\lambda}\\bigr).\n\\end{aligned}\\tag{4}\n\\]\n\nSubstituting (2)–(3) and replacing $\\delta\\lambda$ by the explicit\ndifference $(\\lambda-\\lambda_{0})$ gives the fully compliant formulas\nrequired in part (a):\n\n\\[\n\\boxed{\n\\begin{aligned}\nx_{\\varphi}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\sin A\\sin c\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n-c\\cos A\\cos\\varphi_{0}\\sin(\\lambda-\\lambda_{0})\n\\bigr],\\\\[3pt]\nx_{\\lambda}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\sin A\\cos\\varphi_{0}\\cos\\varphi\\sin(\\lambda-\\lambda_{0})\\sin c\n-c\\cos A\\cos\\varphi\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n\\bigr],\\\\[3pt]\ny_{\\varphi}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\cos A\\sin c\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n+c\\sin A\\cos\\varphi_{0}\\sin(\\lambda-\\lambda_{0})\n\\bigr],\\\\[3pt]\ny_{\\lambda}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\cos A\\cos\\varphi_{0}\\cos\\varphi\\sin(\\lambda-\\lambda_{0})\\sin c\n+c\\sin A\\cos\\varphi\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n\\bigr].\n\\end{aligned}}\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nNumerical evaluation at $P^{\\star}\\;(\\varphi^{\\star}=-5^{\\circ},\\;\n\\lambda^{\\star}=100^{\\circ})$\n--------------------------------------------------------------------\n(All angles are first converted to radians.)\n\n\\[\n\\begin{aligned}\n\\varphi_{0}&=0.61086524,\\qquad &\\lambda_{0}&=0.69813170,\\\\\n\\varphi^{\\star}&=-0.08726646,\\qquad &\\lambda^{\\star}&=1.74532925,\\\\\n\\delta\\lambda^{\\star}&=1.04719755.\n\\end{aligned}\n\\]\n\nCentral angle and azimuth  \n\\[\n\\cos c^{\\star}=0.357667\\;\\Longrightarrow\\;\nc^{\\star}=1.20600\\text{ rad},\\qquad\n\\sin c^{\\star}=0.933799,\n\\]\n\\[\n\\sin A^{\\star}=0.927\\,,\\qquad\n\\cos A^{\\star}=-0.374\\,,\n\\quad\\Longrightarrow\\;A^{\\star}=112.2^{\\circ}.\n\\]\n\nUsing (2)–(4) (or directly (5)) one finds\n\n\\[\n\\frac{1}{R}\n\\begin{bmatrix}\nx_{\\varphi} & x_{\\lambda}\\\\\ny_{\\varphi} & y_{\\lambda}\n\\end{bmatrix}_{P^{\\star}}\n=\n\\begin{bmatrix}\n-0.22507 & 1.01837\\\\\n\\;1.15392 & 0.48209\n\\end{bmatrix}\\!.\\tag{6}\n\\]\n\n--------------------------------------------------------------------\nb)  Principal scales and indicatrix orientation\n--------------------------------------------------------------------\n(i)  Singular values (principal scales).  \n\\[\n\\frac{J^{\\mathsf T}J}{R^{2}}=\n\\begin{bmatrix}\n1.38219 & 0.32652\\\\\n0.32652 & 1.26949\n\\end{bmatrix},\n\\quad\n\\lambda_{1}=1.65375,\\;\n\\lambda_{2}=0.99794.\n\\]\nHence  \n\\[\nk_{\\max}=\\sqrt{\\lambda_{1}}=1.286,\\qquad\nk_{\\min}=\\sqrt{\\lambda_{2}}=0.999.\n\\]\n\n(ii)  Azimuth of the major axis with respect to \\emph{true North}.  \n\nThe image of the true-north tangent vector is  \n\\[\nJ\\,e_{N}=J(1,0)^{\\mathsf T}\n        =\\bigl(x_{\\varphi},y_{\\varphi}\\bigr)^{\\mathsf T}\n        =R\\,(-0.22507,\\,1.15392)^{\\mathsf T}.\n\\]\nHence the direction of true north in the map plane makes the angle  \n\\[\n\\theta_{N}=\\operatorname{atan2}(-0.22507,\\,1.15392)\n          =-11.0^{\\circ}\n\\]\nwith the positive $y$-axis.\n\nLet $v_{\\max}$ be a unit eigen-vector of $J^{\\mathsf T}J$ for\n$\\lambda_{1}$; one finds $v_{\\max}\\propto(1,0.8323)^{\\mathsf T}$.\nIts image in the map plane is  \n\\[\nw_{\\max}=Jv_{\\max}\\propto(0.622,\\,1.555)^{\\mathsf T},\n\\qquad\n\\theta_{w}=\\operatorname{atan2}(0.622,\\,1.555)=21.8^{\\circ}.\n\\]\n\nTherefore the azimuth of the major axis measured clockwise from \\emph{true\nnorth} is  \n\n\\[\n\\boxed{\\;\\theta_{w}-\\theta_{N}=21.8^{\\circ}-(-11.0^{\\circ})\n      =32.8^{\\circ}\\;}.\n\\]\n\n(The earlier version erroneously took $\\theta_{w}$ itself as the azimuth,\nimplicitly presuming that the $+y$-axis corresponds to true north\neverywhere, which is not the case for an oblique projection.)\n\n--------------------------------------------------------------------\nc)  Directional scales\n--------------------------------------------------------------------\nFor small displacements the spherical line element is\n${\\mathrm d}s^{2}=R^{2}\\bigl({\\mathrm d}\\varphi^{2}+\n\\cos^{2}\\varphi\\,{\\mathrm d}\\lambda^{2}\\bigr)$.\nUnit tangent vectors in $(\\varphi,\\lambda)$ space are therefore  \n\n\\[\ne_{N}=(1,0),\\qquad\ne_{E}=\\bigl(0,1/\\cos\\varphi^{\\star}\\bigr),\\qquad\ne_{NE}=\\frac{1}{\\sqrt{2}}\\Bigl(1,\\;1/\\cos\\varphi^{\\star}\\Bigr).\n\\]\n\nUsing (6)\n\n\\[\nk_{N}=||(J/R)e_{N}||=1.176,\\qquad\nk_{E}=||(J/R)e_{E}||=1.131,\\qquad\nk_{NE}=||(J/R)e_{NE}||=1.288.\n\\]\n\nThus, to four significant digits,\n\\[\n\\boxed{\\;\nk_{E}/k_{N}=0.9619,\\qquad\nk_{E}/k_{NE}=0.8780\n\\;}.\n\\]\n(Numerically $k_{NE}$ is virtually equal to $k_{\\max}$, even though its\ndirection differs from the exact major-axis azimuth by\nabout $23^{\\circ}$; the distortion ellipse is almost, but not quite,\ncircular.)\n\n--------------------------------------------------------------------\nd)  Clipping test\n--------------------------------------------------------------------\n\\[\n\\rho^{\\star}=R\\,c^{\\star}=6371\\text{ km}\\times 1.20600=7686\\text{ km}\n<13\\,000\\text{ km}.\n\\]\nHence $P^{\\star}$ lies inside the designer’s cut-off circle and\n\\emph{does} appear on the logo.\n\n--------------------------------------------------------------------\ne)  Length of a $100$-km NE–SW geodesic\n--------------------------------------------------------------------\nThe relevant local scale is $k_{NE}=1.288$.  \nTherefore the segment is drawn with length\n\\[\n\\text{map length}\\approx k_{NE}\\times100\\text{ km}=129\\text{ km}\n\\]\n(to the nearest kilometre).\n\n--------------------------------------------------------------------\nAnswer summary\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nk_{\\max}&\\approx1.286,\\qquad\nk_{\\min}\\approx0.999,\\\\\n\\text{major-axis azimuth (clockwise from N)}&\\approx32.8^{\\circ},\\\\\nk_{E}&\\approx1.131,\\;\nk_{N}\\approx1.176,\\;\nk_{NE}\\approx1.288,\\\\\nk_{E}/k_{N}&\\approx0.9619,\\;\nk_{E}/k_{NE}\\approx0.8780,\\\\\n\\rho^{\\star}&\\approx7686\\text{ km}\\ (<13\\,000\\text{ km}),\\\\\n\\text{100-km NE–SW segment}&\\longrightarrow\\;129\\text{ km (map).}\n\\end{aligned}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.445276",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional data:  the point of interest is not on a\n  polar circle; the projection is fully oblique, so both c and A depend\n  on ϕ and λ.  \n• Additional constraints: the problem demands principal scales,\n  Tissot-indicatrix orientation, several specific directional scales,\n  and a geometric trimming test.  \n• Mathematical sophistication: solving requires differential geometry\n  (Jacobian, first fundamental form), eigen-analysis, and careful use of\n  spherical trigonometry—not just the single ratio used in the original\n  problem.  \n• Interacting concepts: azimuthal-equidistant geometry, matrix calculus,\n  numerical evaluation, and navigation bearings all appear and must be\n  combined coherently.  \n• Workload: five substantial sub-tasks replace the single computation of\n  the earlier versions; each step builds on the previous ones and\n  significant algebraic as well as numerical effort is required."
      }
    },
    "original_kernel_variant": {
      "question": "A commemorative logo for an international cartographic congress depicts the Earth by means of the oblique azimuthal–equidistant projection whose centre  \n$C$ has geographic latitude $\\varphi_{0}=35^{\\circ}\\text{ N}$ and longitude $\\lambda_{0}=40^{\\circ}\\text{ E}$.  \nFor any point $P(\\varphi,\\lambda)$ on the spherical Earth (radius $R=6371\\text{ km}$) put  \n\\[\nc(\\varphi,\\lambda)=\\arccos\\!\\bigl[\\sin\\varphi_{0}\\sin\\varphi+\n\\cos\\varphi_{0}\\cos\\varphi\\cos\\bigl(\\lambda-\\lambda_{0}\\bigr)\\bigr],\\qquad\n0\\le c\\le\\pi,\n\\]\n\\[\nA(\\varphi,\\lambda)=\\operatorname{atan2}\\!\\Bigl(\n\\cos\\varphi\\,\\sin\\bigl(\\lambda-\\lambda_{0}\\bigr),\\;\n\\cos\\varphi_{0}\\sin\\varphi-\\sin\\varphi_{0}\\cos\\varphi\\cos\\bigl(\\lambda-\\lambda_{0}\\bigr)\n\\Bigr).\n\\]\n\nThe map–plane coordinates are  \n\\[\nx   =Rc\\sin A,\\qquad\ny   =Rc\\cos A.\\tag{1}\n\\]\n\na)  Obtain fully–simplified expressions for the four partial derivatives  \n\\[\nx_{\\varphi},\\;x_{\\lambda},\\;y_{\\varphi},\\;y_{\\lambda}\\qquad\n(\\text{i.e.\\ for the Jacobian }J=\\partial(x,y)/\\partial(\\varphi,\\lambda))\n\\]\nin terms only of the elementary functions of\n$\\varphi,\\lambda,\\varphi_{0},\\lambda_{0},c$ and $A$\n— \\emph{no extra shorthand symbols such as $s_{0}$, $\\bar c$, \n$\\Delta\\lambda$, etc.\\ may appear in the final formulas.}\n\nb)  At the point  \n\\[\nP^{\\star}:\\;\n\\varphi^{\\star}=-5^{\\circ}\\;(5^{\\circ}\\text{ S}),\\qquad\n\\lambda^{\\star}=100^{\\circ}\\text{ E},\n\\]\nevaluate $J$ numerically and compute  \n• the principal scale factors $k_{\\max}\\ge k_{\\min}$ \n(the singular values of $J/R$);  \n• the azimuth, measured clockwise from true North, of the major\naxis of the Tissot indicatrix at $P^{\\star}$.\n\nc)  From $J$ derive the local linear scales  \n\\[\nk_{E}\\quad(\\text{true East–West}),\\qquad\nk_{N}\\quad(\\text{true North–South}),\\qquad\nk_{NE}\\quad(\\text{bearing }45^{\\circ}\\text{ east of North}),\n\\]\nthen give the ratios $k_{E}/k_{N}$ and $k_{E}/k_{NE}$ correct to\nfour significant digits.\n\nd)  The designer removes every map point whose planar distance\n$\\rho=\\sqrt{x^{2}+y^{2}}$ from $C$ exceeds $13\\,000\\text{ km}$.\nWill $P^{\\star}$ still appear on the logo? Give a quantitative argument.\n\ne)  A $100$-km geodesic segment at $P^{\\star}$ is oriented NE–SW.\nTo the nearest kilometre, what length will it have on the map?\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout write $\\delta\\lambda=\\lambda-\\lambda_{0}$ for convenience\n\\emph{inside the calculations}; the final answers of part (a) will be\nre-expressed without this abbreviation.\n\nStep 1 – derivatives of $c$.  \nPut\n\\[\nu(\\varphi,\\lambda)=\\sin\\varphi_{0}\\sin\\varphi+\n\\cos\\varphi_{0}\\cos\\varphi\\cos\\delta\\lambda=\\cos c.\n\\]\nBecause $c=\\arccos u$,  \n\\[\n\\partial c=-\\frac{\\partial u}{\\sin c}.\n\\]\nDifferentiating $u$ gives\n\\begin{align}\nc_{\\varphi}&=\n\\frac{\\cos\\varphi_{0}\\sin\\varphi\\cos\\delta\\lambda-\\sin\\varphi_{0}\\cos\\varphi}\n{\\sin c},\\tag{2a}\\\\\nc_{\\lambda}&=\n\\frac{\\cos\\varphi_{0}\\cos\\varphi\\sin\\delta\\lambda}{\\sin c}.\\tag{2b}\n\\end{align}\n\nStep 2 – derivatives of $A$.  \nDefine\n\\[\nB=\\cos\\varphi\\,\\sin\\delta\\lambda,\\qquad\nD=\\cos\\varphi_{0}\\sin\\varphi-\\sin\\varphi_{0}\\cos\\varphi\\cos\\delta\\lambda,\n\\qquad A=\\operatorname{atan2}(B,D).\n\\]\nBecause $B^{2}+D^{2}=(\\sin c)^{2}$, straightforward differentiation\nyields\n\\begin{align}\nA_{\\varphi}&=\n-\\frac{\\cos\\varphi_{0}\\sin\\delta\\lambda}{\\sin^{2}c},\\tag{3a}\\\\\nA_{\\lambda}&=\n\\frac{\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos\\delta\\lambda-\n\\sin\\varphi_{0}\\cos\\varphi\\bigr)\\cos\\varphi}{\\sin^{2}c}.\\tag{3b}\n\\end{align}\n\nStep 3 – the Jacobian $J$.  \nFrom (1)\n\\[\n\\begin{aligned}\nx_{\\varphi}&=R\\bigl(c_{\\varphi}\\sin A+c\\cos A\\,A_{\\varphi}\\bigr),\\\\\nx_{\\lambda}&=R\\bigl(c_{\\lambda}\\sin A+c\\cos A\\,A_{\\lambda}\\bigr),\\\\\ny_{\\varphi}&=R\\bigl(c_{\\varphi}\\cos A-c\\sin A\\,A_{\\varphi}\\bigr),\\\\\ny_{\\lambda}&=R\\bigl(c_{\\lambda}\\cos A-c\\sin A\\,A_{\\lambda}\\bigr).\n\\end{aligned}\\tag{4}\n\\]\n\nSubstituting (2)–(3) and replacing $\\delta\\lambda$ by the explicit\ndifference $(\\lambda-\\lambda_{0})$ gives the fully compliant formulas\nrequired in part (a):\n\n\\[\n\\boxed{\n\\begin{aligned}\nx_{\\varphi}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\sin A\\sin c\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n-c\\cos A\\cos\\varphi_{0}\\sin(\\lambda-\\lambda_{0})\n\\bigr],\\\\[3pt]\nx_{\\lambda}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\sin A\\cos\\varphi_{0}\\cos\\varphi\\sin(\\lambda-\\lambda_{0})\\sin c\n-c\\cos A\\cos\\varphi\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n\\bigr],\\\\[3pt]\ny_{\\varphi}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\cos A\\sin c\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n+c\\sin A\\cos\\varphi_{0}\\sin(\\lambda-\\lambda_{0})\n\\bigr],\\\\[3pt]\ny_{\\lambda}&=\\frac{R}{\\sin^{2}c}\n\\bigl[\n\\cos A\\cos\\varphi_{0}\\cos\\varphi\\sin(\\lambda-\\lambda_{0})\\sin c\n+c\\sin A\\cos\\varphi\\bigl(\\cos\\varphi_{0}\\sin\\varphi\\cos(\\lambda-\\lambda_{0})\n-\\sin\\varphi_{0}\\cos\\varphi\\bigr)\n\\bigr].\n\\end{aligned}}\n\\tag{5}\n\\]\n\n--------------------------------------------------------------------\nNumerical evaluation at $P^{\\star}\\;(\\varphi^{\\star}=-5^{\\circ},\\;\n\\lambda^{\\star}=100^{\\circ})$\n--------------------------------------------------------------------\n(All angles are first converted to radians.)\n\n\\[\n\\begin{aligned}\n\\varphi_{0}&=0.61086524,\\qquad &\\lambda_{0}&=0.69813170,\\\\\n\\varphi^{\\star}&=-0.08726646,\\qquad &\\lambda^{\\star}&=1.74532925,\\\\\n\\delta\\lambda^{\\star}&=1.04719755.\n\\end{aligned}\n\\]\n\nCentral angle and azimuth  \n\\[\n\\cos c^{\\star}=0.357667\\;\\Longrightarrow\\;\nc^{\\star}=1.20600\\text{ rad},\\qquad\n\\sin c^{\\star}=0.933799,\n\\]\n\\[\n\\sin A^{\\star}=0.927\\,\\; ,\\qquad\n\\cos A^{\\star}=-0.374\\,\\; ,\n\\; \\Longrightarrow A^{\\star}=112.2^{\\circ}.\n\\]\n\nUsing (2)–(4) (or directly (5)) one finds\n\n\\[\n\\frac{1}{R}\n\\begin{bmatrix}\nx_{\\varphi} & x_{\\lambda}\\\\\ny_{\\varphi} & y_{\\lambda}\n\\end{bmatrix}_{P^{\\star}}\n=\n\\begin{bmatrix}\n-0.22507 & 1.01837\\\\\n\\;1.15392 & 0.48209\n\\end{bmatrix}\\!.\\tag{6}\n\\]\n\n--------------------------------------------------------------------\nb)  Principal scales and indicatrix orientation\n--------------------------------------------------------------------\n(i)  Singular values.  \n\\[\n\\frac{J^{\\mathsf T}J}{R^{2}}=\n\\begin{bmatrix}\n1.38219 & 0.32652\\\\\n0.32652 & 1.26949\n\\end{bmatrix},\n\\quad\n\\lambda_{1}=1.65375,\\;\n\\lambda_{2}=0.99794.\n\\]\nHence  \n\\[\nk_{\\max}=\\sqrt{\\lambda_{1}}=1.286\\;,\\qquad\nk_{\\min}=\\sqrt{\\lambda_{2}}=0.999.\n\\]\n\n(ii)  Orientation on the map plane.  \nAn eigen-vector of $J^{\\mathsf T}J$ for $\\lambda_{1}$ is\n$v_{\\max}\\propto(1,0.8323)^{\\mathsf T}$.\nMapping it to the plane,\n\\[\nw_{\\max}=Jv_{\\max}\\propto(0.622,\\,1.555)^{\\mathsf T}.\n\\]\nThe azimuth (clockwise from the $+y$–axis, i.e.\\ true North) is  \n\\[\n\\theta^{\\star}=\\operatorname{atan2}(w_{x},w_{y})\n              =\\operatorname{atan2}(0.622,1.555)\n              =21.8^{\\circ}.\n\\]\nA direct dot-product check confirms that $w_{\\max}$ is orthogonal\nto the radial direction $(\\sin A^{\\star},\\cos A^{\\star})$ on the map,\nas is characteristic of the azimuthal–equidistant projection.\n\n--------------------------------------------------------------------\nc)  Directional scales\n--------------------------------------------------------------------\nFor small displacements the spherical line element is\n${\\mathrm d}s^{2}=R^{2}\\bigl({\\mathrm d}\\varphi^{2}+\n\\cos^{2}\\varphi\\,{\\mathrm d}\\lambda^{2}\\bigr)$.\nUnit tangent vectors in $(\\varphi,\\lambda)$ space are therefore  \n\n\\[\ne_{N}=(1,0),\\qquad\ne_{E}=\\bigl(0,1/\\cos\\varphi^{\\star}\\bigr),\\qquad\ne_{NE}=\\frac{1}{\\sqrt{2}}\\Bigl(1,\\;1/\\cos\\varphi^{\\star}\\Bigr).\n\\]\n\nUsing (6)\n\n\\[\nk_{N}=||(J/R)e_{N}||=1.176,\\qquad\nk_{E}=||(J/R)e_{E}||=1.131,\\qquad\nk_{NE}=||(J/R)e_{NE}||=1.288.\n\\]\n\nThus, to four significant digits,\n\\[\n\\boxed{\\;\nk_{E}/k_{N}=0.9619,\\qquad\nk_{E}/k_{NE}=0.8780\n\\;}.\n\\]\n(Numerically $k_{NE}$ is virtually equal to $k_{\\max}$, even though its\ndirection differs from the exact major-axis azimuth by\nabout $23^{\\circ}$; the distortion ellipse is almost, but not quite,\ncircular.)\n\n--------------------------------------------------------------------\nd)  Clipping test\n--------------------------------------------------------------------\n\\[\n\\rho^{\\star}=Rc^{\\star}=6371\\text{ km}\\times 1.20600=7686\\text{ km}\n<13\\,000\\text{ km}.\n\\]\nHence $P^{\\star}$ lies inside the designer’s cut-off circle and\n\\emph{does} appear on the logo.\n\n--------------------------------------------------------------------\ne)  Length of a $100$-km NE–SW geodesic\n--------------------------------------------------------------------\nThe relevant local scale is $k_{NE}=1.288$.  \nTherefore the segment is drawn with length\n\\[\n\\text{map length}\\approx k_{NE}\\times100\\text{ km}=129\\text{ km}\n\\]\n(to the nearest kilometre).\n\n--------------------------------------------------------------------\nAnswer summary\n--------------------------------------------------------------------\n\\[\n\\begin{aligned}\nk_{\\max}&\\approx1.286,\\qquad\nk_{\\min}\\approx0.999,\\\\\n\\text{major-axis azimuth}&\\approx21.8^{\\circ}\\text{ E of N},\\\\\nk_{E}&\\approx1.131,\\;\nk_{N}\\approx1.176,\\;\nk_{NE}\\approx1.288,\\\\\nk_{E}/k_{N}&\\approx0.9619,\\;\nk_{E}/k_{NE}\\approx0.8780,\\\\\n\\rho^{\\star}&\\approx7686\\text{ km}\\ (<13\\,000\\text{ km}),\\\\\n\\text{100-km NE–SW segment}&\\longrightarrow\\;129\\text{ km (map).}\n\\end{aligned}\n\\]\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.384252",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional data:  the point of interest is not on a\n  polar circle; the projection is fully oblique, so both c and A depend\n  on ϕ and λ.  \n• Additional constraints: the problem demands principal scales,\n  Tissot-indicatrix orientation, several specific directional scales,\n  and a geometric trimming test.  \n• Mathematical sophistication: solving requires differential geometry\n  (Jacobian, first fundamental form), eigen-analysis, and careful use of\n  spherical trigonometry—not just the single ratio used in the original\n  problem.  \n• Interacting concepts: azimuthal-equidistant geometry, matrix calculus,\n  numerical evaluation, and navigation bearings all appear and must be\n  combined coherently.  \n• Workload: five substantial sub-tasks replace the single computation of\n  the earlier versions; each step builds on the previous ones and\n  significant algebraic as well as numerical effort is required."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}