{
  "index": "1978-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem B-1\nFind the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form \\( r+s \\sqrt{t} \\) with \\( r, s \\), and \\( t \\) positive integers.",
  "solution": "B-1.\nThe area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure \\( 3 \\pi / 4 \\) and one can augment the octagon into a square with sides of length \\( 3+2 \\sqrt{2} \\) by properly placing a \\( \\sqrt{2}, \\sqrt{2}, 2 \\) isosceles right triangle on each of the sides of length 2 . Hence the desired area is\n\\[\n(3+2 \\sqrt{2})^{2}-(4 \\sqrt{2} \\cdot \\sqrt{2} / 2)=13+12 \\sqrt{2} .\n\\]\n\nA second solution follows. Let \\( r \\) be the radius of the circle and let \\( \\alpha \\) and \\( \\beta \\) be half of the central angles for the chords of lengths 3 and 2 , respectively. Then \\( 8 \\alpha+8 \\beta=2 \\pi \\) and so \\( \\beta=(\\pi / 4)-\\alpha \\). Also\n\\[\n\\begin{array}{c}\n\\frac{3}{2 r}=\\sin \\alpha, \\quad \\frac{1}{r}=\\sin \\beta=\\sin \\left(\\frac{\\pi}{4}-\\alpha\\right)=\\frac{\\cos \\alpha-\\sin \\alpha}{\\sqrt{2}}, \\\\\n\\frac{2}{3}=\\frac{2 r}{3} \\cdot \\frac{1}{r}=\\frac{\\cos \\alpha-\\sin \\alpha}{\\sqrt{2} \\sin \\alpha}=\\frac{\\cot \\alpha-1}{\\sqrt{2}} .\n\\end{array}\n\\]\n\nNow \\( \\cot \\alpha=(3+2 \\sqrt{2}) / 3=[(3+2 \\sqrt{2}) / 2] /(3 / 2) \\) and hence the distance from the center of the circle to a chord of length 3 is \\( h_{3}=(3+2 \\sqrt{2}) / 2 \\). Similarly the distance to a chord of length 2 is \\( h_{2}=(2+3 \\sqrt{2}) / 2 \\). Finally, the desired area is\n\\[\n4\\left(3 h_{3}+2 h_{2}\\right) / 2=(9+6 \\sqrt{2})+(4+6 \\sqrt{2})=13+12 \\sqrt{2} .\n\\]",
  "vars": [
    "r",
    "s",
    "t",
    "\\\\alpha",
    "\\\\beta",
    "h_3",
    "h_2"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "radiallen",
        "s": "coeffval",
        "t": "radicand",
        "\\alpha": "primeangle",
        "\\beta": "secondangle",
        "h_3": "heightthree",
        "h_2": "heighttwo"
      },
      "question": "Problem B-1\nFind the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form \\( radiallen+coeffval \\sqrt{radicand} \\) with \\( radiallen, coeffval \\), and \\( radicand \\) positive integers.",
      "solution": "B-1.\nThe area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure \\( 3 \\pi / 4 \\) and one can augment the octagon into a square with sides of length \\( 3+2 \\sqrt{2} \\) by properly placing a \\( \\sqrt{2}, \\sqrt{2}, 2 \\) isosceles right triangle on each of the sides of length 2 . Hence the desired area is\n\\[\n(3+2 \\sqrt{2})^{2}-(4 \\sqrt{2} \\cdot \\sqrt{2} / 2)=13+12 \\sqrt{2} .\n\\]\n\nA second solution follows. Let \\( radiallen \\) be the radius of the circle and let \\( primeangle \\) and \\( secondangle \\) be half of the central angles for the chords of lengths 3 and 2 , respectively. Then \\( 8 primeangle+8 secondangle=2 \\pi \\) and so \\( secondangle=(\\pi / 4)-primeangle \\). Also\n\\[\n\\begin{array}{c}\n\\frac{3}{2 radiallen}=\\sin primeangle, \\quad \\frac{1}{radiallen}=\\sin secondangle=\\sin \\left(\\frac{\\pi}{4}-primeangle\\right)=\\frac{\\cos primeangle-\\sin primeangle}{\\sqrt{2}}, \\\\\n\\frac{2}{3}=\\frac{2 radiallen}{3} \\cdot \\frac{1}{radiallen}=\\frac{\\cos primeangle-\\sin primeangle}{\\sqrt{2} \\sin primeangle}=\\frac{\\cot primeangle-1}{\\sqrt{2}} .\n\\end{array}\n\\]\n\nNow \\( \\cot primeangle=(3+2 \\sqrt{2}) / 3=[(3+2 \\sqrt{2}) / 2] /(3 / 2) \\) and hence the distance from the center of the circle to a chord of length 3 is \\( heightthree=(3+2 \\sqrt{2}) / 2 \\). Similarly the distance to a chord of length 2 is \\( heighttwo=(2+3 \\sqrt{2}) / 2 \\). Finally, the desired area is\n\\[\n4\\left(3 heightthree+2 heighttwo\\right) / 2=(9+6 \\sqrt{2})+(4+6 \\sqrt{2})=13+12 \\sqrt{2} .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "paperbloom",
        "s": "cobblestone",
        "t": "dragonfly",
        "\\alpha": "pebblestone",
        "\\beta": "moonlilac",
        "h_3": "buttercup",
        "h_2": "marigold"
      },
      "question": "Problem B-1\nFind the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form \\( \\mathrm{paperbloom}+\\mathrm{cobblestone} \\sqrt{\\mathrm{dragonfly}} \\) with \\( \\mathrm{paperbloom}, \\mathrm{cobblestone} \\), and \\( \\mathrm{dragonfly} \\) positive integers.",
      "solution": "B-1.\nThe area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure \\( 3 \\pi / 4 \\) and one can augment the octagon into a square with sides of length \\( 3+2 \\sqrt{2} \\) by properly placing a \\( \\sqrt{2}, \\sqrt{2}, 2 \\) isosceles right triangle on each of the sides of length 2. Hence the desired area is\n\\[\n(3+2 \\sqrt{2})^{2}-(4 \\sqrt{2} \\cdot \\sqrt{2} / 2)=13+12 \\sqrt{2} .\n\\]\n\nA second solution follows. Let \\( \\mathrm{paperbloom} \\) be the radius of the circle and let \\( \\mathrm{pebblestone} \\) and \\( \\mathrm{moonlilac} \\) be half of the central angles for the chords of lengths 3 and 2, respectively. Then \\( 8 \\mathrm{pebblestone}+8 \\mathrm{moonlilac}=2 \\pi \\) and so \\( \\mathrm{moonlilac}=(\\pi / 4)-\\mathrm{pebblestone} \\). Also\n\\[\n\\begin{array}{c}\n\\dfrac{3}{2 \\mathrm{paperbloom}}=\\sin \\mathrm{pebblestone}, \\quad \\dfrac{1}{\\mathrm{paperbloom}}=\\sin \\mathrm{moonlilac}=\\sin \\left(\\dfrac{\\pi}{4}-\\mathrm{pebblestone}\\right)=\\dfrac{\\cos \\mathrm{pebblestone}-\\sin \\mathrm{pebblestone}}{\\sqrt{2}}, \\\\\n\\dfrac{2}{3}=\\dfrac{2 \\mathrm{paperbloom}}{3} \\cdot \\dfrac{1}{\\mathrm{paperbloom}}=\\dfrac{\\cos \\mathrm{pebblestone}-\\sin \\mathrm{pebblestone}}{\\sqrt{2} \\sin \\mathrm{pebblestone}}=\\dfrac{\\cot \\mathrm{pebblestone}-1}{\\sqrt{2}} .\n\\end{array}\n\\]\n\nNow \\( \\cot \\mathrm{pebblestone}=(3+2 \\sqrt{2}) / 3=[(3+2 \\sqrt{2}) / 2] /(3 / 2) \\) and hence the distance from the center of the circle to a chord of length 3 is \\( \\mathrm{buttercup}=(3+2 \\sqrt{2}) / 2 \\). Similarly the distance to a chord of length 2 is \\( \\mathrm{marigold}=(2+3 \\sqrt{2}) / 2 \\). Finally, the desired area is\n\\[\n4\\left(3 \\mathrm{buttercup}+2 \\mathrm{marigold}\\right) / 2=(9+6 \\sqrt{2})+(4+6 \\sqrt{2})=13+12 \\sqrt{2} .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "perimeter",
        "s": "absence",
        "t": "simplicity",
        "\\alpha": "obtuseangle",
        "\\beta": "acuteangle",
        "h_3": "closeness",
        "h_2": "adjacency"
      },
      "question": "Problem B-1\nFind the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form \\( perimeter+absence \\sqrt{simplicity} \\) with \\( perimeter, absence \\), and \\( simplicity \\) positive integers.",
      "solution": "B-1.\nThe area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure \\( 3 \\pi / 4 \\) and one can augment the octagon into a square with sides of length \\( 3+2 \\sqrt{2} \\) by properly placing a \\( \\sqrt{2}, \\sqrt{2}, 2 \\) isosceles right triangle on each of the sides of length 2 . Hence the desired area is\n\\[\n(3+2 \\sqrt{2})^{2}-(4 \\sqrt{2} \\cdot \\sqrt{2} / 2)=13+12 \\sqrt{2} .\n\\]\n\nA second solution follows. Let \\( perimeter \\) be the radius of the circle and let \\( obtuseangle \\) and \\( acuteangle \\) be half of the central angles for the chords of lengths 3 and 2 , respectively. Then \\( 8 obtuseangle+8 acuteangle=2 \\pi \\) and so \\( acuteangle=(\\pi / 4)-obtuseangle \\). Also\n\\[\n\\begin{array}{c}\n\\frac{3}{2 perimeter}=\\sin obtuseangle, \\quad \\frac{1}{perimeter}=\\sin acuteangle=\\sin \\left(\\frac{\\pi}{4}-obtuseangle\\right)=\\frac{\\cos obtuseangle-\\sin obtuseangle}{\\sqrt{2}}, \\\\\n\\frac{2}{3}=\\frac{2 perimeter}{3} \\cdot \\frac{1}{perimeter}=\\frac{\\cos obtuseangle-\\sin obtuseangle}{\\sqrt{2} \\sin obtuseangle}=\\frac{\\cot obtuseangle-1}{\\sqrt{2}} .\n\\end{array}\n\\]\n\nNow \\( \\cot obtuseangle=(3+2 \\sqrt{2}) / 3=[(3+2 \\sqrt{2}) / 2] /(3 / 2) \\) and hence the distance from the center of the circle to a chord of length 3 is \\( closeness=(3+2 \\sqrt{2}) / 2 \\). Similarly the distance to a chord of length 2 is \\( adjacency=(2+3 \\sqrt{2}) / 2 \\). Finally, the desired area is\n\\[\n4\\left(3 closeness+2 adjacency\\right) / 2=(9+6 \\sqrt{2})+(4+6 \\sqrt{2})=13+12 \\sqrt{2} .\n\\]"
    },
    "garbled_string": {
      "map": {
        "r": "qzmdvktp",
        "s": "lkjsdmfa",
        "t": "rnbxovqc",
        "\\alpha": "hzgxqpwe",
        "\\beta": "wtnpqsla",
        "h_3": "kldjsmno",
        "h_2": "vpsndkqe"
      },
      "question": "Problem B-1\nFind the area of a convex octagon that is inscribed in a circle and has four consecutive sides of length 3 units and the remaining four sides of length 2 units. Give the answer in the form \\( qzmdvktp+lkjsdmfa \\sqrt{rnbxovqc} \\) with \\( qzmdvktp, lkjsdmfa \\), and \\( rnbxovqc \\) positive integers.",
      "solution": "B-1.\nThe area is the same as for an octagon inscribed in a circle and with sides alternately 3 units and 2 units in length. For such an octagon, all angles measure \\( 3 \\pi / 4 \\) and one can augment the octagon into a square with sides of length \\( 3+2 \\sqrt{2} \\) by properly placing a \\( \\sqrt{2}, \\sqrt{2}, 2 \\) isosceles right triangle on each of the sides of length 2 . Hence the desired area is\n\\[\n(3+2 \\sqrt{2})^{2}-(4 \\sqrt{2} \\cdot \\sqrt{2} / 2)=13+12 \\sqrt{2} .\n\\]\n\nA second solution follows. Let \\( qzmdvktp \\) be the radius of the circle and let \\( hzgxqpwe \\) and \\( wtnpqsla \\) be half of the central angles for the chords of lengths 3 and 2 , respectively. Then \\( 8 hzgxqpwe+8 wtnpqsla=2 \\pi \\) and so \\( wtnpqsla=(\\pi / 4)-hzgxqpwe \\). Also\n\\[\n\\begin{array}{c}\n\\frac{3}{2 qzmdvktp}=\\sin hzgxqpwe, \\quad \\frac{1}{qzmdvktp}=\\sin wtnpqsla=\\sin \\left(\\frac{\\pi}{4}-hzgxqpwe\\right)=\\frac{\\cos hzgxqpwe-\\sin hzgxqpwe}{\\sqrt{2}}, \\\\\n\\frac{2}{3}=\\frac{2 qzmdvktp}{3} \\cdot \\frac{1}{qzmdvktp}=\\frac{\\cos hzgxqpwe-\\sin hzgxqpwe}{\\sqrt{2} \\sin hzgxqpwe}=\\frac{\\cot hzgxqpwe-1}{\\sqrt{2}} .\n\\end{array}\n\\]\n\nNow \\( \\cot hzgxqpwe=(3+2 \\sqrt{2}) / 3=[(3+2 \\sqrt{2}) / 2] /(3 / 2) \\) and hence the distance from the center of the circle to a chord of length 3 is \\( kldjsmno=(3+2 \\sqrt{2}) / 2 \\). Similarly the distance to a chord of length 2 is \\( vpsndkqe=(2+3 \\sqrt{2}) / 2 \\). Finally, the desired area is\n\\[\n4\\left(3 kldjsmno+2 vpsndkqe\\right) / 2=(9+6 \\sqrt{2})+(4+6 \\sqrt{2})=13+12 \\sqrt{2} .\n\\]"
    },
    "kernel_variant": {
      "question": "Let a convex octagon be inscribed in a circle.  Suppose that, in order around the octagon, the first four consecutive sides have length 5 and the remaining four consecutive sides have length 3.  Find the area of the octagon and express your answer in the form \\(r+s\\sqrt{t}\\) with positive integers \\(r,s,t.\\)",
      "solution": "Because the student’s key step (that one may reorder the sides arbitrarily) is false, we rederive from first principles.  Let R be the circumradius, and let chords of length 5 subtend central angle 2α, those of length 3 subtend 2β.  Then\n  4(2α)+4(2β)=2π ⇒ α+β=π/4,\n  5=2R sin α,   3=2R sin β,\nand sin β=sin(π/4−α)=(cos α−sin α)/√2.  Solving gives\n  cot α=(5+3√2)/5,\n  cos α=(5+3√2)/√(68+30√2),\n  cos β=(5√2+3)/√(68+30√2),\n  R=5/(2 sin α)=√(68+30√2)/2.\n\nSplit the octagon into eight triangles apexed at the center.  Four have bases 5 and height R cos α, four have bases 3 and height R cos β.  Thus\n  Area = 4·½·5·(R cos α) + 4·½·3·(R cos β)\n= 10·(R cos α) + 6·(R cos β).\nSince R cos α=(5+3√2)/2 and R cos β=(5√2+3)/2, this becomes\n  10·((5+3√2)/2) + 6·((5√2+3)/2)\n= 5(5+3√2) + 3(5√2+3)\n= 25+15√2 +15√2+9\n= 34 + 30√2.\n\nHence the area is 34 + 30√2.",
      "_meta": {
        "core_steps": [
          "Replace the given ordering of sides by an alternating 3-,2-,3-,2-… pattern; a cyclic polygon’s area depends only on its side lengths, not their order.",
          "Show that a cyclic octagon whose sides alternate between two lengths is equiangular, with each interior angle equal to 3π⁄4 (135°).",
          "Because every interior angle is 135°, each side of length 2 is the hypotenuse of an external isosceles right triangle; attaching those four triangles turns the octagon into a square.",
          "Compute:  square-area  = (3+2√2)²,  triangles-area = 4·(√2·√2)/2;  subtract to obtain 13+12√2."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Length of the longer set of four sides.",
            "original": "3"
          },
          "slot2": {
            "description": "Length of the shorter set of four sides.",
            "original": "2"
          },
          "slot3": {
            "description": "Initial positioning of the equal sides (stated as 'four consecutive' in the problem; can be rearranged to alternate without affecting the argument).",
            "original": "four consecutive 3’s followed by four consecutive 2’s"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}