{
  "index": "1969-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "\\text { B-4. Show that any curve of unit length can be covered by a closed rectangle of area } 1 / 4 \\text {. }",
  "solution": "B-4 Place the curve so that its endpoints lie on the \\( x \\)-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be \\( a \\) and \\( b \\) respectively. Let \\( P_{0} \\) and \\( P_{6} \\) be the endpoints of the curve, and let \\( P_{1}, P_{2}, P_{3} \\), and \\( P_{4} \\) be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line \\( P_{0} P_{1} P_{2} P_{3} P_{4} P_{5} \\). This line has length at most one. The horizontal components of the segments of this broken line add up at least to \\( a \\), since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least \\( 2 b \\) since we start and finish on the \\( x \\)-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least \\( \\left(a^{2}+4 b^{2}\\right)^{1 / 2} \\).\n\nWe now have that \\( a \\) and \\( b \\) both lie between 0 and 1 and that \\( a^{2}+4 b^{2} \\leqq 1 \\). Under these conditions the product \\( a b \\) is a maximum for \\( a=\\frac{1}{2} \\sqrt{ } 2, b=\\frac{1}{a} \\sqrt{ } 2 \\) and so the maximum of \\( a b \\) is \\( \\frac{1}{4} \\). Thus the area of the rectangle we have constructed is at most \\( \\frac{1}{4} \\).",
  "vars": [
    "x",
    "a",
    "b",
    "P_0",
    "P_1",
    "P_2",
    "P_3",
    "P_4",
    "P_5",
    "P_6"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horzaxis",
        "a": "widthdim",
        "b": "heightdm",
        "P_0": "pointzero",
        "P_1": "pointonee",
        "P_2": "pointtwoo",
        "P_3": "pointthree",
        "P_4": "pointfourr",
        "P_5": "pointfivee",
        "P_6": "pointsixxx"
      },
      "question": "\\text { B-4. Show that any curve of unit length can be covered by a closed rectangle of area } 1 / 4 \\text {. }",
      "solution": "B-4 Place the curve so that its endpoints lie on the \\( horzaxis \\)-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be \\( widthdim \\) and \\( heightdm \\) respectively. Let \\( pointzero \\) and \\( pointsixxx \\) be the endpoints of the curve, and let \\( pointonee, pointtwoo, pointthree, \\) and \\( pointfourr \\) be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line \\( pointzero\\ pointonee\\ pointtwoo\\ pointthree\\ pointfourr\\ pointfivee \\). This line has length at most one. The horizontal components of the segments of this broken line add up at least to \\( widthdim \\), since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least \\( 2 heightdm \\) since we start and finish on the \\( horzaxis \\)-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least \\( \\left(widthdim^{2}+4 heightdm^{2}\\right)^{1 / 2} \\).\n\nWe now have that \\( widthdim \\) and \\( heightdm \\) both lie between 0 and 1 and that \\( widthdim^{2}+4 heightdm^{2} \\leqq 1 \\). Under these conditions the product \\( widthdim heightdm \\) is a maximum for \\( widthdim=\\frac{1}{2} \\sqrt{ } 2, heightdm=\\frac{1}{widthdim} \\sqrt{ } 2 \\) and so the maximum of \\( widthdim heightdm \\) is \\( \\frac{1}{4} \\). Thus the area of the rectangle we have constructed is at most \\( \\frac{1}{4} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "harmonic",
        "a": "vaportrail",
        "b": "lanternfly",
        "P_0": "sunflower",
        "P_1": "thunderclap",
        "P_2": "moonlitpath",
        "P_3": "crystalwave",
        "P_4": "emberglade",
        "P_5": "shadowbrook",
        "P_6": "mistyforest"
      },
      "question": "\\text { B-4. Show that any curve of unit length can be covered by a closed rectangle of area } 1 / 4 \\text {. }",
      "solution": "B-4 Place the curve so that its endpoints lie on the \\( harmonic \\)-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be \\( vaportrail \\) and \\( lanternfly \\) respectively. Let \\( sunflower \\) and \\( mistyforest \\) be the endpoints of the curve, and let \\( thunderclap, moonlitpath, crystalwave \\), and \\( emberglade \\) be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line \\( sunflower\\, thunderclap\\, moonlitpath\\, crystalwave\\, emberglade\\, shadowbrook \\). This line has length at most one. The horizontal components of the segments of this broken line add up at least to \\( vaportrail \\), since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least \\( 2 lanternfly \\) since we start and finish on the \\( harmonic \\)-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least \\( \\left(vaportrail^{2}+4 lanternfly^{2}\\right)^{1 / 2} \\).\n\nWe now have that \\( vaportrail \\) and \\( lanternfly \\) both lie between 0 and 1 and that \\( vaportrail^{2}+4 lanternfly^{2} \\leqq 1 \\). Under these conditions the product \\( vaportrail \\, lanternfly \\) is a maximum for \\( vaportrail=\\frac{1}{2} \\sqrt{ } 2, lanternfly=\\frac{1}{vaportrail} \\sqrt{ } 2 \\) and so the maximum of \\( vaportrail \\, lanternfly \\) is \\( \\frac{1}{4} \\). Thus the area of the rectangle we have constructed is at most \\( \\frac{1}{4} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "a": "narrowheight",
        "b": "widewidth",
        "P_0": "endlesszero",
        "P_1": "endlessone",
        "P_2": "endlesstwo",
        "P_3": "endlessthree",
        "P_4": "endlessfour",
        "P_5": "endlessfive",
        "P_6": "endlesssix"
      },
      "question": "\\text { B-4. Show that any curve of unit length can be covered by a closed rectangle of area } 1 / 4 \\text {. }",
      "solution": "B-4 Place the curve so that its endpoints lie on the \\( verticalaxis \\)-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be \\( narrowheight \\) and \\( widewidth \\) respectively. Let \\( endlesszero \\) and \\( endlesssix \\) be the endpoints of the curve, and let \\( endlessone, endlesstwo, endlessthree \\), and \\( endlessfour \\) be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line \\( endlesszero endlessone endlesstwo endlessthree endlessfour endlessfive \\). This line has length at most one. The horizontal components of the segments of this broken line add up at least to \\( narrowheight \\), since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least \\( 2 widewidth \\) since we start and finish on the \\( verticalaxis \\)-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least \\( \\left(narrowheight^{2}+4 widewidth^{2}\\right)^{1 / 2} \\).\n\nWe now have that \\( narrowheight \\) and \\( widewidth \\) both lie between 0 and 1 and that \\( narrowheight^{2}+4 widewidth^{2} \\leqq 1 \\). Under these conditions the product \\( narrowheight widewidth \\) is a maximum for \\( narrowheight=\\frac{1}{2} \\sqrt{ } 2, widewidth=\\frac{1}{narrowheight} \\sqrt{ } 2 \\) and so the maximum of \\( narrowheight widewidth \\) is \\( \\frac{1}{4} \\). Thus the area of the rectangle we have constructed is at most \\( \\frac{1}{4} \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "mcfpluoq",
        "P_0": "zdbrtlef",
        "P_1": "vqmsnhdz",
        "P_2": "lkpgrtae",
        "P_3": "rwcfxmbo",
        "P_4": "typqsdni",
        "P_5": "swhzmkjo",
        "P_6": "gphlrtva"
      },
      "question": "\\text { B-4. Show that any curve of unit length can be covered by a closed rectangle of area } 1 / 4 \\text {. }",
      "solution": "B-4 Place the curve so that its endpoints lie on the \\( qzxwvtnp \\)-axis. Then take the smallest rectangle with sides parallel to the axes which covers the curve. Let its horizontal and vertical dimensions be \\( hjgrksla \\) and \\( mcfpluoq \\) respectively. Let \\( zdbrtlef \\) and \\( gphlrtva \\) be the endpoints of the curve, and let \\( vqmsnhdz, lkpgrtae, rwcfxmbo, \\) and \\( typqsdni \\) be the points on the curve, in the order named, which lie one on each of the four sides of the rectangle. Draw the broken line \\( zdbrtlef vqmsnhdz lkpgrtae rwcfxmbo typqsdni swhzmkjo \\). This line has length at most one. The horizontal components of the segments of this broken line add up at least to \\( hjgrksla \\), since one of the vertices of the broken line lies on the left end of the rectangle and one on the right end. The vertical segments add to at least \\( 2 mcfpluoq \\) since we start and finish on the \\( qzxwvtnp \\)-axis and go to both the top and bottom sides. This implies that the total length of the broken line is at least \\( \\left(hjgrksla^{2}+4 mcfpluoq^{2}\\right)^{1 / 2} \\).\n\nWe now have that \\( hjgrksla \\) and \\( mcfpluoq \\) both lie between 0 and 1 and that \\( hjgrksla^{2}+4 mcfpluoq^{2} \\leqq 1 \\). Under these conditions the product \\( hjgrksla mcfpluoq \\) is a maximum for \\( hjgrksla=\\frac{1}{2} \\sqrt{ } 2, mcfpluoq=\\frac{1}{hjgrksla} \\sqrt{ } 2 \\) and so the maximum of \\( hjgrksla mcfpluoq \\) is \\( \\frac{1}{4} \\). Thus the area of the rectangle we have constructed is at most \\( \\frac{1}{4} \\)."
    },
    "kernel_variant": {
      "question": "Let  \\(\\Gamma\\) be a plane curve of total length \\(2\\).  Prove that there exists a rigid motion of the plane (i.e. a rotation followed by a translation) which moves \\(\\Gamma\\) into a closed rectangle of area at most \\(1\\) whose sides are parallel to the two fixed lines \\(y=x\\) and \\(y=-x\\).",
      "solution": "We keep the spirit of the classical solution to Putnam B–4, but we never alter the shape of the given curve – every step uses the very same curve after a single rigid motion.\n\nStep 1.  Choose convenient axes.\nIntroduce the orthogonal coordinates\n\\[\n   u=\\tfrac{1}{\\sqrt2}(x+y), \\qquad v=\\tfrac{1}{\\sqrt2}(x-y).\n\\]\nThe \\(u\\!\\)-axis is the line \\(y=x\\) and the \\(v\\!\\)-axis the line \\(y=-x\\).  A rectangle whose edges are parallel to those axes is therefore exactly a rectangle whose edges are parallel to the two prescribed lines.\n\nStep 2.  A single rigid motion that fixes the two end–points at the same height.\nLet \\(A\\) and \\(B\\) be the two end–points of a parametrisation of \\(\\Gamma\\).  (If the curve is closed, simply choose any point on it as both the starting and ending point; the argument below still works.)\n\n•  If \\(A\\ne B\\), rotate the plane so that the segment \\(AB\\) becomes parallel to the \\(u\\)-axis (i.e. to the vector \\((1,1)\\)).  In the rotated position the \\(v\\)-coordinates of \\(A\\) and \\(B\\) coincide, because the \\(v\\)-direction is perpendicular to \\(AB\\).\n\n•  If \\(A=B\\) (a closed curve), perform any rotation – its choice is irrelevant – the two coincident end–points automatically share the same \\(v\\)-coordinate.\n\nFinally translate the rotated plane parallel to the \\(v\\)-axis until that common \\(v\\)-coordinate becomes zero.  The composition of the rotation and this translation is a single rigid motion.  From now on we assume that this rigid motion has already been applied and therefore\n\\[\n   v(A)=v(B)=0.\\tag{1}\n\\]\n\nStep 3.  The minimal enclosing rectangle.\nLet\n\\[\n   u_{\\min}=\\min_{P\\in\\Gamma} u(P), \\; u_{\\max}=\\max_{P\\in\\Gamma} u(P),\n\\qquad\n   v_{\\min}=\\min_{P\\in\\Gamma} v(P), \\; v_{\\max}=\\max_{P\\in\\Gamma} v(P).\n\\]\nSet\n\\[\n   a:=u_{\\max}-u_{\\min}\\ge 0,\\qquad b:=v_{\\max}-v_{\\min}\\ge 0.\n\\]\nThe closed rectangle\n\\[\n   R=[u_{\\min},u_{\\max}]\\times[v_{\\min},v_{\\max}]\n\\]\nwith sides parallel to the \\(u\\)- and \\(v\\)-axes is the smallest such rectangle that contains \\(\\Gamma\\).  Observe that, because of (1),\n\\[\n   v_{\\min}\\le 0\\le v_{\\max}.\\tag{2}\n\\]\n\nStep 4.  A broken-line contained in \\(\\Gamma\\).\nTraverse \\(\\Gamma\\) once, starting at \\(A=P_{0}\\) and ending at \\(B=P_{5}\\).  Define successively\n• \\(P_{1}\\) = the first point encountered on the left side \\(u=u_{\\min}\\),\n• \\(P_{2}\\) = the first point thereafter on the top side \\(v=v_{\\max}\\),\n• \\(P_{3}\\) = the first point thereafter on the right side \\(u=u_{\\max}\\),\n• \\(P_{4}\\) = the first point thereafter on the bottom side \\(v=v_{\\min}\\).\n\nThe polygonal path\n\\[\n   P_{0}P_{1}P_{2}P_{3}P_{4}P_{5}\n\\]\nlies entirely on \\(\\Gamma\\); hence its length does not exceed the length of \\(\\Gamma\\), namely\n\\[\n   \\sum_{i=0}^{4}|P_{i}P_{i+1}| \\le 2.\\tag{3}\n\\]\nDenote the coordinate differences along each segment by\n\\[\n   \\Delta u_{i}=u(P_{i+1})-u(P_{i}),\\quad \\Delta v_{i}=v(P_{i+1})-v(P_{i}).\n\\]\nBecause the path reaches both vertical sides, its total horizontal excursion satisfies\n\\[\n   \\sum_{i=0}^{4}|\\Delta u_{i}|\\;\\ge\\;a.\\tag{4}\n\\]\nSimilarly, starting and finishing at the level \\(v=0\\) and visiting both the top and the bottom sides forces a total vertical travel of at least\n\\[\n   \\underbrace{v_{\\max}}_{0\\to v_{\\max}}+\\underbrace{(v_{\\max}-v_{\\min})}_{v_{\\max}\\to v_{\\min}}+\\underbrace{(-v_{\\min})}_{v_{\\min}\\to 0}=2(v_{\\max}-v_{\\min})=2b,\n\\]\nso\n\\[\n   \\sum_{i=0}^{4}|\\Delta v_{i}|\\;\\ge\\;2b.\\tag{5}\n\\]\nApplying the triangle (or Minkowski) inequality in \\(\\mathbb R^{2}\\) to the five vectors \\((\\Delta u_{i},\\,\\Delta v_{i})\\) gives\n\\[\n   \\sum_{i=0}^{4}\\sqrt{\\,\\Delta u_{i}^{2}+\\Delta v_{i}^{2}\\,}\n     \\;\\ge\\; \\sqrt{\\left(\\sum|\\Delta u_{i}|\\right)^{2}+\\left(\\sum|\\Delta v_{i}|\\right)^{2}}\n     \\;\\ge\\; \\sqrt{\\,a^{2}+4b^{2}\\,}.\\tag{6}\n\\]\nCombining (3) and (6) we obtain the fundamental inequality\n\\[\n   a^{2}+4b^{2}\\;\\le\\;4.\\tag{7}\n\\]\n\nStep 5.  Maximising the area under the constraint.\nThe area of the rectangle is \\(A=ab\\).  With \\(a,b\\ge0\\) and (7)\n\\[\n   2ab \\le \\frac{a^{2}+4b^{2}}{2} \\le \\frac{4}{2}=2, \\qquad\\Longrightarrow\\qquad ab\\le1.\n\\]\nThus the rectangle \\(R\\) that contains the moved curve \\(\\Gamma\\) has area at most \\(1\\), exactly what had to be shown.\n\nTherefore, after one suitable rigid motion, the curve of length 2 fits inside a rectangle of area \\(1\\) whose sides are parallel to the fixed lines \\(y=x\\) and \\(y=-x\\).",
      "_meta": {
        "core_steps": [
          "Cover the curve with the smallest axis-aligned rectangle; call its width a and height b.",
          "Pick one point of the curve on each side and join them in order to form a broken line lying on the curve (length ≤ total curve length).",
          "Horizontal projections of that line sum to ≥ a, vertical projections to ≥ 2b, hence (a^2 + 4b^2)^{1/2} ≤ curve length.",
          "Maximize ab subject to a^2 + 4b^2 ≤ (curve length)^2; the optimum occurs at b = a/2, giving ab ≤ (curve length)^2 / 4.",
          "Therefore the covering rectangle has area ≤ (curve length)^2 / 4 ( = 1/4 for a unit-length curve)."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Total length of the curve; every inequality and the final bound scale with this value.",
            "original": "1"
          },
          "slot2": {
            "description": "Resulting maximal rectangle area, which is the square of slot1 divided by 4.",
            "original": "1/4"
          },
          "slot3": {
            "description": "Choice of coordinate reference so that the rectangle’s sides are parallel to the chosen axes.",
            "original": "“parallel to the (x,y)-axes”"
          },
          "slot4": {
            "description": "Placing both endpoints of the curve on one chosen reference line to guarantee a vertical travel of 2b.",
            "original": "“on the x-axis”"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}