{
  "index": "1963-A-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "1. (i) Show that a regular hexagon, six squares, and six equilateral triangles can be assembled without overlapping to form a regular dodecagon.\n(ii) Let \\( P_{1}, P_{2}, \\ldots, P_{12} \\) be the successive vertices of a regular dodecagon. Explain how the three diagonals \\( P_{1} P_{9}, P_{2} P_{11} \\), and \\( P_{4} P_{12} \\) intersect.",
  "solution": "Solution. (i) Let squares be drawn on the outside of each side of a regular hexagon of side \\( s \\). Then the wedge-shaped indentations have vertex angle \\( 60^{\\circ}\\left(=360^{\\circ}-90^{\\circ}-120^{\\circ}-90^{\\circ}\\right) \\), and adjacent sides of length \\( s \\). Hence they can be filled in with equilateral triangles of side \\( s \\). The resulting dodecagon has all sides of length \\( s \\) and all angles \\( 150^{\\circ}\\left(=90^{\\circ}+60^{\\circ}\\right) \\); hence it is regular.\n(ii) Letter the vertices as shown in the diagram. Since the regular dodecagon can be inscribed in a circle, \\( \\angle P_{1} P_{12} P_{4} \\) is half the central angle of arc \\( P_{1} P_{4} \\), which is clearly a quarter circle. Hence \\( \\angle P_{1} P_{12} P_{4}=45^{\\circ} \\), so \\( P_{12} P_{4} \\) is the diagonal of the square \\( P_{1} A B P_{12} \\) and passes through \\( A \\). Similarly \\( P_{1} P_{9} \\) passes through \\( B \\). Hence \\( P_{1} P_{9} \\) and \\( P_{4} P_{12} \\) both pass through the center of the square \\( P_{1} A B P_{12} \\). Now the hexagon \\( P_{1} P_{2} A B P_{11} P_{12} \\) is evidently symmetric about the line \\( P_{2} P_{11} \\), so \\( P_{2} P_{11} \\) also passes through the midpoint of the square \\( P_{1} A B P_{12} \\). Thus the three diagonals mentioned are concurrent.",
  "vars": [
    "P_1",
    "P_2",
    "P_4",
    "P_9",
    "P_11",
    "P_12",
    "A",
    "B"
  ],
  "params": [
    "s"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P_1": "vertexone",
        "P_2": "vertextwo",
        "P_4": "vertexfour",
        "P_9": "vertexnine",
        "P_11": "vertexeleven",
        "P_12": "vertextwelve",
        "A": "pointalpha",
        "B": "pointbeta",
        "s": "sidelength"
      },
      "question": "1. (i) Show that a regular hexagon, six squares, and six equilateral triangles can be assembled without overlapping to form a regular dodecagon.\n(ii) Let \\( vertexone, vertextwo, \\ldots, vertextwelve \\) be the successive vertices of a regular dodecagon. Explain how the three diagonals \\( vertexone vertexnine, vertextwo vertexeleven \\), and \\( vertexfour vertextwelve \\) intersect.",
      "solution": "Solution. (i) Let squares be drawn on the outside of each side of a regular hexagon of side \\( sidelength \\). Then the wedge-shaped indentations have vertex angle \\( 60^{\\circ}\\left(=360^{\\circ}-90^{\\circ}-120^{\\circ}-90^{\\circ}\\right) \\), and adjacent sides of length \\( sidelength \\). Hence they can be filled in with equilateral triangles of side \\( sidelength \\). The resulting dodecagon has all sides of length \\( sidelength \\) and all angles \\( 150^{\\circ}\\left(=90^{\\circ}+60^{\\circ}\\right) \\); hence it is regular.\n(ii) Letter the vertices as shown in the diagram. Since the regular dodecagon can be inscribed in a circle, \\( \\angle vertexone vertextwelve vertexfour \\) is half the central angle of arc \\( vertexone vertexfour \\), which is clearly a quarter circle. Hence \\( \\angle vertexone vertextwelve vertexfour=45^{\\circ} \\), so \\( vertextwelve vertexfour \\) is the diagonal of the square \\( vertexone pointalpha pointbeta vertextwelve \\) and passes through \\( pointalpha \\). Similarly \\( vertexone vertexnine \\) passes through \\( pointbeta \\). Hence \\( vertexone vertexnine \\) and \\( vertexfour vertextwelve \\) both pass through the center of the square \\( vertexone pointalpha pointbeta vertextwelve \\). Now the hexagon \\( vertexone vertextwo pointalpha pointbeta vertexeleven vertextwelve \\) is evidently symmetric about the line \\( vertextwo vertexeleven \\), so \\( vertextwo vertexeleven \\) also passes through the midpoint of the square \\( vertexone pointalpha pointbeta vertextwelve \\). Thus the three diagonals mentioned are concurrent."
    },
    "descriptive_long_confusing": {
      "map": {
        "P_{1}": "marigold",
        "P_{2}": "chrysanthemum",
        "P_{4}": "buttercup",
        "P_{9}": "snowflake",
        "P_{11}": "dawnlight",
        "P_{12}": "lavender",
        "A": "tangerine",
        "B": "pinecone",
        "s": "peppercorn"
      },
      "question": "Problem:\n<<<\n1. (i) Show that a regular hexagon, six squares, and six equilateral triangles can be assembled without overlapping to form a regular dodecagon.\n(ii) Let \\( marigold, chrysanthemum, \\ldots, lavender \\) be the successive vertices of a regular dodecagon. Explain how the three diagonals \\( marigold snowflake, chrysanthemum dawnlight \\), and \\( buttercup lavender \\) intersect.\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. (i) Let squares be drawn on the outside of each side of a regular hexagon of side \\( peppercorn \\). Then the wedge-shaped indentations have vertex angle \\( 60^{\\circ}\\left(=360^{\\circ}-90^{\\circ}-120^{\\circ}-90^{\\circ}\\right) \\), and adjacent sides of length \\( peppercorn \\). Hence they can be filled in with equilateral triangles of side \\( peppercorn \\). The resulting dodecagon has all sides of length \\( peppercorn \\) and all angles \\( 150^{\\circ}\\left(=90^{\\circ}+60^{\\circ}\\right) \\); hence it is regular.\n(ii) Letter the vertices as shown in the diagram. Since the regular dodecagon can be inscribed in a circle, \\( \\angle marigold lavender buttercup \\) is half the central angle of arc \\( marigold buttercup \\), which is clearly a quarter circle. Hence \\( \\angle marigold lavender buttercup=45^{\\circ} \\), so \\( lavender buttercup \\) is the diagonal of the square \\( marigold tangerine pinecone lavender \\) and passes through \\( tangerine \\). Similarly \\( marigold snowflake \\) passes through \\( pinecone \\). Hence \\( marigold snowflake \\) and \\( buttercup lavender \\) both pass through the center of the square \\( marigold tangerine pinecone lavender \\). Now the hexagon \\( marigold chrysanthemum tangerine pinecone dawnlight lavender \\) is evidently symmetric about the line \\( chrysanthemum dawnlight \\), so \\( chrysanthemum dawnlight \\) also passes through the midpoint of the square \\( marigold tangerine pinecone lavender \\). Thus the three diagonals mentioned are concurrent.\n>>>\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "P_1": "irregularnode",
        "P_2": "skewedvertex",
        "P_4": "crookedcorner",
        "P_9": "offcenterpoint",
        "P_11": "misalignedspot",
        "P_12": "warpedjunction",
        "A": "emptiness",
        "B": "fuzziness",
        "s": "immaterial"
      },
      "question": "1. (i) Show that a regular hexagon, six squares, and six equilateral triangles can be assembled without overlapping to form a regular dodecagon.\n(ii) Let \\( irregularnode, skewedvertex, \\ldots, warpedjunction \\) be the successive vertices of a regular dodecagon. Explain how the three diagonals \\( irregularnode offcenterpoint, skewedvertex misalignedspot \\), and \\( crookedcorner warpedjunction \\) intersect.",
      "solution": "Solution. (i) Let squares be drawn on the outside of each side of a regular hexagon of side \\( immaterial \\). Then the wedge-shaped indentations have vertex angle \\( 60^{\\circ}\\left(=360^{\\circ}-90^{\\circ}-120^{\\circ}-90^{\\circ}\\right) \\), and adjacent sides of length \\( immaterial \\). Hence they can be filled in with equilateral triangles of side \\( immaterial \\). The resulting dodecagon has all sides of length \\( immaterial \\) and all angles \\( 150^{\\circ}\\left(=90^{\\circ}+60^{\\circ}\\right) \\); hence it is regular.\n(ii) Letter the vertices as shown in the diagram. Since the regular dodecagon can be inscribed in a circle, \\( \\angle irregularnode \\, warpedjunction \\, crookedcorner \\) is half the central angle of arc \\( irregularnode \\, crookedcorner \\), which is clearly a quarter circle. Hence \\( \\angle irregularnode \\, warpedjunction \\, crookedcorner=45^{\\circ} \\), so \\( warpedjunction \\, crookedcorner \\) is the diagonal of the square \\( irregularnode \\, emptiness \\, fuzziness \\, warpedjunction \\) and passes through \\( emptiness \\). Similarly \\( irregularnode \\, offcenterpoint \\) passes through \\( fuzziness \\). Hence \\( irregularnode \\, offcenterpoint \\) and \\( crookedcorner \\, warpedjunction \\) both pass through the center of the square \\( irregularnode \\, emptiness \\, fuzziness \\, warpedjunction \\). Now the hexagon \\( irregularnode \\, skewedvertex \\, emptiness \\, fuzziness \\, misalignedspot \\, warpedjunction \\) is evidently symmetric about the line \\( skewedvertex \\, misalignedspot \\), so \\( skewedvertex \\, misalignedspot \\) also passes through the midpoint of the square \\( irregularnode \\, emptiness \\, fuzziness \\, warpedjunction \\). Thus the three diagonals mentioned are concurrent."
    },
    "garbled_string": {
      "map": {
        "P_1": "qzxwvtnp",
        "P_2": "hjgrksla",
        "P_4": "mbcuvyeq",
        "P_9": "dafplori",
        "P_11": "nsutyvek",
        "P_12": "wprliadz",
        "A": "ksyuomaz",
        "B": "venjildr",
        "s": "uxomretq"
      },
      "question": "Problem:\n<<<\n1. (i) Show that a regular hexagon, six squares, and six equilateral triangles can be assembled without overlapping to form a regular dodecagon.\n(ii) Let \\( qzxwvtnp, hjgrksla, \\ldots, wprliadz \\) be the successive vertices of a regular dodecagon. Explain how the three diagonals \\( qzxwvtnp dafplori, hjgrksla nsutyvek \\), and \\( mbcuvyeq wprliadz \\) intersect.\n>>>",
      "solution": "Solution. (i) Let squares be drawn on the outside of each side of a regular hexagon of side \\( uxomretq \\). Then the wedge-shaped indentations have vertex angle \\( 60^{\\circ}\\left(=360^{\\circ}-90^{\\circ}-120^{\\circ}-90^{\\circ}\\right) \\), and adjacent sides of length \\( uxomretq \\). Hence they can be filled in with equilateral triangles of side \\( uxomretq \\). The resulting dodecagon has all sides of length \\( uxomretq \\) and all angles \\( 150^{\\circ}\\left(=90^{\\circ}+60^{\\circ}\\right) \\); hence it is regular.\n(ii) Letter the vertices as shown in the diagram. Since the regular dodecagon can be inscribed in a circle, \\( \\angle qzxwvtnp wprliadz mbcuvyeq \\) is half the central angle of arc \\( qzxwvtnp mbcuvyeq \\), which is clearly a quarter circle. Hence \\( \\angle qzxwvtnp wprliadz mbcuvyeq=45^{\\circ} \\), so \\( wprliadz mbcuvyeq \\) is the diagonal of the square \\( qzxwvtnp ksyuomaz venjildr wprliadz \\) and passes through \\( ksyuomaz \\). Similarly \\( qzxwvtnp dafplori \\) passes through \\( venjildr \\). Hence \\( qzxwvtnp dafplori \\) and \\( mbcuvyeq wprliadz \\) both pass through the center of the square \\( qzxwvtnp ksyuomaz venjildr wprliadz \\). Now the hexagon \\( qzxwvtnp hjgrksla ksyuomaz venjildr nsutyvek wprliadz \\) is evidently symmetric about the line \\( hjgrksla nsutyvek \\), so \\( hjgrksla nsutyvek \\) also passes through the midpoint of the square \\( qzxwvtnp ksyuomaz venjildr wprliadz \\). Thus the three diagonals mentioned are concurrent."
    },
    "kernel_variant": {
      "question": "Let  A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}  be a regular hexagon whose sides all have length 2.  On every side  A_{i}A_{i+1}\\;(\\text{indices taken modulo }6) construct externally the square  A_{i}B_{i}C_{i}A_{i+1};  that is, each square lies completely outside the hexagon and has the segment  A_{i}A_{i+1}  as one of its sides.\n\n(i)  Prove that the six gaps left between neighbouring squares are equilateral triangles of side-length 2.  Deduce that the outer boundary of the figure obtained by adjoining the six squares and the six triangles is a regular dodecagon.\n\n(ii)  Label the successive vertices of this dodecagon, in clockwise order, by\n        Q_{1},Q_{2},\\dots ,Q_{12},\nchoosing Q_{1}=B_{1}.  Show that the three diagonals \n        Q_{1}Q_{9},\\;Q_{2}Q_{11},\\;Q_{4}Q_{12}\nare concurrent.",
      "solution": "Throughout, indices of the A–vertices are understood modulo 6, and those of the Q–vertices modulo 12.\n\n(i)  Shape of the gaps and the outer boundary\n-------------------------------------------\nFix the vertex A_{i+1} of the original hexagon.  Three plane angles meet there:\n • the interior angle 120^{\\circ} of the regular hexagon, and\n • two right angles originating from the two squares that share that vertex.\n\nIndeed, the square built on the side A_{i}A_{i+1} contains the right angle \\widehat{C_{i}A_{i+1}A_{i}}, so one of its edges leaving A_{i+1} is A_{i+1}C_{i}.  The square on the neighbouring side A_{i+1}A_{i+2} contains the right angle \\widehat{A_{i+2}A_{i+1}B_{i+1}}, so its edge A_{i+1}B_{i+1} also leaves A_{i+1}.  Both segments A_{i+1}C_{i} and A_{i+1}B_{i+1} have length 2, since they are sides of squares of side 2.\n\nThe total angle around a point is 360^{\\circ}.  Subtracting the three known angles gives the angle of the gap between the two squares:\n360^{\\circ}-\\bigl(120^{\\circ}+90^{\\circ}+90^{\\circ}\\bigr)=60^{\\circ} .\n\nThus the gap at A_{i+1} is a triangle whose two sides issuing from the vertex have equal length 2 and include an angle of 60^{\\circ}.  Consequently the third side also has length 2, and the triangle is equilateral.\n\nDoing this at every vertex produces six congruent equilateral triangles which exactly fill the wedges between neighbouring squares.  Let \\mathcal P be the polygon which is the outer boundary of the union of the six squares and the six triangles.\n\nWalking once around \\mathcal P one meets alternately an edge of a square (length 2) and an edge of one of the triangles (also length 2).  Hence all twelve sides of \\mathcal P have the common length 2.  Whenever a square-edge meets a triangle-edge, the interior angle of \\mathcal P equals 90^{\\circ}+60^{\\circ}=150^{\\circ}.  Because \\mathcal P has twelve equal sides and all interior angles equal, it is a regular dodecagon whose side length is 2.\n\n(ii)  Concurrency of Q_{1}Q_{9}, Q_{2}Q_{11}, Q_{4}Q_{12}\n---------------------------------------------------------\nOnly the regular dodecagon from part (i) is used in this part; the hexagon and squares no longer play a role.\n\n1.  A complex-number model\nPlace the centre O of the dodecagon at the origin of the complex plane and put\n                 q_{k}=e^{i\\pi(k-1)/6}\\quad(k=1,2,\\dots ,12)\nfor the affix of Q_{k}.  Hence\n  Q_{1}=1,\\;Q_{2}=e^{i\\pi/6},\\;Q_{4}=e^{i\\pi/2}=i,\\;Q_{9}=e^{4i\\pi/3},\\;Q_{11}=e^{5i\\pi/3},\\;Q_{12}=e^{11i\\pi/6}.\n(Using the circum-radius 1 is merely a convenience; scaling does not affect concurrence.)\n\n2.  Intersection of Q_{1}Q_{9} and Q_{4}Q_{12}\nWrite the two diagonals in parametric form:\n      z=1+t\\bigl(e^{4i\\pi/3}-1\\bigr),\\qquad   (t\\in\\mathbb R),\n      z=i+s\\bigl(e^{11i\\pi/6}-i\\bigr),\\qquad (s\\in\\mathbb R).\nSeparating real and imaginary parts and solving yields the unique common point\n      z_{0}=\\dfrac{1+\\sqrt3}{4}+i\\,\\dfrac{1-\\sqrt3}{4},\nwith\n      t=\\dfrac{3-\\sqrt3}{6},\\qquad s=\\dfrac{1+\\sqrt3}{2\\sqrt3}=\\dfrac{3+\\sqrt3}{6}.  \\qquad(1)\n\n3.  Verifying that z_{0} lies on Q_{2}Q_{11}\nThe line through Q_{2} and Q_{11} has slope\n      m=\\frac{\\operatorname{Im}(e^{5i\\pi/3})-\\operatorname{Im}(e^{i\\pi/6})}\n            {\\operatorname{Re}(e^{5i\\pi/3})-\\operatorname{Re}(e^{i\\pi/6})}\n        =2+\\sqrt3.\nIts equation is therefore\n      y-\\tfrac12=(2+\\sqrt3)\\Bigl(x-\\tfrac{\\sqrt3}{2}\\Bigr).\nSubstituting x_{0}=\\dfrac{1+\\sqrt3}{4} and y_{0}=\\dfrac{1-\\sqrt3}{4} gives\n      y_{0}-\\tfrac12=(2+\\sqrt3)\\bigl(x_{0}-\\tfrac{\\sqrt3}{2}\\bigr),\nso z_{0} indeed lies on Q_{2}Q_{11}.\n\nBecause the point z_{0} belongs to each of the three lines, the diagonals\nQ_{1}Q_{9}, Q_{2}Q_{11}, and Q_{4}Q_{12} are concurrent.  The concurrency point is the interior point with Cartesian coordinates\n      \\bigl(\\tfrac{1+\\sqrt3}{4},\\;\\tfrac{1-\\sqrt3}{4}\\bigr).",
      "_meta": {
        "core_steps": [
          "Attach an exterior square to every side of a regular hexagon; the leftover wedges are isosceles with 60° vertex angles.",
          "Insert an equilateral triangle into each wedge; the resulting 12-gon has equal side-lengths and 150° interior angles → it is a regular dodecagon.",
          "Inscribe the dodecagon in its circumcircle; an inscribed-angle argument shows P4P12 (and analogously P1P9) passes through the center of the square on side P1P12.",
          "A symmetry line of the composite hexagon (through P2 and P11) also passes through that same square-center, so P2P11 meets the other two diagonals there; hence the three diagonals are concurrent."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Overall scale of the figure (common edge-length of hexagon, squares, and triangles).",
            "original": "s"
          },
          "slot2": {
            "description": "Direction and starting point of the vertex labeling around the dodecagon (clockwise vs. counter-clockwise, or which vertex is called P1).",
            "original": "Vertices labeled P1, P2, …, P12 consecutively counter-clockwise starting at an arbitrary vertex"
          },
          "slot3": {
            "description": "Choice of drawing the six squares outward or inward relative to the hexagon, producing the same wedges and final dodecagon up to reflection.",
            "original": "Squares drawn on the outside of each hexagon side"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}